Class 6

Expert Teachers at HSSLive.Guru has created Kerala Syllabus Std 6 Basic Science Notes Pdf Malayalam Medium English Medium of SCERT Class 6 Basic Science Notes Textbook Solutions, Basic Science Class 6 Question Answer are part of Kerala Syllabus 6th Standard Textbooks Solutions. Here we have given Basic Science Class 6 Kerala Syllabus Notes Pdf of SCERT Class 6 Basic Science Solutions Part 1 and Part 2.

SCERT Class 6 Basic Science Notes Textbook Solutions

Basic Science Class 6 Question Answer Solutions

Basic Science Class 6 Kerala Syllabus Notes Pdf Part 1

• Class 6 Basic Science Chapter 1 Question Answer – Food for Health Class 6 Notes
• Class 6 Basic Science Chapter 2 Question Answer – Marvel of the Magnetic Realm Class 6 Notes
• Class 6 Basic Science Chapter 3 Question Answer – Let’s Stand Straight Class 6 Notes
• Class 6 Basic Science Chapter 4 Question Answer – Flowering and Fruiting Class 6 Notes
• Class 6 Basic Science Chapter 5 Question Answer – Association of Substances Class 6 Notes

Basic Science for Class 6 Questions and Answers Part 2

• Class 6 Basic Science Chapter 6 Question Answer – Tiny Chambers of Life Class 6 Notes
• Class 6 Basic Science Chapter 7 Question Answer – The World of Changes Class 6 Notes
• Class 6 Basic Science Chapter 8 Question Answer – Motion in Daily Life Class 6 Notes
• Class 6 Basic Science Chapter 9 Question Answer – Better Lifestyle for a Brighter Tomorrow Class 6 Notes
• Class 6 Basic Science Chapter 10 Question Answer – For Shape and Strength Class 6 Notes

SCERT Class 6 Basic Science Solutions (Extra Questions)

1. Food for Health Class 6 Important Questions
2. Marvel of the Magnetic Realm Class 6 Important Questions
3. Let’s Stand Straight Class 6 Important Questions
4. Flowering and Fruiting Class 6 Important Questions
5. Association of Substances Class 6 Important Questions
6. Tiny Chambers of Life Class 6 Important Questions
7. The World of Changes Class 6 Important Questions
8. Motion in Daily Life Class 6 Important Questions
9. Better Lifestyle for a Brighter Tomorrow Class 6 Important Questions

Class 6 Basic Science Previous Year Question Papers English Medium

• 6th Standard Basic Science First Term Question Paper 2022-23
• 6th Standard Basic Science Second Term Question Paper 2022-23
• 6th Standard Basic Science Annual Exam Question Paper 2022-23
• 6th Standard Basic Science Annual Exam Question Paper 2021-22

Kerala Syllabus 6th Standard Basic Science Question Paper English Medium

• 6th Standard Basic Science Model Question Paper Set 1
• Class 6 Basic Science Model Question Paper Set 2

Std 6 Basic Science Notes Pdf Malayalam Medium

SCERT Class 6 Basic Science Solutions Malayalam Medium Part 1

• Class 6 Basic Science Chapter 1 Malayalam Medium – ജീവന്റെ ചെപ്പുകൾ
• Class 6 Basic Science Chapter 2 Malayalam Medium – മാറ്റത്തിന്റെ പൊരുൾ
• Class 6 Basic Science Chapter 3 Malayalam Medium – പൂവില്‍നിന്ന് പൂവിലേക്ക്
• Class 6 Basic Science Chapter 4 Malayalam Medium – ചലനത്തിനൊപ്പം
• Class 6 Basic Science Chapter 5 Malayalam Medium – ആഹാരം ആരോഗ്യത്തിന്

Std 6 Basic Science Notes Pdf Malayalam Medium Part 2

• Class 6 Basic Science Chapter 6 Malayalam Medium – ഒന്നിച്ചു നിലനിൽക്കാം
• Class 6 Basic Science Chapter 7 Malayalam Medium – ആകർഷിച്ചും വികർഷിച്ചും
• Class 6 Basic Science Chapter 8 Malayalam Medium – തിങ്കളും താരങ്ങളും
• Class 6 Basic Science Chapter 9 Malayalam Medium – ചേർക്കാം പിരിക്കാം
• Class 6 Basic Science Chapter 10 Malayalam Medium – രൂപത്തിനും ബലത്തിനും

Class 6 Basic Science Previous Year Question Papers Malayalam Medium

• Basic Science Class 6 First Term Question Paper 2023-24
• Basic Science Class 6 Second Term Question Paper 2023-24
• Basic Science Class 6 Annual Exam Question Paper 2023-24
• Basic Science Class 6 Annual Exam Question Paper 2021-22

6th Standard Basic Science Question Paper Malayalam Medium

• SCERT Class 6 Basic Science Model Question Paper Set 1
• 6th Std Basic Science Model Question Paper Set 2

We hope the given Std 6 Basic Science Notes Pdf English Medium Kerala Syllabus of Basic Science for Class 6 Questions and Answers, SCERT Class 6 Basic Science Solutions will help you. If you have any queries regarding Kerala Syllabus 6th Standard Basic Science Notes Pdf of 6th Class Basic Science Notes Part 1 and Part 2, drop a comment below and we will get back to you at the earliest.

Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 5 Decimal Forms Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 6 Maths Chapter 5 Solutions Decimal Forms

Class 6 Kerala Syllabus Maths Solutions Chapter 5 Decimal Forms Questions and Answers

Decimal Forms Class 6 Questions and Answers Kerala Syllabus

Decimal Places (Page No. 69)

Question 1.
Split the numbers below according to place value:
(i) 4.5
(ii) 4.57
(iii) 4.572
(iv) 45.72
(v) 457.2
Answer:

Textbook Page No. 71

Question 1.
Now try to write 4 kilograms and 55 grams as kilograms in decimal form.
Answer:
4 kilograms 55 grams
55 grams means \(\frac {55}{1000}\) kilograms.
So, 4 kilograms 55 grams = 4\(\frac {55}{1000}\) kilograms
Splitting 4\(\frac {55}{1000}\) according to place value

So we can write the decimal form of 4\(\frac {55}{1000}\) as 4.055

Question 2.
Convert the measures below into the measures specified, using fractions and decimal forms.

Answer:

Decimals and Fractions (Page No. 74)

Question 1.
The decimal form of some numbers is given below. Write each of them as a fraction with a denominator of 10, 100, or 1000.
(i) 3.7
(ii) 3.07
(iii) 30.7
(iv) 3.72
(v) 37.2
(vi) 3.072
(vii) 30.72
Answer:
(i) \(\frac {37}{10}\)
(ii) \(\frac {307}{100}\)
(iii) \(\frac {307}{10}\)
(iv) \(\frac {372}{100}\)
(v) \(\frac {372}{10}\)
(vi) \(\frac {3072}{1000}\)
(vii) \(\frac {3072}{100}\)

Question 2.
Write the decimal form of the fractions given below.
(i) \(\frac {51}{100}\)
(ii) \(\frac {513}{10}\)
(iii) \(\frac {513}{100}\)
(iv) \(\frac {513}{1000}\)
(v) \(\frac {5130}{1000}\)
Answer:
(i) 5.1
(ii) 51.3
(iii) 5.13
(iv) 0.513
(v) 5.13

Addition and Subtraction (Page No. 79)

Question 1.
Anu made an 8.5 metre long festoon and Sarah made a 7.8 metre long one to decorate their classroom for the school anniversary. What is the total length of the festoon they made?
Answer:
Length of the festoon Anu made = 8.5 metres
Length of the Festoon Sarah made = 7.8 metres
Removing the measures and converting into a fraction
8.5 = \(\frac {85}{10}\)
7.8 = \(\frac {78}{10}\)
Adding the fraction
\(\frac{85}{10}+\frac{78}{10}=\frac{163}{10}\)
Converting to decimals
\(\frac {163}{10}\) = 16.3
Total length of the festoon = 16.3 metres

Question 2.
Amal needs 2.25 metres of cloth and Sagar, 1.85 metres for a school uniform. How many metres of cloth in all?
Answer:
Length of the cloth Amal needs for the school uniform = 2.25 metres
Length of the cloth Sagar needs for the school uniform = 1.85 metres
Removing measures and converting to fractions
2.25 = \(\frac {225}{100}\)
1.85 = \(\frac {185}{100}\)
Adding these fractions
\(\frac{225}{100}+\frac{185}{100}=\frac{200+100+25+85}{100}=\frac{410}{100}\)
Converting this to decimals \(\frac {410}{100}\) = 4.1
Total length of cloth in all is 4.1 metres.

Question 3.
A tin weighs 2.85 kilograms, and it is filled with 12.5 kilograms of rice. What is the total weight?
Answer:
Weight of the tin = 2.85 kilograms
Amount of rice filled in the tin = 12.5 kilograms
Removing measures and converting to fractions
2.85 = \(\frac {285}{100}\)
12.5 = \(\frac {125}{10}\)
Adding these fractions
\(\frac{285}{100}+\frac{125}{10}\)
To add these changing \(\frac {125}{10}\) to a form with denominator
\(\frac{125 \times 10}{10 \times 10}=\frac{1250}{100}\)
Adding these fractions,
\(\frac{1250}{100}+\frac{285}{100}\)
1000 + 200 + 250 + 85 = 1200 + 335 = 1535
\(\frac{1250}{100}+\frac{285}{100}=\frac{1535}{100}\)
Converting the fractions to decimals
\(\frac {1535}{100}\) = 15.35
Total weight = 15.35 kilograms

Question 4.
Bakul walks 2.25 kilometres in the morning and 1.5 kilometres in the evening every day. What is the total distance she walks each day?
Answer:
Distance Bakul walks in the morning = 2.25 kilometres
Distance Bakul walks in the evening = 1.5 kilometres
Total distance she walks each day = 2.25 kilometres + 1.5 kilometres
Removing measures and converting to fractions
2.25 = \(\frac {225}{100}\)
1.5 = \(\frac {15}{10}\)
To add these, change \(\frac {15}{10}\) to a form with denominator 100
\(\frac{15 \times 10}{10 \times 10}=\frac{150}{100}\)
Adding the fractions
\(\frac{225}{100}+\frac{150}{100}\)
200 + 100 + 25 + 50 = 375
\(\frac{225}{100}+\frac{150}{100}=\frac{375}{100}\)
Converting the fractions to decimals
\(\frac {375}{100}\) = 3.75
Total distance she walks each day = 3.75 kilometres

Question 5.
Two small bottles contain 0.850 litres and 0.375 litres of honey. If both the bottles are emptied into a large bottle, how much honey does it contain?
Answer:
Amount of honey in the first bottle = 0.850 litres
Amount of honey in the second bottle = 0.375 litres
Amount of honey in the large bottle = 0.850 litre + 0.375 litre
Removing measures and making into fractions
0.375 = \(\frac {375}{1000}\)
0.850 = \(\frac {850}{1000}\)
Adding these \(\frac{375}{1000}+\frac{850}{1000}\)
375 + 850 = 300 + 800 + 75 + 50 = 1225
\(\frac{375}{1000}+\frac{850}{1000}=\frac{1225}{1000}\)
Converting the fractions into decimals
\(\frac {1225}{1000}\) = 1.225
Amount of honey in the large bottle = 1.225 litres.

Textbook Page No. 82

Question 1.
From a rod 14.7 metres long, a piece 7.75 metres long is cut off. What is the length of the remaining piece?
Answer:
Length of the long rod = 14.7 metres
Length of the piece cut off from the long rod = 7.75 metres
Length of the remaining piece = 14.7 metre – 7.75 metres
Changing into fractions
14.7 = \(\frac {147}{10}\)
7.75 = \(\frac {775}{100}\)
To subtract change \(\frac {147}{10}\) to a form with denominator 100
\(\frac{147 \times 10}{10 \times 10}=\frac{1470}{100}\)
Subtracting \(\frac{1470}{100}-\frac{775}{100}=\frac{695}{100}\)
Changing back to decimals
\(\frac {695}{100}\) = 6.95
Length of the remaining piece = 6.95 metres.

Question 2.
There were 38.7 kilograms of rice in a sack, and 12.350 kilograms of this were used up. How much rice remains in the sack?
Answer:
Amount of rice in the sack = 38.7 kilograms
Amount of rice used up = 12.350 kilograms
Amount of rice remaining in the sack = 38.7 kilograms – 12.350 kilograms
38.7 = \(\frac {387}{10}\)
12.350 = \(\frac {12350}{1000}\)
\(\frac{387}{10}-\frac{12350}{1000}\)
\(\frac{387 \times 100}{10 \times 100}=\frac{38700}{1000}\)
Subtracting \(\frac{38700}{1000}-\frac{12350}{1000}=\frac{26350}{1000}\)
Changing back to decimals
\(\frac {26350}{1000}\) = 26.35
Amount of rice remaining in the sack = 26.35 kilograms

Question 3.
The perimeter of a rectangle is 24 centimetres and the length of one side is 6.4 centimetres. What is the length of the other side?
Answer:
Perimeter of rectangle = 2(length + breadth) = 24 centimetres
Length of one side = 6.4 centimetres
Length of the other side = 12 – 6.4 = 5.6 centimetres

Question 4.
There were 2.50 litres of oil in a bottle, and 0.475 litres of this were used for cooking. How much oil is left in the bottle?
Answer:
Amount of oil in the bottle = 2.50 litres
Amount of oil used for cooking = 0.475 litres
Amount of oil left in the bottle = 2.50 litres – 0.475 litres
2.50 = \(\frac {250}{100}\)
0.475 = \(\frac {475}{1000}\)
\(\frac{250}{100}-\frac{475}{1000}\)
\(\frac{250 \times 10}{100 \times 10}=\frac{2500}{1000}\)
Subtracting \(\frac{2500}{1000}-\frac{475}{1000}=\frac{2025}{1000}\)
Amount of oil remaining = 2.025 litres

Question 5.
What number must we add to 14.32 to get 16.43?
Answer:
Number to be added to 14.32 to get 16.43 = 16.43 – 14.32
16.43 = \(\frac {1643}{100}\)
14.32 = \(\frac {1432}{100}\)
Subtracting \(\frac{1643}{100}-\frac{1432}{100}=\frac{211}{100}\)
Converting back to decimals
\(\frac {211}{100}\) = 2.11

Class 6 Maths Chapter 5 Kerala Syllabus Decimal Forms Questions and Answers

Class 6 Maths Decimal Forms Questions and Answers

Question 1.
Split the numbers below according to place value.
(i) 3.6
(ii) 3.64
(iii) 3.641
(iv) 36.41
(v) 364.1
Answer:

Question 2.
Convert the following measures into the specified forms, using both fractions and decimal forms.

Answer:

Question 3.
The decimal form of some numbers is given below. Write each of them as a fraction with a denominator of 10, 100, or 1000.
(i) 4.2
(ii) 4.02
(iii) 40.2
(iv) 4.25
(v) 42.5
(vi) 4.025
(vii) 40.25
Answer:
(i) \(\frac {42}{10}\)
(ii) \(\frac {402}{100}\)
(iii) \(\frac {402}{10}\)
(iv) \(\frac {425}{100}\)
(v) \(\frac {4025}{1000}\)
(vi) \(\frac {4025}{100}\)

Question 4.
Write the decimal form of the fractions given below.
(i) \(\frac {9}{10}\)
(ii) \(\frac {47}{100}\)
(iii) \(\frac {381}{1000}\)
(iv) \(\frac {15}{10}\)
(v) \(\frac {245}{100}\)
(vi) \(\frac {7}{100}\)
(vii) \(\frac {82}{1000}\)
(viii) \(\frac {3456}{1000}\)
Answer:
(i) 0.9
(ii) 0.47
(iii) 0.381
(iv) 1.5
(v) 2.45
(vi) 0.07
(vii) 0.082
(viii) 3.456

Question 5.
John ran a distance of 4.25 kilometres and then walked another 2.5 kilometres. What is the total distance he covered?
Answer:
Distance John ran = 4.25 kilometres
Distance he walked = 2.5 kilometres
Total distance he covered = 4.25 kilometres + 2.5 kilometres
Converting into fractions
4.25 = \(\frac {425}{100}\)
25 = \(\frac{25}{10}=\frac{25 \times 10}{10 \times 10}=\frac{250}{100}\)
Adding \(\frac{425}{100}+\frac{250}{100}=\frac{675}{100}\)
Converting to decimals
\(\frac {675}{100}\) = 6.75
Total distance John covered = 6.75 kilometres

Question 6.
A baker has two bags of flour, one with 1.75 kilograms of flour and the other with 2.5 kilograms. If the baker combines all the flour into a single container, what is the total weight of the flour in the container?
Answer:
Total weight of the flour in the container = Amount of flour in Bag 1 + Amount of flour in Bag 2 = 1.75 kilograms + 2.5 kilograms
1.75 = \(\frac {175}{100}\)
2.5 = \(\frac{25}{10}=\frac{25 \times 10}{10 \times 10}=\frac{250}{100}\)
Adding \(\frac{175}{100}+\frac{250}{100}=\frac{425}{100}\)
Converting into decimals
\(\frac {425}{100}\) = 4.25
Total weight of the flour in the container = 4.25 kilograms

Question 7.
A ribbon was 15.8 metres long. If a piece measuring 4.25 metres was cut from it, how much ribbon is left?
Answer:
Length of the ribbon = 15.8 metres
Length of the piece cut off from the ribbon = 4.25 metres
Length of the ribbon left = 15.8 metres – 4.25 metres
15.8 = \(\frac{158}{10}=\frac{158 \times 10}{10 \times 10}=\frac{1580}{100}\)
4.25 = \(\frac {425}{100}\)
Subtracting \(\frac{1580}{100}-\frac{425}{100}=\frac{1155}{100}\)
Converting back to decimals
\(\frac {1155}{100}\) = 11.55
Length of the ribbon left = 11.55 metres

Question 8.
A water tank holds 50.5 litres of water. If 25.5 litres are used for gardening, how much water is left in the tank?
Answer:
Amount of water left in the tank = Amount of water that the water tank can hold – Amount of water used for gardening
50.5 = \(\frac {505}{10}\)
255 = \(\frac {255}{10}\)
Subtracting \(\frac{505}{10}-\frac{255}{10}=\frac{250}{10}\)
Converting to decimals
\(\frac {250}{10}\) = 2.5

Question 9.
A carpenter uses a 2.75-metre board and a 1.5-metre board for a project. What is the total length of the wood used?
Answer:
Total length of the wood used = 2.75 metre + 1.5 metre
2.75 = \(\frac {275}{100}\)
1.5 = \(\frac{15}{10}=\frac{15 \times 10}{10 \times 10}=\frac{150}{100}\)
Adding \(\frac{275}{100}+\frac{150}{100}=\frac{425}{100}\)
Converting to decimals
\(\frac {425}{100}\) = 4.25
Total length of the wood used = 4.25 metres

Question 10.
A plant is 15.6 centimetres tall. How many more centimetres must it grow to reach a height of 20.1 centimetres?
Answer:
Height of the plant = 15.6 centimetres
Target height = 20.1 centimetres
Height the plant must grow = 20.1 centimetres – 15.6 centimetres
20.1 = \(\frac {201}{10}\)
15.6 = \(\frac {156}{10}\)
Subtracting \(\frac{201}{10}-\frac{156}{10}=\frac{45}{10}\)
Converting to decimals
\(\frac {45}{10}\) = 4.5
The plant must grow 4.5 centimetres to reach the target height.

Class 6 Maths Chapter 5 Notes Kerala Syllabus Decimal Forms

→ In the decimal form of a number, the dot (decimal point) separates the whole number part and the fractional part.

→ Digits to the left of the decimal point represent ones, tens, hundreds, and so on;

→ The digits to the right represent tenths, hundredths, thousandths, and so on.

→ Decimals allow us to represent quantities that are not whole numbers with greater precision, making them essential for measurements, money, and other real-world applications.

→ To convert a measurement from centimetres to metres in decimal form, divide the number of centimetres by 100. This is equivalent to moving the decimal point two places to the left.

→ If you have a combination of metres and centimetres, first convert the centimetres to metres as a decimal and then add them to the whole number of metres.

→ To convert a measurement from centimetres to millimetres, multiply the number by 10.

→ To convert from millimetres to centimetres, divide the number by 10.

→ To add decimal numbers representing measurements (like 4.3 cm and 2.5 cm), align the decimal points and add them directly.

→ Alternatively, you can convert the measurements to a smaller unit (like millimetres) and complete the addition, then convert the result back to the original unit.

→ To subtract a decimal from another (like subtracting 3.2 cm from 8.5 cm), you can convert both decimals to fractions with a common denominator, subtract them, and then convert the result back into a decimal.

This chapter comprehensively covers the representation and manipulation of numbers beyond whole units. Key topics include understanding decimal places to denote fractional parts, establishing the crucial relationship and conversion techniques between decimals and fractions, and mastering the fundamental operations of addition and subtraction of decimal numbers to build foundational arithmetic skills.

Decimal Places
The length of a pencil can be said in different ways:

• 5 centimetres 7 millimetres
• 5\(\frac {7}{10}\) centimetres
• 5.7 centimetres

We can write other measures also like this:
5\(\frac {7}{10}\) litres = 5.7 litres
5\(\frac {7}{10}\) kilograms = 5.7 kilograms

We can drop all references to measures and simply say that 5.7 is the decimal form of the number 5\(\frac {7}{10}\).
5\(\frac {7}{10}\) = 5.7
Similarly, 4.29 is the decimal form of 4\(\frac {29}{100}\)
4\(\frac {29}{100}\) = 4.29

We write natural numbers using ones, tens, hundreds, and so on.
For example: 247 = 2 hundreds + 4 tens + 7 ones
Splitting 247.3
Split it as the sum of a whole number and a fraction.
247.3 = 247\(\frac {3}{10}\) = 247 + \(\frac {3}{10}\)
First, we split 247.3 as the sum of a whole number and a fraction, as
247.3 = 247\(\frac {3}{10}\) = 247 + \(\frac {3}{10}\)
The \(\frac {3}{10}\) here can be written as
\(\frac{3}{10}=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}\)
That is, 3 tenths. So, we can write 247.3 in terms of hundreds, tens, ones, and tenths:
247.3 = 2 hundreds + 4 tens + 7 ones + 3 tenths

Splitting 247.39
First, we write it as
247.39 = 247\(\frac {39}{100}\)= 247 + \(\frac {39}{100}\)
Then, we can split \(\frac {39}{100}\) as
\(\frac{39}{100}=\frac{30+9}{100}=\frac{30}{100}+\frac{9}{100}=\frac{3}{10}+\frac{9}{100}\)
The \(\frac {3}{10}\) here is 3 tenths: and \(\frac {9}{100}\) is 9 hundredths.
So 247.39 = 2 hundreds + 4 tens + 7 ones + 3 tenths + 9 hundredths

In the decimal form of a number, the dot separates the whole number part and the fractional part. Digits to the left of the dot show the multiples of ones, tens, hundreds, and so on; the digits to the right show the multiples of tenths, hundredths, thousandths, and so on.
For example, the two numbers used in the above examples can be split according to place value like this:

Question 1.
What is the decimal form of 23 metres 40 centimetres?
Answer:
Method 1
23 metres 40 centimetres = 23\(\frac {40}{100}\) metres = 23.40 metres
Taking only the numbers, we have
23\(\frac {40}{100}\) = 23.40
We can write the \(\frac {40}{100}\) here as
\(\frac{40}{100}=\frac{4}{10}\)
So, we get 23\(\frac {40}{100}\) = 23\(\frac {4}{10}\) = 23.4
This means 23.40 = 23.4

Method 2
Using place value

Thus, we can write 23 metres and 40 centimetres in two different ways:
23 metres 40 centimetres = 23.40 metres
23 metres 40 centimetres = 23.4 metres

Question 2.
What is the decimal form of 23 metres 4 centimetres?
Answer:
4 centimetres = \(\frac {4}{100}\) metre.
23 metres 4 centimetres =23\(\frac {4}{100}\) metres
Split 23\(\frac {4}{100}\) according to place value:
23\(\frac {4}{100}\) = 2 tens + 3 ones + 4 hundredths

The decimal form of 23\(\frac {4}{100}\) = 23.04

Question 3.
What is the decimal form of 23 metres and 4 millimetres?
Answer:
4 millimetres means \(\frac {4}{1000}\) metres.
So 23 metres 4 millimetres = 23\(\frac {4}{1000}\) metres
Split 23\(\frac {4}{1000}\) according to place value.

So, we can write the decimal form of 23\(\frac {4}{1000}\) as
23\(\frac {4}{1000}\) = 23.004

Converting Centimetres to a Decimal Form of Metres
To convert centimetres to a decimal form of metres, first, write the number of metres as the whole number part of your decimal.
Next, express the number of centimetres as a fraction of a metre. Since there are 100 centimetres in 1 metre, the denominator of your fraction will be 100. For example, 40 centimetres would be written as \(\frac {40}{100}\) metres.
Combine the whole number and the fraction to create a mixed number, such as 23\(\frac {40}{100}\).
To get the decimal form, divide the numerator of your fraction by 100. This is the same as moving the decimal point two places to the left. So, \(\frac {40}{100}\) becomes 0.40.
Finally, add the decimal to the whole number.
For example, 23 + 0.40 = 23.40.
The result is the length in meters.

Decimals and Fractions
Conversion of decimals into fractions
Start with the decimal number you want to convert.

Example: 7.3 centimetres
Write this in millimetres
7 centimetres = 70 millimetres
To convert \(\frac {3}{10}\) centimetres into millimetres:
\(\frac {3}{10}\) is three \(\frac {1}{10}\)
\(\frac {3}{10}\) centimetres = 3 millimetres
7.3 centimetres = 70 millimetres + 3 millimetres

Converting 73 millimetres into centimetres:
Divide it by 10
73 millimetres = \(\frac {73}{10}\) centimetres
Removing measure and writing as just numbers 7.3 = \(\frac {73}{10}\)

Question 4.
How do we write 7.31 metres as a fraction?
Answer:
Write it as a whole number and a fraction
7.31 metres = 7\(\frac {31}{100}\) metres
7 metres = 700 centimetres
\(\frac {31}{100}\) metres = 31 centimetres
7.31 metres = 700 centimetres + 31 centimetres = 731 centimeters
Converting this into metres
731 centimetres = \(\frac {731}{100}\) metres
7.31 metres = \(\frac {731}{100}\) metres

Question 5.
Convert 7.319 litres as a fraction.
Answer:
7.319 litres = 7\(\frac {319}{1000}\) litres
7 litres = 7000 millilitres
\(\frac {319}{1000}\) litres = 319 millilitres
7.319 litres = 7319 millilitres
Converting back to litres
7319 millilitres = \(\frac {7319}{1000}\) litres
7.319 litres = \(\frac {7319}{1000}\) litres

Converting 12.03 to a Fraction

• Step 1: Count the digits after the decimal point in 12.03. There are two digits (0 and 3).
• Step 2: The denominator is 100 because there are two digits after the decimal.
• Step 3: Remove the decimal point from 12.03 to get the numerator, which is 1203.
• Step 4: The final fraction is \(\frac {1203}{100}\).

Question 6.
What is the decimal form of \(\frac {1203}{1000}\)?
Answer:
Looking at the denominator, we can say there will be three digits after the decimal point.
\(\frac {1203}{1000}\) = 1.203

Addition and Subtraction

Addition of Decimal Numbers
A line 4.3 centimetres long was drawn and then extended by another 2.5 centimetres:

To find the total length of the line we have to add 4.3 centimetres and 2.5 centimetres

Method 1
Convert these to centimetres and millimetres.
4.3 centimetres = 4 centimetres 3 millimetres
2.5 centimetres = 2 centimetres 5 millimetres
And add the centimetres and millimetres separately.
4 centimetres + 2 centimetres = 6 centimetres
3 millimetres + 5 millimetres = 8 millimetres
The length of the line is 6 centimetres, 8 millimetres
Convert back to centimetres.
6 centimetres 8 millimetres = \(\frac {68}{10}\) centimetres = 6.8 centimetres

Method 2
Write the lengths in millimetres.
4.3 centimetres = 43 millimetres
2.5 centimetres = 25 millimetres
Add 43 and 25
43 + 25 = 40 + 20 + 3 + 5 = 68
Thus, the length of the line is 68 millimetres
Write as centimetres in decimal form.
68 millimetres = 6 centimetres 8 millimetres = 6.8 centimetres

Method 3
Remove the measures and write the numbers as fractions.
4.3 = \(\frac {43}{10}\)
2.5 = \(\frac {25}{10}\)
And these fractions we can add like this:
\(\frac{43}{10}+\frac{25}{10}=\frac{43+25}{10}=\frac{68}{10}\)
Write the fraction as a decimal number
\(\frac {68}{10}\) = 6.8
The length of the line is 6.8 centimetres.

Question 7.
Add 4.3 centimetres and 2.8 centimetres.
Answer:
Changing the lengths into millimetres
4.3 centimetres = 43 millimetres
2.8 centimetres = 28 millimetres
Adding 43 and 28
43 + 28 = 40 + 20 + 3 + 8 = 60 + 11 = 71
Thus, the length of this line is 71 millimetres.
Write in centimetres as a decimal:
71 millimetres = 7 centimetres 1 millimetre = 7.1 centimetres
Convert the numbers to fractions
4.3 = \(\frac {43}{10}\)
2.8 = \(\frac {28}{10}\)
Add the fractions:
\(\frac{43}{10}+\frac{28}{10}=\frac{43+28}{10}=\frac{71}{10}\)
Convert the fraction back to the decimal form.
\(\frac {71}{10}\) = 7.1
The length of the line is 7.1 centimetres.

Question 8.
A jar contains 3.5 litres of oil, and 6.25 litres more is poured into it. How much oil does the jar contain now?
Answer:
Convert just the numbers to fractions:
3.5 = \(\frac {35}{10}\)
6.25 = \(\frac {625}{100}\)
We can write \(\frac {35}{10}\) also as a fraction with denominator 100:
\(\frac{35}{10}=\frac{35 \times 10}{10 \times 10}\) = \(\frac {350}{100}\)
Now we can add like this:
\(\frac{35}{10}+\frac{625}{100}=\frac{350}{100}+\frac{625}{100}=\frac{300+600+50+25}{100}\)
= \(\frac{350+625}{100}\)
= \(\frac {975}{100}\)
Amount of oil the jar contains = 9.75 litres

Question 9.
A person bought 2.5 kilograms of rice and 3.125 kilograms of vegetables. What is the total weight?
Answer:
Converting into fractions
2.5 = \(\frac {25}{10}\)
3.125 = \(\frac {3125}{1000}\)
To add these, we change \(\frac {25}{10}\) to a form with a denominator of 1000.
\(\frac{25}{10}=\frac{25 \times 100}{10 \times 100}\) = \(\frac {2500}{1000}\)
Now, we add the fractions:
\(\frac{25}{10}+\frac{3125}{1000}=\frac{2500}{1000}+\frac{3125}{1000}\) = \(\frac{2500+3125}{1000}\)
One way of adding 2500 and 3125 is this:
2500 + 3125 = 2000 + 3000 + 500 + 125 = 5000 + 625 = 5625
So, we can continue our addition of fractions:
\(\frac{25}{10}+\frac{3125}{1000}=\frac{2500+3125}{1000}\) = \(\frac {5625}{1000}\)
Converting this to decimals:
\(\frac {5625}{1000}\) = 5.625
Thus, the total weight is 5.625 kilograms.

Subtraction of Decimal Numbers
Example: From an 8.5 centimetres long eerkkil, a 3.2 centimetres long piece is broken off. What is the length of the remaining piece?
Thinking in terms of numbers alone, what we need is to subtract 3.2 from 8.5.
We change the numbers to fractions.
8.5 = \(\frac {85}{10}\)
3.2 = \(\frac {32}{10}\)
Now we can subtract:
\(\frac{85}{10}-\frac{32}{10}=\frac{85-32}{10}\)
One way to subtract 32 from 85 is this:
85 – 32 = (80 – 30) + (5 – 2) = 50 + 3 = 53
So we get \(\frac{85}{10}-\frac{32}{10}=\frac{53}{10}\)
Finally, we switch back to decimals:
\(\frac {53}{10}\) = 5.3
Thus, the length of the remaining piece of eerkkil is 5.3 centimetres.

Question 10.
From an 8.5 centimetres long eerkkil, a 3.7 centimetres long piece is broken off. What is the length of the remaining piece?
Answer:
We start as before by converting the decimals to fractions:
8.5 = \(\frac {85}{10}\)
3.7 = \(\frac {37}{10}\)
And then subtract
\(\frac{85}{10}-\frac{37}{10}=\frac{85-37}{10}\)
We have seen in earlier classes that subtraction like 85 different ways.
For example,
85 – 37 = (85 – 35) – 2 = 50 – 2 = 48
85 – 37 = (87 – 37) – 2 = 50 – 2 = 48
85 – 37 = (85 – 40) + 3 = 45 + 3 = 48
Anyway, we find
\(\frac{85}{10}-\frac{37}{10}=\frac{48}{10}\)
Changing back to decimals,
\(\frac {48}{10}\) = 4.8
The remaining piece of eerkkil is 4.8 centimetres long.

Question 11.
There are 15 kilograms of rice in a sack. 4.25 kilograms from this are put in a bag. How much rice remains in the sack?
Answer:
Thinking just in terms of numbers, what we have to do is subtract 4.25 from 15.
Write 4.25 as a fraction:
4.25 = \(\frac {425}{100}\)
Write 15 as a fraction with a denominator of 100.
15 = \(\frac{15}{1}=\frac{15 \times 100}{1 \times 100}=\frac{1500}{100}\)
Subtracting \(\frac{1500}{100}-\frac{425}{100}=\frac{1500-425}{100}\)
We can do 1500 – 425 in several ways:
1500 – 425 = 1000 + 500 – 425 = 1000 + 75 = 1075
1500 – 425 – 1425 – 425 + 75 = 1000 + 75 = 1075
1500 – 425 = 1500 – 500 + 75 = 1000 + 75 = 1075
Thus, we have:
\(\frac{1500}{100}-\frac{425}{100}=\frac{1075}{100}\)
Changing back to decimals:
\(\frac {1075}{100}\) = 10.75
So, there are 10.75 kilograms of rice still in the sack.

Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 6 Multiples and Factors Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 6 Maths Chapter 6 Solutions Multiples and Factors

Class 6 Kerala Syllabus Maths Solutions Chapter 6 Multiples and Factors Questions and Answers

Multiples and Factors Class 6 Questions and Answers Kerala Syllabus

Multiples of Multiples (Page No. 85)

Question 1.
For each of the multiples given below, find the other numbers they are multiples of:
(i) Multiples of 8
(ii) Multiples of 10
(iii) Multiples of 12
Answer:
(i) The factors of 8 are 2 and 4
8 is a multiple of 2 and 4
Therefore, the multiples of 8 are also the multiples of 2 and 4

(ii) The factors of 10 are 2 and 5
10 is a multiple of 2 and 5
Therefore, the multiples of 10 are also the multiples of 2 and 5

(iii) The factors of 12 are 2, 3, 4, and 6
12 is a multiple of 2, 3, 4, and 6
Therefore, the multiples of 12 are also the multiples of 2, 3, 4, and 6.

Question 2.
Check whether each of the statements below is true or false. For true statements, explain why they are so. For the false statements, give an example in which it is not true.
(i) All multiples of 20 are multiples of 10
(ii) All multiples of 10 are multiples of 2
(iii) All multiples of 15 are multiples of 5
(iv) All multiples of 15 are multiples of 3
(v) All multiples of 5 are multiples of 15
(vi) All multiples of 3 are multiples of 15
Answer:
(i) True
Since 20 = 2 × 10,
So every multiple of 20 is a multiple of 10.

(ii) True
Since 10 = 2 × 5,
So every multiple of 10 is a multiple of 2.
Or
Since 10 is an even number, and every multiple of 10 ends in 0, it is divisible by 2
That is, every multiple of 10 is also a multiple of 2.

(iii) True
Since 15 = 3 × 5,
So every multiple of 15 is a multiple of 5.

(iv) True
Since 15 = 3 × 5,
So every multiple of 15 is a multiple of 3.

(v) False
Not all multiples of 5 are divisible by 15.
E.g.: 10 is a multiple of 5 but not of 15.

(vi) False
Not all multiples of 3 are divisible by 15.
E.g.: 6 is a multiple of 3, but not a multiple of 15.

Primary Factors (Page No. 88)

Question 1.
Can you write the numbers below as a product of primes?
(i) 24
(ii) 35
(iii) 36
(iv) 60
(v) 100
Answer:
(i) 24 = 2 × 12
= 2 × 2 × 6
= 2 × 2 × 2 × 3
Therefore 24 = 2 × 2 × 2 × 3

(ii) 35 = 5 × 7

(iii) 36 = 2 × 18
= 2 × 2 × 9
= 2 × 2 × 3 × 3
Therefore 36 = 2 × 2 × 3 × 3

(iv) 60 = 2 × 30
= 2 × 2 × 15
= 2 × 2 × 3 × 5
Therefore 60 = 2 × 2 × 3 × 5

(v) 100 = 2 × 50
= 2 × 2 × 25
= 2 × 2 × 5 × 5
Therefore 100 = 2 × 2 × 5 × 5

Textbook Page No. 89

Question 1.
Write each of the numbers below as a product of primes.
(i) 72
(ii) 105
(iii) 144
(iv) 330
(v) 900
Answer:
(i) 72 = 12 × 6
12 = 2 × 2 × 3
6 = 2 × 3
Therefore, 72 = 2 × 2 × 2 × 3 × 3

(ii) 105 = 21 × 5
21 = 3 × 7
5 = 5
Therefore 105 = 3 × 5 × 7

(iii) 144 = 12 × 12
12 = 2 × 2 × 3
Therefore 144 = 2 × 2 × 3 × 2 × 2 × 3

(iv) 330 = 10 × 33
10 = 2 × 5
33 = 3 × 11
Therefore, 330 = 2 × 3 × 5 × 11

(v) 900 = 30 × 30
30 = 2 × 3 × 5
Therefore, 900 = 2 × 3 × 5 × 2 × 3 × 5

All Factors (Page No. 89)

Question 1.
Find all the factors of the number below:
(i) 35
(ii) 77
(iii) 26
(iv) 51
(v) 95
Answer:
(i) 35 = 1 × 35 = 5 × 7
Factors of 35 are: 1, 5, 7, and 35

(ii) 77 = 1 × 77 = 11 × 7
Factors of 77 are: 1, 7, 11, and 77

(iii) 26 = 1 × 26 = 2 × 13
Factors of 26 are: 1, 2, 13, and 26

(iv) 51 = 1 × 51 = 3 × 17
Factors of 51 are: 1, 3, 17, and 51

(v) 95 = 1 × 95 = 5 × 19
Factors of 95 are: 1, 5, 19, and 95

Textbook Page No. 90

Question 1.
Write each of the numbers below as a product of three primes and find all its factors:
(i) 66
(ii) 70
(iii) 105
(iv) 110
(v) 130
Answer:
(i) 66 is the product of three prime numbers.
66 = 2 × 3 × 11
Factors of 66 are:
1
2, 3, 11
2 × 3 = 6
2 × 11 = 22
3 × 11 = 33
Therefore, the factors are: 1, 2, 3, 6, 11, 22, 33, and 66

(ii) 70 as the product of three prime numbers.
70 = 2 × 5 × 7
Factors of 70 are:
1
2, 5, 7
2 × 5 = 10
2 × 7 = 14
5 × 7 = 35
Therefore, the factors are: 1, 2, 5, 7, 10, 14, 35, and 70

(iii) 105 is the product of three prime numbers.
105 = 3 × 5 × 7
Factors of 105 are:
1
3, 5, 7
3 × 5 = 15
3 × 7 = 21
5 × 7 = 35
Therefore, the factors are: 1, 3, 5, 7, 15, 21, 35, and 105

(iv) 110 as the product of three prime numbers.
110 = 2 × 5 × 11
Factors of 110 are:
1
2, 5, 11
2 × 5 = 10
2 × 11 = 22
5 × 11 = 55
Therefore, the factors are: 1, 2, 5, 10, 11, 22, 55, and 110

(v) 130 is the product of three prime numbers.
130 = 2 × 5 × 13
Factors of 130 are:
1
2, 5, 13
2 × 5 = 10
2 × 13 = 26
5 × 13 = 65
Therefore, the factors are: 1, 2, 5, 10, 13, 26, 65, and 130

Prime Numbers (Page No. 92)

Question 1.
Find all primes less than 100. Find the primes that differ by 2 among these.
Answer:
Prime numbers less than 100 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
Prime numbers that differ by 2 are:
(3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73)

Question 2.
Can the product of two natural numbers be a prime?
Answer:
The product of two natural numbers can only be a prime number if one of the numbers is 1 and the other is a prime.
If both numbers are greater than one, their product will always have more than two factors, so the result cannot be a prime.

Question 3.
Can the sum of two prime numbers be prime?
Answer:
Yes, sometimes, but not always.
If the sum of two prime numbers is a prime only when one of the numbers is 2 (the only even prime).
If 2 is added to an odd prime, the result is strange and may be prime.
But if two odd primes are added, the result is even and will never be a prime (except 2 + 2 = 4, which is not a prime number).

Class 6 Maths Chapter 6 Kerala Syllabus Multiples and Factors Questions and Answers

Class 6 Maths Multiples and Factors Questions and Answers

Question 1.
For each of the multiples given below, find the other numbers they are multiples of:
(i) Multiples of 15
(ii) Multiples of 21
(iii) Multiples of 33
Answer:
(i) The factors of 15 are 3 and 5.
15 is a multiple of 3 and 5.
Therefore, the multiples of 15 are also the multiples of 3 and 5.

(ii) The factors of 21 are 3 and 7.
21 is a multiple of 3 and 7.
Therefore, the multiples of 21 are also the multiples of 3 and 7.

(iii) The factors of are 3 and 11.
33 is a multiple of 3 and 11.
Therefore, the multiples of 33 are also the multiples of 3 and 11.

Question 2.
Write the numbers below as a product of primes?
(i) 18
(ii) 40
(iii) 150
(iv) 210
(v) 300
Answer:
(i) 18 = 2 × 9 = 2 × 3 × 3
Therefore 18 = 2 × 3 × 3

(ii) 40 = 2 × 20
= 2 × 2 × 10
= 2 × 2 × 2 × 5
Therefore 40 = 2 × 2 × 2 × 5

(iii) 150 = 2 × 75
= 2 × 3 × 25
= 2 × 3 × 5 × 5
Therefore 150 = 2 × 3 × 5 × 5

(iv) 210 = 2 × 105
= 2 × 3 × 35
= 2 × 3 × 5 × 7
Therefore 210 = 2 × 3 × 5 × 7

(v) 300 = 2 × 150
= 2 × 2 × 75
= 2 × 2 × 3 × 25
= 2 × 2 × 3 × 5 × 5
Therefore 300 = 2 × 2 × 3 × 5 × 5

Question 3.
Write the following numbers as the product of three prime numbers, and find all the factors of them.
(i) 174
(ii) 385
(iii) 182
Answer:
(i) 174 is the product of three prime numbers.
174 = 2 × 3 × 29
Factors of 174 are:
1
2, 3, 29
2 × 3 = 6
2 × 29 = 39
3 × 29 = 78
Therefore, the factors are: 1, 2, 3, 6, 29, 58, 87, 174

(ii) 385 is the product of three prime numbers.
385 = 5 × 7 × 11
Factors of 385 are:
1
5, 7, 11
5 × 7 = 35
5 × 11 = 55
7 × 11 = 77
Therefore, the factors are: 1, 5, 7, 11, 35, 55, 77, 88, and 385.

(iii) 182 as the product of three prime numbers.
182 = 2 × 7 × 13
Factors of 182 are:
1
2, 7, 13
2 × 7 = 14
2 × 13 = 26
7 × 13 = 91
Therefore, the factors are: 1, 2, 7, 13, 14, 26, 91, and 182.

Class 6 Maths Chapter 6 Notes Kerala Syllabus Multiples and Factors

→ The multiples of a natural number are the product of that number with the natural numbers 1, 2, 3,…

→ All multiples of the multiple of a number are also multiples of that number.

→ All multiples of a number are also multiples of any of its factors.

→ A natural number greater than 1, which has no factors other than 1 and itself, is called a prime number.

→ Any composite number can be written as a product of primes.

→ The only even number among the prime numbers is 2.

Multiples and factors are fundamental concepts in mathematics that help us understand the relationships between numbers. A multiple of a number is the result of multiplying that number by any natural number, while a factor is a number that divides another number exactly, without leaving a remainder.

For example, 4 is a multiple of 2, and 2 is a factor of 4. Learning about multiples and factors is essential for solving problems involving divisibility, simplifying fractions, finding the greatest common factor (GCF), the least common multiple (LCM), and much more. In this chapter, we discuss multiples of multiples, primary factors, and prime numbers.

Multiples of Multiples
The multiples of a natural number are the product of that number with the natural numbers 1, 2, 3,…
For example:
The multiples of 2 are the numbers 2, 4, 6,… obtained by multiplying the natural number by 2.
The multiples of 4 are the numbers 4, 8, 12,… obtained by multiplying the natural number by 4.
Here, all the multiples of 4 can be written as multiples of 2 also:
1 × 4 = 4
2 × 4 = 8
3 × 4 = 12
4 × 4 = 16
5 × 4 = 20
……………

2 × 2 = 4
4 × 2 = 8
6 × 2 = 12
8 × 2 = 16
10 × 2 = 20
……………….

Similarly, if 6 is a multiple of 2 and 3.
So all multiples of 6 can be written as multiples of 2 and 3.

1 × 6 = 6
2 × 6 = 12
3 × 6 = 18
4 × 6 = 24
5 × 6 = 30
………………..

3 × 2 = 6
6 × 2 = 12
9 × 2 = 18
12 × 2 = 24
15 × 2 = 30
………………….

2 × 3 = 6
4 × 3 = 12
6 × 3 = 18
8 × 3 = 24
10 × 3 = 30
………………..

In general, all multiples of the multiple of a number are also multiples of that number.

The multiples of 15 can be written as the multiples of what numbers?
Answer:
15 is a multiple of 3 and 5.
So the multiples of 10 can be written as the multiples of 2 and 5.

We have seen that multiples can also be put in terms of factors.
For example:
4 is the multiple of 2 can also be written as 2 is a factor of 4.
6 is the multiple of 2 and 3 can also be written as 2 and 3 are two factors of 6.
12 is a multiple of 3, and 4 can also be written as 3 and 4 are factors of 12.

In general, we can say that all multiples of a number are also multiples of any of its factors.

14 is a multiple of 2 and 7. Express it in the form of factors.
Answer:
14 is a multiple of 2 and 7.
Since, 2 × 7 = 14
So 2 and 7 are two factors of 14.

70 is a multiple of 2, 5, and 7. Express it in the form of factors.
Answer:
70 is a multiple of 2, 5, and 7.
Since, 2 × 5 × 7 = 70
So 2, 5, and 7 are the factors of 70.

If 2, 3, and 7 are factors of a number. Then what is that number?
Answer:
2, 3, and 7 are factors of a number.
That means 2 × 3 × 7 = 42
Therefore, the number is 42, and multiples of 42 are 2, 3, and 7.

Primary Factors
Any number can be written as the product of its factors in different ways.
For example, consider the number 70,
1 × 70 = 70
2 × 35 = 70
5 × 14 = 70
10 × 7 = 70
We can write 70 as a product of three factors, without using 1:
That is 70 = 2 × 5 × 7
The only factors of each of the numbers 2, 5, and 7 are 1 and the number itself.
For any number 1, the number itself is are factor.

The numbers below 20 with factors 1 and itself are: 1, 2, 3, 5, 7, 11, 13, 17, 19.
Such numbers, excluding 1, are said to be prime numbers.

A natural number greater than 1, which has no factors other than 1 and itself, is called a prime number.

Numbers greater than 1, which are not primes, are called composite numbers.
For example, 4 is a composite number.
Since 4 = 2 × 2

A composite number can be written as the product of a prime number.

When a number is written as the product of two factors and any one of them is not a prime, then that factor can be written as the product of two factors. This can continue till all factors are prime.

Write 48 as the product of primes.
Answer:
48 = 2 × 24
= 2 × 2 × 12
= 2 × 2 × 2 × 6
= 2 × 2 × 2 × 2 × 3
Therefore 48 = 2 × 2 × 2 × 2 × 3

Write the following as the product of primes.
(i) 30
(ii) 45
(iii) 64
Answer:
(i) 30 = 2 × 15 = 2 × 3 × 5
Therefore 30 = 2 × 3 × 5

(ii) 45 = 5 × 9 = 5 × 3 × 3
Therefore 45 = 5 × 3 × 3

(iii) 64 = 2 × 32
= 2 × 2 × 16
= 2 × 2 × 2 × 8
= 2 × 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2 × 2
Therefore 64 = 2 × 2 × 2 × 2 × 2 × 2

Product of Primes:
Once we write two numbers as a product of primes, it is easy to write the product of these numbers also as a product of primes.
For example: 12 × 24 can be split like this:
12 = 2 × 2 × 3
24 = 2 × 2 × 2 × 3
Split 12 × 24 as shown below:
288 = 12 × 24
= (2 × 2 × 3) × (2 × 2 × 2 × 3)
= 2 × 2 × 3 × 2 × 2 × 2 × 3

To split a number into a product of primes, we first split it into the product of any two factors, then split each of these factors into a product of primes, and finally put these prime factors together.
For example: Split 140 into a product of primes,
140 = 14 × 10
Next, write 14 and 10 as products of primes
14 = 2 × 7
10 = 2 × 5
We can write 140 like this;
140 = 14 × 10
= (2 × 7) × (2 × 5)
= 2 × 2 × 5 × 7

Split the following numbers into a product of primes.
(i) 420
(ii) 180
(iii) 336
Answer:
(i) 420 = 15 × 28
15 = 3 × 5
28 = 2 × 2 × 7
Therefore, 420 = 2 × 2 × 3 × 5 × 7

(ii) 180 = 12 × 15
12 = 2 × 2 × 3
15 = 3 × 5
Therefore 180 = 2 × 2 × 3 × 3 × 5
(iii) 336 = 14 × 24
14 = 2 × 7
24 = 2 × 2 × 2 × 3
Therefore, 336 = 2 × 2 × 2 × 2 × 3 × 7

All Factors
If we know the prime factors of a number, we can find all its factors.
For example, the factors of 6 are: 1, 2, 3, and 6.

Write the prime factors of 15. And what are its other factors?
Answer:
Prime factors of 15 are: 3 and 5
Factors of 15 are: 1, 3, 5, and 15

Find all the factors of the numbers given below:
(i) 42
(ii) 54
(iii) 63
Answer:
(i) 42 = 1 × 42
= 2 × 21
= 3 × 14
= 6 × 7
Therefore, factors of 42 are: 1, 2, 3, 6, 7, 14, 21, and 42.

(ii) 54 = 1 × 54
= 2 × 27
= 3 × 18
= 6 × 9
Therefore, factors of 54 are: 1, 2, 3, 6, 9, 18, 27, and 54.

(iii) 63 = 1 × 63
= 3 × 21
= 7 × 9
Therefore, factors of 63 are: 1, 3, 7, 9, 21, and 63.

Product of Three Prime Numbers:
Now let’s look at the product of three different primes.

Write 42 as the product of three prime numbers and find all the factors?
Answer:
42 as the product of three prime numbers,
42 = 2 × 3 × 7
Factors of 42 are:
1
2, 3, 7
2 × 3 = 6
2 × 7 = 14
3 × 7 = 21
Therefore, the factors are: 1, 2, 3, 6, 7, 14, 21, and 42

Write the following numbers as the product of three prime numbers, and find all the factors of it?
(i) 102
(ii) 154
(iii) 195
Answer:
(i) 102 is the product of three prime numbers.
102 = 2 × 3 × 17
Factors of 102 are:
1
2, 3, 17
2 × 3 = 6
2 × 17 = 34
3 × 17 = 51
Therefore, the factors are: 1, 2, 3, 6, 17, 34, 51, and 102

(ii) 154 is the product of three prime numbers.
154 = 2 × 7 × 11
Factors of 154 are:
1
2, 7, 11
2 × 7 = 14
2 × 11 = 22
7 × 11 = 77
Therefore, the factors are: 1, 2, 7, 11, 14, 22, 77, and 154.

(iii) 195 as the product of three prime numbers.
195 = 3 × 5 × 13
Factors of 195 are:
1
3, 5, 13
1 × 5 = 15
3 × 13 = 39
5 × 13 = 65
Therefore, the factors are: 1, 3, 5, 13, 15, 39, 65, and 195

Prime Numbers
The only even number among the prime numbers, 2, 3, 5, 7, 11,… is 2.
All primes afterwards are odd numbers. But not all odd numbers are primes;
For example: 9 = 3 × 3, 15 = 3 × 5,……. They are not prime numbers.

There is no definite pattern for the odd primes.
For example, after 3, 5, 7 are consecutive primes that differ by 2, the next prime is not 9 (which is not a prime), but 11. Thus, the difference between 7 and 11 is 4. Similarly, after the prime 31, the next prime is 37, and their difference is 6; the prime after 89 is 97, with a difference of 8. But even as such consecutive primes drift further apart, there are consecutive primes like 41 and 43 or 71 and 73 in between, which are only 2 apart. There is a technique to list all primes less than a specified number. That is, first write all numbers up to 50 in rows and columns like this:

Strike off 1 from this. Then strike off all multiples of 2, except 2:

Keep 3 and strike off all multiples of 3:

Strike all multiples of 5, except 5 itself.
If we remove the multiples of 7 other than itself also, we can see that there are no multiples, except themselves, of the other numbers that remain:

Now the numbers not struck off are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.
These are the prime numbers less than 50.

Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 4 Arithmetic of Parts Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 6 Maths Chapter 4 Solutions Arithmetic of Parts

Class 6 Kerala Syllabus Maths Solutions Chapter 4 Arithmetic of Parts Questions and Answers

Arithmetic of Parts Class 6 Questions and Answers Kerala Syllabus

Intext Questions (Page No. 50)

In each of the pictures below, write the parts of each colour and the total coloured parts as fractions. Write the sum of the fractions obtained from each picture in the lowest terms.

Question 1.

Answer:
The fraction of circle coloured in blue = \(\frac {3}{8}\)
The fraction of circle coloured in green = \(\frac {1}{8}\)
The circle is divided into 8 equal parts.
Sum of fractions is \(\frac{1}{8}+\frac{3}{8}=\frac{4}{8}=\frac{1}{2}\)

Question 2.

Answer:
The fraction of circle coloured in orange = \(\frac {3}{8}\)
The fraction of circle coloured in brown = \(\frac {3}{8}\)
The circle is divided into 8 equal parts.
Sum of fractions is \(\frac{3}{8}+\frac{3}{8}=\frac{6}{8}=\frac{3}{4}\)

Question 3.

Answer:
The fraction coloured in yellow = \(\frac {1}{10}\)
The fraction coloured in red = \(\frac {3}{10}\)
The ribbon is divided into 10 equal parts.
Sum of fractions is \(\frac{1}{10}+\frac{3}{10}=\frac{4}{10}=\frac{2}{5}\)

Question 4.

Answer:
The fraction coloured in yellow = \(\frac {5}{16}\)
The fraction coloured in green = \(\frac {7}{16}\)
The ribbon is divided into 10 equal parts.
Sum of fractions is \(\frac{5}{16}+\frac{7}{16}=\frac{12}{16}=\frac{3}{4}\)

Addition of Fractions (Page No. 56)

Question 1.
In each pair of pictures below, find the fraction of the circle we get by cutting up the coloured pieces of both circles and putting them together:

Answer:

Question 2.
Calculate the sums given below:
(i) \(\frac{1}{4}+\frac{1}{8}\)
(ii) \(\frac{3}{4}+\frac{1}{6}\)
(iii) \(\frac{1}{3}+\frac{2}{5}\)
(iv) \(\frac{1}{2}+\frac{2}{5}\)
(v) \(\frac{2}{3}+\frac{1}{5}\)
Answer:

Question 3.
There are two taps to fill a tank with water. If the first tap alone is opened, the tank would fill up in 10 minutes. If the second tap alone is opened, it would take 15 minutes to fill up the tank.
(i) If the first tap alone is opened, what fraction of the tank would be filled in one minute?
(ii) If the second tap alone is opened, what fraction of the tank would be filled in one minute?
(iii) If both the taps are opened, what fraction of the tank would be filled in one minute?
(iv) If both the taps are opened, how much time would it take for the tank to be filled up?
Answer:
(i) The first tap fills the whole tank in 10 minutes.
Therefore, in 1 minute it fills upto \(\frac {1}{10}\) of the tank.

(ii) The second tap fills the whole tank in 15 minutes.
Therefore, in 1 minute it fills upto \(\frac {1}{15}\) of the tank.

(iii) If both taps are opened,
First tap fills \(\frac {1}{10}\)
Second tap fills \(\frac {1}{15}\)
So together they fill = \(\frac{1}{10}+\frac{1}{15}=\frac{1 \times 3}{10 \times 3}+\frac{1 \times 2}{15 \times 2}\) = \(\frac{3}{30}+\frac{2}{30}=\frac{5}{30}=\frac{1}{6}\)
That means both taps together fill \(\frac {1}{6}\) of the tank in 1 minute.

(iv) The time taken to fill the tank when both taps are opened.
\(1 \div \frac{1}{6}=\frac{1}{\frac{1}{6}}=1 \times \frac{6}{1}\) = 6 minutes

Some Other Sums (Page No. 57)

Question 1.
For each fraction given below, can you mentally calculate the fraction to be added to make it 1?
(i) \(\frac {2}{7}\)
(ii) \(\frac {4}{7}\)
(iii) \(\frac {3}{8}\)
(iv) \(\frac {3}{10}\)
Answer:
(i) \(\frac {5}{7}\) is added with \(\frac {2}{7}\) to get 1.
That is, \(\frac{2}{7}+\frac{5}{7}=\frac{7}{7}\) = 1

(ii) \(\frac {3}{7}\) is added with \(\frac {4}{7}\) to get 1.
That is, \(\frac{3}{7}+\frac{4}{7}=\frac{7}{7}\) = 1

(iii) \(\frac {5}{8}\) is added with \(\frac {3}{8}\) to get 1.
That is, \(\frac{3}{8}+\frac{5}{8}=\frac{8}{8}\) = 1

(iv) \(\frac {7}{10}\) is added with \(\frac {3}{10}\) to get 1.
That is, \(\frac{7}{10}+\frac{3}{10}=\frac{10}{10}\) = 1

Textbook Page No. 60

Question 1.
Calculate the sums below:
(i) \(\frac{5}{6}+\frac{1}{3}\)
(ii) \(\frac{7}{8}+\frac{1}{4}\)
(iii) \(\frac{5}{6}+\frac{1}{4}\)
(iv) \(\frac{5}{8}+\frac{3}{4}\)
(v) \(2 \frac{1}{3}+3 \frac{1}{2}\)
Answer:

Question 2.
One jar contains one and a half litres of milk, and another contains two and three-quarters litres of milk. How much milk is in both jars together?
Answer:
Milk contained in first jar = 1\(\frac {1}{2}\) litres
Second jar contained in second jar = 2\(\frac {3}{4}\) litres
Milk contained in both jars is,

Question 3.
Two strings of lengths one and a half metres are joined end to end. What is the total length?
Answer:
Lengths of strings = 1\(\frac {1}{2}\)
Two strings are joined together,
\(1 \frac{1}{2}+1 \frac{1}{2}=(1+1)+\left(\frac{1}{2}+\frac{1}{2}\right)\)
= 2 + \(\frac {2}{2}\)
= 2 + 1
= 3 metres

Question 4.
Ahirath bought one and a half kilograms of beans and three-quarters kilograms of yams. What is the total weight?
Answer:
Quantity of beans Ahirath bought = 1\(\frac {1}{2}\) kg
Quantity of yam Ahirath bought = \(\frac {3}{4}\) kg

Removing Parts (Page No. 61)

Question 1.
(i) \(\frac{1}{2}-\frac{1}{8}\)
Answer:

(ii) \(\frac{3}{4}-\frac{1}{8}\)
Answer:

(iii) \(\frac{1}{3}-\frac{1}{5}\)
Answer:

(iv) \(\frac{2}{5}-\frac{1}{3}\)
Answer:

(v) \(\frac{2}{3}-\frac{1}{5}\)
Answer:

Textbook Page No. 64

Question 1.
Natasha drew a circle and coloured \(\frac {5}{12}\) of it. What fraction of the circle remains to be coloured?
Answer:
Coloured portion of the circle = \(\frac {5}{12}\)
Portion of the circle to be coloured is,
\(1-\frac{5}{12}=\frac{12}{12}-\frac{5}{12}\)
= \(\frac{12-5}{12}\)
= \(\frac {7}{12}\)

Question 2.
A bucket can hold 10 litres of water, and it contains 3\(\frac {3}{4}\) litres. How much more is needed to fill it?
Answer:
Total quantity of water that can be held in the bucket = 10 litres
Quantity of water contained in the bucket = 3\(\frac {3}{4}\) litres
The quantity of water needed to fill the bucket is,
10 – 3\(\frac {3}{4}\) = \(\left(9 \frac{1}{4}+\frac{3}{4}\right)-\left(3 \frac{3}{4}\right)\)
= \(\left(9+\frac{1}{4}+\frac{3}{4}\right)-\left(3+\frac{3}{4}\right)\)
= (9 – 3) + \(\left(\frac{1}{4}+\frac{3}{4}-\frac{3}{4}\right)\)
= 6 + \(\frac {1}{4}\)
= 6\(\frac {1}{4}\) litres
OR
If \(\frac {1}{4}\) added to 3\(\frac {3}{4}\), we get 4
And if 6 is added to 4 will results 10
That means, 6 + \(\frac {1}{4}\) = 6\(\frac {1}{4}\) litres

Question 3.
From a string, one and three-quarters of a metre long, a piece half a metre long is cut off. What is the length of the remaining piece?
Answer:
Length of the original string = 1\(\frac {3}{4}\) metre
Piece cut off = \(\frac {1}{2}\) metre
Length of the remaining piece is

Question 4.
A panchayat constructed a new road, 14\(\frac {3}{4}\) kilometres long last year. This year, a road 16\(\frac {1}{4}\) kilometres long was constructed. How much more was constructed this year?
Answer:
Road constructed last year = 14\(\frac {3}{4}\) km
Road constructed this year = 16\(\frac {1}{4}\) km
The more roads constructed this year are,

Question 5.
Ashadev bought 20 metres of string. He cut off a piece 5\(\frac {3}{4}\) metres long first, and then a piece 6\(\frac {1}{2}\) metres long later. What is the length of the string left?
Answer:
Total length of the string = 20 metres
First cut piece = 5\(\frac {3}{4}\) metre
Second cut piece = 6\(\frac {1}{2}\) metres
Total length of the string cut is,

Length of the remaining string is,

Question 6.
The milk society got 75\(\frac {1}{4}\) litres in the morning and 55\(\frac {1}{2}\) litres in the evening. Of this 85\(\frac {3}{4}\) litres of milk is sold. How much milk is left?
Answer:
Milk received in the morning = 75\(\frac {1}{4}\) litres
Milk received in the evening = 55\(\frac {1}{2}\) litres
Milk sold is = 85\(\frac {3}{4}\) litres
Remaining milk is,

Large and Small (Page No. 66)

Question 1.
Find the larger and smaller of each pair of fractions below and write this using the < or > symbol:
(i) \(\frac{2}{5}, \frac{3}{5}\)
(ii) \(\frac{2}{5}, \frac{2}{3}\)
(iii) \(\frac{2}{5}, \frac{3}{4}\)
(iv) \(\frac{3}{7}, \frac{2}{9}\)
(v) \(\frac{2}{7}, \frac{3}{8}\)
(vi) \(\frac{4}{9}, \frac{3}{8}\)
Answer:
(i) Here, the denominators are the same.
So compare the numerators.
That is 2 < 3,
Therefore \(\frac{2}{5}<\frac{3}{5}\)

(ii) Here, the numerators are the same.
So the smaller denominator is larger.
Therefore \(\frac{2}{3}>\frac{2}{5}\)

(iii) Convert the fractions into forms with the same denominators
\(\frac{2}{5}=\frac{2 \times 4}{5 \times 4}=\frac{8}{20}\) and \(\frac{3}{4}=\frac{3 \times 5}{4 \times 5}=\frac{15}{20}\)
Therefore \(\frac{2}{5}<\frac{3}{4}\)

(iv) Convert the fractions into forms with the same denominators
\(\frac{3}{7}=\frac{3 \times 9}{7 \times 9}=\frac{27}{63}\) and \(\frac{2}{9}=\frac{2 \times 7}{9 \times 7}=\frac{14}{63}\)
Therefore \(\frac{3}{7}>\frac{2}{9}\)

(v) Convert the fractions into forms with the same denominators
\(\frac{2}{7}=\frac{2 \times 8}{7 \times 8}=\frac{16}{56}\) and \(\frac{3}{8}=\frac{3 \times 7}{8 \times 7}=\frac{21}{56}\)
Therefore \(\frac{2}{7}<\frac{3}{8}\)

(vi) Convert the fractions into forms with the same denominators
\(\frac{4}{9}=\frac{4 \times 8}{9 \times 8}=\frac{32}{72}\) and \(\frac{3}{8}=\frac{3 \times 9}{8 \times 9}=\frac{27}{72}\)
Therefore \(\frac{4}{9}>\frac{3}{8}\)

Question 2.
Arrange each triple of fractions below from the smallest to the largest and write it using the < symbol:
(i) \(\frac{2}{5}, \frac{3}{4}, \frac{3}{5}\)
(ii) \(\frac{3}{7}, \frac{2}{9}, \frac{2}{7}\)
(iii) \(\frac{1}{2}, \frac{1}{3}, \frac{2}{3}\)
Answer:
(i) Convert the fractions into forms with the same denominators
\(\frac{2}{5}=\frac{2 \times 4}{5 \times 4}=\frac{8}{20}, \quad \frac{3}{4}=\frac{3 \times 5}{4 \times 5}=\frac{15}{20}, \quad \frac{3}{5}=\frac{3 \times 4}{5 \times 4}=\frac{12}{20}\)
\(\frac{8}{20}<\frac{12}{20}<\frac{15}{20}\)
Therefore \(\frac{2}{5}<\frac{3}{5}<\frac{3}{4}\)

(ii) Convert the fractions into forms with the same denominators
\(\frac{3}{7}=\frac{3 \times 9}{7 \times 9}=\frac{27}{63}, \quad \frac{2}{9}=\frac{2 \times 7}{9 \times 7}=\frac{14}{63}, \frac{2}{7}=\frac{2 \times 9}{7 \times 9}=\frac{18}{63}\)
\(\frac{14}{63}<\frac{18}{63}<\frac{27}{63}\)
Therefore \(\frac{2}{9}<\frac{2}{7}<\frac{3}{7}\)

(iii) Convert the fractions into forms with the same denominators
\(\frac{1}{2}=\frac{1 \times 3}{2 \times 3}=\frac{3}{6}, \quad \frac{1}{3}=\frac{1 \times 2}{3 \times 2}=\frac{2}{6} \quad, \frac{2}{3}=\frac{2 \times 2}{3 \times 2}=\frac{4}{6}\)
\(\frac{2}{6}<\frac{3}{6}<\frac{4}{6}\)
Therefore \(\frac{1}{3}<\frac{1}{2}<\frac{2}{3}\)

Class 6 Maths Chapter 4 Kerala Syllabus Arithmetic of Parts Questions and Answers

Class 6 Maths Arithmetic of Parts Questions and Answers

Question 1.
Compare \(\frac {5}{8}\) and \(\frac {3}{4}\)?
Answer:
Convert the fractions into forms with the same denominators
\(\frac{5}{8}<\frac{6}{8}\)
Therefore \(\frac{5}{8}<\frac{3}{4}\)

Question 2.
Arrange from smallest to largest: \(\frac{1}{6}, \frac{1}{2}, \frac{5}{12}\)
Answer:
Convert the fractions into forms with the same denominators
\(\frac{1}{6}=\frac{1 \times 2}{6 \times 2}=\frac{2}{12}, \quad \frac{1}{2}=\frac{1 \times 6}{2 \times 6}=\frac{6}{12} \quad, \frac{5}{12}\)
\(\frac{2}{12}<\frac{5}{12}<\frac{6}{12}\)
Therefore \(\frac{1}{6}<\frac{5}{12}<\frac{1}{2}\)

Question 3.
Calculate: \(\frac{3}{5}+\frac{1}{2}\)
Answer:

Question 4.
A bottle contains 2\(\frac {1}{4}\) litres of juice, and another bottle contains 1\(\frac {2}{3}\) litres. What is the total amount of juice?
Answer:
Juice in first bottle = 2\(\frac {1}{4}\) litres
Juice in second bottle = 1\(\frac {2}{3}\) litres
Total amount of the juice is;

Question 5.
A ribbon is 2\(\frac {1}{2}\) metres long. Another ribbon is 3\(\frac {1}{3}\) metres long. What is their total length?
Answer:
First ribbon = 2\(\frac {1}{2}\) metres
Second ribbon = 3\(\frac {1}{3}\) metres
Total length is:

Question 6.
Reshma had 12\(\frac {1}{2}\) metres of ribbon. She used 4\(\frac {3}{4}\) metres for a project. How much ribbon is left?
Answer:
Total ribbon = 12\(\frac {1}{2}\) metres
Ribbon used = 4\(\frac {3}{4}\) metres
The remaining ribbon is,

Question 7.
A container holds 18 litres of oil. 6\(\frac {2}{3}\) litres are used in the morning and 5\(\frac {1}{6}\) litres in the evening. How much oil is left?
Answer:
Total quantity of oil held in the container = 18 litres
Oil used is,

The quantity of oil remaining is,

Question 8.
A pipe is 9\(\frac {1}{2}\) metres long. A piece 3\(\frac {3}{4}\) metres is cut from it. How much pipe remains?
Answer:
Length of the original pipe = 9\(\frac {1}{2}\) metres
Length of the cut piece = 3\(\frac {3}{4}\) metres
Length of the remaining pipe is,

Question 9.
A recipe needs 1\(\frac {1}{4}\) cups of sugar. Aliya only has \(\frac {3}{4}\) cup. How much more does she need?
Answer:
Required sugar = 1\(\frac {1}{4}\) cups
Available sugar = \(\frac {3}{4}\) cups
Sugar needed is,

Question 10.
Meera walked 2\(\frac {2}{5}\) kilometre in the morning and 3\(\frac {3}{5}\) kilometre in the evening. How much did she walk in total?
Answer:
Meera walked in the morning = 2\(\frac {2}{5}\) km
Walked in the evening = 3\(\frac {3}{5}\) km
Total she walked:

Class 6 Maths Chapter 4 Notes Kerala Syllabus Arithmetic of Parts

→ To find the sum of two fractional measures put together, we must see them as a set of the same equal pieces.

→ Convert both fractional measures to forms with the same denominator.

→ Increasing the numerator alone makes a fraction larger.

→ Of two fractions with the same denominator, the one with the larger numerator is the larger and the one with the smaller numerator is the smaller.

→ If the denominator alone of a fraction is increased, the fraction becomes smaller.

→ Of two fractions with the same numerator, the one with the larger denominator is smaller and the one with the smaller denominator is larger.

→ To compare two fractions with the same denominator, we need only compare the numerator; to compare two fractions with the same numerator, we need only compare the denominators.

In our everyday life, we often come across situations where we need to work with parts of a whole, like dividing a chocolate bar between friends, measuring ingredients for a recipe, or cutting ribbon into equal pieces. This idea of working with parts is what we study in this chapter, Arithmetic of Parts. This chapter mainly deals with understanding fractions – numbers that represent parts of a whole, dividing object quantities into equal parts, performing operations such as addition, subtraction, multiplication, and division with fractions, comparing and converting fractions, especially simplifying them to their lowest terms. This chapter helps build a strong foundation in dealing with fractions and their applications in real-life situations.

Joining Parts
A circle is divided into 4 equal parts, and two of them are joined together to get half of the circle.

The quarter circle joined to a quarter circle makes half a circle.
That means, two quarters make a half.
We can write like this: \(\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\)

Suppose a circle is divided into eight equal parts ,and two of them are joined together.
2 of 8 equal parts is \(\frac {2}{8}\); also, \(\frac{2}{8}=\frac{1 \times 2}{4 \times 2}=\frac{1}{4}\)
So we have \(\frac{1}{8}+\frac{1}{8}=\frac{2}{8}=\frac{1}{4}\)

Question 1.
Join \(\frac {1}{8}\) of a circle and \(\frac {3}{8}\) of the circle, what fraction of the circle would we get?
Answer:
We took 1 + 3 = 4 parts of 8 equal parts, that is \(\frac {4}{8}\)
Reducing the numerator and denominator,
\(\frac{1}{8}+\frac{3}{8}=\frac{4}{8}=\frac{1}{2}\)

Question 2.
Look at the picture given below;

What sum of fractions of the ribbon do we get from this?
Answer:
The ribbon is divided into 8 equal parts
The fraction of the ribbon coloured in red = \(\frac {1}{8}\)
The fraction of the ribbon coloured in green = \(\frac {5}{8}\)
Sum of the fraction is, \(\frac{1}{8}+\frac{5}{8}=\frac{6}{8}=\frac{3}{4}\)

Addition of Fractions
If a circle is cut into four equal pieces and two of them joined together, we get half a circle:

If one more piece is joined to this, we get three-quarters of a circle.

That is half and a quarter make three quarters, \(\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)

Question 1.
What sum do we get from this picture below?

Answer:
Here, the circle is divided into 8 equal pieces
The pieces coloured in red = 1
The pieces coloured in yellow = 4
Fraction of the coloured portion = \(\frac{1}{8}+\frac{4}{8}=\frac{5}{8}\)
The lowest form of \(\frac{4}{8}=\frac{1}{2}\)
Therefore the sum is \(\frac{1}{8}+\frac{1}{2}=\frac{5}{8}\)

Question 2.
Look at the picture below.

\(\frac {1}{4}\) of a circle and \(\frac {3}{8}\) of another circle of the same size are cut out, and these pieces are put together. What fraction of the full circle is this?
Answer:
In the first figure, the circle is divided into 4 equal parts, and 1 is coloured.
Here one one-fourth of the circle can be considered as two one-eighths joined together.
That is \(\frac{1}{4}=\frac{2}{8}\)
In the second figure, the circle is divided into 8 equal parts, and 3 is coloured.
That is \(\frac {3}{8}\)

Therefore, the fraction of the full circle is \(\frac{1}{4}+\frac{3}{8}=\frac{2}{8}+\frac{3}{8}=\frac{5}{8}\)

Question 3.
Two ribbons of length \(\frac {3}{10}\) metre and \(\frac {2}{5}\) metre are joined end to end. What is the total length?
Answer:
The first ribbon is 3 of 10 equal parts of a 1 metre long ribbon:

The second ribbon is 2 of 5 equal parts of a 1 metre long ribbon:

Considering \(\frac {2}{5}\) metre as 4 of 10 equal parts of a metre:

The sum of fraction is \(\frac{3}{10}+\frac{2}{5}=\frac{3}{10}+\frac{4}{10}=\frac{7}{10}\)
Therefore, the total length is \(\frac {7}{10}\) metre.

To find the sum of two fractional measures put together, we must see them as a set of the same equal pieces.
Convert both fractional measures to forms with the same denominator.

Question 4.
Two ribbons of length \(\frac {1}{2}\) metre and \(\frac {2}{5}\) metre are joined end to end. What is the total length?
Answer:
For the same denominators we want for both must be a multiple of 2 and 5.
That is \(\frac{1}{2}=\frac{1 \times 5}{2 \times 5}=\frac{5}{10}\) and \(\frac{2}{5}=\frac{2 \times 2}{5 \times 2}=\frac{4}{10}\)
Total length of the ribbon is \(\frac{1}{2}+\frac{2}{5}=\frac{5}{10}+\frac{4}{10}=\frac{9}{10}\)

Some Other Sums
Of the two fractions added, one is some parts of one divided into equal parts taken together, and the other is the remaining parts taken together. When both these are taken together, we get all the parts or the whole.
\(\frac{1}{2}+\frac{1}{2}\) = 1
\(\frac{1}{3}+\frac{2}{3}\) = 1
\(\frac{1}{4}+\frac{3}{4}\) = 1
\(\frac{1}{5}+\frac{4}{5}\) = 1
\(\frac{2}{5}+\frac{3}{5}\) = 1

Question 1.
What fraction to be added to get 1?
(i) \(\frac {3}{7}\)
(ii) \(\frac {5}{6}\)
(iii) \(\frac {5}{9}\)
Answer:
(i) \(\frac{3}{7}+\frac{4}{7}=\frac{7}{7}\) = 1
(ii) \(\frac{5}{6}+\frac{1}{6}=\frac{6}{6}\) = 1
(iii) \(\frac{5}{9}+\frac{4}{9}=\frac{9}{9}\) = 1

Question 2.
Draw two circles of the same size and colour half of one and two-thirds of the other. If we cut out these pieces, can we put them together?

What if we cut them like this?

Answer:
Figure 1
The first circle is divided into 2 equal parts, and 1 is considered. In the second circle, it is divided into 3 equal parts, and 2 is considered from it.
If we join these two circles together, we get,

Figure 2
If we join these two circles together, we get,

Removing Parts
Here we can do this subtraction just as we did the addition of fractions:
If a half metre piece is cut off from a three-quarter metre long ribbon, then the length of the remaining portion is;

Answer:
Length of the ribbon is 1 metre.
Here, a half metre piece is cut off from a three-quarter metre long ribbon,

If one-third of a metre is removed from half a metre;
Answer:

Therefore, the remaining portion is \(\frac {1}{6}\) metres.

We have seen some pairs of fractions that add up to 1.
We can rewrite them as subtractions:

Question 3.
From one liter of milk, a quarter liter was used up. How much milk is left?
Answer:
The remaining portion of milk is,
\(1-\frac{1}{4}=\frac{4}{4}-\frac{1}{4}\)
= \(\frac{4-1}{4}\)
= \(\frac {3}{4}\) litres

Question 4.
From two and a half kilograms of yams, one and a quarter kilograms are cut off. What is the weight of the remaining piece?
Answer:
Weight of yam = 2\(\frac {1}{2}\)
The weight cut off from the yam = 1\(\frac {1}{4}\)
Weight of the remaining piece

Large and Small
Case 1: To find the largest fraction with the same denominator,
Increasing the numerator alone makes a fraction larger.
For example:
\(\frac{1}{7}<\frac{2}{7}<\frac{3}{7}<\frac{4}{7}<\frac{5}{7}<\frac{6}{7}\)

Which is larger, \(\frac {2}{5}\) or \(\frac {3}{5}\)?
Answer:
\(\frac {2}{5}\) is 2 parts of 5 equal parts taken together, and \(\frac {3}{5}\) is 3 parts of 5 equal parts taken together.

Therefore \(\frac {2}{5}\) is less than the number \(\frac {3}{5}\).
We can write like this:
\(\frac {2}{5}\) < \(\frac {3}{5}\) Or \(\frac {3}{5}\) is greater than \(\frac {2}{5}\).
That is, \(\frac {3}{5}\) > \(\frac {2}{5}\)

In general, we can say “Of two fractions with the same denominator, the one with a larger numerator is the larger and the one with a smaller numerator is the smaller.”

Case 2
If the denominator of a fraction is increased, the fraction becomes smaller.
For example:
\(\frac{7}{2}>\frac{7}{3}>\frac{7}{4}>\frac{7}{5}>\frac{7}{6}\)

Which is larger \(\frac {3}{4}\) and \(\frac {3}{5}\)?
Answer:
4 equal parts are larger than each of 5 equal parts.

3 of the first kind of pieces taken together is larger than 3 of the second kind of pieces.

This means \(\frac {3}{4}\) > \(\frac {3}{5}\)
In general, of two fractions with the same numerator, the one with the larger denominator is smaller and the one with the smaller denominator is larger.

In conclusion,
To compare two fractions with the same denominator, we need only compare the numerators
To compare two fractions with the same numerator, we need only compare the denominators.
If both the numerator and denominator are different, convert the fractions into forms with the same denominators.

Which of \(\frac {1}{2}\) and \(\frac {3}{5}\) are larger?
Answer:
\(\frac{1 \times 5}{2 \times 5}=\frac{5}{10}\) and \(\frac{3 \times 2}{5 \times 2}=\frac{6}{10}\)
That is, \(\frac{5}{10}<\frac{6}{10}\)
Therefore \(\frac{1}{2}<\frac{3}{5}\)

Which is larger, \(\frac {3}{5}\) or \(\frac {7}{5}\)?
Answer:
\(\frac {3}{5}\) < \(\frac {7}{5}\)

Which is larger, \(\frac {4}{5}\) or \(\frac {4}{6}\)?
Answer:
\(\frac {4}{5}\) > \(\frac {4}{6}\)

Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 3 Volume Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 6 Maths Chapter 3 Solutions Volume

Class 6 Kerala Syllabus Maths Solutions Chapter 3 Volume Questions and Answers

Volume Class 6 Questions and Answers Kerala Syllabus

Size as Number (Page No. 37)

Question 1.
All blocks shown below are made up of cubes of side 1 centimetre. Calculate the violume of each.


Answer:
Figure 1
Length = 4 small cube = 4 cm
Width = 4 small cube = 4 cm
Height = 4 small cube = 4 cm
Therefore volume = 4 × 4 × 4 = 64 cubic cm

Figure 2
Length = 2 small cube = 2 cm
Width = 2 small cube = 2 cm
Height = 3 small cube = 3 cm
Therefore volume = 2 × 2 × 3 = 12 cubic cm

Figure 3
Length = 4 small cube = 4 cm
Width = 3 small cube = 3 cm
Height = 3 small cube = 3 cm
Therefore volume = 4 × 3 × 3 = 36 cubic cm

Figure 4
Length = 7 small cube = 7 cm
Width = 2 small cube = 2 cm
Height = 3 small cube = 3 cm
Therefore volume = 7 × 2 × 3 = 42 cubic cm

Figure 5
Length = 3 small cube = 3 cm
Width = 2 small cube = 2 cm
Height = 2 small cube = 2 cm
Therefore volume = 3 × 2 × 2 = 12 cubic cm

Intext Questions (Page No. 40)

Question 1.
Calculate the volume of each of the rectangular blocks shown below:

Answer:
Figure 1
Length = 7 small cube = 7 cm
Width = 4 small cube = 4 cm
Height = 1 small cube = 1 cm
Therefore volume = length × width × height
= 7 × 4 × 1
= 28 cubic cm

Figure 2
Length = 6 small cube = 6 cm
Width = 3 small cube = 3 cm
Height = 3 small cube = 3 cm
Therefore volume = length × width × height
= 6 × 3 × 3
= 54 cubic cm

Figure 3
Length = 5 small cube = 5 cm
Width = 5 small cube = 5 cm
Height = 5 small cube = 5 cm
Therefore volume = length × width × height
= 5 × 5 × 5
= 125 cubic cm

Figure 4
Length = 5 small cube = 5 cm
Width = 4 small cube = 4 cm
Height = 5 small cube = 5 cm
Therefore volume = length × width × height
= 5 × 4 × 5
= 100 cubic cm

Volume Calculation (Page No. 40)

Question 1.
The length, width, and height of a brick are 21 centimetres, 15 centimetres, and 7 centimetres. What is its volume?
Answer:
Length of a brick = 21 cm
Width = 15 cm
Height = 7 cm
Volume = 21 × 15 × 7
= 315 × 7
= 2205 cubic cm

Question 2.
An iron cube is of side 8 centimetres. What is its volume? 1 cubic centimetre of iron weighs 8 grams. What is the weight of this cube?
Answer:
Volume = 8 × 8 × 8 = 512 cubic cm.
Weight of this cube = 512 × 8 = 4096 cubic cm

Volume and Length (Page No. 41)

Question 1.
The table shows the measurements of some rectangular blocks. Calculate the missing measures.

Answer:

New Shapes (Page No. 42)

Question 1.
Calculate the volumes of the shapes shown below. All lengths are in centimetres.

Answer:
Figure 1

(i) 8 × 4 × 2 = 64 cubic cm.
(ii) 8 × 4 × 2 = 64 cubic cm.
(iii) 20 × 4 × 2 = 160 cubic cm.
(iv) 4 × 4 × 2 = 32 cubic cm.
Total volume = 64 + 64 + 160 + 32 = 320 cubic cm.

Figure 2

(i) 16 × 3 × 4 = 192 cubic cm.
(ii) 16 × 3 × 4 = 192 cubic cm.
(iii) 4 × 3 × 4 = 48 cubic cm.
Total volume = 192 + 192 + 48 = 432 cubic cm.

Figure 3

(i) 16 × 3 × 4 = 192 cubic cm.
(ii) 11 × 3 × 4 = 132 cubic cm.
Total volume = 192 + 132 = 324 cubic cm

Large Measures (Page No. 43)

Question 1.
A truck is loaded with sand, 4 metres long, 2 metres wide, and 1 metre high. The price of 1 cubic metre of sand is 1000 rupees. What is the price of this truckload?
Answer:
Volume of the truck = 4 × 2 × 1 = 8 cubic metre
The price of 1 cubic metre of sand = 1000 rupees.
The price of 8 cubic metre sand = 8 × 1000 = 8000 rupees

Question 2.
What is the volume in cubic centimetres of a platform 6 metres long, 1 metre wide, and 50 centimetres high?
Answer:
Volume of the platform = 6 m × 1 m × 50 cm
= 600 × 100 × 50
= 3,000,000 cubic cm

Question 3.
What is the volume of a piece of wood that is 5 metres long, 1 metre wide, and 25 centimetres high? The price of 1 cubic metre of wood is 60000 rupees. What is the price of this piece of wood?
Answer:
Volume of apiece of wood = 5 m × 1 m × 25 cm
= 500 cm × 100 cm × 25 cm
= 1250000 cubic cm
= 1250000 ÷ 1000000
= 1 cubic metre 250000 cubic cm
= 1\(\frac {1}{4}\) cubic metre
The price of 1 cubic metre of wood = 60000 rupees.
The price of \(\frac {1}{4}\) cubic metre of wood = 15000 rupees.
Therefore the total price = 60000 + 15000 = 75000 rupees

Capacity & Liquid Measures (Page No. 45)

Question 1.
The inner sides of a cubical box are of length 4 centimetres. What is its capacity? How many cubes of side 2 centimetres can be stacked inside it?
Answer:
Inner length of the cubical box = 4 cm
Capacity of the box = 4 × 4 × 4 = 64 cubic cm.
Volume of one small cube = 2 × 2 × 2 = 8 cubic cm.
Number of small cubes that can fit inside = 64 ÷ 8 = 8 cubes

Question 2.
The inner sides of a rectangular tank are 70 centimetres, 80 centimetres, and 90 centimetres. How many litres of water can it contain?
Answer:
Capacity of the water tank = 70 × 80 × 90
= 504000 cubic cm
= 504 litres (Since 1000 cubic cm = 1 liter)

Question 3.
The length and width of a rectangular box are 90 centimetres and 40 centimetres. It contains 180 litres of water. How high is the water level?
Answer:
1 litre = 1000 cubic cm.
180 litres = 180 × 1000 = 180,000 cubic cm.
Volume = length × width × height
180,000 = 90 × 40 × height
height = 180,000 ÷ (90 × 40) = 50 cm

Question 4.
The inner length, width, and height of a tank are 80 centimetres, 60 centimetres, and 50 centimetres, and it contains water 15 centimetres high. How much more water is needed to fill it?
Answer:
Inner dimensions of the tank:
Length = 80 cm
Width = 60 cm
Height = 50 cm
Current water height = 15 cm
Capacity of the tank,
Total volume = 80 × 60 × 50 = 240,000 cubic cm.
Capacity = 240 litres
Volume of the water already in the tank = 80 × 60 × 15 = 72,000 cubic cm
Current Capacity = 72 litres
Remaining water needed = 240 – 72 = 168 litres
Or
Now the water is at 15 cm height, and the remaining height is 35 cm.
The water can fill in the remaining portion = 80 × 60 × 35
= 168,000 millilitres
= 168 litres

Question 5.
The panchayat decided to make a rectangular pond. The length, width, and depth were decided to be 20 metres, 15 metres, and 2 metres. How many litres of water are needed to fill this pond to a height of one and a half metres?
Answer:
Length = 20 m = 2000 cm
Width = 15 m = 1500 cm
Height = 1\(\frac {1}{2}\) m = 150 cm
Volume = 2000 × 1500 × 150 = 450,000,000 cubic cm.
= 450 cubic metres
= 450000 litre
Or
Volume = 20 × 15 × 1\(\frac {1}{2}\)
= 300 × 1\(\frac {1}{2}\)
= 300 + 150
= 450 cubic metres
= 450000 litre

Question 6.
The inner length and width of an aquarium are 60 centimetres and 30 centimetres. It is half-filled with water. When a stone is immersed in it, the water level rises by 10 centimetres. What is the volume of the stone?
Answer:
When a stone is immersed in the aquarium, the water level rises by 10 centimetres
Volume of the water roses = length × width × 10 cm
= 60 × 30 × 10
= 18000 cubic metres.
Volume of the stone = 18000 cubic cm.

Question 7.
A rectangular iron block has a length of 20 centimetres, a width of 10 centimetres, and a height of 5 centimetres. It is melted and recast into a cube. What is the length of a side of this cube?
Answer:
Volume = 20 × 10 × 5 = 1000 cubic cm
The side of the cube = 10 cm
10 × 10 × 10 = 1000 cubic cm

Question 8.
A tank 2 metres long and 1 metre wide is to contain 10000 litres of water. What should be the height of the tank?
Answer:
1 cubic metre = 1000 litres
10,000 litres = 10 cubic metres
Volume = length × width × height
10 = 2 × 1 × height
height = 10 ÷ (2 × 1)
= 10 ÷ 2
= 5 metres

Question 9.
From the four corners of a square piece of paper of side 12 centimetres, small squares of side 1 centimetre are cut off. The edges of this are bent up and joined to form a container of height 1 centimetre. What is the capacity of this container? If squares of side 2 centimetres are cut off. What would be the capacity?
Answer:

Original side of the square = 12 cm
After cutting 1 cm from each corner, the length and width were reduced by 2 cm (1 cm from each end)
New length = 12 – 2 = 10 cm
New width = 12 – 2 = 10 cm
Height = 1 cm
Capacity = 10 × 10 × 1 = 100 cubic cm.
If 2 cm is cut off from each side,
New length = 12 – 4 = 8 cm
New width = 12 – 4 = 8 cm
Height = 2 cm
Volume = 8 × 8 × 2 = 128 cubic cm.

Class 6 Maths Chapter 3 Kerala Syllabus Volume Questions and Answers

Class 6 Maths Volume Questions and Answers

Question 1.
Which of the following rectangular blocks has a volume of 30 cubic centimetres?
(a) 3 cm, 4 cm, 5 cm
(b) 5 cm, 3 cm, 2 cm
(c) 4 cm, 7 cm, 2 cm
(d) 10 cm, 2 cm, 2 cm
Answer:
(b) 5 cm, 3 cm, 2 cm
Volume = 5 × 3 × 2 = 30 cubic cm.

Question 2.
Which of the following is correct for the volume of a rectangular block?
(a) Volume is the product of its length and height.
(b) Unit of volume is the square centimetre.
(c) If the length of a rectangular block is doubled, then its volume is also doubled.
(d) 100 cubic centimetres is equal to 1 cubic metre.
Answer:
(a) False.
Reason: volume = length × width × height
(b) False.
Reason: unit of volume is cubic centimetre or cubic metre.
(c) If the length of a rectangular block is doubled, then its volume is also doubled.
(d) False.
Reason: 1000 cubic centimetres is 1 litre
100 cubic centimetres is 100 millilitres.

Question 3.
Match the following:
(a) 1000 cubic centimetre – 10 litre
(b) 1000000 cubic centimetre – 1 millilitre
(c) 1 cubic centimetre – 1 cubic metre
(d) 10000 millilitre – 1 litre
Ans:
(a) 1000 cubic centimetre – 1 litre
(b) 1000000 cubic centimetre – 1 cubic metre
(c) 1 cubic centimetre – 1 millilitre
(d) 10000 millilitre – 10 litre

Question 4.
Find the volume of the rectangular block whose length is 12 centimetres, width is 6 centimetres, and height is 4 centimetres.
Answer:
Volume = 12 × 6 × 4 = 288 cubic centimetres

Question 5.
A rectangular block of length 50 centimetres and width 20 centimetres has a volume of 3000 cubic centimetres. What is the height of the rectangular block?
Answer:
Volume = length × width × height
3000 = 50 × 20 × height
height = 3000 ÷ (50 × 20)
= 3000 ÷ 1000
= 3 cm

Question 6.
A wooden block of length 15 metres, width 4 metres, and height 2 metres. What is its volume?
Answer:
Volume of the wodden block = 15 × 4 × 2 = 120 cubic metres

Question 7.
A tank 4 metres long and 2 metres wide is to contain 16000 litres of water. What should be the height of the tank?
Answer:
Volume = 16000 litre = 16 cubic metre
Volume = length × width × height
16 = 4 × 2 × height
height = 16 ÷ (4 × 2)
= 16 ÷ 8
= 2 metres

Question 8.
Complete the following given below:
(a) 180 cubic metre = _____________ litre
(b) 3000 cubic centimetre = _____________ litre
(c) 2000 litre = _____________ cubic metre
(d) 15 cubic centimetre = _____________ millilitre
(e) 75 cubic metre = _____________ litre
Answer:
(a) 180 cubic metre = 180,000 litre
(b) 3000 cubic centimetre = 3 litre
(c) 2000 litres = 2 cubic metres
(d) 15 cubic centimetre = 15 millilitre
(e) 75 cubic metre = 75,000 litre

Class 6 Maths Chapter 3 Notes Kerala Syllabus Volume

→ To compare the size of two rectangular blocks, we must consider their length, width, and height.

→ The volume of a rectangular block can be calculated by multiplying its length, width, and height.

→ If the measurements are in centimetres, then the volume will be in cubic centimetres.

→ If measurements are in metres, then the volume should be in cubic metres.

→ The volume of a rectangular block is the product of its length, width, and height.

→ For a rectangular block,

• If the length is doubled, the volume becomes twice.
• If the length and width are doubled, the volume becomes four times.
• If the length, width, and height are all doubled, the volume becomes eight times.

→ The volume of a rectangular block is 1 cubic metre if its length is 1 metre, width is 1 metre, and height is 1 metre.

→ 1 cubic metre = 100 × 100 × 100 = 1000000 cubic centimetre

→ The volume is called the capacity of the box or a vessel.

→ The capacity of a rectangular box or a vessel is calculated by multiplying its inner length, width, and height.

→ 1 cubic centimeter = 1 millilitre

→ 1000 cubic centimeter = 1 litre

→ 1 cubic metre =1000 litre

→ 1 cubic metre = 1000000 cubic centimetre

In our everyday life, we often come across objects that occupy spaces – like a water bottle, a box, a tank, or even a cupboard. The amount of space an object takes up is called its volume. Volume helps us understand how much space is inside a three-dimensional object or how much it can hold. In this chapter, we will learn how to measure the volume of different solid shapes, such as cubes, cuboids. We will also explore the formulas used to calculate volume, and understand the units in which volume is measured, like cubic centimetres or liters.

Large and Small & Rectangular Blocks
In the previous classes, we have already learned how to compare the lengths of objects, to find their perimeter. The size of an object can be determined by measuring its length of the objects. Measuring only the length of a rectangular-shaped object is not sufficient to determine its size. Instead, we need to measure both its length and breadth to find its area. To compare the size of two rectangular blocks, we must consider their length, width, and height.

Size of a Rectangular Block & Size as a Number
Look at the rectangular block given below.

The length, width, and height of the rectangular block are 10 cm, 5 cm, and 10 cm, respectively. Also, consider that it is made by stacking smaller blocks, each having length, width, and height as 1 cm, 1 cm, 1 cm.

Let us find out how many small rectangular blocks are contained within the large rectangular block. In the bottom row, there are 10 numbers in length and 5 numbers in width. So the total number of small rectangular blocks is 50.

The height of the larger rectangular block is 10 cm, so we can place 10 smaller blocks along the height. If we stack 50 such columns, each with 10 blocks in height, we get a total of 50 × 10 = 500 blocks. So the size of the larger rectangular block is the same as the size of the small 500 rectangular blocks, each measuring 1 cm in length, 1 cm in width, and 1 cm in height.

We learned that the area of a rectangle with a length of 1 cm and a width of 1 cm is 1 square centimeter. Similarly, the volume of a rectangular block with a length of 1 cm, a width of 1 cm, and a height of 1 cm is 1 cubic centimeter. Therefore, the volume of the above rectangular block is 10 × 5 × 10 = 500 cubic centimeters. That means the volume of a rectangular block can be calculated by multiplying its length, width, and height. If the measurements are in centimeters, then the volume will be in cubic centimeters. If measurements are in metres, then the volume should be in cubic metres.

Question 1.
Find the volume of the given rectangular block.

Answer:
Length of the rectangular block = 20 cm
Width =10 cm
Height = 8 cm
Therefore, Volume = 20 × 10 × 8 = 1600 cubic cm

Question 2.
How many small cubes are used to make this large cube? If one small block is removed from each corner of the large block, how many would be left?

Answer:
The large cube is made by stacking small cubes.
Length = 4 small cubes
Width = 4 small cubes
Height = 4 small cubes
Total number of small cubes = 4 × 4 × 4 = 64
A cube has 8 corners. Removing 1 cube from each corner means a total of 8 cubes are removed.
Therefore the cubes left = 64 – 8 = 56
The number of cubes remaining after removing corners is 56.

Question 3.
All sides of the large cube are painted. How many small cubes would have no paint at all?

Answer:
Total number of small cubes = 3 × 3 × 3 = 27
Only the cubes that are completely inside the large cube will have no paint on them.
The top layer and bottom layer have 9 cubes each, and all have at least one side painted.
In the middle layer, only the outer edge cubes have paint on any of their sides, which means 8 cubes have at least one side painted.
So the total number of cubes with at least one side painted is = 9 + 9 + 8 = 26
Therefore, the small cube with no paint at all = 27 – 26 = 1 cube

Volume Calculation
How do we calculate its volume?

For calculating the volume, we must find out how many cubes of side 1 cm we need to make it.

30 small cubes are needed to make it.
So the volume of the cube is 30 cubic cm.
Or
Volume of the cube = 5 × 3 × 2 = 30 cubic cm.
The volume of a rectangular block is the product of its length, width, and height.

Question 1.
The dimensions of a rectangular block are given below. Find its volume.
(i) Length = 10 cm, Width = 7 cm, Height = 4 cm.
(ii) Length = 5 cm, Width = 5 cm, Height = 5 cm.
(iii) Length = 8 cm, Width = 4 cm, Height = 6 cm.
(iv) Length = 20 cm, Width = 10 cm, Height = 10 cm.
(v) Length = 40 cm, Width = 10 cm, Height = 10 cm.
Answer:
(i) 10 × 7 × 4 = 280 cubic cm
(ii) 5 × 5 × 5 = 125 cubic cm
(iii) 8 × 4 × 6 = 192 cubic cm
(iv) 20 × 10 × 10 = 2000 cubic cm
(v) 40 × 10 × 10 = 4000 cubic cm

Volume and Length
Volume can be calculated when the length, width, and height are given. If any three of these four terms are given, we can find the fourth term.

Question 1.
The volume of a rectangular block is 160 cubic centimetres. Its height and width are 5 centimetres and 4 centimetres respectively. What is its length?
Answer:
Volume = length × breadth × height
160 = length × 5 × 4
160 = length × 20
To find the length, divide 160 by 20
Length = 160 ÷ 20 = 8 cm

Question 2.
A rectangular block of length 23 centimetres and height 6 centimetres has a volume of 1380 cubic centimetres. What is its width?
Answer:
Volume = length × breadth × height
1380 = 23 × width × 6
1380 = width × 138
Width = 1380 ÷ 138 = 10 cm

New Shapes
We can make shapes other than rectangular blocks by stacking cubes.

Question 1.
It is made by stacking cubes of side 1 centimetre. Calculate its volume?

Answer:
1st row (bottom row) = 9 × 9 = 81
2nd row = 7 × 9 = 63
3rd row = 5 × 9 = 45
4th row = 3 × 9 = 27
Total = 81 + 63 + 45 + 27 = 216
Therefore, volume = 216 cubic cm.
Or
If the length and width are 9 in every row, and the height is 4.
Then total number of blocks = 9 × 9 × 4 = 324
Now consider the missing blocks.
1st row (bottom row) = 2 × 9 = 18
2nd row = 4 × 9 = 36
3rd row = 6 × 9 = 54
Total missing blocks = 18 + 36 + 54 = 108
Total number of blocks = 324 – 108 = 216
Therefore, volume = 216 cubic cm.

Question 2.
It is made by stacking square blocks. The bottom block is of side 9 centimetres. As we move up, the sides decrease by 2 centimetres at each step. All blocks are of height 1 centimetre. What is the volume of this figure?

Answer:
The blocks are arranged in 5 rows.
Volume of the first square block = 9 × 9 × 1 = 81 cubic cm.
Volume of the second square block = 7 × 7 × 1 = 49 cubic cm.
Volume of the third square block = 5 × 5 × 1 = 25 cubic cm.
Volume of the fourth square block = 3 × 3 × 1 = 9 cubic cm.
Volume of the fifth (top) square block = 1 × 1 × 1 = 1 cubic cm.
Toatal volume of the square blocks = 81 + 49 + 25 + 9 + 1 = 165 cubic cm.

Question 3.
What is the volume of a rectangular block of length 4 centimetres, width 3 centimetres, and height 1 centimetre? If the length, width, and height are doubled, what happens to the volume?
Answer:
Length of the rectangular block = 4 cm
Width = 3 cm
Height = 1 cm
Volume of the rectangular block = 4 × 3 × 1 = 12 cubic cm.
If the length, width, and height are doubled.
Length = 8 cm
Width = 6 cm
Height = 2 cm
Volume = 8 × 6 × 2 = 96 cubic cm.

For a rectangular block,
If the length is doubled, the volume becomes twice.
If the length and width are doubled, the volume becomes four times.
If the length, width, and height are all doubled, the volume becomes eight times.

Large Measures
The volume of a rectangular block is 1 cubic centimetre if its length is 1 centimetre, its width is 1 centimetre, and its height is 1 centimetre.
The volume of a rectangular block is 1 cubic metre if its length is 1 metre, width is h metres, and height is 1 metre.
1 cubic metre = 100 × 100 × 100 = 1000000 cubic centimetre
Then, the volume of a rectangular block, if its length is 5 metres, width 3 metres, and height 2 metres, is 5 × 3 × 2 = 30 cubic metres
If it is converted into cubic centimetres, 30 × 1000000 = 30000000 cubic cm

Question 1.
What is the volume of a platform if its length is 12 metres, width is 8 metres, and height is 75 centimetres?
Answer:
Here, the measurements are both in centimetres and metres.
Convert every value into centimetres.
1200 cm × 800 cm × 75 cm = 72,000,000 cubic cm
If converting it into cubic metres, 72,000,000 ÷ 1000000 = 72 cubic metres

Capacity & Liquid Measures
Capacity refers to the amount a box or a vessel can hold. So it is closely related to volume. That is, the volume is called the capacity of the box or a vessel. The capacity of a rectangular box or a vessel is calculated by multiplying its inner length, width, and height. While considering the thickness of a rectangular box or vessel, the inner and outer dimensions are different.

For this rectangular box, the outer dimensions are greater than the inner dimensions. For calculating the capacity, we have to consider the inner dimensions.

Question 1.
Calculate the capacity when its inner dimensions are of length 20 centimetres, width 10 centimetres, and height 5 centimetres.
Answer:
Capacity = 20 × 10 × 5 = 1000 cubic cm

If we are discussing the quantity of water contained in a rectangular box, it can be expressed in millilitres or litres.
Consider a rectangular box with dimensions of length 1 centimetre, width 1 centimetre, and height 1 centimetre.
Its capacity is calculated as = 1 × 1 × 1 = 1 cubic cm. This is equal to 1 millilitre

If we combine 1000 such boxes, their total capacity becomes 1000 cubic cm, which is equal to 1 litre.
1000 millilitre = 1 litre
Therefore, the capacity of the above-mentioned rectangular box is 1000 cubic cm = 1 litre.

Consider a rectangular box with dimensions of length 10 centimetres, width 10 centimetres, and height 10 centimetres.
Its capacity is calculated as = 10 × 10 × 10 = 1000 cubic cm. This is equal to 1 litre. These relationships can be written like this:

• 1 cubic centimeter = 1 millilitre
• 1000 cubic centimeter = 1 litre
• 1 cubic metre = 1000 litre
• 1 cubic metre = 1000000 cubic centimetre

Question 2.
What is the capacity of a box whose inner length, width, and height are 15 centimetres, 10 centimetres, and 8 centimetres?
Answer:
Capacity = 15 × 10 × 8
= 1200 cubic cm.
= 1000 cubic cm + 200 cubic cm
= 1 litre + 200 millilitre
= 1 litre 200 millilitre

Question 3.
A vessel is filled with water. If a cube of side 1 centimetre is immersed in it, how many cubic centimetres of water would overflow? What if 20 such cubes are immersed?
Answer:
If a cube of side 1 centimetre is immersed in a vessel filled with water, the amount of water that overflows will be equal to the volume (or capacity) of the cube.
Volume of the cube = 1 cm × 1 cm × 1 cm = 1 cubic centimeter
Since 1 cubic centimetre = 1 millilitre
1 millilitre of water will overflow.
If 20 such cubes are immersed.
Each cube displaces 1 millilitre of water.
So, 20 × 1 = 20 millilitres of water will overflow.

Question 4.
A water tank of length 4 metres, width 3 metres, and height 2 metres. How many litres of water does it contain?
Answer:
Capacity of the water tank = 4 × 3 × 2 = 24 cubic metre
Since 1 cubic metre = 1000 litre
24 cubic metre = 24000 litre
Therefore, the tank can hold 24,000 litres of water.

Question 5.
A rectangular tank of length 8 metres and width 6 metres. It contains 96000 litres of water. Calculate the height of the water in the tank?
Answer:
1 cubic metre = 1000 litres
96000 litres = \(\frac {96000}{10000}\) = 96 cubic metres
Volume = length × width × height
96 = 8 × 6 × height
height = 96 ÷ 48 = 2 metre

Question 6.
A swimming pool is 25 metres long, 10 metres wide, and 2 metres deep. It is half-filled. How many litres of water does it contain now?
25 × 10 × 1 = 250 cubic metres = 250000 litres
Suppose the water level is increased by 1 centimetre. How many more litres of water does it contain now?
Answer:
If it is half-filled
Full depth = 2 metre, Half depth = 1 metre
Volume = 25 × 10 × 1 = 250 cubic metres
250 cubic metres = 250 × 1000 = 250,000 litres
If the water level rises by 1 cm
Additional volume = 2500 × 1000 × 1 = 2500000 cubic cm
2500000 cubic cm = \(\frac {2500000}{1000}\) = 2500 litres

Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 2 One Fraction Many Forms Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 6 Maths Chapter 2 Solutions One Fraction Many Forms

Class 6 Kerala Syllabus Maths Solutions Chapter 2 One Fraction Many Forms Questions and Answers

One Fraction Many Forms Class 6 Questions and Answers Kerala Syllabus

Numerator and Denominator (Page No. 24)

Question 1.
Find the fractions specified below:
(i) The form of \(\frac {1}{2}\) with denominator 24
(ii) The form of \(\frac {1}{2}\) with numerator 24
(iii) The form of \(\frac {1}{3}\) with denominator 24
(iv) The form of \(\frac {1}{3}\) with numerator 24
(v) The form of \(\frac {1}{4}\) with numerator 100
Answer:
(i) \(\frac{1}{2}=\frac{12}{24}\)
(ii) \(\frac{1}{2}=\frac{24}{48}\)
(iii) \(\frac{1}{3}=\frac{8}{24}\)
(iv) \(\frac{1}{3}=\frac{24}{72}\)
(v) \(\frac{1}{4}=\frac{100}{400}\)

Question 2.
Write three different forms of \(\frac {1}{4}\).
Answer:
Three different forms of \(\frac {1}{4}\) is
\(\frac{1}{4}=\frac{2}{8}=\frac{20}{80}=\frac{50}{200}\)

Question 3.
For each pair of fractions below, find three forms with the same denominator:
(i) \(\frac{1}{2}, \frac{1}{3}\)
(ii) \(\frac{1}{2}, \frac{1}{4}\)
(iii) \(\frac{1}{3}, \frac{1}{4}\)
Answer:

Question 4.
Does \(\frac {1}{3}\) have another form with a denominator of 10, 100, or 1000? Give reasons.
Answer:
No, \(\frac {1}{3}\) does not have another form with a denominator 10, 100, or 1000, because none of these numbers is divisible by 3.

Textbook Page No. 26

Question 1.
Now, can’t you fill up the following table by multiplying the numerator and denominator by a number?

Answer:

Textbook Page No. 28

Question 1.
Write each of the following fractions in lowest terms:
(i) \(\frac {32}{64}\)
(ii) \(\frac {27}{81}\)
(iii) \(\frac {30}{45}\)
(iv) \(\frac {12}{21}\)
(v) \(\frac {45}{54}\)
Answer:
(i) \(\frac{32}{64}=\frac{16}{32}=\frac{8}{16}=\frac{4}{8}=\frac{2}{4}=\frac{1}{2}\)
The largest common factor of 32 and 64 is 32
So divide 32 and 64 by 32, we get \(\frac{32}{64}=\frac{1}{2}\)

(ii) \(\frac {27}{81}\)
3, 9, 27 are the common factors of 27 and 81
\(\frac{27}{81}=\frac{9 \times 3}{27 \times 3} ; \quad \frac{27}{81}=\frac{3 \times 9}{9 \times 9} ; \quad \frac{27}{81}=\frac{1 \times 27}{3 \times 27}\)
So divide it by the largest common factor, 27, to get \(\frac {1}{3}\)
Therefore, the lowest form is \(\frac {1}{3}\).

(iii) \(\frac {30}{45}\)
Common factors are 5 and 15
So divide it by 15
\(\frac{30}{45}=\frac{2}{3}\)

(iv) \(\frac {12}{21}\)
3 is the common factor of 12 and 21
\(\frac{12}{21}=\frac{3 \times 4}{3 \times 7}=\frac{4}{7}\)

(v) \(\frac {45}{54}\)
3, 9 are the common factors of 45 and 54.
\(\frac{45}{54}=\frac{15 \times 3}{18 \times 3}=\frac{15}{18}, \quad \frac{15}{18}=\frac{5 \times 3}{6 \times 3}=\frac{5}{6}\)
First, removing the factor 9 from it will result in \(\frac {5}{6}\)
\(\frac{45}{54}=\frac{9 \times 5}{9 \times 6}=\frac{5}{6}\)

Fraction as Division (Page No. 32)

Question 1.
20 litres of water are used to fill 8 identical bottles. How many litres of water are there in each bottle?
Answer:
20 ÷ 8 = \(\frac {20}{8}\)
\(\frac {16}{8}\) = 2
The remaining 4 is divided into 8 equal parts, that is,
\(\frac{4}{8}=\frac{1}{2}\)
Therefore the quantity of water contained in one bottle is 2\(\frac {1}{2}\) liters
Or \(\frac{20}{8}=\frac{10}{4}=\frac{5}{2}=2 \frac{1}{2}\)

Question 2.
A rope of length 140 centimetres is cut into 16 equal pieces. What is the length of each piece?
Answer:
140 ÷ 16 = \(\frac {140}{16}\)
= \(\frac {70}{8}\)
= \(\frac {35}{4}\)
= 8\(\frac {3}{4}\) metre

Question 3.
If 215 kilograms of rice is divided equally among 15 people, how many kilograms of rice would each get?
Answer:
215 ÷ 15 = \(\frac{215}{15}=\frac{43 \times 5}{3 \times 5}=\frac{43}{3}\)
If 42 is divided by 3 we get 14
The remainder 1 is divided into 3 parts, we get \(\frac {1}{3}\)
The kilogram of rice each get 14\(\frac {1}{3}\) kg

Class 6 Maths Chapter 2 Kerala Syllabus One Fraction Many Forms Questions and Answers

Class 6 Maths One Fraction Many Forms Questions and Answers

Question 1.
Which is the correct statement given below:
(a) The form of \(\frac {1}{3}\) with denominator 9 is \(\frac {2}{9}\)
(b) The form of \(\frac {1}{3}\) with numerator 9 is \(\frac {9}{27}\)
(c) The form of \(\frac {1}{3}\) with denominator 6 is \(\frac {4}{6}\)
(d) The form of \(\frac {1}{3}\) with numerator 2 is \(\frac {2}{6}\)
Answer:
(b) The form of \(\frac {1}{3}\) with numerator 9 is \(\frac {9}{27}\)
(d) The form of \(\frac {1}{3}\) with numerator 2 is \(\frac {2}{6}\)

Question 2.
Which is the wrong statement given below:
(a) Another form of \(\frac {1}{4}\) is \(\frac {5}{20}\)
(b) Another form of \(\frac {3}{7}\) is \(\frac {9}{21}\)
(c) Another form of \(\frac {5}{8}\) is \(\frac {30}{40}\)
(d) Another form of \(\frac {9}{10}\) is \(\frac {27}{30}\)
Answer:
(c) Another form of \(\frac {5}{8}\) is \(\frac {30}{40}\)
Multiplying 8 by 5 we get 40, and multiplying 5 by 5 we get 25.
So the another form of \(\frac {5}{8}\) is \(\frac {25}{40}\)

Question 3.
Match the following:

Answer:

Question 4.
The form of \(\frac {3}{5}\) with denominator 25 is _____________
Answer:
\(\frac{3}{5}=\frac{15}{25}\) (3 × 5 = 15, 5 × 5 = 25)

Question 5.
The form of \(\frac {1}{8}\) with numerator 10 is _____________
Answer:
\(\frac{1}{8}=\frac{10}{80}\) (1 × 10 = 10, 8 × 10 = 80)

Question 6.
Write 3 different forms of \(\frac {3}{10}\)?
Answer:
\(\frac{3}{10}=\frac{6}{20}=\frac{12}{40}=\frac{24}{80}\)

Question 7.
Write 3 different forms of the pair \(\frac {2}{5}\), \(\frac {1}{6}\) with same denominator.
Answer:
\(\frac {2}{5}\), \(\frac {1}{6}\)
Denominator 30: \(\frac{12}{30}, \frac{5}{30}\)
Denominator 60: \(\frac{24}{60}, \frac{10}{60}\)
Denominator 120: \(\frac{48}{120}, \frac{20}{120}\)

Question 8.
If 10 litres of milk are filled into 3 identical bottles. How many litres of milk are contained in one bottle?
Answer:
Milk contained in one bottle is 10 ÷ 3 = \(\frac {10}{3}\) = 3\(\frac {1}{3}\) litres

Question 9.
Dividing 12 by which number is equal to 24 divided by 18.
Answer:
24 ÷ 18 = 12 ÷ 9 (\(\frac{24}{18}=\frac{12}{9}\))

Question 10.
Write the lowest term of \(\frac {60}{90}\).
Answer:
Lowest term is \(\frac{60}{90}=\frac{2 \times 30}{3 \times 30}=\frac{2}{3}\)

Class 6 Maths Chapter 2 Notes Kerala Syllabus One Fraction Many Forms

→ Multiplying the numerator and denominator of a fraction by the same natural number gives a form of the same fraction.

→ If the numerator and denominator of a fraction have a common factor, then dividing them by this common factor gives a form of this fraction.

→ The form of a fraction in the lowest terms is obtained by removing all common factors of the numerator and denominator by division.

→ If we remove common factors of the numerator and denominator of a fraction, then we get another form of the same fraction.

→ In writing divisions with remainders as fractions, we can remove common factors of the numerator and denominator.

Fractions are numerical representations used to describe parts of a whole or the ratio between quantities. A fraction consists of two components: the numerator (the top number), which shows how many parts are being considered, and the denominator (the bottom number), which indicates the total number of equal parts the whole is divided into. Fractions are commonly used in everyday life to divide objects or quantities evenly, express proportions, and make comparisons. In this chapter, we will explore the concept of fractions in detail and learn how a single fraction can be expressed in various forms.

Numerator and Denominator

→ A portion of two equal parts is called a half. It is also known as one by two portion and is written as \(\frac {1}{2}\).

→ If an object is divided into 4 equal parts and 2 parts are taken, or it is divided into 6 equal parts and 3 parts are taken, or even if it is divided into 8 equal parts and 4 parts are taken. In all these cases, the portion taken is the same as \(\frac {1}{2}\) of the whole.

→ If an object is divided into 3 equal parts, then 1 part is \(\frac {1}{3}\) (one-third) and 2 parts are \(\frac {2}{3}\) (two-thirds) of the whole.

→ If an object is divided into 4 equal parts, then 1 part represents \(\frac {1}{4}\) (one-fourth or a quarter), and 3 parts represent \(\frac {3}{4}\) (three-fourths or three-quarters) of the whole.

→ Like this, fractions are the numbers used to represent parts of an object being considered.

→ If 2 objects are divided into 3 equal parts, then one part will be \(\frac {2}{3}\) of it.

→ If 3 objects are divided into 2 equal parts, each part will be \(\frac {3}{2}\), which is equal to one and a half.
\(\frac{3}{2}=1+\frac{1}{2}=1 \frac{1}{2}\)

Now let’s check the different forms of a fraction in detail:

→ Different forms of the fraction \(\frac {1}{2}\) are \(\frac{2}{4}, \frac{3}{6}, \frac{4}{8}, \frac{5}{10}\)

→ Similarly different forms of \(\frac {1}{3}\) are \(\frac{2}{6}, \frac{3}{9}, \frac{4}{12}, \frac{5}{15}\)

→ If an object is divided into 100 equal parts and we take 50 of them or divide it into 50 equal parts and we take 25 of them will be half \(\frac {1}{2}\). So \(\frac{1}{2}, \frac{25}{50}, \frac{50}{100}\) are all different forms of a fraction.

→ Similarly, we can see the different forms of the fraction \(\frac {1}{3}\) as \(\frac{10}{30}, \frac{20}{60}, \frac{100}{300}\)

→ For the fraction \(\frac {1}{2}\), its numerator is 1 and denominator is 2.

→ For the fraction \(\frac {2}{3}\), its numerator is 2 and denominator is 3.

To get the different forms of a fraction when the numerator is 1, we can write the numerator as the number used to multiply the denominator.
Another form of \(\frac {1}{2}\)
If we multiply 2 by 8, we get 16
So another form of \(\frac {1}{2}\) is \(\frac {8}{16}\).

Another form of \(\frac {1}{5}\)
If we multiply 5 by 10, we get 50
Numerator = 10, Denominator = 50
So another form of \(\frac {1}{5}\) is \(\frac {10}{50}\).

The form of \(\frac {1}{7}\) with denominator 21
If we multiply 7 by 3, we get 21
The numerator of another form is 3
So the another form is \(\frac {3}{21}\)

The form of \(\frac {1}{4}\) with denominator 10 is \(\frac {10}{40}\)
Because the denominator is multiplied by 10 is its numerator.

Write the different forms of \(\frac {1}{7}\)
Write the numerator as the number multiplied by 7
7 × 4 = 28 → \(\frac {4}{28}\)
7 × 6 = 42 → \(\frac {6}{42}\)
7 × 10 = 70 → \(\frac {10}{70}\)

Complete the following given below.

Question 1.
The form of \(\frac {1}{6}\) with denominator 12 is _____________
Answer:
The denominator of \(\frac {1}{6}\) is multiplied by 2, we get 12.
So its numerator is 12
Therefore the form is \(\frac {2}{12}\)

Question 2.
The form of \(\frac {1}{10}\) with denominator 40 is _____________
Answer:
The denominator of \(\frac {1}{10}\) is multiplied by 4, we get 40.
So its numerator is 4
Therefore the form is \(\frac {4}{40}\)

Question 3.
The form of \(\frac {1}{8}\) with numerator 4 is _____________
Answer:
The denominator of \(\frac {1}{8}\) is multiplied by 4, we get 32.
So its numerator is 4
That is \(\frac{1}{8}=\frac{4}{32}\)
(Since the numerator will be the number used to multiply the denominator)

Question 4.
The form of \(\frac {1}{12}\) with numerator 5 is _____________
Answer:
The denominator of \(\frac {1}{12}\) is multiplied by 5, we get 60.
So its numerator is 5
Therefore the form is \(\frac{1}{12}=\frac{5}{60}\)

Question 5.
The form of \(\frac {1}{5}\) with denominator 50 is _____________
Answer:
The denominator of \(\frac {1}{5}\) is multiplied by 10, we get 50.
So its numerator is 10
Therefore the form is \(\frac{1}{5}=\frac{10}{50}\)

Question 6.
The form of \(\frac {1}{7}\) with numerator 20 is _____________
Answer:
The denominator of \(\frac {1}{7}\) is multiplied by 20, we get 140.
So its numerator is 20
Therefore the form is \(\frac{1}{7}=\frac{20}{140}\)

Question 7.
Any 3 different forms of \(\frac {1}{4}\) is _____________
Answer:
Different forms of \(\frac {1}{4}\) is,

Question 8.
Any 3 different forms of \(\frac {1}{8}\) is _____________
Answer:

Different forms of fractions:
So far, we have discussed the different forms of fractions with a numerator of 1.
To find the different forms of fractions in this form, we can take the numerator as the number that is used to multiply the denominator.
\(\frac{1}{4}=\frac{1 \times 10}{4 \times 10}=\frac{10}{40}\)

How do we find the different forms of \(\frac {2}{5}\)?
For this, we have to multiply both the numerator and denominator by the same number.
In the fraction \(\frac {2}{5}\), multiply 5 by 2, we get 10.
Here, we multiply the denominator by 2, so we need to multiply the numerator also by 2.
That is, \(\frac{2}{5}=\frac{2 \times 2}{5 \times 2}=\frac{4}{10}\)
Another form, \(\frac{2}{5}=\frac{2 \times 5}{5 \times 5}=\frac{10}{25}\)
That means instead of taking 2 from 5 equal parts, take 4 from 10 equal parts (\(\frac{2}{5}=\frac{4}{10}\))
Similarly, instead of taking 2 from 5 equal parts, take 10 from 25 equal parts (\(\frac{2}{5}=\frac{10}{25}\))

Now check the different forms of \(\frac {3}{7}\).
\(\frac{3}{7}=\frac{6}{14}\) (multiply 7 by 2, we get 14, and multiply 3 by 2, we get 6)
\(\frac{3}{7}=\frac{30}{70}\) (multiply 7 by 10 we get 70, and multiply 3 by 10 we get 30)

Complete the following given below:

Question 1.
The form of \(\frac {2}{3}\) with denominator 15 is _____________
Answer:
Multiply 3 by 5 to get the denominator as 15
So, multiply 2 by 5, and we get the numerator as 10
\(\frac{2}{3}=\frac{2 \times 5}{3 \times 5}=\frac{10}{15}\)

Question 2.
The form of \(\frac {3}{4}\) with denominator 100 is _____________
Answer:
Multiply 4 by 25 to get the denominator as 100
So, multiply 3 by 25, and  we get the numerator as 75
\(\frac{3}{4}=\frac{3 \times 25}{4 \times 25}=\frac{75}{100}\)

Question 3.
The form of \(\frac {5}{8}\) with numerator 25 is _____________
Answer:
Multiply 5 by 5 to get the numerator as 25
So, multiply 8 by 5, and we get the denominator as 40
\(\frac{5}{8}=\frac{5 \times 5}{8 \times 5}=\frac{25}{40}\)

Question 4.
The form of \(\frac {7}{10}\) with numerator 70 is _____________
Answer:
Multiply 7 by 10 to get the numerator as 70
So, multiply 10 by 10, we get the denominator as 100
\(\frac{7}{10}=\frac{7 \times 10}{10 \times 10}=\frac{70}{100}\)

Question 5.
The 3 forms of \(\frac {3}{8}\) is _____________
Answer:
Different forms is,

Question 6.
The 3 forms of \(\frac {5}{7}\) is _____________
Answer:
Different forms is,

Till now, we have discussed the fractions with a numerator of 1 and different numbers separately.
In general, we can say this: “Multiplying the numerator and denominator of a fraction by the same natural number gives a form of the same fraction.”
For example: \(\frac{1}{5}=\frac{1 \times 6}{5 \times 6}=\frac{6}{30}, \quad \frac{2}{5}=\frac{2 \times 6}{5 \times 6}=\frac{12}{30}\)

Lowest terms:
We can form the different forms of a fraction by multiplying the numerator and denominator by the same number.

Let’s look at another form of \(\frac {20}{25}\);
We can divide 20 and 25 by their factor 5.
Divide 20 by 5, and we get 4
Divide 25 by 5, we get 5
Therefore \(\frac{20}{25}=\frac{4}{5}\)
Now 1 is the only number that can divide 4 and 5.
So the lowest form of \(\frac {20}{25}\) is \(\frac {4}{5}\).

Now let’s look at another form of \(\frac {30}{36}\);
Divide 30 and 36 by their common factor 2, we get
\(\frac{30}{36}=\frac{15}{18}\)
Now, divide 15 and 18 by their common factor 3, and we get
\(\frac{15}{18}=\frac{5}{6}\) (15 ÷ 3 = 5, 18 ÷ 3 = 6)
There is no common factor for 5 and 6.
So the lowest form of \(\frac {30}{36}\) is \(\frac {5}{6}\).
(Instead of dividing the numerator and denominator of the fraction \(\frac {30}{36}\) by 2 and 3, it is enough to divide both 30 and 36 by their common factor 6)
\(\frac{30}{36}=\frac{15}{18}=\frac{5}{6}\)

Like this, find another form of \(\frac {40}{50}\) by removing their common factor;
Divide both 40 and 50 by their common factor 5,
40 ÷ 5 = 8, 50 ÷ 5 = 10
\(\frac{40}{50}=\frac{8}{10}\)
2 is the factor of 8 and 10.
So, divide both 8 and 10 by 2, we get
8 ÷ 2 = 4, 10 ÷ 2 = 5
\(\frac{8}{10}=\frac{4}{5}\)
There is no common factor for 4 and 5, so the lowest form of \(\frac {40}{50}\) is \(\frac {4}{5}\).
(When the numerator and denominator of \(\frac {40}{50}\) is divided by 10 we get \(\frac {4}{5}\))
By removing the common factor of the numerator and denominator of any fraction, we get the lowest form of this fraction.
In general, if the numerator and denominator of a fraction have a common factor, then dividing them by this common factor gives a form of this fraction.

Write each of the following fractions in lowest terms:
(i) \(\frac {21}{28}\)
Answer:
7 is a factor of 21 and 28
Divide 21 by 7 we get 3
Divide 28 by 7 we get 4
\(\frac{21}{28}=\frac{3}{4}\) (3 and 4 do not have any common factor)

(ii) \(\frac {35}{50}\)
Answer:
5 is a factor of 35 and 50
35 ÷ 5 = 7, 50 ÷ 5 = 10
\(\frac{35}{50}=\frac{7}{10}\)

(iii) \(\frac {6}{60}\)
Answer:
2 is the factor of 6 and 60
So, \(\frac{6}{60}=\frac{3}{30}\)
3 is the factor of 3 and 30
\(\frac{3}{30}=\frac{1}{10}\)

(iv) \(\frac {40}{70}\)
Answer:
5 is a factor of 40 and 70
So, \(\frac{40}{70}=\frac{8}{14}\)
2 is a factor of 8 and 14
\(\frac{8}{14}=\frac{4}{7}\)
4 and 7 do not have any common factor.
So the lowest form is \(\frac {4}{7}\)
(Also 40 and 70 can be divided by 10 to get \(\frac {4}{7}\))

(v) \(\frac {7}{70}\)
Answer:
7 is the factor of 7 and 70
7 ÷ 7 = 1, 70 ÷ 10 = 7
\(\frac{7}{70}=\frac{1}{10}\)

Fraction as Division
In class 5, we have seen that the fraction \(\frac {2}{3}\) can be expressed in two different ways.

• If one is divided into 3 equal parts and 2 fares are taken from it.
• If two is divided into 3 equal parts and 1 is removed from it.

Any division can be written as a fraction like this.
For example 8 ÷ 4 = \(\frac {8}{4}\)
The lowest form is
\(\frac{8}{4}=\frac{2}{1}\) = 2
10 ÷ 5 = \(\frac {10}{5}\) = 2
24 ÷ 6 = \(\frac {24}{6}\) = 4
We can write like this.
\(\frac {4}{1}\) means 4 itself.
Like this, any natural number with a denominator of 1 can be expressed as a fraction.
\(\frac {5}{1}\) = 5, \(\frac {50}{1}\) = 50, \(\frac {100}{1}\) = 100
The remaining divisions can also be written as fractions.

If a 3 metre ribbon is divided equally among 2 people. Then what is the length of the ribbon each will get?
Answer:
One will get 3 ÷ 2.
Each person will get 1 metre of ribbon and 1 metre as a remainder.
Divide the remainder of 1 metre equally among the 2 people.
Each will get \(\frac {1}{2}\) metre.
So one person will get, 1 metre + \(\frac {1}{2}\) metre = 1\(\frac {1}{2}\) metre.

Like this, what is 8 ÷ 3?
8 ÷ 3 = \(\frac {8}{3}\)
Divide 6 into 3 equal parts, we get 2
Divide the remaining 2 into 3 equal parts, that is \(\frac {2}{3}\)
So, \(\frac{8}{3}=2 \frac{2}{3}\)

\(\frac {15}{4}\)
Divide 12 by 4, and we get 3
Divide the remaining 3 into 4 equal parts, we get \(\frac {3}{4}\)
So, \(\frac{15}{4}=3 \frac{3}{4}\)

If 14 is divided into 4 parts
\(\frac{14}{4}=\frac{7}{2}\)
That means dividing 14 into 4 parts is the same as dividing 7 into 2 parts.
\(\frac{14}{4}=3 \frac{2}{4}=3 \frac{1}{2}\)
\(\frac{7}{2}=3 \frac{1}{2}\)

Write the following in the form of fractions:
15 ÷ 3 = \(\frac {15}{3}\) = 5
21 ÷ 7 = \(\frac {21}{7}\) = 3
24 ÷ 4 = \(\frac {24}{4}\) = 6
32 ÷ 16 = \(\frac {32}{16}\) = 2
7 ÷ 3 = \(\frac {7}{3}\)
If 7 is divided into 3 parts, we get 2, and the remaining 1 is divided into 3 parts
\(\frac{7}{3}=2 \frac{1}{3}\)

25 ÷ 7 = \(\frac {25}{7}\)
If 21 is divided into 7 equal parts, we get 3, and the remaining 4 is divided into 7 parts
\(\frac{25}{7}=3 \frac{4}{7}\)

Answer the following questions:

Question 1.
If a 12 metre long ribbon is divided among 5 people. How long will a person get?
Answer:
12 ÷ 5 = \(\frac {12}{5}\)
If 10 is divided into 5 equal parts, we get 2, and the remaining 2 is divided into 5 parts, that is \(\frac {2}{5}\)
One person will get 2\(\frac {2}{5}\)

Question 2.
If a 37 metre iron rod is divided into 5 equal parts. What is the length of the rod in metres?
Answer:
37 ÷ 5 = \(\frac {37}{5}\)
If 35 is divided into 5 equal parts, we get 7, and the remaining 2 is divided into 5 parts, that is \(\frac {2}{5}\)
The length of one rod is 7\(\frac {2}{5}\) metre

Question 3.
If 18 kg of sugar is packed equally in 7 identical packets. What is the quantity of sugar in one packet?
Answer:
18 ÷ 7 = \(\frac {18}{7}\)
If 14 is divided into 7 equal parts, we get 2, and the remaining 4 is divided into 7 parts, that is \(\frac {4}{7}\)
Sugar in one packet is 2\(\frac {4}{7}\)

Question 4.
If 20 litres of coconut oil are filled in 6 bottles, then how many liters are in each bottle contains.
Answer:
20 ÷ 6 = \(\frac {20}{6}\)
If 18 is divided into 6 equal parts, we get 3, and the remaining 2 is divided into 6 parts, that is \(\frac{2}{6}=\frac{1}{3}\).
One bottle contain 3\(\frac {1}{3}\)
Or dividing 20 into 6 parts is the same as dividing 10 into 3 parts.
\(\frac{20}{6}=\frac{10}{3}, \quad \frac{10}{3}=3 \frac{1}{3}\)

Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 1 Angles Questions and Answers Notes Pdf to clear their doubts.

SCERT Class 6 Maths Chapter 1 Solutions Angles

Class 6 Kerala Syllabus Maths Solutions Chapter 1 Angles Questions and Answers

Angles Class 6 Questions and Answers Kerala Syllabus

Drawing Angles (Page 13)

Question 1.
Measure and mark each angle below:


Answer:
(i) 60°
(ii) 135°
(iii) 60°
(iv) 130°

Question 2.
Draw the pictures below in the notebook:

Answer:
(i)

• Draw PQ of length 4 cm.
• From P measure 50° (right measure).
• From Q measure 50° (left).
• Join the two lines.

(ii)

• Make a line of 5 cm.
• From the right end, measure 90° and make a line.
• From the left end, draw a line making a 30° angle.
• Meet the two lines to get the required figure.

(iii)

• Draw a line of length 5 cm.
• From one end, draw a line of 3 cm at an angle of 60°.
• From the other end, draw a line of 3 cm, making 120°.
• Join the angles of the two lines.

(iv)

• Draw a line (dotted) of 6 cm.
• From both ends, draw a line to make a 20° angle on both sides.
• Join the lines together to get the required figure.

Circle Division (Pages 17 & 18)

Question 1.
In each of the pictures below, calculate what fractions of the circle are the yellow and green parts:


Answer:
The measure of an angle inside a circle is 360°
Figure-1
Angle measure of the yellow portion is 20°
Total fractions of the circle is 360° ÷ 20 = 18
The fraction of yellow portion is \(\frac {1}{18}\)
The green portion is \(\frac {17}{18}\)

Figure-2
Angle measure of the yellow portion is 24°
Total fractions of the circle is 360° ÷ 24 = 15
The fraction of yellow portion is \(\frac {1}{15}\)
The green portion is \(\frac {14}{15}\)

Figure-3
Angle measure of the yellow portion is 54°
But for 18° its fraction is, \(\frac {1}{20}\). So for 54° it is \(\frac {3}{20}\).
Therefore the fraction of yellow portion is \(\frac {3}{20}\)
The green portion is \(\frac {17}{20}\)

Figure-4
The angle measure of the yellow portion is 80°
Total fractions of the circle when it is 40°;
360° ÷ 40 = 9. So for 40° it is \(\frac {1}{9}\)
The yellow portion is 80°.
The fraction of yellow portion is \(\frac {2}{9}\)
The green portion is \(\frac {7}{9}\)

Figure-5
Angle measure of the yellow portion is 108°
If it is 36° its fraction is \(\frac {1}{10}\).
That means 36 × 3 = 108
So for 108° it is \(\frac {3}{10}\)
Therefore the fraction of yellow portion is \(\frac {3}{10}\)
The green portion is \(\frac {7}{10}\)

Figure-6
Angle measure of the yellow portion is 150°
If it is 30° its fraction is, \(\frac {1}{12}\).
So for 150° it is \(\frac {5}{12}\).
Therefore the fraction of yellow portion is \(\frac {5}{12}\)
The green portion is \(\frac {7}{12}\)

Question 2.
Mark each of the fractions below as part of a circle and colour the pictures.
(i) \(\frac {3}{8}\)
(ii) \(\frac {2}{5}\)
(iii) \(\frac {4}{9}\)
(iv) \(\frac {5}{12}\)
(v) \(\frac {5}{24}\)
Answer:
(i) \(\frac {3}{8}\)

Angle inside a circle is 360°.
360 is divided into 8 equal parts.
\(\frac {360}{8}\) = 45°
Each part measures 45°.
Angle of shaded region = 3 × 45 = 135°

(ii) \(\frac {2}{5}\)

\(\frac {360}{5}\) = 72°
Each part measures 72°.
Shaded angle = 72 × 2 = 144°

(iii) \(\frac {4}{9}\)

\(\frac {360}{9}\) = 40°
Each part measures 40°.
Shaded angle = 4 × 40° = 160°

(iv) \(\frac {5}{12}\)

\(\frac {360}{12}\) = 30°
Each part measures 30°.
Shaded angle = 30 × 5 = 150°

(v) \(\frac {5}{24}\)

\(\frac {360}{24}\) = 15°
Each part measures 15°.
Shaded angle = 15 × 5 = 75°

Question 3.
Draw the pictures below:

Answer:
Figure-1
Draw a circle and mark 72° in the centre of the circle, and complete the pattern. From the corner-1, draw a line to the corners 3 and 4 similarly, from corner-5 draw a line to the corners 2 and 3, and from corner-2 draw lines to the corners 4 and 5. And colour the picture.

Figure-2
To draw the second figure, make some changes to the above second figure.

Figure-3
Draw an 8-sided figure and draw lines inside the figure to get a square.

Figure-4
Draw it like this.

Class 6 Maths Chapter 1 Kerala Syllabus Angles Questions and Answers

Class 6 Maths Angles Questions and Answers

Question 1.
Draw the angles on the lines that are marked below.

Answer:
Place the point which the bottom line and the perpendicular line joins together in the protractor at the end of the line where the angle should be drawn. For drawing the figures 4 and 5 place the protractor downwards and mark the angles and complete the angles.

Question 2.
Measure and mark each angle below:

Answer:
(i) 50°
(ii) 130°
(iii) 90°
(iv) 105°
(v) 103°

Question 3.
From the angles given below, without measuring it classify them into angles right angle, less than and greater than right angles.

Answer:
Less than right angle: a, d, h
Greater than right angle: b, e, f, g
Right angle: c

Question 4.
Measure the angles below.

Answer:
(a) 120°
(b) 60°
(c) 58°
(d) 125°
(e) 90°
(f) 22°

Question 5.
Divide the circle into 6 equal parts and draw different figures.
Answer:

Question 6.
Draw a picture with 5 and 8 sides.
Answer:
Draw a circle and mark a horizontal line (radius), and mark a 72° angle on it.
Again, mark a 72° angle with each line.

Draw a circle and mark a horizontal line (radius), and mark a 45° angle on it.
Again, mark a 45° angle with each line.

Question 7.
Measure all the angles.

Answer:
(a) 1 – 90°, 2 – 90°, 3 – 130°, 4 – 50°
(b) 1 – 150°, 2 – 120°, 3 – 60°, 4 – 30°

Question 8.
Draw these pictures with the same measurements.

Answer:

Question 9.
Calculate what fraction of the circle is the shaded portion.

Answer:
Figure-1
Angle measure of the shaded portion is 36°
Total fractions of the circle is 360° ÷ 36 = 10
The fraction of shaded portion is \(\frac {1}{10}\)

Figure-2
Angle measure of the shaded portion is 120°
Total fractions of the circle is 360° ÷ 120 = 3
The fraction of shaded portion is \(\frac {1}{3}\)

Class 6 Maths Chapter 1 Notes Kerala Syllabus Angles

→ A figure formed by two lines meeting at a point is called an angle.

→ A protractor is a simple measuring instrument that is used to measure angles.

→ It is in the shape of a semicircle with an inner scale and an outer scale and with markings from 0° to 180° on it.

→ The angle at a square corner is 90°. It is also called a right angle.

→ The measure of a circle is 360°.

An angle is formed when two lines meet at a point. The space between these lines is called an angle. Angles are measured in degrees or radians, and they are grouped based on their size. Angles are very useful in our daily lives. For example, engineers use angles to build houses, bridges, and buildings. Athletes use angles in sports to improve their movements. Carpenters use angles to make things like doors, tables, and chairs. In this chapter, we will learn how to measure angles using a protractor, and how to divide a circle into equal parts.

When Lines Join
Remember how we drew various shapes joining lines straight up and slanted in the section Line Math of the lesson Lines and Circles in the class 5 textbook. Here we draw different patterns using the set square.

If we draw two lines upward of the same length on the two ends of a line and join the top ends using another line, the lengths of the lines in the top and bottom are of the same length.

Instead of drawing the straight-up lines, two slanted lines can be drawn using the set square. We know that the length of the top and bottom lines is are same.

We learn more about the topic straight up and slanted in this unit.
A figure formed by two lines meeting at a point is called an angle.

For example: Arrange these angles based on their size.

Write the numbers of the cones in ascending order.
The angle with less spread out is 4, and the more spread out is 6.
Therefore, the smallest angle is 4 and the largest angle is 6.
Thu,s the arrangement is of the form, 4, 1, 3, 2, 5, 6.

Protractor:
A protractor is a simple measuring instrument that is used to measure angles. It is in the shape of a semicircle with an inner scale and an outer scale, and with markings from 0° to 180° on it.

Considering the inner scale, starting with the bottom line marked 0, there are other lines upward; and as they move up, the angles between them and the bottom line become larger and larger. The numbers at their ends show the sizes of these angles.

How do you measure these angles using a protractor?

Place the protractor at the corner of each angle, as shown below:

Here we can see that the angle on the second figure is 60°.
In the first figure, the slant line of this angle passes between 50 and 60 on the protractor.
We can see that the small lines in the protractor divide the gaps between the multiples of 10 into ten equal parts. Each of them shows a difference of 1°.
Here, in the first figure, the slant line goes through the fifth (slightly larger) line between 50 and 60. That means the angle is 55°.

The angle at a square corner is 90°. It is also called a right angle.

A line making an angle of 90° with another line is said to be perpendicular to the first line.

For example, from the following figures, find out the perpendicular lines.

The lines in Figures 1, 6, and 8 are perpendiculars.
The other figures the angles does not form 90°; therefore, the lines are not perpendicular.

Drawing Angles
Draw an angle of 40° using a protractor.
First, draw a horizontal line. Then place the protractor at its left end as shown below and mark a point at the number 40 in it.

Now remove the protractor and join this point and the left endpoint of the first line to get a 40° angle:

For example: Draw an angle measure of 80°.
First, draw a horizontal line. Then, place the point where the bottom line and the perpendicular line join together in the protractor at the end of the line where the angle should be drawn.

Then mark the point 80° and join this point.

Circle Division
A protractor is a semicircle with 180° measure. Then for a circle, its measure is 360°.
That is, the measure of the angle around the centre of a circle is 360°.
In a circle, 360 equally spaced radii are drawn, and then the angle between any two nearby radii is 1°.
Here, the circle is divided into 360 equal parts.

What if we divide a circle into 4 equal parts?
Its one part is, 360° ÷ 4 = 90°

What if we draw radii 10° apart?
We get 36 radii, which divide the circle into 36 equal parts.

For example, divide a circle into 8 equal parts.
360° ÷ 8 = 45°
Here we get 8 radii, which are 45° apart.

The angle measures of some other portions are;

Expert Teachers at HSSLive.Guru has created Kerala Syllabus 6th Standard Maths Textbook Solutions Notes Pdf Download English & Malayalam Medium of SCERT Class 6 Maths Solutions are part of Kerala Syllabus 6th Standard Textbooks Solutions. Here HSSLive.Guru has given SCERT 6th Class Maths Textbook Solutions of Class 6 Maths Solutions Kerala Syllabus Part 1 and Part 2.

SCERT Class 6 Maths Solutions Pdf

Kerala Syllabus 6th Standard Maths Solutions Pdf Part 1

• Class 6 Maths Chapter 1 Kerala Syllabus – Angles Class 6 Questions and Answers
• Class 6 Maths Chapter 2 Kerala Syllabus – One Fraction Many Forms Class 6 Questions and Answers
• Class 6 Maths Chapter 3 Kerala Syllabus – Volume Class 6 Questions and Answers
• Class 6 Maths Chapter 4 Kerala Syllabus – Arithmetic of Parts Class 6 Questions and Answers
• Class 6 Maths Chapter 5 Kerala Syllabus – Decimal Forms Class 6 Questions and Answers
• Class 6 Maths Chapter 6 Kerala Syllabus – Multiples and Factors Class 6 Questions and Answers

Class 6 Maths SCERT Solutions Part 2

• Class 6 Maths Chapter 7 Kerala Syllabus – Money Math Class 6 Questions and Answers
• Class 6 Maths Chapter 8 Kerala Syllabus – Unchanging Relations Class 6 Questions and Answers
• Class 6 Maths Chapter 9 Kerala Syllabus – Lines and Angles Class 6 Questions and Answers
• Class 6 Maths Chapter 10 Kerala Syllabus – Number Relations Class 6 Questions and Answers
• Class 6 Maths Chapter 11 Kerala Syllabus – Letter Math Class 6 Questions and Answers
• Class 6 Maths Chapter 12 Kerala Syllabus – Data Pictures Class 6 Questions and Answers

6th Standard Maths Question Paper with Answers Kerala Syllabus

• 6th Standard Maths Annual Exam Question Paper 2023-24
• 6th Standard Maths First Term Question Paper 2022-23
• 6th Standard Maths Second Term Question Paper 2022-23
• 6th Standard Maths Annual Exam Question Paper 2022-23
• 6th Standard Maths Annual Exam Question Paper 2021-22
• 6th Standard Maths First Term Model Question Paper
• 6th Standard Maths Mid Term Model Question Paper
• 6th Standard Maths Annual Exam Model Question Paper

SCERT Class 6 Maths Solutions Malayalam Medium

6th Std Maths Textbook Solutions Pdf Malayalam Medium Part 1

• Class 6 Maths Chapter 1 Malayalam Medium – കോണുകൾ
• Class 6 Maths Chapter 2 Malayalam Medium – ശരാശരി
• Class 6 Maths Chapter 3 Malayalam Medium – ഭിന്നസംഖ്യകൾ
• Class 6 Maths Chapter 4 Malayalam Medium – വ്യാപ്തം
• Class 6 Maths Chapter 5 Malayalam Medium – ദശാംശരൂപങ്ങൾ

6th Class Maths Textbook Solutions Pdf Malayalam Medium Part 2

• Class 6 Maths Chapter 6 Malayalam Medium – സംഖ്യകൾ
• Class 6 Maths Chapter 7 Malayalam Medium – ദശാംശരീതി
• Class 6 Maths Chapter 8 Malayalam Medium – കോണുകൾ ചേരുമ്പോൾ
• Class 6 Maths Chapter 9 Malayalam Medium – നൂറിൽ എത്ര
• Class 6 Maths Chapter 10 Malayalam Medium – അക്ഷരഗണിതം
• Class 6 Maths Chapter 11 Malayalam Medium – സ്ഥിതിവിവരക്കണക്കുകൾ

Class 6 Maths Previous Year Question Paper Kerala Syllabus Malayalam Medium

• 6th Standard Maths First Term Question Paper 2023-24
• 6th Standard Maths Second Term Question Paper 2023-24
• 6th Standard Maths Annual Exam Question Paper 2023-24
• 6th Standard Maths First Term Question Paper 2022-23
• 6th Standard Maths Second Term Question Paper 2022-23
• 6th Standard Maths Annual Exam Question Paper 2022-23
• 6th Standard Maths Annual Exam Question Paper 2021-22
• 6th Standard Maths Second Term Question Paper 2019-20
• 6th Standard Maths First Term Model Question Paper
• 6th Standard Maths Mid Term Model Question Paper
• 6th Standard Maths Annual Exam Model Question Paper

We hope the given Class 6 Maths SCERT Solutions, Class 6 Kerala Syllabus Maths Solutions English Medium will help you. If you have any queries regarding 6th Std Maths Textbook Solutions Kerala Syllabus, 6th Standard Maths Textbook Answers Solutions Notes Pdf Part 1 and Part 2, drop a comment below and we will get back to you at the earliest.

Practicing with SCERT Kerala Syllabus 6th Standard Hindi Textbook Solutions Unit 3 Chapter 2 चलें Chalen Question Answer Notes Summary in Malayalam & Hindi improves language skills.

Chalen Class 6 Question Answer Notes Summary

SCERT Class 6 Hindi Unit 3 Chapter 2 Question Answer Kerala Syllabus चलें

Chalen Question Answer

चलें पाठ के आधार पर दिए प्रश्नों के उत्तर

प्रश्न 1.
क्रम में लिखें।
ക്രമമായി എഴുതാം.

अम्मा ने सपना देखा वो चुक्कू को स्कूटी पर बिठाए स्कूल छोड़ने जा रही है।
• अम्मा ने टुक्कू के लिए साइकिल और अपने लिए एक स्कूटी खरीद लीं।
• अम्मा की कोहनी और घुटने छिल गए।
• अम्मा और टुक्कू एक-दूसरे के गले में बाँहें डालकर सो गए।
• साइकिल, टुक्कू दोनों गिर पड़े।
• सुबह टुक्कू ने अपनी साइकिल निकाली और अम्मा ने अपनी स्कूटी।

• अम्मा ने टुक्कू के लिए साइकिल और अपने लिए एक स्कूटी खरीद लीं।
• …………………………………………………………………………
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उत्तर:
• अम्मा ने टुक्कू के लिए साइकिल और अपने लिए एक स्कूटी खरीद लीं ।
അമ്മ ടുക്കുവിനായി ഒരു സൈക്കിളും തനി ക്കായി ഒരു സ്കൂട്ടറും വാങ്ങി.

• सुबह टुक्कू ने अपनी साइकिल निकाली और अम्मा ने अपनी स्कूटी।
രാവിലെ ടുക്കു അവന്റെ സൈക്കിളും അമ്മ തന്റെ സ്കൂട്ടറും എടുത്തു.

• साइकिल, टुक्कू दोनों गिर पड़े।
സൈക്കിളും ടുക്കുവും രണ്ടുപേരും വീണു.

• अम्मा की कोहनी और घुटने छिल गए ।
അമ്മയുടെ കൈമുട്ടിലേയും കാൽ മുട്ടിലേയും തൊലി ഉരിഞ്ഞുപോയി.

• अम्मा और टुक्कू एक-दूसरे के गले में बाँहें डालकर सो गए।
അമ്മയും ടുക്കുവും പരസ്പരം കെട്ടിപ്പിടിച്ചു കിടന്ന് ഉറങ്ങി.

• अम्मा ने सपना देखा वो चुक्कू को स्कूटी पर बिठाए स्कूल छोड़ने जा रही है।
അമ്മ ടുവിനെ സ്കൂട്ടറിൽ ഇരുത്തി സ്കൂളിൽ വിടുന്നതായി സ്വപ്നം കണ്ടു.

प्रश्न 2.
समर्थन करें [✓] सही वाक्य चुनकर लिखें ।
സമർത്ഥിക്കാം. ഏതാണ് ശരിയെന്നും, ശരി
യായ വാക്യം തിരഞ്ഞെടുത്ത് എഴുതാം.

उत्तर:
1.
बाइक चलाते समय हेलमेट पहलें ।
ബൈക്ക് ഓടിക്കുന്ന സമയത്ത് ഹെൽമറ്റ് ധരി ക്കാം.

2.
कार चलाते समय सीट बैल्ट पहनें।
കാർ ഓടിക്കുന്ന സമയത്ത് സീറ്റ് ബെൽറ്റ് ധരിക്കാം.

3.
एक बाइक में दो ही सवार हों ।
ഒരു ബൈക്കിൽ രണ്ട് യാത്രക്കാർ മാത്രമേ ഉണ്ടാ കാൻ പാടുള്ളൂ.

4.
गाड़ी चलाते समय मोबाइल का इस्तेमाल न करें।
വാഹനം ഓടിക്കുമ്പോൾ മൊബൈൽ ഫോൺ ഉപയോഗിക്കാൻ പാടില്ല.

प्रश्न 3.
टुक्कू की डायरी लिखें।
ടുക്കുവിന്റെ ഡയറി എഴുതാം.

उत्तर:
मंगलवार
12/06/2025

आज एक मज़ेदार दिन है। कई दिनों से मेरे मन में एक आशा थी कि एक साईकिल खरीदने के लिए। अगले दिन अम्मा ने मेरे लिए साइकिल और अपने लिए एक स्कूटी खरीद लीं। उस रात हम दोनों को नींद न आई क्योंकि हम दोनों को यह चलाना नहीं आता। अगले दिन हम दोनों ने यह चलाने के लिए कोशिश किये। लेकिन हम दोनों स्कूटर और साइकिल से नीचे गिर पड़ा। मेरे सिर पर गूमड़ निकल आया और अम्मा की कोहनी और घुटने छिल गए। रात में एक स्वप्न देखा कि अम्मा को साइकिल पर बिठाकर बाज़ार घुमाने की और अम्मा ने स्वप्न देखा कि वह मुझे स्कूटी पर बिठाए स्कूल छोड़ने जा रही है। सचमुच में बहुत मज़ा आया ।

ചൊവ്വ
12/6/2025

ഇന്ന് നല്ല രസകരമായ ഒരു ദിവസമായിരുന്നു. കുറെ ദിവസങ്ങളായി എന്റെ മനസ്സിൽ ഒരു ആഗ്രഹം ഉണ്ടായിരുന്നു ഒരു സൈക്കിൾ വാങ്ങു ന്നതിനായി. അടുത്ത ദിവസം തന്നെ അമ്മ എനി ക്കായി ഒരു സൈക്കിളും അമ്മയ്ക്കായി ഒരു സ്കൂട്ടറും വാങ്ങി. ആ രാത്രി ഞങ്ങൾ രണ്ടു പേർക്കും ഉറക്കം വന്നില്ല. കാരണം ഞങ്ങൾക്ക് രണ്ടുപേർക്കും സൈക്കിൾ ഓടിക്കാൻ അറിയി ല്ലായിരുന്നു. അടുത്ത ദിവസം ഞങ്ങൾ രണ്ടു പേരും അത് ഓടിക്കാൻ പരിശ്രമിച്ചു. എന്നാൽ ഞാൻ സൈക്കിളിൽ നിന്നും, അമ്മ സ്കൂട്ടറിൽ നിന്നും താഴെ വീണു. എന്റെ തല മുഴച്ചു പൊങ്ങി വന്നു, അമ്മയുടെ കൈമുട്ടിലേയും കാൽമുട്ടി ലേയും തൊലി ഇളകി വന്നു. രാത്രിയിൽ ഞാൻ ഒരു സ്വപ്നം കണ്ടു. അമ്മയെ സൈക്കിളിൽ ഇരുത്തി ഞാൻ ചന്തയിൽ ചുറ്റിക്കറങ്ങുന്നതായും, അമ്മ സ്വപ്നം കണ്ടത് എന്നെ സ്കൂട്ടറിൽ ഇരുത്തി സ്കൂളിൽ കൊണ്ടു വിടുന്നതുമാണ്. വാസ്തവത്തിൽ വളരെ രസകരമായിരുന്നു.

प्रश्न 4.
लिखें, चित्र क्या कह रहा है?
ചിത്രം എന്താണ് പറയുന്നതെന്ന് എഴുതാം.

उत्तर:
• टुक्कू अम्मा को साइकिल पर बिठाकर जाता है।
ടു അമ്മയെ സൈക്കിളിൽ ഇരുത്തി കൊണ്ടുപോകുന്നു.

• टुक्कू के चारों ओर उसके दोस्त खड़े हैं।
ടുക്കുവിന്റെ നാലു വശത്തും അവന്റെ സുഹൃ ത്തുക്കൾ നിൽപ്പുണ്ടായിരുന്നു.

• स्कूल के सामने अम्मा और टुक्कू स्कूटर पर बैठे है ।
സ്കൂളിന്റെ മുൻപിൽ അമ്മയും ടുക്കുവും സ്കൂട്ടറിൽ ഇരിക്കുന്നു.

• साइकिल बैठे टुक्कू को अम्मा टाटा देती है ।
സൈക്കിളിൽ ഇരിക്കുന്ന ടുക്കുവിന് അമ്മ റ്റാറ്റ കൊടുക്കുന്നു.

• स्कूटर पर बैठी अम्मा को टुक्कू टिफिन बॉक्स देती है ।
സ്കൂട്ടറിൽ ഇരിക്കുന്ന അമ്മയ്ക്ക് ടുക്കു ടിഫിൻ ബോക്സ് നൽകുന്നു.

चलें पाठ के अन्य प्रश्न और उत्तर

प्रश्न 1.
टुक्कू कई दिनों से किस को माँगने के लिए जिद कर रहा था?
ടുക്കു കുറെ ദിവസമായി എന്തു വാങ്ങുന്നതി നായിട്ടാണ് നിർബന്ധം പിടിച്ചിരുന്നത്?
उत्तर:
साइकिल माँगने के लिए
സൈക്കിൾ വാങ്ങുന്നതിനായി

प्रश्न 2.
अम्मा उसके लिए क्या खरीदी ?
അമ്മ അവൾക്കുവേണ്ടി എന്ത് വാങ്ങി?
उत्तर:
स्कूटी खरीदी
സ്കൂട്ടർ വാങ്ങി

प्रश्न 3.
“उस रात दोनों को नींद न आई।” क्यों ?
ആ രാത്രി രണ്ടുപേർക്കും ഉറക്കം വന്നില്ല. എന്തു കൊണ്ട്?
उत्तर:
क्योंकि दोनों को साइकिल और स्कूटर चलाना नहीं आता। साथ में दोनों बहुत उत्साहित थे ।
കാരണം രണ്ടുപേർക്കും സൈക്കിളും സ്കൂട്ടറും ഓടിക്കാൻ അറിയില്ലായിരുന്നു. കൂടാതെ അവർ വളരെ ആവേശഭരിതനായിരുന്നു.

प्रश्न 4.
“ज़्यादा दूर मत जाना।” अम्मा ने अम्मा ने टुक्कू को समझाया। “आप भी ज़्यादा दूर मत जाना” – टुक्कु ने अम्मा को समझाया। यहाँ दोनों का कौन – सा मनोभाव प्रकट होता है ?
“ദൂരെ ഒന്നും പോകരുത്”. അമ്മ ടുക്കുവിനെ മനസ്സിലാക്കി കൊടുത്തു. “അമ്മയും ദൂരെ ഒന്നും പോകരുത്” – ടുക്കുവും മനസ്സിലാക്കി കൊടുത്തു. ഇവിടെ രണ്ടുപേരുടേയും എന്ത് മനോഭാവമാണ് പ്രകടമാകുന്നത്?
उत्तर:
यहाँ दोनों का आपसी देखभाल प्रकट होता है। क्योंकि दोनों को स्कूटी और साइकिल चलाने को नहीं आता।
ഇവിടെ രണ്ടുപേരുടേയും പരസ്പരമുള്ള കരുത ലാണ് പ്രകടമാകുന്നത്. കാരണം രണ്ടുപേർക്കും സ്കൂട്ടറും സൈക്കിളും ഓടിക്കാൻ അറിയില്ലായി രുന്നു.

प्रश्न 5.
टुक्कू के सिर पर गूमड़ निकल आया । कैसे ?
ടുക്കുവിന്റെ തലയിൽ മുഴച്ചു പൊങ്ങി വന്നു. എങ്ങനെ?
उत्तर:
टुक्कू साइकिल चलाने को कोशिश करने लगा । इसी वक्त वह साइकिल से नीचे गिर पड़ा। इसी कारण
उसके सिर पर गूमड़ निकल आया ।
ടുക്കു സൈക്കിൾ ഓടിക്കുന്നതിനായി പരിശ്രമി ക്കാൻ തുടങ്ങി. ഈ സമയത്ത് അവൻ സൈക്കി ളിൽ നിന്ന് താഴെ വീണു. ഇക്കാരണത്താലാണ് അവന്റെ തലയിൽ മുഴച്ചു പൊങ്ങിവന്നത്.

प्रश्न 6.
साइकिल गिरकर अम्मा को क्या हुई ?
സൈക്കിളിൽ നിന്ന് വീണിട്ട് അമ്മയ്ക്ക് എന്ത് സംഭവിച്ചു?
उत्तर:
अम्मा को कोहनी और घुटने छिल गए ।
അമ്മയുടെ കാൽമുട്ടിലെയും കൈമുട്ടിലേയും തൊലി ഇളകി.

प्रश्न 7.
अम्मा बोली, “अब नहीं चलाऊँगी स्कूटी ।” अम्मा क्यों ऐसे कहा ?
അമ്മ പറഞ്ഞു, “ഇനിയൊരിക്കലും ഞാൻ സ്കൂട്ടർ ഓടിക്കില്ല.” അമ്മ എന്തുകൊണ്ടാണ് അങ്ങനെ പറഞ്ഞത്?
उत्तर:
अम्मा स्कूटर से गिरता देखकर लोग हँस रहे थे और उसकी कोहनी और घुटने छिल गए।
അമ്മ സ്കൂട്ടറിൽ നിന്ന് വീഴുന്നതു കണ്ട് ആൾക്കാർ ചിരിച്ചു കൊണ്ടിരുന്നു. അമ്മയുടെ കാൽ മുട്ടിലേയും കൈമുട്ടിലേയും തൊലി ഇളകു കയും ചെയ്തു.

प्रश्न 8.
“ गिर गए तो क्या । मज़ा तो आया ।” ऐसा कोई अनुभव आप को है? बताएँ ।
“വീണെങ്കിൽ എന്താ, നല്ല രസമല്ലായിരുന്നോ” ഇങ്ങനെയുള്ള അനുഭവം നിങ്ങൾക്കുണ്ടായി ട്ടുണ്ടോ? ഉണ്ടെങ്കിൽ പറയാം.
उत्तर:
हाँ, मुझे भी हुआ है। मैं अपने भाई के साथ साइकिल चलाते समय हमेशा नीचे गिरता है । क्योंकि मुझे साइकिल चलाने के लिए अच्छे से नहीं आता। लेकिन मुझे भी तो उसी वक्त मज़ा आया।
ഉണ്ട്, എനിക്കും അങ്ങനെയുള്ള അനുഭവം ഉണ്ടാ യിട്ടുണ്ട്. ഞാൻ എന്റെ സഹോദരനോടൊപ്പം
സൈക്കിൾ ചവിട്ടുന്ന സമയത്ത് എപ്പോഴും താഴെ വീഴാറുണ്ട്. കാരണം എനിക്കും സൈക്കിൾ ചവി ട്ടാൻ നന്നായിട്ട് അറിയില്ല. എന്നാൽ എനിക്കും ആ സമയത്ത് വളരെ രസകരമായി തോന്നിയിട്ടുണ്ട്.

प्रश्न 9.
स्कूटी पर बैठकर अम्मा को कैसे लगी?
സ്കൂട്ടറിൽ ഇരുന്നപ്പോൾ അമ്മയ്ക്ക് എങ്ങനെ യാണ് തോന്നിയത്?
उत्तर:
वह हवा में तैर रही है।
അമ്മ വായുവിൽ നീന്തുവാണെന്ന് തോന്നി.

प्रश्न 10.
अम्मा और टुक्कू क्या तय किये ?
അമ്മയും ടുക്കുവും എന്താണ് തീരുമാനിച്ചത്?
उत्तर:
कल फिर साइकिल और स्कूटी चलाने की कोशिश करेंगे।
നാളെ വീണ്ടും സൈക്കിളും സ്കൂട്ടറും ഓടിക്കാൻ പരിശ്രമിക്കുമെന്ന്.

प्रश्न 11.
सड़क सुरक्षा पर पोस्टर बनाइए ।
റോഡ് സുരക്ഷയെപ്പറ്റി ഒരു പോസ്റ്റർ തയ്യാറാ ക്കുക.
उत्तर:

“अब नहीं चलाऊँगी स्कूटी।”

इस वाक्य में ‘चलाऊँगी भविष्यतकाल है।
भविष्यतकाल, जिसे अंग्रेज़ी में ‘future tense’ कहते हैं, हिन्दी व्याकरण में क्रिया का वह रूप है जो आने वाले समय में होने वाली क्रिया का घटन को दर्शाता है। दूसरे शब्दों में, यह बताता है कि कोई काम अभी नहीं हुआ है, बल्कि भविष्य में होगा ।

ഇംഗ്ലീഷിൽ ‘future tense’ എന്ന് വിളിക്കപ്പെടുന്ന भविष्यतकाल ഹിന്ദി വ്യാകരണത്തിലെ ക്രിയയുടെ ഒരു രൂപമാണ്. അത് വരാനിരിക്കുന്ന സമയത്ത് സംഭവിക്കുന്ന പ്രവൃത്തിയേയോ സംഭവത്തേയോ കാണിക്കുന്നു. മറ്റൊരു വിധത്തിൽ പറഞ്ഞാൽ, ചില പ്രവൃത്തികൾ ഇതുവരെ സംഭവിച്ചിട്ടി ല്ലെന്നും ഭാവിയിൽ സംഭവിക്കുമെന്നും അത് പറ യുന്നു.

उदाहरण के लिए,

• कल स्कूल जाऊँगा।
ഞാൻ നാളെ സ്കൂളിൽ പോകും.

• वह आज रात खाना खाएगी।
അവൾ ഇന്ന് രാത്രി അത്താഴം കഴിക്കും.

• हम सब फिल्म देखने जाएँगे ।
നമ്മളെല്ലാവരും സിനിമ കാണാൻ പോകും.

• तुम कब आओगे ?
നീ എപ്പോൾ വരും?

Chalen Summary in Malayalam

चलें Summary in Malayalam

चलें (പോകാം)
കഥയുടെ മലയാള പരിഭാഷ

ടുക്കു കുറെ ദിവസമായി സൈക്കിളിനായി വാശി പിടിച്ചുകൊണ്ടിരിക്കുകയായിരുന്നു. അമ്മ ടുക്കുവിനു വേണ്ടി ഒരു സൈക്കിളും തനിക്കുവേണ്ടി ഒരു സ്കൂട്ടറും വാങ്ങി. “പക്ഷേ അമ്മയ്ക്ക് സ്കൂട്ടർ ഓടിക്കാൻ അറിയില്ലല്ലോ?”- ടുക്കു അമ്മയോട് ചോദിച്ചു. “നിനക്കും സൈക്കിൾ ഓടിക്കാൻ അറിയില്ലല്ലോ’ അമ്മ ചോദിച്ചു. “ഞാൻ പതിയെ പതിയെ പഠിക്കും’ – ടുക്കു പറഞ്ഞു. “ഞാനും പഠിക്കും പതിയെ പതിയെ. അമ്മ ചിരിച്ചു കൊണ്ടു പറഞ്ഞു. ആ രാത്രി രണ്ടുപേർക്കും ഉറക്കം വന്നില്ല.

രാവിലെ ടുക്കു തന്റെ സൈക്കിൾ എടുത്തു. അമ്മ തന്റെ സ്കൂട്ടറും. “ദൂരെ ഒന്നും പോകരുത്” അമ്മ ടുക്കു വിനെ പറഞ്ഞു മനസ്സിലാക്കി കൊടുത്തു. “അമ്മയും ദൂരെ എങ്ങും പോകരുത്” ടുക്കുവും പറഞ്ഞു കൊടുത്തു.

ടുക്കു ആദ്യം പകുതി പെഡൽ ചവിട്ടി. പകുതി ചവിട്ടി ചവിട്ടി അവൻ സീറ്റിലിരി ക്കാൻ ശ്രമിച്ചു. പക്ഷെ എങ്ങനെയൊക്കെ സീറ്റിലിരിക്കുമ്പോഴും ഹാൻഡിൽ തിരി യും. കാല് പെഡൽ വരെ എത്തില്ല. ബാലൻസ് തെറ്റും. സൈക്കിളും, ടുക്കുവം രണ്ടുപേരും താഴെ വീഴും. പൊത്തോ അവിടെ അമ്മ സ്കൂട്ടർ ഓടിക്കാൻ പഠി ക്കുകയാണ്. അവർ രണ്ടു കൈയും കൊണ്ട് ഹാൻഡിലിൽ മുറുകെ പിടിച്ചു. ഇടത് കൈ കൊണ്ട് ബ്രേക്ക് അമർത്തി.

വലതു കൈ കൊണ്ട് ആക്സിലേറ്റർ തിരിച്ചു. എന്നാൽ സ്കൂട്ടർ അവിടെ തന്നെ നിന്നും. പിന്നീട് അമ്മ ബ്രേക്കിൽ നിന്ന് പിടിവിട്ടു. ബ്രേക്ക് വിട്ടപ്പോഴേക്കും സ്കൂട്ടർ കുതിച്ചു. എന്നിട്ട് പിൻചക്രത്തിൽ നിന്നു. അമ്മയും സ്കൂട്ടറും രണ്ടുപേരും താഴെ വീണു. പൊത്തോ!

ടുക്കുവിന്റെ തല മുഴച്ചു പൊങ്ങി. അമ്മയുടെ കൈമുട്ടിലേയും കാൽമുട്ടിലെയും തൊലി ഉരഞ്ഞ് ഇളകി. രാത്രിയിൽ രണ്ടുപേരും കിടക്കയിൽ കിടക്കുകയായിരുന്നു. അമ്മ പറഞ്ഞു “ഇനി മേലിൽ ഞാൻ സ്കൂട്ടർ ഓടിക്കില്ല. ആളുകൾ ഞാൻ വീണത് കണ്ടിട്ട് ചിരിച്ചു കൊണ്ടിരിക്കുകയായിരുന്നു.” “വീണെങ്കിൽ എന്താ? നല്ല രസമല്ലായിരുന്നോ!” ടുക്കു പറഞ്ഞു.

“കാര്യമൊക്കെ ശരിയാണ്” – സ്കൂട്ടറിൽ ഇരുന്നപ്പോൾ എനിക്ക് തോന്നിയത് ഞാൻ വായുവിൽ നീന്തു വായിരുന്നു എന്നാണ്” – രണ്ടുപേരും നാളെ വീണ്ടും ശ്രമിക്കും എന്ന് തീരുമാനമെടുത്തു. രണ്ടുപേരും പരസ്പരം കെട്ടിപിടിച്ച് ഉറങ്ങി. അമ്മ അന്നു രാത്രി അവൾ ടുക്കുവിനെ സ്കൂട്ടറിൽ ഇരുത്തി സ്കൂളിൽ വിടുന്നതായി സ്വപ്നം കണ്ടു. അതേ സമയം ടുക്കുവും ഒരു സ്വപ്നം കാണുകയായിരുന്നു. അവൾ അമ്മ യേയും സൈക്കിളിൽ ഇരുത്തി ചന്തയിലൊക്കെ ചുറ്റിക്കറങ്ങുന്നതായിട്ട്.

चलें लेखक परिचय

लेखक परिचय

श्रीमती गुलसारिका ठाकुर एक हिन्दी साहित्यकार है, जिन्होंने “कुछ बूढ़ी उदास औरतें” नामक कविता संग्रह के लिए युवा साहित्य अकादमी पुरस्कार जीता है। उनका जन्म सन् 1970 में हुआ था । सारिका ठाकुर ने अपनी पढ़ाई के दौरान हिन्दी साहित्य में रुचि विकसित की और लेखन शुरू किया। उनकी कविताएँ विभिन्न पत्रिकाओं में प्रकाशित हुई ।

वे हिन्दी साहित्य के बाल साहित्यकार, अनुवादक, पत्रकार, पटकथा लेखक आदि भी है। उनका जन्म स्थान बिहार के दरभंगा में था। उनकी शिक्षा के बारे में कहें तो, उनको तीन विषयों में ( दर्शन शास्त्र, समाज शास्त्र एवं पत्रकारिता) में स्नातकोत्तर की उपाधी थी। उनकी प्रमुख रचनाएँ हैं मेरी बीरहखड़ी, पर्यटन संदर्भ अद्वितीय मध्यप्रदेश, पर्यटन संदर्भ अतुलनीय छत्तीसगढ़ आदि । उल्लेखनीय कार्य यह है कि वे बालपत्रिका ‘चकमक’ का सह संपादक है।

ശ്രീമതി. ഗുൽസാരിക ഠാക്കൂർ ഹിന്ദിയിലെ ഒരു പ്രസിദ്ധയായ സാഹിത്യകാരിയാണ്. അവർക്ക് അവരുടെ കവിതാ സമാഹാരമായ “കുച് ബൂഢി ഉദാസ് ഔരതേം’-ന് യുവസാഹിത്യ അക്കാദമി അവാർഡ് ലഭിച്ചി ട്ടുണ്ട്. അവരുടെ ജന്മം 1970-ൽ ആയിരുന്നു. അവർക്ക് തന്റെ പഠനം തുടരുന്നതോടൊപ്പം തന്നെ ഹിന്ദി സാഹിത്യത്തോട് താല്പര്യം തോന്നുകയും ലേഖനങ്ങൾ എഴുതാൻ തുടങ്ങുകയും ചെയ്തു. അവർ ഹിന്ദി സാഹിത്യത്തിലെ മികച്ച ബാല സാഹിത്യകാരിയും, വിവർത്തകയും, പത്രപ്രവർത്തകയും, തിരക്ക ഥാകൃത്തുമാണ്. അവരുടെ ജന്മസ്ഥലം ബീഹാറിലെ ദർഭംഗയിലാണ്. അവരുടെ വിദ്യാഭ്യാസത്തെ സംബ ന്ധിച്ച് പറയുകയാണെങ്കിൽ അവർക്ക് മൂന്ന് വിഷയത്തിൽ ബിരുദാനന്തര ബിരുദം ലഭിച്ചിട്ടുണ്ട്. (ഫിലോ സഫിയിലും, സാമൂഹിക ശാസ്ത്രത്തിലും, പത്രപ്രവർത്തനത്തിലും). അവരുടെ പ്രധാന രചനകൾ ഇവ യാണ് – മേരി ബീഹര്ഖടി, പര്യടൻ സംദർഭ് അദ്വതീയ് മധ്യപ്രദേശ്, പര്യടൻ സന്ദർഭ് അതുലനീയ് ഛത്തീ ഗഢ്. അവരിൽ ശ്രദ്ധേയമായ കാര്യം ഇതാണ് – അവർ ബാലമാസികയായ ചകിന്റെ അസോസി യേറ്റ് എഡിറ്ററാണ്.

चलें शब्दार्थ

कई दिन – കുറെ ദിവസം
साइकिल – സൈക്കിൾ
जिद करना – വാശി പിടിക്കുക
खरीदना – വാങ്ങുക
साइकिल चलाना – സൈക്കിൾ ഓടിക്കുക
पूछना – ചോദിക്കുക
सीखना – പഠിക്കുക
धीरे धीरे – പതിയെ പതിയെ
बोलना – പറയുക
हँसना – ചിരിക്കുക
रात – രാത്രി
दोनों – രണ്ടുപേരും
नींद – ഉറക്കം
सुबह – ചെലവഴിക്കുക
आधा – പകുതി
पैडल मारा – പെഡൽ
बैठना – ഇരിക്കുക
पहूँचना – എത്തുക
बैलेंस बिगड़ना – ബാലൻസ് തെറ്റുക
गिर पड़ना – താഴെ വീഴുക
हैंडल – ഹാൻഡിൽ
बाएँ हाथ – ഇടതു കൈ
ब्रेक दबाया – ബ്രേക്ക് പിടിക്കുക / ബൈക്കിൽ അമർത്തുക
उछलना – കുതിക്കുക
पहिए – ചക്രം
बिस्तर – കിടക്ക
हँसना – ചിരിക്കുക
गूमड़ निकल आना – തല മുഴച്ചു പൊങ്ങി വരുക
कोहनी और घुटने छिल गए – കൈമുട്ടിലേയും കാൽമുട്ടി ലേയും തൊലി ഇളകി വരുക
सही – ശരി
तय करना – തീരുമാനിക്കുക
निकालना – എടുക്കുക
ज़्यादा – കുറച്ചധികം
दूर – ദൂരെ
मत जाना – പോകരുത്
साझाया – മനസ്സിലാക്കി കൊടുക്കുക
पहले – ആദ്യം
कोशिश करना – പരിശ്രമിക്കുക
घूमना – തിരിയുക
पाँव – കാൽ
कसकर थामना – മുറുകെ പിടിക്കുക
पिछले – പുറകിലെ
सिर – തല
लेटना – കിടക്കുക
मज़ा आया – രസം ആകുക
तैरना – നീന്തുക
एक दूसरे के गले में बाँहें डालकर – പരസ്പരം കെട്ടിപ്പിടിക്കുക
सपना – സ്വപ്നം
सोना – ഉറങ്ങുക
स्कूल छोडना – സ്കൂളിൽ വിടുക
वक्त – സമയം
बिठाकर – ഇരുത്തികൊണ്ട്
बाज़ार – ചന്ത
घुमाना – ചുറ്റിക്കറങ്ങുക

Practicing with SCERT Kerala Syllabus 6th Standard Hindi Textbook Solutions Unit 3 Chapter 1 मीठा संतरा Meeta santara Question Answer Notes Summary in Malayalam & Hindi improves language skills.

Meeta santara Class 6 Question Answer Notes Summary

SCERT Class 6 Hindi Unit 3 Chapter 1 Question Answer Kerala Syllabus मीठा संतरा

Meeta santara Question Answer

चित्र में एक दूकान है। उस दूकान में शीशे में तरह-तरह के मिठाइयाँ रखते हैं । दूकान के सामने लड़के-लड़कियाँ खड़े है। ऐसा लग रहा है कि स्कूल जाते समय वे मिठाइयाँ खरीदने के लिए खड़े रहे हैं । दूकान में विभिन्न प्रकार के पकवान रखे हुए हैं। दूकान के सामने बड़े-बड़े पेड़ और एक घर भी देखने को मिलता है।

ചിത്രത്തിൽ ഒരു കടയുണ്ട്. കടയിൽ വിവിധ തരത്തിലുള്ള മിഠായികൾ കുപ്പിയിൽ അടച്ചു സൂക്ഷിച്ചിരി ക്കുന്നു. കടയുടെ മുൻപിൽ ആൺകുട്ടികളും പെൺകുട്ടികളും നിൽപുണ്ട്. സ്കൂളിൽ പോകുന്ന സമയത്ത് അവർ മിഠായി വാങ്ങാനായി നിൽക്കുകയാണെന്ന് തോന്നുന്നു. കടയിൽ വിവിധ തരത്തിലുള്ള പലഹാ രങ്ങൾ വച്ചിട്ടുണ്ട്. കടയുടെ മുൻപിൽ വലിയ വൃക്ഷങ്ങളും ഒരു വീടും കാണാൻ സാധിക്കുന്നുണ്ട്.

मीठा संतरा पाठ के आधार पर दिए प्रश्नों के उत्तर

प्रश्न 1.
लिखें, क्या – क्या हैं?
എഴുതാം. ചിത്രത്തിൽ എന്തെല്ലാം ഉണ്ട്?

उत्तर:

प्रश्न 2.
लिखें, किसने कहा (ആര് പറഞ്ഞു)

उत्तर:

प्रश्न 3.
कहानी से गुज़रें, प्रस्ताव पर [✗/✓] लगाएँ ।
കഥയിലൂടെ ഒന്ന് കടന്നുപോകാം, പ്രസ്താവന യിൽ ശരിയോ തെറ്റോ ഇടാം.

उत्तर:
बीस साल बाद 20 വർഷങ്ങൾക്കു ശേഷം

• लेखक शिलांग गया। [✓]
അതെ, വളരെയധികം മധുരമുള്ളത്.

• गुमटी सूपर मार्कट बन गया। [✗]
പെട്ടിക്കട സൂപ്പർ മാർക്കറ്റ് ആയി മാറി.

• युवती के पास उसकी छोटी-सी लड़की भी बैठी थी । [✓]
യുവതിയുടെ അടുത്ത് അവളുടെ ചെറിയ പെൺകുട്ടിയും ഇരിപ്പുണ്ടായിരുന്നു.

• लेखक को बूढ़ी कांग की याद आई। [✓]
ലേഖകന് വൃദ്ധയായ കച്ചവടക്കാരിയെപ്പറ്റി ഓർമ്മ വന്നു.

• गुमटी में वृद्धा नज़र आई। [✗]
പെട്ടിക്കടയിൽ വൃദ്ധയെ കാണാൻ സാധിച്ചു.

• गुमटी में युवती अकेली बैठी थी । [✗]
പെട്ടിക്കടയിൽ യുവതി തനിച്ചിരിക്കുകയായി രുന്നു.

प्रश्न 4.
एक दूकान का चित्र खींचें, रंग दें, नाम दें।
ഒരു കടയുടെ ചിത്രം വരയ്ക്കാം, നിറം കൊടു ക്കാം, പേര് കൊടുക്കാം.

दूकान के बारे में लिखें।
കടയെക്കുറിച്ച് എഴുതാം.
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उत्तर:
चित्र में एक फल का दूकान है। दूकान का नाम ताज़ा फल दूकान है। दूकान में तरह-तरह के फल बेचने के लिए रखता है। संतरा, तरबूजा, अमरूद, आम, सेब, केला, अंगूर ये सब इस दूकान में है । दूकान में फलों को वज़न करने के लिए तराजू भी है।

ചിത്രത്തിൽ ഒരു പഴക്കടയാണ് കാണുന്നത്. കട യുടെ പേര് ‘താജാ ഫൽ ദൂകാൻ’ എന്നാണ്. കട യിൽ പല തലത്തിലുള്ള പഴങ്ങൾ വിൽപ്പന യ്ക്കായി വച്ചിരിക്കുന്നു. ഓറഞ്ച്, തണ്ണിമത്തൻ, പേരക്ക, മാങ്ങ, ആപ്പിൾ, വാഴപ്പഴം, മുന്തിരി ഇതെല്ലാം ഈ കടയിൽ ഉണ്ട്. കടയിൽ പഴങ്ങൾ അളന്നു കൊടുക്കുന്നതിനായിട്ടുള്ള യന്ത്രവുമുണ്ട്.

प्रश्न 5.
पाठभाग से उचित शब्द ढूँढें और लिखें।
പാഠഭാഗത്തിൽ നിന്ന് ഉചിതമായ വാക്കുകൾ കണ്ടെത്തി എഴുതാം.

उत्तर:


മുകളിൽ അടിവരയിട്ട വാക്കുകളെല്ലാം അടുത്ത വാക്കിനെ വിശേഷിപ്പിക്കുന്ന വാക്കുകൾ ആണ്.
संज्ञा की विशेषता बतलाने वाला विकारी शब्द को विशेषण कहते हैं।

നാമത്തിന്റെ വിശേഷതകൾ പറയുന്ന വികാരി ശബ്ദങ്ങളെ വിശേഷണം എന്നുപറയുന്നു. മുകളിൽ തന്നിരിക്കുന്ന അടിവര ഇടാത്ത വാക്കുകളെല്ലാം നാമങ്ങളാണ്. അടിവര ഇട്ടി രിക്കുന്ന വാക്കുകൾ വിശേഷണവും.

ആരെയാണോ വിശേഷിപ്പിച്ചിരിക്കുന്നത് അതിനെ ‘विशेष्य’ എന്നു പറയും.

उदा: नीला आकाश नीला → विशेषण

प्रश्न 6.
सब्जियों और फलों के चित्र इकट्ठा करें, तालिका में नाम लिखें।
പച്ചക്കറികളുടേയും, പഴങ്ങളുടേയും ചിത്രങ്ങൾ ശേഖരിച്ച് താഴെ തന്നിരിക്കുന്ന ടേബിളിൽ പേരു കൾ എഴുതാം.

उत्तर:

फल (പഴങ്ങൾ)	सब्जी (പച്ചക്കറികൾ)
आम (മാങ്ങ)	गाजर (കാരറ്റ്)
संतरा (ഓറഞ്ച്)	प्याज (സബോള)
तरबूजा (തണ്ണിമത്തൻ)	आलू (ഉരുളക്കിഴങ്ങ്)
अंगूर (മുന്തിരി)	टमाटर (തക്കാളി)
अमरूद (പേരയ്ക്ക)	बैंगन (വഴുതനങ്ങ)
सेब (ആപ്പിൾ)	चुकंदर (ബീറ്റ്റൂട്ട്)
अनन्नास (പൈനാപ്പിൾ)	मटर (പയർ)
जामुन (ഞാവൽപ്പഴം)	फूलगोभी (കോളിഫ്ളവർ)
पपीता (പപ്പായ)	कद्दू (മത്തങ്ങ)
अनार (മാതളം)	पालक (ചീര)

एलबम तैयार करें।
താഴെ തന്നിരിക്കുന്ന പഴങ്ങളുടേയും, പച്ചക്കറികളുടേയും ചിത്രങ്ങൾ കോർത്തിണക്കി ആൽബം തയ്യാറാക്കാം.


उत्तर:

मीठा संतरा पाठ के अन्य प्रश्न और उत्तर

प्रश्न 1.
शिलांग में लेखकर कहाँ रहता था ?
ഷില്ലോംഗിൽ ലേഖകൻ എവിടെയാണ് താമസി ച്ചിരുന്നത്?
उत्तर:
मल्की हिल्स में

प्रश्न 2.
उस छोटी-सी गुमटा में क्या बचता था ?
ആ ചെറിയ പെട്ടിക്കടയിൽ എന്താണ് വിറ്റിരു ന്നത്?
उत्तर:
मौसमी फल
അതാത് സീസണിലെ പഴങ്ങൾ

प्रश्न 3.
बूढ़ी कांग के साथ कभी-कभी कौन बैठी होती ?
വൃദ്ധയായ കച്ചവടക്കാരിക്ക് ഒപ്പം ചിലപ്പോ ഴൊക്കെ ആരാണ് ഇരുന്നിരുന്നത്?
उत्तर:
उसकी पोती
അവരുടെ കൊച്ചുമകൾ

प्रश्न 4.
लेखक और बूढ़ी कांग की गपशप आपको कैसे बूढ़ी कांग के साथ कभी-कभी कौन बैठी होती ?
ലേഖകന്റെയും വൃദ്ധയായ കച്ചവടക്കാരിയു ടേയും വർത്തമാനം നിങ്ങൾക്ക് എങ്ങനെ തോന്നി? എന്തുകൊണ്ട്?
उत्तर:
लेखक की गपशप बहुत मज़ाक से भरा हुआ है। इसलिए मुझे यह गपशप बहुत अच्छी लगी ।
ലേഖകന്റെ വർത്തമാനം വളരെ തമാശ നിറഞ്ഞ തായിരുന്നു. അതുകൊണ്ട് എനിക്ക് ഈ വർത്ത മാനം വളരെയധികം നന്നായി തോന്നി.

प्रश्न 5.
युवती ने पैसा लेने से इनकार किया । यहाँ युवती की कौन-सी मनोभाव प्रकट होता है ?
യുവതി പൈസ വാങ്ങാനായി വിസമ്മതിച്ചു. ഇവിടെ യുവതിയുടെ ഏതു മനോഭാവമാണ് പ്രക ടമാകുന്നത്?
उत्तर:
शायद युवती ने लेखक को पहचान लिया था। वर्षों पहले उसकी नानी के गुमटी के ग्राहक थे लेखक । युवती के मन में उनके प्रति कृतज्ञता का भाव प्रकट होता है।
ഒരു പക്ഷേ, യുവതി ലേഖകനെ തിരിച്ചറിഞ്ഞിട്ടു ണ്ടാവണം. വർഷങ്ങൾക്കു മുമ്പേ അവളുടെ മുത്ത ശ്ശിയുടെ പെട്ടിക്കടയിലെ കസ്റ്റമർ ആയിരുന്നു ലേഖകൻ. യുവതിയുടെ മനസ്സിൽ ലേഖകനോ ടുള്ള കൃതജ്ഞതാ ഭാവമാണ് ഇവിടെ പ്രകടമാ കുന്നത്.

प्रश्न 6.
“हाँ, तब तो ज़रूर मीठा होगा”. – कब ?
ശരി, അപ്പോൾ തീർച്ചയായും മധുരമുണ്ടായി രിക്കും. എപ്പോൾ?
उत्तर:
संतरा को मीठा बनाने के लिए दो पत्वा चीनी डालने के बारे में बूढ़ी कांग कहे तो तब लेखक ने इस प्रकार कहा।
ഓറഞ്ച് മധുരമുള്ളതാക്കാൻ വേണ്ടി അരക്കിലോ പഞ്ചസാര ചേർത്തു എന്ന് വൃദ്ധയായ കച്ചവട ക്കാരി പറഞ്ഞപ്പോഴാണ് ലേഖകൻ ഇപ്രകാരം പറഞ്ഞത്.

प्रश्न 7.
छोटी-सी गुमटी के नियमित ग्राहक कौन थे ?
ചെറിയ പെട്ടിക്കടയിലെ സ്ഥിര ഉപഭോക്താക്കൾ ആരായിരുന്നു?
उत्तर:
स्कूली बच्चे
സ്കൂൾ കുട്ടികൾ

प्रश्न 8.
पूरा कीजिए ।
പൂർത്തീകരിക്കാം.
1. शिलांग में लेखक ……………………. में रहता था ।
2. संतरा ………………………. है।
3. वह समझ गई कि मैं ……………………….. कर रहा था।
4. लगभग …………………….. साल बाद मैं फिर शिलांग गया।
5. उसने पैसा लेने से …………………… कर दिया।
उत्तर:
1. मल्की हिल्स
2. मीठा
3. मज़ाक
4. बीस
5. इनकार

प्रश्न 9.
एक ग्राहक सब्जी खरीदने दूकान पर जाता है तथा मोलभाव करके सब्जियाँ खरीदता है, दोनों की बातचीत लिखें।
ഒരു കസ്റ്റമർ പച്ചക്കറി വാങ്ങുന്നതിനായി കട യിലേക്ക് പോയി. അവിടെ വില വിവരങ്ങൾ
ചോദിച്ച് പച്ചക്കറികൾ വാങ്ങുന്നു. രണ്ടുപേരും തമ്മിലുള്ള സംഭാഷണം എഴുതൂ.
उत्तर:
ग्राहक: ये मटर कैसे दिए है भाई ?
सब्जीवाला: ले लो बाबूजी ! बहुत अच्छे मटर है, एकदम ताज़ा।
ग्राहक: भाव तो बताओ।
सब्जीवाला: बेचे तो पचास रुपये किलो हैं पर आपसे पैतालीस रुपये ही लेंगे।
ग्राहक: बहुत महँगे है भाई।
सब्जीवाला: क्या बताऊँ बाबूजी ! मण्डी में सब्जी के भाव आसमान छू रहे हैं।
ग्राहक: फिर भी ….। कुछ तो कम करो।
सब्जीवाला: आप एक रुपया कम दे देना बाबूजी ! कहिए कितने तोल हूँ।
ग्राहक: एक किलो मटर दे दो। और ………… एक किलो आलू भी।
सब्जीवाला: टमाटर भी ले जाइए, साहब । बहुत रास्तें हैं।
ग्राहक: कैसे?
सब्जीवाला: तीस रुपए किलो दे रहा हूँ। माल लुटा दिया बाबू जी ।
ग्राहक: अच्छा! दे दो एक किलो टमाटर भी । …………. और चार नींबू भी डाल देना ।
सब्जीवाला: यह लो बाबू जी । धनिया और हरी मिर्च भी रख दी है।
ग्राहक: कितने पैसे हुए?
सब्जीवाला: सिर्फ सतहत्तर रुपये ।
ग्राहक: लो भाई पैसे ?
सब्जीवाला: धन्यवाद बाबूजी।

കസ്റ്റമർ: ഈ പയർ എത്ര രൂപയ്ക്ക് തരും, സഹോദരാ ?
പച്ചക്കറി: എടുത്തോളൂ സാർ, നല്ലയിനം
കടക്കാരൻ പയർ ആണ് ഏറ്റവും ഫ്രഷ്
കസ്റ്റമർ: വിലയിൽ നിന്ന് രക്ഷപ്പെടുമോ?
പച്ചക്കറി: വിൽക്കുന്നത് ഒരു കിലോക്ക് 50
കടക്കാരൻ രൂപയ്ക്കാണ് എന്നാൽ താങ്കളിൽ നിന്ന് 45 രൂപയേ എടുക്കൂ.
കസ്റ്റമർ: വളരെ വിലയാണല്ലോ സഹോദരാ.
പച്ചക്കറി: എന്തുപറയാനാ സാർ? പച്ചക്കറി
കടക്കാരൻ വില മാർക്കറ്റിൽ കുതിച്ചുയരുക യാണ്.
കസ്റ്റമർ: എന്നാലും ……………………. കുറച്ച് കുറച്ച് തരൂ
പച്ചക്കറി: താങ്കൾ ഒരു രൂപ കൂടി കുറച്ചു
കടക്കാരൻ തന്നാൽ മതി. പറഞ്ഞാലും എത കിലോ വേണം?
കസ്റ്റമർ: ഒരു കിലോ പയർ തരൂ; പിന്നെ ഒരു കിലോ ഉരുളക്കിഴങ്ങും.
പച്ചക്കറി: തക്കാളി കൂടി കൊണ്ടു പോകൂ
കടക്കാരൻ സർ, വളരെ വിലക്കുറവാണ്.
കസ്റ്റമർ: എങ്ങനെ?
പച്ചക്കറി: കിലോ 30 രൂപയ്ക്ക് തരാം,
കടക്കാരൻ സാധനം കൊണ്ടുപൊയ്ക്കോ.
കസ്റ്റമർ: ശരി, അങ്ങനെയാണെങ്കിൽ ഒരു കിലോ തക്കാളി കൂടി തരൂ…….. പിന്നെ നാല് നാരങ്ങ കൂടി ഇടണേ
പച്ചക്കറി: ഇതാ സാർ. മല്ലിയിലയും പച്ച
കടക്കാരൻ മുളകും കൂടി വച്ചിട്ടുണ്ട്.
കസ്റ്റമർ: എത്ര രൂപയായി?
പച്ചക്കറി: 97 രൂപ മാത്രം.
കടക്കാരൻ
കസ്റ്റമർ: ഇതാ സഹോദരാ.
പച്ചക്കറി: നന്ദി സാർ.
കടക്കാരൻ

प्रश्न 10.
फल विक्रेता और ग्राहक के बीच संवाद ।
പഴം വിൽപ്പനക്കാരനും കസ്റ്റമറും തമ്മിലുള്ള സംഭാഷണം.
उत्तर:
ग्राहक: (फल विक्रेता से) संतरे किस भाव है ?
फल विक्रेता: एक किलो के लिए पचास रुपए ।
ग्राहक: एक किलो सेब और आधा किलो केला चाहिए।
फल विक्रेता: जी, सेब 100 रुपये किलो और के 40 रुपए किलो हैं ।
ग्राहक: ठीक है, मुझे दे दीजिए ।
फल विक्रेता: ये लीजिए । कुल 190 रुपये।
ग्राहक: ये लीजिए 190 रुपये। धन्यवाद !
फल विक्रेता: धन्यवाद ! फिर आयेगा ।
കസ്റ്റമർ: (പഴക്കടക്കാരനോട്) ഓറഞ്ചി നെന്താ വില?
പഴക്കടക്കാരൻ: ഒരു കിലോക്ക് അമ്പത് രൂപ.
കസ്റ്റമർ: ഒരു കിലോ ആപ്പിളും അരക്കിലോ ഏത്തക്കയും വേണം.
പഴക്കടക്കാരൻ: സാർ ആപ്പിളിന് കിലോക്ക് 100 രൂപയും ഏത്തയ്ക്കക്ക് കിലോയ്ക്ക് 40 രൂപയുമാണ്.
കസ്റ്റമർ: ശരി, എനിക്കു തരൂ.
പഴക്കടക്കാരൻ: ഇതാ എടുത്താലും, എല്ലാം കൂടി 190 രൂപയായി.
കസ്റ്റമർ: ഇതാ 190 രൂപ. നന്ദി.
പഴക്കടക്കാരൻ: നന്ദി. വീണ്ടും വന്നാലും,

प्रश्न 11.
जोड़ा बनाइए ।
ചേരുംപടി ചേർക്കുക.
उत्तर:
उबले – कांग
मौसमी – गुमटी
थोड़ी – शकरकंद
बूढ़ी – फल
छोटी सी – गपशप
उत्तर:
उबले – शकरकंद
मौसमी – फल
थोड़ी – गपशप
बूढ़ी – कांग
छोटी सी – गुमटी

Meeta santara Summary in Malayalam

मीठा संतरा Summary in Malayalam

मीठा संतरा മധുരമുളള ഓറഞ്ച്
ലേഖനത്തിന്റെ മലയാള പരിഭാഷ

ഞാൻ എപ്പോഴൊക്കെ ഷില്ലോംഗിൽ ചെലവഴിച്ച ദിനങ്ങളെപ്പറ്റി ഓർക്കുമോ അപ്പോഴൊക്കെ ഒരു വൃദ്ധ യായ കച്ചവടക്കാരിയെപ്പറ്റി ഓർമ്മിക്കും. അവിടെ ഞാൻ മൽക്കി ഹിൽസിലായിരുന്നു താമസിച്ചിരുന്നത്. അവർ എന്റെ വീടിന്റെ അടുത്തു തന്നെ ഒരു ചെറിയ പെട്ടിക്കടയിലിരുന്ന് അതത് സീസണിലെ പഴങ്ങൾ ആയിരുന്നു വിറ്റു കൊണ്ടിരുന്നത്. ഓറഞ്ച്, വാഴപ്പഴങ്ങൾ, പ്ലം, പൈനാപ്പിൾ കൂടാതെ വേവിച്ച് മധുരക്ക ങ്ങും, ബിസ്ക്കറ്റും ചോക്ലേറ്റും പെപ്പർമിന്റും എല്ലാമായിരുന്നു അവർ വിൽക്കുന്നത്. സ്കൂൾ കുട്ടികൾ ആയിരുന്നു അവരുടെ സ്ഥിര ഉപഭോക്താക്കൾ.

ഞാനും അതുവഴി പോകുമ്പോൾ ആ കടയിൽ നിന്ന് എന്തെങ്കിലും ഒക്കെ വാങ്ങുമായിരുന്നു. കുറച്ച് വർത്തമാനങ്ങളും പറയുമായിരുന്നു. ചിലപ്പോഴൊക്കെ അവരുടെ കൂടെ അവരുടെ ചെറുമകളും കാണുമാ യിരുന്നു.

ഒരു ദിവസം വീട്ടിലേക്ക് മടങ്ങുന്ന സമയം അതുവഴി കടന്നുപോയപ്പോൾ അവരോട് വർത്തമാനം പറയു ന്നതിനായി അവിടെ അൽപ്പനേരം നിന്നും. ഞാൻ അവരോട് ചോദിച്ചു “ഓറഞ്ച് മധുരമുള്ളതാണോ?” അവർ പറഞ്ഞു, “അതെ, വളരെ മധുരമുള്ളത്.” എന്നാൽ എപ്പോഴാണോ ഞാൻ ചോദ്യം വീണ്ടും ആവർത്തി ച്ചത് അപ്പോൾ അവർ പറഞ്ഞു “മധുരമുള്ളതാണെങ്കിൽ എന്തുകൊണ്ട് മധുരമില്ല?” “എത്രത്തോളം പഞ്ച സാര ഇട്ടു?” ഞാൻ ചോദിച്ചു. അവർക്ക് മനസ്സിലായി ഞാൻ തമാശ പറയുകയാണെന്ന്. അവർ പറഞ്ഞു “കാൽ കിലോ”. ഞാൻ പറഞ്ഞു “കാൽ കിലോ പഞ്ചസാരയിൽ അധികം മധുരമുണ്ടാവില്ല!” അവർ ചിരിച്ചു കൊണ്ട് രണ്ട് വിരലുകൾ കാണിച്ചു പറ ഞ്ഞു. “അരക്കിലോ”. “ശരി, അപ്പോൾ തീർച്ച യായും മധുരമുണ്ടാകും.” ഞാൻ അവരുടെ അടുത്തു നിന്ന് കുറച്ച് ഓറഞ്ചുകൾ വാങ്ങി വീട്ടി ലേക്ക് തിരികെ പോയി.

കുറച്ച് നാളുകൾക്കു ശേഷം ഞാൻ ഷില്ലോംഗ് ഉപേക്ഷിച്ചു പോയി. പിന്നീട് വർഷങ്ങളോളം അവിടേക്ക് ഞാൻ പോയില്ല.

ഏകദേശം ഇരുപത് വർഷത്തിനു ശേഷം ഞാൻ വീണ്ടും ഷില്ലോംഗിലേക്ക് പോയി. എനിക്ക് ആ വൃദ്ധയായ കച്ചവടക്കാരിയെ ഓർമ്മ വന്നു. അവരെ കാണുന്നതിനായി ഞാൻ വീണ്ടും മൽക്കി
ഹിൽസിൽ എത്തി. ആ സ്ഥലത്ത്, എവിടെയാണോ ആ ചെറിയ പെട്ടിക്കടയിലിരുന്ന് പഴങ്ങൾ വിറ്റിരു ന്നത് അവിടെ. കട അവിടെ തന്നെ ഉണ്ടായിരുന്നു. എന്നാൽ വൃദ്ധയെ അവിടെ എങ്ങും കാണാൻ സാധിച്ചില്ല. അവിടെ ഒരു യുവതി ഇരിപ്പുണ്ടായിരുന്നു. അവ ളുടെ അടുത്ത് ഒരു ചെറിയ പെൺകുട്ടി ഇരിപ്പുണ്ടായി രുന്നു. ഇരുപത് വർഷത്തിന് മുൻപ് കട് എങ്ങനെയാ യിരുന്നുവോ അങ്ങനെ തന്നെ കട ഇപ്പോഴും ഉണ്ടാ യിരുന്നു.

ഞാൻ ആ കച്ചവടക്കാരിയോട് ആ വൃദ്ധയായ സ്ത്രീയെപ്പറ്റി ചോദിച്ചു. അവർ അവളുടെ മുത്തശ്ശി യായിരുന്നുവെന്നും അവർ ഇപ്പോൾ ജീവിച്ചിരിപ്പി ല്ലാന്നും എനിക്ക് മനസ്സിലാക്കാൻ സാധിച്ചു. കുറച്ച് സമയം വർത്തമാനം പറഞ്ഞതിനുശേഷം ഓറഞ്ച് എടു ക്കാനുള്ള ഉദ്ദേശത്തോടെ എപ്പോഴും ചോദിക്കുന്നതു പോലെ ചോദിച്ചു, “മധുരമുള്ളതാണോ?” അവൾ പറ ഞ്ഞു. “വളരെ മധുരമുള്ളതാണ്. എന്നാൽ എപ്പോഴാണോ ഈ ചോദ്യം ഞാൻ ആവർത്തിച്ചത് അവൾ കുറച്ച് ദേഷ്യത്തോടെ പറഞ്ഞു “മധുരമുള്ളതാണ ങ്കിൽ എന്തുകൊണ്ട് മധുരമില്ല?” ഞാൻ വർഷങ്ങൾ പഴക്കമുള്ള ചോദ്യം വീണ്ടും ചോദിച്ചു “എത്ര കിലോ പഞ്ചസാര ഇട്ടു?” അവർ ചിരിച്ചു കൊണ്ടു പറഞ്ഞു “കാൽ കിലോ.”

ഞാൻ പറഞ്ഞു “അത് അത്ര മധുരമായിരിക്കില്ല.” അവൾ ചിരിക്കാൻ തുടങ്ങി. എന്നിൽ രണ്ട് വിരലുകൾ കാണിച്ചു പറഞ്ഞു, “അരക്കിലേലോ. ഇത് കേട്ട് എനിക്കു കൂടി ചിരി വന്നു. “എങ്കിൽ തീർച്ചയായും മധുരം കാണും.” ഞാൻ ഓറഞ്ച് വാങ്ങി പൈസ അവൾക്ക് നേരെ നീട്ടി. എന്നാൽ അവൾ പൈസ വാങ്ങാൻ വിസമ്മതിച്ചു. ഒരു പക്ഷേ അവൾക്ക് എന്നെ മനസ്സിലായിട്ടുണ്ടാവും. ഞാനും ഒരു മാലാഖയെപ്പോലുള്ള ആ കൊച്ചുപെൺകുട്ടിയെ മനസ്സിലാക്കിയിരുന്നു. അവളായിരുന്നു അവളുടെ മുത്തശ്ശിയുടെ സമീപം കട യിൽ ഇരുന്നിരുന്നത്.

मीठा संतरा लेखक परिचय

यायावर हिन्दी के प्रसिद्ध साहित्यकार थे । वे एक प्रसिद्ध लेखक एवं कवि थे । उनका जन्म 1949 में उत्तर प्रदेश के फिरोज़ाबाद में हुआ था । यायावर शब्द का अर्थ है घुमंतू या घुमक्कड़ जिसका मतलब है एक स्थान पर न टिकने वाला और लगातार यात्रा करने वाला व्यक्ति। यह शब्द उन लोगों के लिए इस्तेमाल होता है जो एक जगह से दूसरी जगह घूमते रहते हैं, जिनका कोई स्थायी निवास नहीं होता । यायावर भी ऐसा एक साहित्यकार थे। वे कई देशों को भ्रमण करते थे । वहाँ के सब कार्य ही उनकी रचनाओं में हम देख सकते हैं। ‘मीठी’ संतरा उनकी प्रसिद्ध रचना है। वे एक बाल साहित्यकार, कहानीकार और संस्मरणकार थे।

യായാവർ ഹിന്ദിയിലെ പ്രസിദ്ധനായ സാഹിത്യകാരനാണ്. അദ്ദേഹം ഒരു പ്രസിദ്ധനായ ലേഖകനും കവിയുമായിരുന്നു. അദ്ദേഹത്തിന്റെ ജനനം ഉത്തർപ്രദേശിലെ ഫിറാസാബാദിൽ 1949-ൽ ആയിരുന്നു. യായാവർ എന്ന വാക്കിന്റെ അർത്ഥം നാടോടി അല്ലെങ്കിൽ അലഞ്ഞു തിരിയുന്നവൻ എന്നാണ്. അതാ യത് ഒരു സ്ഥലത്ത് താമസിക്കാതെ തുടർച്ചയായി സഞ്ചരിക്കുന്ന വ്യക്തി. സ്ഥിരമായ താമസസ്ഥലം ഇല്ല. ഒരു സ്ഥലത്ത് നിന്ന് മറ്റൊരിടത്തേക്ക് നിരന്തരം സഞ്ചരിക്കുന്ന ആളുകളെ സൂചിപ്പിക്കാൻ ഈ വാക്ക് ഉപ യോഗിക്കുന്നു. യായാവരും അങ്ങനെയുള്ള ഒരു സാഹിത്യകാരനായിരുന്നു. അദ്ദേഹം ഒരു പാട് ദേശ ങ്ങൾ ചുങ്ങി സഞ്ചരിച്ചിട്ടുണ്ട്. അവിടുത്തെ കാര്യങ്ങൾ ഒക്കെ തന്നെയായിരുന്നു അദ്ദേഹത്തിന്റെ രചനക ളിൽ കാണാൻ സാധിച്ചിരുന്നത്. മീഠാ സംതരാ’ അദ്ദേഹത്തിന്റെ പ്രസിദ്ധ രചനയാണ്. അദ്ദേഹം ഒരു ബാലസാഹിത്യകാരനും, കഥാകാരനും, ഓർമ്മക്കുറിപ്പുകാരനുമായിരുന്നു.

मीठा संतरा शब्दार्थ

बिताना – ചെലവഴിക്കുക
याद – ഓർമ്മ
रहना – താമസിക്കുക
छोटी-सी – ചെറിയ
मौसमी – കാലാവസ്ഥയ്ക്കനുസൃതമായ
बेचना – വിൽക്കുക
केला – വാഴപ്പഴം
दिन – ദിവസം
बूढ़ी कांग – വൃദ്ധയായ കച്ചവടക്കാരി
घर – വീട്
गुमटी – പെട്ടിക്കട
फल – പഴങ്ങൾ
संतरा – ഓറഞ്ച്
प्लम – Plum
पाइनेपल – പൈനാപ്പിൾ
पीपरमेंट – പെപ്പർമെന്റ്
नियमिक ग्राहक – സ്ഥിര ഉപഭോക്താക്കൾ
उधर – അവിടെ
थोडी – കുറച്ച്
कभी-कभी – ചിലപ്പോഴൊക്കെ
लौटना – തിരികെ വരുക
गुज़रा – കടന്നുപോവുക
मगर – എന്നാൽ
सवाल – ചോദ്യം
दुहराया – ആവർത്തിച്ചു
पव्वा – കാൽകിലോ
हँसना – ചിരിക്കുക
ज़रूर – തീർച്ചയായും
बाद – ശേഷം
लगभग – ഏകദേശം
साल – വർഷം
नज़र नहीं आना – കണ്ടില്ല
शायद – ഒരു പക്ഷേ
उबले शकरकंद – പുഴുങ്ങിയ മധുരക്കിഴങ്ങ്
स्कूली बच्चे – സ്കൂളിൽ പോകുന്ന കുട്ടികൾ
खरीदना – വാങ്ങുക
गपशप – വർത്തമാനം
पोती – കൊച്ചുമകൾ
ठहराना – നിൽക്കുക
मीठा – മധുരമുള്ള
चीनी – പഞ്ചസാര
डालना – ഇടുക
मज़ाक – തമാശ
ज़्यादा – കുറെ അധികം
ऊँगली – വിരൽ
खरीदना – വാങ്ങുക
बरसों – വർഷങ്ങളോളം
बीस – ഇരുപത്
पहूँचना – എത്തുക
इनकार करना – വിസമ്മതിക്കുക