Students often refer to Kerala State Syllabus SCERT Class 6 Maths Solutions and Class 6 Maths Chapter 8 Unchanging Relations Questions and Answers Notes Pdf to clear their doubts.
SCERT Class 6 Maths Chapter 8 Solutions Unchanging Relations
Class 6 Kerala Syllabus Maths Solutions Chapter 8 Unchanging Relations Questions and Answers
Unchanging Relations Class 6 Questions and Answers Kerala Syllabus
Unchanging Sum (Page Number 116)
Question 1.
Mentally calculate the sums below:
(i) 137 + 199
(ii) 375 + 298
(iii) 697 + 174
(iv) 1489 + 2363
Answer:
(i) 137 + 199 = 136 + 200 = 336 (1 subtracted, 1 added)
(ii) 375 + 298 = 373 + 300 = 673 (2 subtracted, 2 added)
(iii) 697 + 174 = 700 + 171 = 871 (3 added, 3 subtracted)
(iv) 1489 + 2363 = 1500 + 2352 = 3852 (11 added, 11 subtracted)
Question 2.
(i) Take any four consecutive natural numbers. Calculate the sum of the first and the last, and also the sum of the middle two numbers. Take the other four consecutive natural numbers and check. Why are the sums the same every time?
Answer:
1 + 2 + 3 + 4 → 1 + 4 = 5; 2 + 3 = 5
7 + 8 + 9 + 10 → 7 + 10 = 17; 8 + 9 = 17
10 + 11 + 12 + 13 → 10 + 13 = 23; 11 + 12 = 23
The first number and the fourth number represent the sum of the first number and the number that is 3 more than it.
The second number and the third number also mean the first number and the number that is 3 more than it.
Both sums will always be equal. (both will be the first number, add 4)
Adding 3 to 1 gives 4 (1 + 4 = 5)
Adding 1 + 2 to 1 also gives 4 (1 + 4 = 5)
Adding 3 to 10 gives 13 (10 + 13 = 23)
Adding 1 + 2 to 10 also gives 13 (10 + 13 = 23)
It can also be said like this:
If you add 3 to the first number, you get 4.
The middle numbers are the first number plus one, and the fourth number minus one.
When one number decreases by one and the next number increases by one, the total remains unchanged.
(ii) Instead of four consecutive natural numbers, take six consecutive natural numbers. The sum of the first and the last numbers is equal to which other pairs? Why is this so?
Answer:
Six consecutive natural numbers are 3, 4, 5, 6, 7, 8
3 + 8 = 4 + 7 = 5 + 6 = 11
The sixth number is five more than the first number.
Now, if we take the second and fifth numbers:
The second number is one more than the first number, and the fifth number is one less than the sixth number.
(one more than 3 is 4, one less than 8 is 7)
Next, if we take the third and fourth numbers:
The third number is two more than the first number, and the fourth number is two less than the sixth number.
(two more than 3 is 5, and two less than 8 is 6)
So, 3 + 8 = (3 + 1) + (8 – 1) that is 4 + 7 = 11
3 + 8 = (3 + 2) +(8 – 2) that is 5 + 6 = 11
Consecutive counting numbers will always follow this pattern.
(iii) What if we take five consecutive naturalnumbers?
Answer:
Five consecutive natural numbers are 1, 2, 3, 4, 5
1 + 5 = 6, 2 + 4 = 6, 3 + 3 = 6
The second number is one more than the first number, and the fourth number is one less than the fifth number.
Therefore, first number + fifth number = second number + fourth number
Now, if we take the third number:
It is two more than the first number and two less than the fifth number.
So, if we add it twice, it will be equal to the sum of the first and fifth numbers.
Example: 3 + 3 = (1 + 2) + (5 – 2) = 6
(iv) What about three consecutive natural numbers?
Answer:
Three consecutive natural numbers are 4, 5, 6
4 + 6 = 5 + 5
The middle number is one more than the first number and one less than the third number.
Therefore, the sum of the end numbers will always be twice the middle number.
Question 3.
In the calendar of any one month, mark those dates which fall on any one day of the week:
The sum of the first and last numbers is equal to the sum of which other pairs? Why?
Answer:
Let’s take the first 4 Sundays.
5, 12, 19, 26
5 + 26 = 12 + 19 = 31
The second number is 7 more than the first number, and the third number is 7 less than the fourth number.
Therefore, the sums of such pairs will not change.
When we take 5 numbers 1, 8, 15, 22, 29, the same thing happens.
The middle number is 14 less than one end number and 14 more than the other end number.
So, when it is added twice, we get the same sum as the end numbers.
Example: 1 + 29 = 8 + 22 = 15 + 15
Unchanging Difference (Page Number 118)
Question 1.
Calculate these differences in your head:
(i) 453 – 196
(ii) 388 – 189
(iii) 700 – 387
(iv) 1500 – 839
Answer:
(i) 453 – 196 = (453 + 4) – (196 + 4)
= 457 – 200
= 257
(ii) 388 – 189 = (388 + 11) – (189 + 11)
= 399 – 200
= 199
(iii) 700 – 387 = (700 + 13) – (387 + 13)
= 713 – 400
= 313
(iv) 1500 – 839 = (1500 – 1) – (839 – 1)
= 1499 – 838
= 661
(Page Numbers 119 & 120)
Question 1.
Is the difference of any two multiples of 3 again a multiple of 3? Why?
Answer:
The difference of two multiples of 3 is always a multiple of 3.
If you take (3 × one number) – (3 × another number), it can be written as 3 × (bigger number – smaller number).
Since the result is 3 multiplied by some whole number, it will always be a multiple of 3.
Example: 63 – 21
= (21 × 3) – (7 × 3)
= (21 – 7) × 3
= 14 × 3
= 42
Whenever you subtract one multiple of 3.
From another, the answer will always be a multiple of 3.
Question 2.
(i) If two numbers leave a remainder of 1 on division by 3, is their difference a multiple of 3? Why?
Answer:
The difference of two numbers that leave a remainder of 1 when divided by 3 will always be a multiple of 3.
This is because if you subtract 1 from each of those numbers, they become exact multiples of 3.
So, the remaining difference will be a multiple of 3.
3 × 10 + 1 – 3 × 7 + 1
3 × 3 + 0 = 3 × 3, multiple of 3
(ii) If two numbers leave a remainder of 2 on division by 3, is their difference a multiple of 3? Why?
Answer:
A number that leaves a remainder of 2 when divided by 3 can be written in the form 3 × 10 + 2.
Then, 3 × 10 + 2 – 3 × 4 + 2
3 × 6 + 0 = 3 × 6, multiple of 3.
Question 3.
What all things can we say if we take 4 instead of 3 in the last problem?
Answer:
- If you subtract one multiple of 4 from another multiple of 4, the result will always be a multiple of 4.
- When two numbers leave the same remainder when divided by 4, their difference will also be a multiple of 4.
Question 4.
If two numbers leave the same remainder on division by a number, what is the relation between the difference of these numbers and the dividing number?
Answer:
The difference will be the divisor multiplied by some counting number.
Unchanging Product (Page Number 123)
Question 1.
If we start with 1 and double again and again, we get the numbers 1, 2, 4, 8, 16,…
(i) Take any four consecutive numbers from these. Why is the product of the first and last numbers the same as the product of the middle two numbers?
Answer:
The second number will be the first number multiplied by two.
The third number will be the last number divided by two.
1, 2, 4, 8
1 × 8 = 2 × 4
(ii) If we take three consecutive numbers from these, what is the relation between the product of the first and last numbers and the middle number?
Answer:
The product of the first and the last numbers will be equal to the middle number multiplied by itself.
4, 8, 16
4 × 16 = 8 × 8
Question 2.
If we start with 1 and triple it again and again, we get the numbers 1, 3, 9, 27, 81,… For these, are the results of the first problem true? Why?
Answer:
That’s correct. Because the second number is the first number multiplied by 3.
The third number is the last number divided by 3.
1, 3, 9, 27, 81
1 × 27 = 3 × 9
Similarly, the second number is three times the first number.
The second number is the third number divided by 3.
3 × 27 = 9 × 9 = 81
Three multiplied by 3 is nine.
If you divide 81 by 3, you get 27.
Unchanging Quotients (Page Number 126)
Question 1.
Calculate the quotients below:
(i) 345 ÷ 15
(ii) 495 ÷ 45
(iii) 325 ÷ 25
(iv) 975 ÷ 75
(v) 875 ÷ 125
Answer:
(i) 345 ÷ 15 = 690 ÷ 30
= (69 × 10) ÷ (3 × 10)
= 69 ÷ 3
= 23
(ii) 495 ÷ 45 = 990 ÷ 90
= (99 × 10) ÷ (9 × 10)
= 99 ÷ 9
= 11
(iii) 325 ÷ 25 = 650 ÷ 50
= (65 × 10) ÷ (5 × 10)
= 65 ÷ 5
= 13
(iv) 975 ÷ 75 = 3900 ÷ 300
= (390 × 10) ÷ (30 × 10)
= 390 ÷ 30
= 13
(v) 875 ÷ 125 = 3500 ÷ 500
= (350 × 10) ÷ (50 × 10)
= 350 ÷ 50
= 7
Class 6 Maths Chapter 8 Kerala Syllabus Unchanging Relations Questions and Answers
Class 6 Maths Unchanging Relations Questions and Answers
Question 1.
Take any 4 consecutive numbers from the sequence 10, 20, 30, 40, 50, 60, …Find the pairs that give the same sum. Take any 6 numbers. Find the pairs that give the same sum. Explain why? What if we take 7 numbers?
Answer:
4 consecutive numbers are 40, 50,60, 70
40 + 70 = 50 + 60
50 is 10 more than 40, 60 is 10 less than 70.
6 consecutive numbers are 30, 40, 50, 60, 70, 80
30 + 80 = 40 + 70 = 50 + 60
40 is ten more than 30, 70 is ten less than 80.
Similarly, 50 is twenty more than 30, and 60 is twenty less than 80.
7 consecutive numbers are 10, 20, 30, 40, 50, 60, 70
10 + 70 = 20 + 60 = 30 + 50 = 40 + 40
In each case, the numbers on the two sides decrease and increase by the same amount, so the sum does not change.
Question 2.
Write 4 other multiplication expressions that are equal to 36 × 24.
Answer:
36 × 24
18 × 48
72 × 12
9 × 96
144 × 6
Question 3.
Do the following mentally:
(a) 202 + 198
Answer:
202 + 198 = (202 – 2) + (198 + 2) = 400
(b) 515 + 485
Answer:
515 + 485 = (515 – 15) + (485 + 15) = 1000
(c) 394 + 306
Answer:
394 + 306 = (394 + 6) + (306 – 6) = 700
(d) 1204 + 1296
Answer:
1204 + 1296 = (1204 – 4) + (1296 + 4) = 2500
(e) 288 – 178
Answer:
288 – 178 = (288 + 12) – (178 + 12) = 110
(f) 5000 – 1424
Answer:
5000 – 1424 = (5000 – 1) – (1424 – 1) = 3576
(g) 5012 – 3212
Answer:
5012 – 3212 = (5012 – 12) – (3212 – 12) = 1800
(h) 6008 – 1458
Answer:
6008 – 1458 = (6008 – 8) – (1458 – 8) = 4550
Question 4.
For each division problem, write 2 equivalent division problems.
(a) 160 ÷ 40
Answer:
160 ÷ 40
= 80 ÷ 20
= 40 ÷ 10
= 4 ÷ 1
(b) 144 ÷ 12
Answer:
144 ÷ 12
= 288 ÷ 24
= 72 ÷ 6
= 36 ÷ 3
(c) 500 ÷ 150
Answer:
500 ÷ 150
= 1000 ÷ 300
= 100 ÷ 30
= 10 ÷ 3
(d) 750 ÷ 25
Answer:
750 ÷ 25
= 1500 ÷ 50
= 150 ÷ 5
= 300 ÷ 10
Class 6 Maths Chapter 8 Notes Kerala Syllabus Unchanging Relations
→ The sum of two numbers is the same as the sum of the numbers got by adding a number to one of them and subtracting the same number from the other.
→ The difference of two numbers is the same as the difference of the two numbers got by adding or subtracting the same number to each of them.
→ The product of two numbers is the same as the product of the numbers got by multiplying one of them by a number and dividing the other by the same number.
→ The quotient of two numbers is the same as the quotient of these two numbers multiplied or divided by the same number.
We have been learning many types of relationships involving numbers since the lower classes. For example, the sum of two consecutive natural numbers is always an odd number. The product of two odd numbers is always an odd number. We have understood such ideas. In this unit, we explain some other relationships that remain unchanged even when the numbers change.
Unchanging Sum
A rope is 10 metres long. From it, a 2-metre piece is cut off. If we add the cut piece and the remaining piece, we get 2 + 8 = 10.
Now, if the piece cut off is 3 metres, the remaining part is 7 metres: 3 + 7 = 10
If 6 metres is cut off, the remaining part is 4 metres: 6 + 4 = 10
Thus, when the cut length changes, the remaining length also changes, but the total remains 10.
Mayookhi has 100 rupees in her money box. She took 10 rupees from it and bought a pen. How much money is left? What if she had spent 20 rupees?
In the money problems in the textbook, even if the amount spent and the amount remaining change, the total always remains 100.
10 + 90 = 100, 20 + 80 = 100, 30 + 70 = 100, and so on.
Let us look at another example. We need to draw a rectangle with a perimeter of 20 cm. To draw a rectangle, we need the length of one side, right? What should it be?
We have learned that the perimeter of a rectangle is twice the sum of its width and height.
Here, since the perimeter is 20 cm, the sum of the width and height is 10 cm.
So, if we take the width as 6 cm, the height will be 10 – 6 = 4 cm.
With the same perimeter, we can draw many other rectangles too.
Width = 1 cm, height = 9 cm, Sum = 10 cm, Perimeter = 20 cm
Width = 2 cm, height = 8 cm, Sum = 10 cm, Perimeter = 20 cm
Here, the width and height of the rectangle may change, but the total remains the same.
What did we observe here?
When the perimeter remains the same, the more the width increases, the more the height decreases.
When one piece of the rope becomes longer, the other piece becomes shorter by the same amount.
When the amount of money spent increases, the amount remaining decreases by the same amount.
Now look at another situation.
A child put 64 manchadi seeds into a box. Now he has 98 seeds left in his hand. How many seeds did he have altogether?
It is 64 + 98.
If the number put into the box decreases by one, the number left increases by one.
If the number put into the box decreases by two, the number left increases by two.
So, 64 + 98 becomes 66 + 96, which is 162.
Here, when one number increases by two and the other decreases by two, the total remains the same.
Now, consider 128 + 99.
This is safe as 127 + 100, which is 227.
How can we write 97 + 203?
(97 + 3) + (203 – 3) = 100 + 200 = 300
That is, one number increases by a certain amount, and the other number decreases by the same amount.
We can say it like this:
The sum of two numbers = (one number + a certain amount) + (the other number – the same number)
Now look at these examples.
425 + 597 = 422 + 600 (3 subtracted, 3 added) = 1022
3249 + 894 = 3243 + 900 (6 subtracted, 6 added) = 4143
395 + 876 = 400 + 871 (5 added, 5 subtracted) = 1271
Look at the numbers 5, 10, 15, 20
If you add 5 and 20, you get 25 (the end of numbers).
If you add 10 and 15, you also get 25(the middle numbers).
Why are their sums equal? Because 5 + 20 = 10 + 15
In 5 + 20, the number 5 is five less than the 10 in 10 + 15, and the number 20 is five more than the 15 in 10 + 15.
So, when one number decreases by 5 and the other number increases by 5, the total doesn’t change.
Let’s extend the sequence 5, 10, 15, 20, 25, 30.
Here, also 15 + 30 = 20 + 25
Again, one number increases by 5, and the other number decreases by 5.
So, the sum remains the same.
It can also be explained like this:
When the first number and the fourth number are added, it is the same as adding the second number and its four times.
When the second number and the third number are added, it is the same as adding twice the first number and three times the first number.
Both results are equal to five times the same number.
Similarly, take 10, 15, 20, 25: 10 + 25 = 15 + 20.
Here, 10 increases to 15 and 25 decreases to 20.
So, the sum stays the same.
In the sequence 4, 8, 12, 16, 20, 24, if you take any four consecutive numbers, will the sum of the two end numbers equal the sum of the two middle numbers? Why?
Answer:
Yes.
4, 8, 12, 16
4 + 16 = 8 + 12 = 20
If you add 4 to 4, you get 8.
If you subtract 4 from 16, you get 12.
One number increases by 4, and the other decreases by 4.
So, the total does not change.
Unchanging Difference
The age difference between two people is a relationship that never changes.
If one person is 17 years old and the other is 25 years old, the age difference now is 25 – 17 = 8 years.
After 10 years?
Difference = (25 + 10) – (17 + 10) = 35 – 27 = 8.
After 3 years?
Difference = (25 + 3) – (17 + 3) = 28 – 20 = 8 again.
What do we observe here?
When finding the difference between two numbers, if we add or subtract the same number from both the first and the second number, the difference does not change.
Examples:
195 – 98 = (195 + 2) – (98 + 2)
= 197 – 100
= 97
212 – 172 = (212 – 12) – (172 – 12)
= 200 – 160
= 40
487 – 247 = (487 + 13) – (247 + 13)
= 500 – 260
= 240
5000 – 3674 = (5000 – 1) – (3674 – 1)
= 4999 – 3673
= 1326
Now we can do this mentally:
1. 704-304
704 – 304 = (704 – 4) – (304 – 4)
= 700 – 300
= 400
2. 598 – 128
598 – 128 = (598 + 2) – (128 + 2)
= 600 – 130
= 470
3. 312 – 142
312 – 142 = (312 – 12) – (142 – 12)
= 300 – 130
= 170
4. 325 – 135
325 – 135 = (325 – 25) – (135 – 25)
= 300 – 110
= 190
5. 1524 – 324
1524 – 324 = (1524 + 26) – (324 + 26)
= 1550 – 350
= 1200
6. 4000 – 1299
4000 – 1299 = (4000 + 1) – (1299 + 1)
= 4001 – 1300
= 2701
Even and Odd Numbers
The difference between two even numbers will always be an even number. Let’s see why?
When you subtract 8 × 2 from 12 × 2, you get 4 × 2. This is an even number.
In the same way, the difference between two odd numbers is always an even number.
Here is the reason.
An odd number can be written as an even number plus one.
So, 125 – 73 can be written as (125 + 1) – (73 + 1), this difference is an even number.
This is true for any pair of odd numbers.
325 – 293 = (325 + 1) – (293 + 1) = 326 – 294; this difference is an odd number.
Let us now check the difference of multiples of 5 in the same way.
125 – 85 = (25 × 5) – (17 × 5) = (25 – 17) × 5 = 8 × 5
The difference is also a multiple of 5.
Similarly, let us check whether the difference of multiples of 9 is also a multiple of 9.
Explanation: 81 and 63 are multiples of 9
81 = 9 × 9, 63 = 7 × 9
(9 × 9) – (7 × 9) = (9 – 7) × 9 = 2 × 9, the difference is multiple of 9.
Unchanging Product
There are 36 cups. We want to arrange them in rows and columns to make a rectangle.
Possible arrangements:
9 rows and 4 columns; 9 × 4 = 36
12 rows and 3 columns; 12 × 3 = 36
18 rows and 2 columns; 18 × 2 = 36
6 rows and 6 columns; 6 × 6 = 36
Look at these examples.
Even though the numbers have changed, the product has not changed.
18 × 2 = 36
9 × 4 = 36
Look at these. What change has happened to each number?
18 became its half, which is 9.
2 became its double, which is 4
Look at the next pair, 18 × 2 = 36; 6 × 6 = 36.
Here, 18 became on third, which is 6, and 2 became three times, which is 6.
Here, also did not change.
What do we understand from this?
When two numbers are multiplied, if one number is multiplied by a certain number and the other number is divided by the same number, the product remains unchanged.
Example: 24 × 6 = 144
When 24 is multiplied by 2, we get 48.
When 6 is divided by 2, we get 3.
48 × 3 = 144 (24 × 6 = 48 × 3)
When 24 is divided by 3, the result is 8; when 6 is multiplied by 3 is 18
24 × 6 = 8 × 18 = 144
Further, 24 is divided by 12, which is 2; 6 is multiplied by 12, which is 72.
24 × 6 = 2 × 72 = 144
The result can be written like this:
24 × 6 = 48 × 3 = 8 × 18 = 2 × 72
Similarly, let us try to write some multiplication expressions equal to 72 × 40
72 × 40
36 × 80
18 × 160
9 × 320
Write some other multiplication expressions equal to the multiplication given below.
(a) 48 × 32
(b) 60 × 20
(c) 72 × 48
Answer:
(a) 48 × 32 = 24 × 64
= 12 × 128
= 6 × 256
= 3 × 512
(b) 60 × 20 = 30 × 40
= 15 × 80
= 120 × 10
= 240 × 5
(c) 72 × 48 = 36 × 96
= 18 × 192
= 9 × 384
= 144 × 24
1 × 216 and 6 × 36 are equal. Why?
Answer:
Multiplying 1 by 6 gives 6.
Dividing 216 by 6 gives 36.
If you multiply one number in 1 × 216 by 6 and divide the other number by 6, their product will still be the same as 1 × 216.
Answer:
20 × 16 = 40 × 8
42 × 30 = 7 × 180
30 × 15 = 10 × 45
55 × 3 = 11 × 15
Unchanging Quotient
If you put 5 seeds in each of 10 covers, how many seeds are needed?
5 × 10 = 50 seeds
If you put 10 seeds in each of 10 covers, how many seeds are needed?
10 × 10 = 100 seeds
If you put 20 seeds in each of 10 covers, how many seeds are needed?
20 × 10 = 200 seeds
In all these examples, there are 3 quantities: the number of seeds in one packet, the number of covers, and the total number of seeds.
Here, the number of seeds in one packet and the total number of seeds change.
The number of covers does not change.
The total number of seeds divided by the number of seeds in one packet gives the number of covers.
50 ÷ 5 = 10, 100 ÷ 10 = 10, 200 ÷ 20 = 10
Similarly, we can see the same idea in the pen problems in the textbook.
The number of pens and the total cost change.
The price per pen does not change.
40 ÷ 8 = 5, 80 ÷ 16 = 5, 50 ÷ 10 = 5, 150 ÷ 30 = 5
In the same way, in square problems too, when both numbers change in the same multiple, the quotient remains the same.
Eg: 25 ÷ 5 = 5, 50 ÷ 10 = 5, 100 ÷ 20 = 5
We also know that when we divide both numbers by the same number, the quotient does not change.
40 ÷ 10 = 4, 4 ÷ 1 = 4
Here, both numbers are divided by 10; the quotient remains the same.
This is like cancelling common factors.
40 ÷ 10 = 4 × 10 ÷ 1 × 10 = 4
75 ÷ 25 = (5 × 15) ÷ (5 × 5)
If we cancel the common factor: 15 ÷ 5
Again, cancelling the common factor: 15 ÷ 5 = 3 × 5 ÷ 1 × 5 = 3
Thus, the quotient of two numbers and the quotient of their equal multiples or equal parts will always be the same.
Eg: 16 ÷ 4 = 4
If we take the multiples of 16 and 4,
(16 × 5) ÷ (4 × 5) = 80 ÷ 20 = 4
If we take the same part of 16 and 4,
(16 ÷ 2) ÷ (4 ÷ 2) = 8 ÷ 2 = 4, quotient does not change.
Eg: If 175 is divided by 5?
175 ÷ 5 = 350 ÷ 10 = 35
Eg: 495 ÷ 15
495 ÷ 15 = 990 ÷ 30
= 99 ÷ 3
= 33
Match the ones with the same quotient.
Answer:
14 ÷ 2 = 28 ÷ 4
35 ÷ 7 = 350 ÷ 70
150 ÷ 30 = 15 ÷ 3
240 ÷ 8 = 120 ÷ 4
200 ÷ 10 = 100 ÷ 5
