Students often refer to Kerala State Syllabus SCERT Class 7 Maths Solutions and Class 7 Maths Chapter 4 Reciprocals Questions and Answers Notes Pdf to clear their doubts.
SCERT Class 7 Maths Chapter 4 Solutions Reciprocals
Class 7 Maths Chapter 4 Reciprocals Questions and Answers Kerala State Syllabus
Reciprocals Class 7 Questions and Answers Kerala Syllabus
Page 64
Question 1.
Suma has 16 rupees with her. Safeer has 4 rupees.
(i) What part of Suma’s money does Safeer have?
(ii) How many times Safeer’s money does Suma have?
Answer:
Suma’s amount = 16
Safeer’s amount = 4
(i) Part = \(\frac{\text { small number }}{\text { large number }}\)
= \(\frac{4}{16}\)
= \(\frac{1}{4}\)
So, Safeer has of Suma’s money.
(ii) Times = \(\frac{\text { large number }}{\text { small number }}\)
= \(\frac{16}{4}\)
= 4
So, Suma has 4 times that of Safeer’s money.
Question 2.
A large bag contains 9 kilograms of sugar. A small bag contains 6 kilograms.
(i) The weight of sugar in the heavier bag is how much times that in the lighter bag?
(ii) The weight of sugar in the lighter bag is what part of that in the heavier bag?
Answer:
(i) Times = \(\frac{\text { large number }}{\text { small number }}\)
= \(\frac{9}{6}\)
= \(\frac{3}{2}\)
So, the weight of sugar in the heavier bag is \(\frac{3}{2}\) times that in the lighter bag.
(ii) Part = \(\frac{\text { small number }}{\text { large number }}\)
= \(\frac{6}{9}\)
= \(\frac{2}{3}\)
So, the weight of sugar in the lighter bag is \(\frac{2}{3}\) part of that in the heavier bag.
Question 3.
The weight of an iron block is 6 kilograms. The weight of another block is 26 kilograms.
(i) The weight of the lighter block is what fraction of that of the heavier block?
(ii) The weight of the heavier block is how much times that of the lighter block?
Answer:
(i) Part = \(\frac{\text { small number }}{\text { large number }}\)
= \(\frac{6}{26}\)
= \(\frac{3}{13}\)
So, the weight of the lighter block is \(\frac{3}{13}\) fraction of that of the heavier block.
(ii) Times = \(\frac{\text { large number }}{\text { small number }}\)
= \(\frac{26}{6}\)
= \(\frac{13}{3}\)
So, the weight of the heavier block is \(\frac{13}{3}\) times that of the lighter block.
Question 4.
The length of a ribbon is 2 times the length of a smaller ribbon. What part of the length of the large ribbon is the length of the small ribbon?
Answer:
Times = \(\frac{\text { length of the large ribbon }}{\text { length of the smaller ribbon }}\)
2\(\frac{2}{3}=\frac{\text { length of the large ribbon }}{\text { length of the smaller ribbon }}\)
\(\frac{8}{3}=\frac{\text { length of the large ribbon }}{\text { length of the smaller ribbon }}\)
8 × length of the smaller ribbon = 3 × length of the large ribbon
\(\frac{\text { length of the smaller ribbon }}{\text { length of the large ribbon }}=\frac{3}{8}\)
Part = \(\frac{3}{8}\)
So, length of the smaller ribbon is \(\frac{3}{8}\) part of the length of the large ribbon.
Page 67
Question 1.
27 students of a class got A plus in Maths. They form of the entire class. How many students are there in the class?
Answer:
\(\frac{3}{4}\) × entire class = 27
entire class = 27 ÷ \(\frac{3}{4}\)
= 27 × \(\frac{4}{3}\)
= 9 × 4
= 36
Thus, there are 36 students in the class.
Question 2.
\(\frac{2}{3}\) of a bottle was filled with \(\frac{1}{2}\) litre of water. How many litres of water will the bottle hold?
Answer:
\(\frac{2}{3}\) x entire bottle = \(\frac{1}{2}\) litres
entire bottle = \(\frac{1}{2} \div \frac{2}{3}\)
= \(\frac{1}{2} \times \frac{3}{2}\)
= \(\frac{3}{4}\)
Thus, the bottle will hold – litres of water.
Question 3.
\(\frac{3}{4}\) of a vessel holds 1\(\frac{1}{2}\) litres of water. What is the capacity of the vessel in litres if it is completely filled with water?
Answer:
\(\frac{3}{4}\) × entire vessel = 1\(\frac{1}{2}\) = \(\frac{3}{2}\) litres
entire vessel = \(\frac{3}{2} \div \frac{3}{4}\)
= \(\frac{3}{2} \times \frac{4}{3}\)
= \(\frac{4}{2}\)
= 2
Thus, the capacity of the vessel if it is completely filled with water is 2 litres.
Question 4.
Two of the three ribbons of the same length and half the third ribbon were placed end to end. It came to 1 metre. What is the length of a ribbon in centimetres?
Answer:
1 metre = 100 cm
2.5 ribbons = 100 cm
\(\frac{5}{2}\) × length of a ribbon = 100 cm
length of a ribbon = 100 ÷ \(\frac{5}{2}\)
= 100 × \(\frac{2}{5}\)
= 20 × 2
= 40 cm
Page 68
Question 1.
A 16 metres long rod is cut into pieces of length \(\frac{2}{3}\) metre. How many such pieces will be there?
Answer:
Total length of the rod = 16 m
Length of a piece = \(\frac{2}{3}\) m
Number of pieces = Total length of the rod ÷ Length of a piece
= 16 ÷ \(\frac{2}{3}\)
= 16 × \(\frac{3}{2}\)
= 8 × 3
= 24.
Question 2.
How many \(\frac{3}{4}\) litre bottles are needed to fill 5\(\frac{1}{4}\) litres of water?
Answer:
Total water = 5\(\frac{1}{4}=\frac{21}{4}\) litres
Amount of water in a bottle = \(\) litres
Number of bottles = Total water ÷ Amount of water in a bottle
= \(\frac{21}{4} \times \frac{4}{3}\)
= \(\frac{21}{4} \times \frac{4}{3}\)
= \(\frac{21}{3}\)
= 7
Question 3.
13\(\frac{1}{2}\) kilograms of sugar is to be packed into bags with 2\(\frac{1}{4}\) kilograms sugar each. How many bags are needed?
Answer:
Total sugar = 13\(\frac{1}{2}\) = \(\frac{27}{2}\) kg
Amount of sugar in one bag = 2\(\frac{1}{4}=\frac{9}{4}\)kg
Number of bags = Total sugar Amount of sugar in one bag
= \(\frac{27}{2} \div \frac{9}{4}\)
= \(\frac{27}{2} \times \frac{4}{9}\)
= 3 × 2
= 6.
Question 4.
The area of a rectangle is 22\(\frac{1}{2}\) square centimetres, and one side is 3\(\frac{3}{4}\) centimetres long. What is the length of the other side?
Answer:
Area of a rectangle = 22\(\frac{1}{2}\) = \(\frac{45}{2}\)sq.cm
length of one side = 3\(\frac{3}{4}=\frac{15}{4}\) cm
length of the other side = area of a rectangle ÷ length of one side
= \(\frac{45}{2} \div \frac{15}{4}\)
= \(\frac{45}{2} \times \frac{4}{15}\)
= 3 × 2
= 6 cm
Question 5.
How many pieces, each of length 2\(\frac{1}{2}\) metres, can be cut off from a rope of length 11\(\frac{1}{2}\) metres? How many metres of rope will be left?
Answer:
Total length of the rope = 11\(\frac{1}{2}\) = \(\frac{23}{2}\) m
Length of a piece = 2\(\frac{1}{2}=\frac{5}{2}\)m
Number of pieces = Total length of the rope ÷ Length of a piece
= \(\frac{23}{2} \div \frac{5}{2}\)
= \(\frac{23}{2} \times \frac{2}{5}\)
= \(\frac{23}{5}\)
= 4\(\frac{3}{5}\)
So, we can cut 4 pieces of length 2\(\frac{1}{2}\) m.
4 × 2\(\frac{1}{2}\)
= 4 × \(\frac{5}{2}\)
= 2 × 5
= 10 m
Length of the remaining rope =11\(\frac{1}{2}\) – 10 = \(\frac{23}{2}\) – 10 = \(\frac{3}{2}\)m
Page 68
Question 1.
A rod is 36 metres long. How many pieces each of length 2\(\frac{1}{2}\) metres can be cut off from it? What is the length of the rod left?
Appu did the problem this way.
36 ÷ 2\(\frac{1}{2}\) = 36 ÷ \(\frac{5}{2}\)
= 36 ÷ \(\frac{2}{5}\)
= \(\frac{72}{5}\)
When we divide 72 by 5, the quotient is 14 and remainder is 2. So we get 14 pieces. The remaining rod is of length 2 metres.
Ammu used another idea.
2 pieces, each of length 2\(\frac{1}{2}\) metres makes 5 metres.
7 × 5 = 35
So 7 × 2 = 14 pieces can be cut off. The remainder is 36 – 35 = 1 metre. Whose answer is right?
Solution:
In Appu’s case, he find that 14 pieces of length 2\(\frac{1}{2}\) metres can be cut off from it.
This 14 pieces together forms;
14 × 2\(\frac{1}{2}\) = 14 × \(\frac{5}{2}\)
= 7 × 5
= 35 m
So, the remaining length = 36 – 35 = 1 m.
Thus, Ammu’s answer is the right one.
Class 7 Maths Chapter 4 Kerala Syllabus Reciprocals Questions and Answers
Question 1.
Find the reciprocal of \(\frac{3}{7}\)
Answer:
reciprocal of \(\frac{3}{7}=\frac{7}{3}\)
Question 2.
Simplify \(\frac{3}{4} \div \frac{5}{6}\)
Answer:
\(\frac{3}{4} \div \frac{5}{6}=\frac{3}{4} \times \frac{6}{5}=\frac{18}{20}=\frac{9}{10}\)
Question 3.
A recipe requires cup of sugar to make 12 cookies. How much sugar is needed to make 36 cookies?
12 കുക്കീസ് ഉണ്ടാക്കുവാനായി \(\frac{3}{4}\) കപ്പ് പഞ്ചസാര വേണം. അങ്ങനെയങ്കിൽ 36 കുക്കീസ് ഉണ്ടാക്കുവാൻ എത്ര കപ്പ് പഞ്ചസാര വേണം?
Answer:
\(\frac{3}{4}\) cup of sugar gives 12 cookies.
12 × 3 gives 36.
So, for 36 cookies we need \(\frac{3}{4}\) × 3 = \(\frac{9}{4}\) cups of sugar.
Question 4.
If of the cake is eaten, what fraction of the whole cake remains?
ഒരു കേക്കിന്റെ \(\frac{2}{5}\) ഭാഗം കഴിച്ചെങ്കിൽ ബാക്കി എത്ര ഭാഗം ഉണ്ട്?
Answer:
The whole cake is \(\frac{5}{5}\) = 1
Remaining cake = \(\frac{5}{5}-\frac{2}{5}=\frac{3}{5}\)
Thus, \(\frac{3}{5}\) part of the whole cake remains.
Question 5.
A ribbon is cut into 12 equal parts. What fraction of the ribbon is 3 parts?
ഒരു റിബൺ 12 തുല്യ ഭാഗങ്ങളായി മുറിച്ചു. അതിലെ 3 ഭാഗം ആകെയുള്ള റിബണിന്റെ എത്ര ഭാഗമാണ്?
Answer:
Part (or fraction) = \(\frac{\text { small number }}{\text { large number }}\)
= \(\frac{3}{12}\)
= \(\frac{1}{4}\)
Practice Questions
Question 1.
Solve \(\frac{4}{5} \div \frac{5}{4}\)
Answer:
\(\frac{16}{25}\)
Question 2.
Calculate \(\frac{3}{7} \div \frac{7}{3}\)
Answer:
1
Question 3.
A company produces \(\frac{2}{5}\) of its daily output in the morning and \(\frac{3}{8}\) in the afternoon. What fraction of the total daily output is produced in the afternoon?
Answer:
\(\frac{15}{31}\)
Question 4.
Calculate the reciprocal of \(\frac{2}{5}\). Which is larger?
Answer:
\(\frac{5}{2}\), reciprocal is the largest
Question 5.
When a tank is \(\frac{1}{4}\) full, it contains 80 litres of water. What will it contain when it is \(\frac{3}{8}\) full?
Answer:
120 litres
Class 7 Maths Chapter 4 Notes Kerala Syllabus Reciprocals
We are already familiar with a number of mathematical operations, such as addition, subtraction, multiplication and division. This chapter introduces a new type of mathematical operation named “reciprocal”. Following are the topics discussed in this chapter.
Times and parts
If we are given a large number and a small number, we can say the large number is how many times the small number. Always,
Times = \(\frac{\text { large number }}{\text { small number }}\)
If we are given a large number and a small number, we can say the small number is what part of the large number. Always,
Part = \(\frac{\text { small number }}{\text { large number }}\)
Topsy – Turvy
A fraction obtained by interchanging the numerator and denominator of the given fraction is called the reciprocal of the given fraction.
If we multiply a natural number by its reciprocal, we will get one as the result.
Eg:
The reciprocal of \(\frac{2}{3}\) is \(\frac{3}{2}\)
If we multiply a fraction by its reciprocal, we will get one as the result.
Eg:
Consider the fraction \(\frac{3}{4}\). Its reciprocal is \(\frac{4}{3}\). Now, \(\frac{3}{4} \times \frac{4}{3}\) = 1
Zero does not have a reciprocal because any number multiplied by zero is zero, not 1.
Fraction division
Division of natural numbers is the same as multiplication by the reciprocal.
Dividing a fraction (or a number) by another fraction is exactly same as multiplying the first fraction (or number) by the reciprocal of the second fraction.
Eg:
\(\frac{1}{2} \div \frac{3}{4}=\frac{1}{2} \times \frac{4}{3}=\frac{4}{6}=\frac{2}{3}\)
8 ÷ \(\frac{2}{5}\) = 8 × \(\frac{5}{2}\) = 4 × 5 = 20
If we are given a large number and a small number, we can say;
the large number is how many times the small number
the small number is what part of the large number large number
Times = \(\frac{\text { large number }}{\text { small number }}\)
Part = \(\frac{\text { small number }}{\text { large number }}\)