RD Sharma Solutions for Class 7 Maths Chapter 2 – Fractions Exercise 2.3 are the best study materials for those students who are finding difficulties in solving problems. Our subject experts formulate these exercises to help learners with their exam preparation to achieve good marks in Maths. Students can download the PDF of these solutions from the given links. By practising RD Sharma Solutions for Class 7, they will be able to grasp the concepts perfectly. It also helps in boosting their confidence in attempting the annual examination.
RD Sharma Solutions for Class 7 Maths Chapter 2 – Fractions Exercise 2.3
Access answers to Maths RD Sharma Solutions for Class 7 Chapter 2 – Fractions Exercise 2.3
Exercise 2.3 Page No: 2.24
1. Find the reciprocal of each of the following fractions and classify them as proper, improper and whole numbers:
(i) (3/7)
(ii) (5/8)
(iii) (9/7)
(iv) (6/5)
(v) (12/7)
(vi) (1/8)
Solution:
(i) Given (3/7)
Reciprocal of (3/7) is (7/3)
(7/3) is an improper fraction.
(ii) Given (5/8)
Reciprocal of (5/8) is (8/5)
It is an improper fraction.
(iii) Given (9/7)
Reciprocal of (9/7) is (7/9)
It is a proper fraction.
(iv) Given (6/5)
Reciprocal of (6/5) is (5/6)
It is a proper fraction.
(v) Given (12/7)
Reciprocal of (12/7) is (7/12)
It is a proper fraction.
(vi) Given (1/8)
Reciprocal of (1/8) is (8/1) = 8
It is a whole number.
2. Divide:
(i) (3/8) by (5/9)
(ii) 3 1/4 by (2/3)
(iii) (7/8) by 4 1/2
(iv) 6 1/4 by 2 3/5
Solution:
(i) Given (3/8) by (5/9)
From the rule of division of fraction, we know that (a/b) ÷ (c/d) = (a/b) × (d/c)
(3/8)/ (5/9) = (3/8) × (9/5)
= (3 × 9) / (8 × 5)
= (27/40)
(ii) Given 3 1/4 by (2/3)
Converting 3 1/4 to improper fraction, we get (13/4)
From the rule of division of fraction, we know that (a/b) ÷ (c/d) = (a/b) × (d/c)
(13/4)/ (2/3) = (13/4) × (3/2)
= (13 × 3) / (4 × 2)
= (39/8)
= 4 7/8
(iii) Given (7/8) by 4 1/2
Converting 4 1/2 to improper fraction, we get (9/2)
From the rule of division of fraction, we know that (a/b) ÷ (c/d) = (a/b) × (d/c)
(7/8)/ (9/2) = (7/8) × (2/9)
= (7 × 2) / (8 × 9)
= (14/72)
= (7/36)
(iv) Given 6 1/4 by 2 3/5
Converting 6 1/4 and 2 3/5 to improper fraction we get (25/4) and (13/5)
From the rule of division of fraction, we know that (a/b) ÷ (c/d) = (a/b) × (d/c)
(25/4)/ (13/5) = (25/4) × (5/13)
= (25 × 5) / (4 × 13)
= (125/52)
= 2 21/52
3. Divide:
(i) (3/8) by 4
(ii) (9/16) by 6
(iii) 9 by (3/16)
(iv) 10 by (100/3)
Solution:
(i) Given (3/8) by 4
= (3/8)/4
= (3/8 × 4)
= (3/32)
(ii) Given (9/16) by 6
= (9/16)/6
= (9/ 16 × 6)
= (9/ 96)
= (3/32)
(iii) Given 9 by (3/16)
= 9/ (3/16)
= (9 × 16)/3
= 16 × 3
= 48
(iv) Given 10 by (100/3)
= 10/ (100/3)
= (10 × 3)/100
= (3/10)
4. Simplify:
(i) (3/10) ÷ (10/3)
(ii) 4 3/5 ÷ (4/5)
(iii) 5 4/7 ÷ 1 3/10
(iv) 4 ÷ 2 2/5
Solution:
(i) Given (3/10) ÷ (10/3)
= (3 × 3)/ (10 × 10)
= (9/100)
(ii) Given 4 3/5 ÷ (4/5)
First, convert the given mixed fraction into improper fractions.
4 3/5 = (23/5)
(23/5) ÷ (4/5) = (23 × 5)/ (5 × 4)
= (23/4)
= 5 3/4
(iii) Given 5 4/7 ÷ 1 3/10
First, convert the given mixed fractions into improper fractions.
(39/7) and (13/10)
(39/7) ÷ (13/10) = (39 × 10)/ (7 × 13)
= (390/91)
= (30/7)
= 4 2/7
(iv) Given 4 ÷ 2 2/5
First, convert the given mixed fraction into an improper fraction.
2 2/5 = (12/5)
4 ÷ (12/5) = (4 × 5)/12
= (20/12)
= 1 2/3
5. A wire of length 12 1/2 m is cut into 10 pieces of equal length. Find the length of each piece.
Solution:
Given total length of the wire is = 12 1/2 = (25/2) m
It is cut into 10 pieces, so the length of one piece is
= (25/2)/10
= (25/20)
= (5/4)
= 1 1/4 m
6. The length of rectangular plot of area 65 1/3 m2 is 12 1/4 m. What is the width of the plot?
Solution:
Given area of rectangular plot is 65 1/3 m2 = (196/3) m2
Length of the same plot is 12 1/4 m = (49/4) m
Width of the plot is
Area = length × breadth
(196/3) = (49/4) × breadth
Breadth = (196/3)/ (49/4)
= (196 × 4)/ (49 × 3)
= (784/147)
= 5 1/3
7. By what number should 6 2/9 be multiplied to get 4 4/9?
Solution:
Let x be the number which needs to be multiplied by 6 2/9 = (56/9)
x × (56/9) = 4 4/9
x × (56/9) = (40/9)
x = (40/9) × (9/56)
x = (40/56)
x = (5/7)
8. The product of two numbers is 25 5/6. If one of the numbers is 6 2/3, find the other.
Solution:
Given product of two numbers is 25 5/6 = (155/6)
One of the numbers is 6 2/3 = (20/3)
Let the other number be x
(155/6) = x × (20/3)
x = (3/20) × (155/6)
x = (31/8)
x = 3 7/8
9. The cost of 6 1/4 kg of apples is Rs 400. At what rate per kg are the apples being sold?
Solution:
The cost of 6 1/4 kg = (25/4) of apples is Rs 400
Cost of apple per kg is = (25/4) /400
= (4/25) × 400
= Rs 64
10. By selling oranges at the rate of Rs 5 1/4 per orange, a fruit-seller gets Rs 630. How many dozens of oranges does he sell?
Solution:
Given cost of 1 orange is Rs 5 1/4 = (21/4)
He got Rs 630 by selling the oranges
Number of dozens of oranges sold by him for Rs 630 is = (4/21) × 630
= 120 apples
But we know that 1 dozen = 12
120 apples mean 10 dozens
11. In mid-day meal scheme (3/10) litre of milk is given to each student of a primary school. If 30 litres of milk is distributed every day in the school, how many students are there in the school?
Solution:
Given (3/10) litre of milk is given to each student
Number of student given (3/10) litre of milk = 1
Number of students giving 1 litre of milk = (10/3)
Numbers of students giving 30 litres of milk = (10/3) × 30 = 100 students
12. In a charity show Rs 6496 were collected by selling some tickets. If the price of each ticket was Rs 50 3/4, how many tickets were sold?
Solution:
Given amount collected by selling tickets = Rs 6496
The price of each ticket is = 50 3/4 = (203/4)
Number of ticket bought at Rs (203/4) = 1
Number of tickets bought for Rs 6496 is = (4/203) × 6496
= 4 × 32
= 128 tickets
RD Sharma Solutions for Class 7 Maths Chapter – 2 Fractions Exercise 2.3
Exercise 2.3 of RD Sharma Solutions for Class 7 Maths Chapter 2 Fractions explains the fundamental operations of fractions. Some of the topics prior to Exercise 2.3 are listed below.
- Reciprocal of fractions
- Division of fractions
Students can refer to this exercise to improve their conceptual knowledge and understand the different ways used to solve problems.
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