The PDF of RD Sharma Solutions for Exercise 2.2 of Class 7 Maths Chapter 2 Fractions is provided here. The questions present in this exercise have been solved by BYJU’S experts in Maths, and this will help students solve the problems without any difficulties. This exercise contains fifteen questions with many sub-questions, which deal with the multiplication of integers. By practising sincerely, the students can obtain worthy results in the final exam using the RD Sharma Solutions for Class 7.
RD Sharma Solutions for Class 7 Maths Chapter 2 – Fractions Exercise 2.2
Access answers to Maths RD Sharma Solutions for Class 7 Chapter 2 – Fractions Exercise 2.2
Exercise 2.2 Page No: 2.19
1. Multiply:
(i) (7/11) by (3/5)
(ii) (3/5) by 25
(iii) 3 4/15 by 24
(iv) 3 1/8 by 4 10/11
Solution:
(i) Given (7/11) by (3/5)
We have to multiply the given number.
(7/11) × (3/5) = (21/55)
(ii) Given (3/5) by 25
(3/5) × 25 = 15 [dividing 25 by 5]
(iii) Given 3 4/15 by 24
First, convert the given mixed fraction to an improper fraction.
(49/15) × 24 = (1176/15)
= 78 2/5
(iv) Given 3 1/8 by 4 10/11
First, convert the given mixed fractions to improper fractions.
(25/8) × (54/11) = (1350/88) = (675/44)
= 15 15/44
2. Find the product:
(i) (4/7) × (14/25)
(ii) 7 1/2 × 2 4/15
(iii) 3 6/7 × 4 2/3
(iv) 6 11/14 × 3 1/2
Solution:
(i) Given (4/7) × (14/25)
(4/7) × (14/25) = (4 × 14)/ (7 × 25)
= (56/175)
Converting the above fractions into the simplest form
= (8/25)
(ii) Given 7 1/2 × 2 4/15
We have to convert mixed fractions into improper fractions.
Then, we get (15/2) and (34/15)
7 (1/2) × 2 (4/15) = (15/2) × (34/15)
= (15 × 34)/ (2 × 15)
= (510/30)
= 17
(iii) Given 3 6/7 × 4 2/3
We have to convert mixed fractions into improper fractions.
Then we get (27/7) and (14/3)
3 6/7 × 4 2/3 = (27/7) × (14/3)
On simplifying
= 9 × 2
= 18
(iv) Given 6 11/14 × 3 1/2
We have to convert mixed fractions into improper fractions.
Then, we get (95/14) and (7/2)
6 11/14 × 3 1/2 = (95/14) × (7/2)
= (95 × 7)/28
= (665/28)
= 23 3/4
3. Simplify:
(i) (12/25) × (15/28) × (35/36)
(ii) (10/27) × (39/56) × (28/65)
(iii) 2 2/17 × 7 2/9 × 1 33/52
Solution:
(i) Given (12/25) × (15/28) × (35/36)
= (12 × 15 × 35)/ (25 × 28 × 36)
= (6300/25200)
On simplifying, we get
= (1/4)
(ii) Given (10/27) × (39/56) × (28/65)
= (10 × 39 × 28)/ (27 × 56 × 65)
= (10920/98280)
On simplifying, we get
= (1/9)
(iii) Given 2 2/17 × 7 2/9 × 1 33/52
First, convert the given mixed fractions into improper fractions then, we get
= (36/17) × (65/9) × (85/52)
= (36 × 65 × 85)/ (17 × 9 × 52)
= (198900/7956)
On simplifying, we get
= 25
4. Find:
(i) (1/2) of 4 2/9
(ii) (5/8) of 9 2/3
(iii) (2/3) of (9/16)
Solution:
(i) Given (1/2) of 4 2/9
First, convert given mixed fraction into an improper fraction then we get (38/9)
= (1/2) × (38/9)
= (1 × 38)/ (2 × 9)
= (38 /18)
= 2 1/9
(ii) Given (5/8) of 9 2/3
First, convert given mixed fraction into an improper fraction then we get (29/3)
= (5/8) × (29/3)
= (5 × 29)/ (8 × 3)
= (145 /24)
= 6 1/24
(iii) Given (2/3) of (9/16)
= (2/3) × (9/16)
= (2 × 9)/ (3 × 16)
= (18 /48)
= (3/8)
5. Which is greater? (1/2) of (6/7) or (2/3) of (3/7)
Solution:
Given (1/2) of (6/7)
= (1/2) × (6/7)
= (1 × 6)/ (2 × 7)
= (6 /14)
Also given that (2/3) of (3/7)
= (2/3) × (3/7)
= (2 × 3)/ (3 × 7)
= (6 /21)
While comparing two fractions, if the numerators of both fractions are the same, then the denominator having a higher value shows the fraction having a lower value.
Therefore, (6/14) is greater.
Hence, (1/2) of (6/7) is greater.
6. Find:
(i) (7/11) of Rs 330
(ii) (5/9) of 108 meters
(iii) (3/7) of 42 litres
(iv) (1/12) of an hour
(v) (5/6) of an year
(vi) (3/20) of a kg
(vii) (7/20) of a litre
(viii) (5/6) of a day
(ix) (2/7) of a week
Solution:
(i) Given (7/11) of Rs 330
= (7/11) × 330
On dividing by 11 we get
= 7 × 30
= 210
(7/11) of Rs 330 is Rs 210
(ii) Given (5/9) of 108 meters
= (5/9) × 108
Dividing 108 by 9 we get
= 5 × 12
= 60
(5/9) of 108 meters is 60 meters
(iii) Given (3/7) of 42 litres
= (3/7) × 42
Dividing 42 by 7 we get
= 3 × 6
= 18
(3/7) of 42 litres is 18 litres
(iv) Given (1/12) of an hour
An hour = 60 minutes
= (1/12) × 60
Dividing 60 by 12 we get
= 1 × 5
= 5
(1/12) of an hour is 5 minutes
(v) Given (5/6) of an year
1 year = 12 months
= (5/6) × 12
Dividing 12 by 6 we get
= 5 × 2
= 10
(5/6) of an year is 10 months
(vi) Given (3/20) of a kg
1 kg = 1000 grams
= (3/20) × 1000
= 3 × 50
= 150
(3/20) of a kg is 150 grams
(vii) Given (7/20) of a litre
1 liter = 1000 ml
= (7/20) × 1000
= 7 × 50
= 350
(7/20) of a litre is 350ml
(viii) Given (5/6) of a day
1 day = 24 hours
= (5/6) × 24
= 5 × 4
= 20
(5/6) of a day is 20 hours
(ix) Given (2/7) of a week
1 week = 7 days
= (2/7) × 7
= 2 × 1
= 2
(2/7) of a week is 2 days
7. Shikha plants 5 saplings in a row in her garden. The distance between two adjacent saplings is ¾ m. Find the distance between the first and the last sapling.
Solution:
Given that the distance between two adjacent saplings is (3/4) m
There are 4 adjacent spacing for 5 sapling
Therefore, the distance between the first and the last sapling is
= (3/4) × 4
= 3
The distance between them is 3m
8. Ravish reads (1/3) part of a book in 1 hour. How many parts of the book will he read in 2 1/5 hours?
Solution:
Given Ravish takes 1 hour to read (1/3) part of the book
Then we have to calculate how many parts he will read in 2 1/5 hours
First, convert the given mixed fraction into an improper fraction, i.e., (11/5)
Now, let x be the full part of the book
1 hour = (1/3) x
Remaining part of the book, he will read in
= (11/5) × (1/3) x
= (11/15) part of the book
9. Lipika reads a book for 1 3/4 hours every day. She reads the entire book in 6 days. How many hours in all were required by her to read the book?
Solution:
Given time taken by Lipika to read a book per day = 1 3/4 = (7/4) hours
Time taken by Lipika to read a book in 6 days = (7/4) × 6
= (42/4)
= 10 ½ hours
10. Find the area of a rectangular park, which is 41 2/3 m long and 18 3/5 m broad.
Solution:
Given length of rectangular park is = 41 2/3 = (125/3)
Breadth of rectangular park is = 18 3/5 = (93/5)
Area of rectangular park = length × breadth
= (125/3) × (93/5)
= (125 × 93)/15
= (11625/15)
= 775 m2
11. If milk is available at Rs 17 3/4 per litre, find the cost of 7 2/5 litres of milk.
Solution:
Given the cost of milk per liter is = 17 3/4 = Rs (71/4)
And the cost of 7 2/5 = (37/5) is
= (37/ 5) × (71/4)
= (37 × 71)/20
= (2627/20)
= Rs 131 7/20
12. Sharada can walk 8 1/3 km in one hour. How much distance will she cover in 2 2/5 hours?
Solution:
Given distance covered by Sharada in one hour = 8 1/3 = (25/3) km
Distance covered by her in 2 2/5 hours = (12/5) is
= (25/3) × (12/5)
= (25 × 12)/15
= (300/15)
= 20 km
13. A sugar bag contains 30kg of sugar. After consuming (2/3) of it, how much sugar is left in the bag?
Solution:
A sugar bag contains 30kg of sugar.
After consuming, the left sugar in the bag is = 30- (2/3) × 30
= 30 – 2 × 10
= 30 – 20
= 10kg
14. Each side of a square is 6 2/3 m long. Find its area.
Solution:
Side of a square = 6 2/3 = (20/3) m
Area of square = side × side
= (20/3) × (20/3)
= (400/9)
=44 4/9 m2
15. There are 45 students in a class, and (3/5) of them are boys. How many girls are there in the class?
Solution:
Total number of students = 45
Number of boys out of 45 is = (3/5)
Number of girls = 45 – (3/5) × 45
= 45 – 3 × 9
= 45 – 27
= 18 girls
RD Sharma Solutions for Class 7 Maths Chapter – 2 Fractions Exercise 2.2
Exercise 2.2 of RD Sharma Solutions for Class 7 Maths Chapter 2 Fractions explains the fundamental operations of fractions. In this exercise, students shall study how to multiply mixed fractions with improper fractions, proper fractions and mixed fractions.
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