RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7 PDF Download

Q.1. Write down the reciprocal of:
(i) 5/8
(ii) 7
(iii) 1/12
(iv)
Ans. (i) Reciprocal of 5/8 = 8/5 [ ∵ 5/8×8/5=1]
(ii) Reciprocal of  7 =1/7 [ ∵ 7×1/7=1]
(iii) Reciprocal of 1/12= 12 [ ∵ 1/12×12=1]
(iv) Reciprocal of = Reciprocal of 63/5 = 5/63 [∵ 63/5 × 5/63=1]

Q.2. Simplify:
(i)  4/7 ÷ 9/14
(ii) 7/10 ÷ 3/5
(iii) 8/9÷16
(iv) 9÷1/3
(v) 24÷6/7
(vi)÷4/5
(vii)÷8/21
(viii)÷
(ix)
Ans. (i) 4/7÷9/14 = 4/7×14/9 [∵ Reciprocal of 9/14 = 14/9]
= 8/9
(ii) 7/10÷3/5 = 7/10×5/3 [∵ Reciprocal of 3/5 = 5/3]
= 7/6 =
(iii) 8/9÷16 = 8/9×1/16 [∵ Reciprocal of 16 = 1/16]
= 1/18
(iv) 9÷1/3=9×3 [∵ Reciprocal of 1/3 = 3]
= 27
(v) 24÷6/7=24×7/6 [∵ Reciprocal of 6/7 = 7/6]
= 4 × 7 = 28
(vi)÷4/5 = 18/5÷4/5
= 18/5×5/4 [∵ Reciprocal of 4/5 = 5/4]
= 18/4 = 9/2 =
(vii)÷ 8/21 = 24/7÷8/21
= 24/7×21/8 [∵ Reciprocal of 8/21 = 21/8]
= 3  3 = 9
(viii)÷=39/7÷13/10
= 39/7×10/13 [∵ Reciprocal of 13/10 = 10/13]
= 30/7
=
(ix)
=108/7÷72/49
= 108/7×49/72 [∵ Reciprocal of 72/49 = 49/72]
= (9×7) / (1×6)
= (3×7) / (1×2)
= 21/2
=

Q.3. Divide:
(i) 11/24 by 7/8
(ii)by 11/16
(iii)
(iv) 32 by 
(v) 45 by 
(vi) 63 by
Ans. (i) 11/24÷7/8
= 11/24×8/7 [∵ Reciprocal of 7/8 = 8/7]
= 11/21
(ii)÷11/16 = 55/8÷11/16
=55/8×16/11 [∵ Reciprocal of 11/16 = 16/11]
= 5 × 2 = 10
(iii)= 50/9÷10/3
= 50/9×3/10 [∵ Reciprocal of 10/3 = 3/10]
=  5/3 =
(iv) 32÷= 32÷8/5
= 32×5/8 [∵ Reciprocal of 8/5 = 5/8]
= 4 × 5 = 20
(v) 45÷= 45÷9/5
= 45×5/9 [∵ Reciprocal of 9/5 = 5/9]
= 5 × 5 = 25
(vi) 63÷= 63÷9/4
= 63×4/9 [∵ Reciprocal of 9/4 = 4/9]
= 7 × 4 = 28

Q.4. A rope of lengthm has been divided into 9 pieces of the same length. What is the length of each piece?
Ans. Length of the rope =m =27/2 m
Number of equal pieces = 9
∴ Length of each piece = (27/2÷9) m
= (27/2×1/9) m [∵ Reciprocal of 9 = 1/9]
= 3/2 m =m
Hence, the length of each piece of rope ism.

Q.5. 18 boxes of nails weigh equally and their total weight iskg. How much does each box weigh?
Ans. Weight of 18 boxes of nails =kg = 99/2 kg
∴ Weight of 1 box = (99/2÷18) kg
= (99/2×1/18) [∵ Reciprocal of 18 = 1/18]
= (99×1) / (2×18) kg
= (11×1) / (2×2) kg
=11/4 kg
= kg
Hence, the weight of each box iskg.

Q.6. By selling oranges at the rate of Rsper orange, a man gets Rs 378. How many oranges does he sell?
Ans. Selling price of an orange = Rs= Rs 27/4
Total money received after selling oranges = Rs 378
Total no. of oranges = 378/27×4=56
Hence, total no. of oranges = 56

Q.7. Mangoes are sold at Rsper kg. What is the weight of mangoes available for Rs ?
Ans. Selling price of 1kg mango == 87/2
Weight of mangoes at 1305/4 =1305/4×2/87 =435/58 =15/2 =
Hence, the weight of mangoes=kg

Q.8. Vikas can cover a distance ofkm inhours on foot. How many km per hour does he walk?
Ans. Distance covered by Vikas inh =km
∴ Distance covered by him in 1 h = (÷) km
= (62/3÷31/4) km
= (62/3×4/31) km
= ((2×4)/3) km =(8/3) km
=km
Hence, the distance covered by Vikas in 1 h is km.

Q.9. Preeti boughtkg of sugar for Rs. Find the price of sugar per kg.
Ans. Cost of 17/2 kg of sugar = 969/4
Cost of 1 kg of sugar =
= 969/4×2/17=57/2
=
Hence, the cost of sugar is Rsper kg

Q.10. If the cost of a notebook is Rs, how many notebooks can be purchased for Rs ?
Ans. Cost of 1 notebook == 111/4
Number of notebooks purchased for
= 999/4×4/111=9
Hence, the number of notebooks purchased are 9.

Q.11. At a charity show the price of each ticket was Rs. The total amount collected by a boy was Rs. How many tickets were sold by him?
Ans. Total amount collected == 1755/2
Price of 1 ticket =
= 65/2
Number of tickets sold ==1755/2×2/65=27
Hence, the number of tickets sold were 27.

Q.12. A group of students arranged a picnic. Each student contributed Rs.The total contribution was Rs. How many students are there in the group?
Ans. Total contribution = =5753/2
Contribution of each student == 523/2
Number of students = 5753/2÷523/2 = 5753/2 × 2/523=11
Hence, number of students in the group are 11.

Q.13. 24 litres of milk was distributed equally among all the students of a hostel. If each student got 2/5 litre of milk, how many students are there in the hostel?
Ans. Quantity of milk given to each student  = 2/5 L
Total quantity of milk distributed among all the students = 24 L
∴ Number of students = (24÷2/5)
= (24×52)24×52 [∵ Reciprocal of 2525 = 5252]
= (12 × 5) = 60
Hence, there are 60 students in the hostel.

Q.14. A bucket containslitres of water. A small jug has a capacity of 3/4 litre. How many times the jug has to be filled with water from the bucket to get it emptied?
Ans. Capacity of the small jug = 3/4 L
Capacity of the bucket =L = 81/4 L
∴ Required number of small jugs = (81/4 ÷ 3/4)
= (81/4×4/3) [∵ Reciprocal of 3/4 = 4/3]
= (81/3) = 27
Hence, the small jug has to be filled 27 times to empty the water from the bucket.

Q.15. The product of two numbers is. If one of the numbers is, find the other.
Ans. Product of the two numbers = =95/6
One of the numbers ==19/3
∴ The other number = (95/6 ÷ 19/3)
= (95/6 × 3/19) [∵ Reciprocal of 19/3 = 3/19]
= (5/2)
=
Hence, the other number is.

Q.16. By what number shouldbe multiplied to get 42?
Ans. Product of the two numbers = 42
One of the numbers == 49/5
∴ The other number = (42 ÷ 49/5)
=(42 × 5/49) [∵ Reciprocal of 49/5 = 5/49]
=(6 × 5) / 7 = 30/7 =
Hence, the required number is.

Q.17. By what number shouldbe divided to obtain?
Ans. Required number = (÷)
= (56/9 ÷ 14/3)
= (56/9  × 3/14) [ ∵ Reciprocal of 14/3 = 3/14]
= (4/3)
=
Hence, we have to divide byto get.