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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)RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7
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 RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7= 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)RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7÷4/5
(vii)RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7÷8/21

(viii)RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7÷RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7
(ix)RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7
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 =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7
(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)RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7÷4/5 = 18/5÷4/5
= 18/5×5/4 [∵ Reciprocal of 4/5 = 5/4]
= 18/4 = 9/2 =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7
(vii)RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7÷ 8/21 = 24/7÷8/21
= 24/7×21/8 [∵ Reciprocal of 8/21 = 21/8]
= 3  3 = 9
(viii)RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7÷RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7=39/7÷13/10
= 39/7×10/13 [∵ Reciprocal of 13/10 = 10/13]
= 30/7
=RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7
(ix)RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7
=108/7÷72/49
= 108/7×49/72 [∵ Reciprocal of 72/49 = 49/72]
= (9×7) / (1×6)
= (3×7) / (1×2)
= 21/2
=RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7

Q.3. Divide:
(i) 11/24 by 7/8
(ii)RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7by 11/16
(iii)RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7

(iv) 32 byRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7 
(v) 45 byRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7 
(vi) 63 byRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7
Ans.
(i) 11/24÷7/8
= 11/24×8/7 [∵ Reciprocal of 7/8 = 8/7]
= 11/21
(ii)RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7÷11/16 = 55/8÷11/16
=55/8×16/11 [∵ Reciprocal of 11/16 = 16/11]
= 5 × 2 = 10
(iii)RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7= 50/9÷10/3
= 50/9×3/10 [∵ Reciprocal of 10/3 = 3/10]
=  5/3 =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7
(iv) 32÷RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7= 32÷8/5
= 32×5/8 [∵ Reciprocal of 8/5 = 5/8]
= 4 × 5 = 20
(v) 45÷RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7= 45÷9/5
= 45×5/9 [∵ Reciprocal of 9/5 = 5/9]
= 5 × 5 = 25
(vi) 63÷RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7= 63÷9/4
= 63×4/9 [∵ Reciprocal of 9/4 = 4/9]
= 7 × 4 = 28

Q.4. A rope of lengthRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7m has been divided into 9 pieces of the same length. What is the length of each piece?
Ans. 
Length of the rope =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7m =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 =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7m
Hence, the length of each piece of rope isRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7m.

Q.5. 18 boxes of nails weigh equally and their total weight isRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7kg. How much does each box weigh?
Ans. Weight of 18 boxes of nails =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7kg = 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
= RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7kg
Hence, the weight of each box isRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7kg.

Q.6. By selling oranges at the rate of RsRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7per orange, a man gets Rs 378. How many oranges does he sell?
Ans. Selling price of an orange = RsRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7= 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 RsRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7per kg. What is the weight of mangoes available for Rs RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7?
Ans.
Selling price of 1kg mango =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7= 87/2
Weight of mangoes at 1305/4 =1305/4×2/87 =435/58 =15/2 =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7
Hence, the weight of mangoes=RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7kg

Q.8. Vikas can cover a distance ofRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7km inRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7hours on foot. How many km per hour does he walk?
Ans. Distance covered by Vikas inRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7h =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7km
∴ Distance covered by him in 1 h = (RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7÷RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7) km
= (62/3÷31/4) km
= (62/3×4/31) km
= ((2×4)/3) km =(8/3) km
=RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7km
Hence, the distance covered by Vikas in 1 h is RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7km.

Q.9. Preeti boughtRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7kg of sugar for RsRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7. Find the price of sugar per kg.
Ans.
Cost of 17/2 kg of sugar = 969/4
Cost of 1 kg of sugar =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7
= 969/4×2/17=57/2
=RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7
Hence, the cost of sugar is RsRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7per kg

Q.10. If the cost of a notebook is RsRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7, how many notebooks can be purchased for Rs RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7?
Ans. 
Cost of 1 notebook =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7= 111/4
Number of notebooks purchased forRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7
= 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 RsRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7. The total amount collected by a boy was RsRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7. How many tickets were sold by him?
Ans. 
Total amount collected =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7= 1755/2
Price of 1 ticket =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7
= 65/2
Number of tickets sold =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7=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 RsRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7.The total contribution was RsRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7. How many students are there in the group?
Ans.
Total contribution =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7 =5753/2
Contribution of each student =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7= 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 containsRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7litres 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 =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7L = 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 isRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7. If one of the numbers isRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7, find the other.

Ans. Product of the two numbers = RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7=95/6
One of the numbers =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7=19/3
∴ The other number = (95/6 ÷ 19/3)
= (95/6 × 3/19) [∵ Reciprocal of 19/3 = 3/19]
= (5/2)
=RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7
Hence, the other number isRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7.

Q.16. By what number shouldRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7be multiplied to get 42?
Ans. Product of the two numbers = 42
One of the numbers =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7= 49/5
∴ The other number = (42 ÷ 49/5)
=(42 × 5/49) [∵ Reciprocal of 49/5 = 5/49]
=(6 × 5) / 7 = 30/7 =RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7
Hence, the required number isRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7.

Q.17. By what number shouldRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7be divided to obtainRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7?
Ans.
Required number = (RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7÷RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7)
= (56/9 ÷ 14/3)
= (56/9  × 3/14) [ ∵ Reciprocal of 14/3 = 3/14]
= (4/3)
=RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7
Hence, we have to divide RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7byRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7to getRS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7.

The document RS Aggarwal Solutions: Integers (Exercise 2C) | Mathematics (Maths) Class 7 is a part of the Class 7 Course Mathematics (Maths) Class 7.
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FAQs on RS Aggarwal Solutions: Integers (Exercise 2C) - Mathematics (Maths) Class 7

1. What are RS Aggarwal Solutions?
RS Aggarwal Solutions are a set of comprehensive solutions to the exercises and questions provided in the RS Aggarwal textbook. These solutions help students understand and solve mathematical problems effectively.
2. What is the topic covered in RS Aggarwal Solutions: Integers (Exercise 2C) Class 7?
RS Aggarwal Solutions: Integers (Exercise 2C) Class 7 covers the topic of integers and provides solutions to the exercises given in the textbook for Class 7 students.
3. How can RS Aggarwal Solutions for Integers (Exercise 2C) Class 7 help students?
RS Aggarwal Solutions for Integers (Exercise 2C) Class 7 can help students in multiple ways. These solutions provide step-by-step explanations and solutions to the exercises, helping students understand the concepts better. They also serve as a valuable resource for practicing and improving problem-solving skills.
4. Are RS Aggarwal Solutions for Integers (Exercise 2C) Class 7 available online?
Yes, RS Aggarwal Solutions for Integers (Exercise 2C) Class 7 are available online. Many educational websites and platforms offer these solutions for free or with a subscription. Students can access them anytime and anywhere, making it convenient for self-study and revision.
5. How can students use RS Aggarwal Solutions for Integers (Exercise 2C) Class 7 effectively?
To use RS Aggarwal Solutions for Integers (Exercise 2C) Class 7 effectively, students should first read the textbook chapter thoroughly. They can then attempt the exercises on their own and refer to the solutions for guidance and verification. It is important to understand the steps and concepts behind each solution to improve problem-solving skills. Regular practice using these solutions will help students gain confidence in solving integer-related problems.
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