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

Q.1. Find the product:
(i) 3/5×7/11
(ii) 5/8×4/7
(iii) 4/9×15/16
(iv) 2/5×15
(v) 8/15×20
(vi) 5/8×1000
(vii)RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7×16
(viii) RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7×12
(ix) RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7×RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7

(x) RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7×RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7
(xi)RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7
(xii)RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7
Ans.
(i) 3/5×7/11 = (3×7) / (5×11) = 21/55
(ii) 5/8×4/7 = (5×4) / (8×7) = (5×1) / (2×7) =5/14
(iii) 4/9×15/16 = (4×15) / (9×16) = (1×5) / (3×4) = 5/12
(iv) 2/5×15 = 2/5×15/1 = (2×15) / (5×1) = (2×3) / (1×1) =6
(v) 8/15×20 = 8/15×20/1 = (8×20)/(15×1) = (8×4) / (3×1) = 32/3 =RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7
(vi)  5/8×1000 = 5/8×1000/1 = (5×1000) / (8 × 1) = (5×125) / (1×1) = 625
(vii) RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7×16 = 25/8 × 16/1 = (25×16) / (8×1) = (25×2) / (1×1) = 50
(viii)RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7× 12 = 34/15 × 12/1 = (34×12) / (15×1) = (34×4) / (5×1) = 136/5 =RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7
(ix) RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7×RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7= 27/7 × 14/3 = (27×14) / (7×3) = (9×2) / (1×1) =18
(x)RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7= 19/2 × 28/19 = (19×28) / (2×19) = (1×14) / (1×1) = 14
(xi)RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7 = 33/8×32/11 = (33×32) / (8×11) = (3×4) / (1×1) =12
(xii)RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7= 35/6×12/7 = (35×12) / (6×7) = (5×2) / (1×1) =10

Q.2. Simplify:
(i) 2/3×5/44×33/35
(ii) 12/25×15/28×35/36
(iii) 10/27×28/65×39/56
(iv)RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7

(v)RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7
(vi)RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7
Ans.
We have the following:
(i) 2/3×5/44×33/35 = (2×5×33) / (3×44×35)
= (1×1×11) / (1×22×7) = (1×1×1) / (1×2×7) =1/14
(ii)12/25 × 15/28 × 35/36 = (1×3×5) / (5×4×3) = (1×1×1) / (1×4×1) = 1/4
(iii) 10/27 × 28/65 × 39/56 = (10×1×3) / (27×5×2) = (1×1×3) / (27×1×1) = 3/27 = 1/9
(iv)RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7= 11/7 × 35/22 × 16/15 = (11×35×16) / (7×22×15) = (1×5×16) / (1×2×15) = (1×1×8) / (1×1×3) = 8/3 =RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7
(v)RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7= 36/17 × 65/9 × 85/52 = (36×65×85) / (17×9×52) = (4×5×5) / (1×1×4) =(1×5×5) / (1×1×1) = 25
(vi)RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7= 49/16 × 52/7 × 64/39 = (7×4×4) / (1×1×3) = 112/3 =RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7

Q.3. Find:
(i) 1/3 of 24

(ii) 3/4 of 32
(iii) 5/9 of 45
(iv) 7/50 of 1000
(v) 3/20 of 1020
(vi) 5/11 of Rs 220
(vii) 4/9 of 54 metres
(viii) 6/7 of 35 litres
(ix) 1/6 of an hour
(x) 5/6 of an year
(xi) 7/20 of a kg
(xii) 9/20 of a metre
(xiii) 7/8 of a day
(xiv) 3/7 of a week
(xv) 7/50 of a litre
Ans. We have the following:
(i) 1/3 of 24 = 24×1/3 = 24/1 × 1/3 = 24 × 1 / 1×3 =8
(ii) 3/4 of 32 = 32 × 3/4 = 32/1×3/4= (32×3) / (1×4) = (8×3) / (1×1)=24
(iii) 5/9 of 45 = 45×5/9 = 45/1 × 5/9 = (45×5) / (1×9) = (5×5) / (1×1)=25
(iv) 7/50 of 1000 = 1000 × 7/50 = 1000/1×7/50 = (20×7) / (1×1)=140
(v) 3/20 of 1020 = 1020×3/20 = 1020/1×3/20 = (51×3)/(1×1)=153
(vi) 5/11 of Rs 220 = Rs (220×5/11) = Rs (20 × 5 ) = Rs 100
(vii) 4/9 of 54 m = (4/9×54)m = (4 × 6)m = 24 m
(viii) 6/7 of 35 L = (6/7×35)L = (6 × 5) L = 30 L
(ix) 1/6 of 1 h = 1/6 of 60 min = (60×1/6) min = 10 min
(x) 5/6 of an year = 5/6 of 12 months = (12×5/6) months = (2 × 5) months = 10 months
(xi) 7/20 of a kg = 7/20 of 1000 g = (1000×7/20) g = (50 × 7) gm = 350 g
(xii) 9/20 of 1 m = 9/20 of 100 cm = (100×9/20) cm = (5 × 9) cm = 45 cm
(xiii) 7/8 of a day = 7/8 of 24 h = (24×7/8) h = (3 × 7) = 21 h
(xiv) 3/7 of a week = 3/7 of 7 days = (7×3/7) days = 3 days
(xv) 7/50 of 1 L = 7/50 of 1000 ml = (1000×7/50) ml = (20 × 7) ml = 140 ml

Q.4. Apples are sold at RsRS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7per kg. What is the cost of RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7kg of apples?

Ans. Cost of 1 kg apple = RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7= 244/5
Cost of RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7kg apples =RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7× 244/5 = 15/4 × 244/5= 183
Hence, the cost ofRS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7apples is 183.

Q.5. Cloth is being sold at RsRS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7per metre. What is the cost of RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7metres of this cloth?
Ans. Cost of 1 m of cloth = RsRS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7= Rs 85/2
∴ Cost ofRS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7m 535 m of cloth = Rs (85/2 ×RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7)
= Rs (85/2 × 28/5)=Rs ((85×28) / (2×5))
=Rs (17×14)=Rs 238
Hence, the cost ofRS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7of cloth is Rs 238.

Q.6. A car covers a certain distance at a uniform speed ofRS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7km per hour. How much distance will it cover in 9 hours?
Ans. Distance covered by the car in 1h =RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7km
Distance covered by the car in 9h  =(RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7×9) km
=(200/3×9) km =(200×9) / (3×1) km
=(200×3) km=600 km
Hence, the distance covered by the car in 9 h will be 600 km.

Q.7. One tin holdsRS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7litres of oil. How many litres of oil can 26 such tins hold?
Ans.
Capacity of 1 tin =RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7L= 51/4 L
∴ Capacity of 26 such tins =(26×51/4) L
= (26/1×51/4) L =(26×51) / (1×4) L
= (13×51) / (1×2) L = (663/2) L
=RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7L
Hence, 26 such tins can holdRS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7L of oil.

Q.8. For a particular show in a circus, each ticket costs Rs 35123512. If 308 tickets are sold for the  show, how much amount has been collected?
Ans.
Cost of 1 ticket = RsRS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7= Rs 7/12
∴ Cost of 308 tickets = Rs (7/12×308) =Rs (7/12×308/1)
=Rs (71×154)
=Rs 1093
Hence, 308 tickets were sold for Rs 10,934.

Q.9. Nine boards are stacked on the top of each other. The thickness of each board isRS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7cm. How high is the stack?
Ans. Thickness of 1 board =RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7cm
∴ Thickness of 9 boards = (9×RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7) cm
=(9/1×11/3) cm
= (3 × 11) cm
= 33 cm
Hence, the height of the stack is 33 cm.

Q.10. Rohit takesRS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7minutes to make complete round of a circular park. How much time will he take to make 15 rounds?

Ans. Time taken by Rohit to complete one round of the circular park =RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7min = 24/5 min
∴ Time taken to complete 15 rounds =(15×24/5) min
= (3 × 24) min
= 72 min
= 1 h 12 min [∵ 1 hr = 60 min]
Hence, Rohit will take 1 h 12 min to make 15 complete rounds of the circular park.

Q.11. Amit weighs 35 kg. His sister Kavita's weight is 3/5 of Amit's weight. How much does Kavita weigh?
Ans. 
Weight of Amit = 35 kg
Weight of Kavita = 3/5 of Amit's weight
= 35 kg x 3/5
= (35×3/5)kg
=(7×3) kg=21 kg
Hence, Kavita's weight is 21 kg.

Q.12. There are 42 students in a class and 5/7 of the students are boys. How many girls are there in the class?
Ans. 
Number of boys in the class = 5/7 of the total no. of students
=5/7 × 42 = (5×42) / 7
=5×6=30
∴ Number of girls in the class = 42 − 30 = 12
Hence, there are 12 girls in the class.

Q.13. Sapna earns Rs 24000 per month. She spends 7/8 of her income and deposits rest of the money in a bank. How much money does she deposit in the bank each month?
Ans. Total monthly income = 24000
Monthly expenditure = 7/8 of 24000
=24000×7/8
=21000
Monthly savings =24000−21000
= 3000

Q.14. Each side of a square field isRS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7m. Find its area.
Ans.
Side of the square field =RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7m
∴ Area of the square = (side)2
= (RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7m)2
= (14/3 m)2
= 14/3 m ×14/3 m
= (14×14) / (3×3) m2
= 196/9 m2
=RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7m2
Hence, the area of the square field isRS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7m2.

Q.15. Find the area of a rectangular park which isRS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7m long andRS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7m broad.
Ans. 
Length of the rectangular park =RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7m = 125/3 m
Its breadth =RS Aggarwal Solutions: Integers (Exercise 2B) | Mathematics (Maths) Class 7m = 93/5 m
∴ Its area = length × breadth
= (125/3 × 93/5) m2
= (25 × 31) m = 775 m2
Hence, the area of the rectangular park is 775 m2.

The document RS Aggarwal Solutions: Integers (Exercise 2B) | 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 2B) - Mathematics (Maths) Class 7

1. What are RS Aggarwal Solutions?
Ans. RS Aggarwal Solutions are a set of comprehensive solutions provided for various textbooks written by RS Aggarwal. These solutions help students understand and solve the problems given in the textbooks effectively.
2. What is the importance of RS Aggarwal Solutions for Class 7 Integers?
Ans. RS Aggarwal Solutions for Class 7 Integers are important as they provide step-by-step solutions to the exercise questions in the textbook. These solutions help students to understand the concepts better and practice solving problems related to integers.
3. How can RS Aggarwal Solutions for Class 7 Integers help in exam preparation?
Ans. RS Aggarwal Solutions for Class 7 Integers are a valuable resource for exam preparation. By referring to these solutions, students can understand the correct approach to solve different types of integer problems. Regular practice with these solutions can help students improve their problem-solving skills and perform well in exams.
4. Are RS Aggarwal Solutions for Class 7 Integers available online?
Ans. Yes, RS Aggarwal Solutions for Class 7 Integers are available online. Students can access these solutions on educational websites, mobile apps, or download them as PDF files. These solutions provide a convenient way for students to study and revise the integer concepts at their own pace.
5. How can I access RS Aggarwal Solutions for Class 7 Integers?
Ans. To access RS Aggarwal Solutions for Class 7 Integers, you can search for them on educational websites or download mobile apps that provide these solutions. You can also find these solutions in PDF format on various online platforms.
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