CBSE Class 7 Mathematics Operations with Integers Practice Paper Worksheet Set 02 with Answers

This is Class 7 Maths Ganita Prakash Part 2 Chapter 2 Operations with Integers Practice Paper 27, a 40-mark test of 20 questions in Sections A to E, with a step-by-step answer under every question. It practises multiplying and dividing negative integers, the sign rules for products and quotients, solving for an unknown in statements such as x ÷ (−9) × (−4) = 48, and long chains like (−420) ÷ (−7) × (−5) × 2. Word problems on a deep-sea robot, a rescue drone, a mining lift, satellite fuel and warehouse stock show how descents, losses and refills become negative and positive integers.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. A rescue drone descends at 24 m/min for 5 minutes and ascends at 16 m/min for 4 minutes.
(a) Find the total descent.
(b) Find the total ascent.
Answer:
Take downward movement as negative and upward movement as positive.
(a) Total descent = 5 × (−24) = −120 m (120 m down).
(b) Total ascent = 4 × 16 = +64 m (64 m up).
(So the drone ends 120 − 64 = 56 m below where it started.)
In simple words: Rate × time gives each movement: 120 m down and 64 m up.

Teacher's Note:
1. Write the direction as a sign: down is negative, up is positive.
2. Multiply the rate per minute by the number of minutes.

Question 12. The product of two integers is −1008. One of the integers is 42.
(a) Find the other integer.
(b) State the rule of signs used.
Answer:
(a) Other integer = (−1008) ÷ 42.
1008 ÷ 42 = 24, and a negative divided by a positive is negative.
So the other integer is −24. Check: 42 × (−24) = −1008.
(b) Rule used: positive × negative = negative (and so negative ÷ positive = negative).
In simple words: Divide the product by the known number: −1008 ÷ 42 = −24.

Teacher's Note:
1. Division undoes multiplication: product ÷ one factor = other factor.
2. The product is negative, so the two integers must have different signs.

Question 13. Evaluate: (−144) × (−7) ÷ (−12)
Answer:
Work from left to right.
(−144) × (−7) = 1008 (negative × negative = positive).
1008 ÷ (−12) = −84 (positive ÷ negative = negative).
So the value is −84.
In simple words: Three negative numbers, so the answer is negative: −84.

Teacher's Note:
1. Multiplication and division are done in order from left to right.
2. An odd number of negative signs gives a negative result.

Question 14. Find x if x × (−18) ÷ 6 = −48
Answer:
Undo the division: x × (−18) = −48 × 6 = −288.
Undo the multiplication: x = −288 ÷ (−18) = 16.
Check: 16 × (−18) = −288 and −288 ÷ 6 = −48.
So x = 16.
In simple words: Work backwards: multiply by 6, then divide by −18 to get x = 16.

Teacher's Note:
1. Undo the operations in reverse order.
2. A negative divided by a negative is positive.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Determine the sign and value of:
(−15) × 8 ÷ (−4) × (−2)
Answer:
Sign first: there are three negative numbers (−15, −4, −2), an odd number, so the answer is negative.
Value, working left to right:
(−15) × 8 = −120.
−120 ÷ (−4) = 30.
30 × (−2) = −60.
So the sign is negative and the value is −60.
In simple words: Three negatives make the answer negative, and the value is −60.

Teacher's Note:
1. Counting negative signs predicts the sign before you calculate.
2. Use the prediction to check your final answer.

Question 16. Evaluate: (−420) ÷ (−7) × (−5) × 2
Answer:
Work from left to right.
(−420) ÷ (−7) = 60.
60 × (−5) = −300.
−300 × 2 = −600.
So the value is −600.
In simple words: Three negative numbers make the answer negative: −600.

Teacher's Note:
1. The expression has three negative numbers, so the answer must be negative.
2. Do the division first because it comes first from the left.

Question 17. If x × (−24) ÷ (−6) × 5 = 200, find x.
Answer:
Simplify the known numbers: (−24) ÷ (−6) = 4, and 4 × 5 = 20.
So the statement becomes x × 20 = 200.
x = 200 ÷ 20 = 10.
Check: 10 × (−24) = −240; −240 ÷ (−6) = 40; 40 × 5 = 200.
So x = 10.
In simple words: The numbers after x combine to 20, so x = 200 ÷ 20 = 10.

Teacher's Note:
1. Combining the known numbers into one multiplier makes the equation easy.
2. Always substitute back to check.

Section D: Long Answer Question (5 Marks)

Question 18. A mining lift descends 20 m/min for 8 minutes, ascends 14 m/min for 6 minutes, and then descends again 9 m/min for 4 minutes.
(a) Represent each movement using integers.
(b) Calculate the total downward movement.
(c) Calculate the total upward movement.
(d) Find the net movement.
(e) State the final position relative to the starting point.
Answer:
(a) Down is negative and up is positive: first descent −20 m/min, ascent +14 m/min, second descent −9 m/min.
(b) First descent: 8 × (−20) = −160 m. Second descent: 4 × (−9) = −36 m.
Total downward movement = −160 + (−36) = −196 m.
(c) Total upward movement = 6 × 14 = +84 m.
(d) Net movement = −196 + 84 = −112 m.
(e) The lift ends 112 m below its starting point.
In simple words: It went down 196 m and up 84 m, so it ends 112 m below the start.

Teacher's Note:
1. Add all the downward parts together before combining with the upward part.
2. A negative net movement means the final position is below the start.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. Case Study 1: Satellite Fuel
A satellite in low Earth orbit is undergoing a critical fuel efficiency test. During the test, it consumes fuel at a rate of 45 litres per second for 8 seconds to maintain its trajectory. Afterward, due to system calibration, it is refueled at 30 litres per second for 6 seconds. Engineers must calculate the net fuel usage to ensure the satellite remains within operational safety limits. Accurate integer multiplication and division help determine how much fuel is effectively consumed and replenished. Monitoring fuel balance is essential to avoid mission failure and ensure long-term satellite stability.
Answer the following questions:
(a) What integer represents the fuel consumed per second?
(b) What is the total fuel consumed in 8 seconds?
(c) What is the total fuel replenished in 6 seconds?
(d) What is the net fuel change after both operations?
Answer:
Fuel used is a decrease (negative); fuel added is an increase (positive).
(a) Fuel consumed per second = −45 litres.
(b) Total consumed = 8 × (−45) = −360 litres.
(c) Total replenished = 6 × 30 = +180 litres.
(d) Net fuel change = −360 + 180 = −180 litres (the tank has 180 litres less).
In simple words: 360 litres used and 180 litres added leaves the satellite 180 litres down.

Teacher's Note:
1. Decide the sign from the meaning: using up is negative, refilling is positive.
2. The net change is the sum of the two signed totals.

Question 20. Case Study 2: Warehouse Inventory
A warehouse managing electronic equipment reports stock losses and arrivals over a week. It records 15 damaged shipments, each losing 220 units. During the same period, 10 new shipments arrive, each containing 400 units. To maintain accurate inventory records, the manager calculates the net stock change using integer multiplication and division. Proper stock calculation prevents shortages, avoids overstocking, and supports smooth warehouse operations. Accurate inventory control ensures timely delivery, reduces losses, and helps maintain efficiency in supply chain management.
Answer the following questions:
(a) What integer represents the loss per damaged shipment?
(b) What is the total number of units lost?
(c) What is the total number of units added?
(d) What is the net stock change?
Answer:
(a) A loss of 220 units per shipment is written as −220.
(b) Units lost = 15 × (−220) = −3300 (3300 units lost).
(c) Units added = 10 × 400 = +4000.
(d) Net stock change = −3300 + 4000 = +700 units (stock rises by 700).
In simple words: 4000 units came in and 3300 were lost, so stock went up by 700.

Teacher's Note:
1. 15 × 220 = 15 × 200 + 15 × 20 = 3000 + 300 = 3300.
2. A positive net change means the stock increased overall.