CBSE Class 7 Mathematics Operations with Integers Practice Paper Worksheet Set 04 with Answers

Class 7 Maths Ganita Prakash Part 2 Chapter 2 Operations with Integers Practice Paper 31 is a full 40-mark practice test with 20 questions spread over Sections A to E, and every question has a detailed answer. It tests comparing products to find the greatest and smallest values, solving (−x) × 18 = −360, the missing factor in □ × (−18) = 306, when a product of three integers is positive, and dividing in order as in (−480) ÷ 12 ÷ (−4). Word problems about a submarine, a hospital service elevator, a melting Himalayan glacier and a factory's lost and regained production practise net change with integers.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. A submarine descends 24 m per minute for 5 minutes.
(a) Write the integer expression.
(b) Find the total vertical movement.
Answer:
Going down is negative, so the rate is −24 m per minute.
(a) Expression: (−24) × 5.
(b) (−24) × 5 = −120 m, that is, 120 m downward.
In simple words: Down 24 m a minute for 5 minutes is −120 m.

Teacher's Note:
1. Put the sign on the rate, not on the time.
2. 24 × 5 = 120.

Question 12. The product of two integers is −960. One integer is 16.
(a) Write the division statement.
(b) Find the other integer.
Answer:
(a) Other integer = (−960) ÷ 16.
(b) 960 ÷ 16 = 60, and negative ÷ positive is negative.
So the other integer is −60. Check: 16 × (−60) = −960.
In simple words: −960 ÷ 16 = −60.

Teacher's Note:
1. A multiplication fact can be rewritten as a division fact.
2. The product is negative, so the two integers have unlike signs.

Question 13. Evaluate: (−180) ÷ 9 × (−4)
(a) Find the quotient.
(b) Find the final result.
Answer:
(a) (−180) ÷ 9 = −20.
(b) (−20) × (−4) = 80.
In simple words: −180 ÷ 9 = −20, then −20 × −4 = 80.

Teacher's Note:
1. Work from left to right.
2. Two negatives in the expression give a positive answer.

Question 14. Find x if: x × (−12) ÷ 4 = −36
(a) Remove the division.
(b) Find x.
Answer:
(a) Multiply both sides by 4: x × (−12) = −36 × 4 = −144.
(b) x = −144 ÷ (−12) = 12.
Check: 12 × (−12) = −144 and −144 ÷ 4 = −36.
In simple words: Multiply by 4, then divide by −12: x = 12.

Teacher's Note:
1. Multiplication undoes division.
2. Negative ÷ negative = positive.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Evaluate: (−9) × 8 ÷ (−6) × 5
Answer:
Left to right:
(−9) × 8 = −72.
−72 ÷ (−6) = 12.
12 × 5 = 60.
So the value is 60.
In simple words: Two negatives make the answer positive: 60.

Teacher's Note:
1. An even number of negative numbers gives a positive result.
2. Do × and ÷ in order from the left.

Question 16. The product of three integers is −1,080. Two of them are −12 and 15. Find the third integer.
Answer:
Product of the two known integers: (−12) × 15 = −180.
Third integer = (−1080) ÷ (−180).
Negative ÷ negative is positive, and 1080 ÷ 180 = 6.
So the third integer is 6.
Check: (−12) × 15 × 6 = −180 × 6 = −1080.
In simple words: Multiply the two known integers to get −180, then −1080 ÷ (−180) = 6.

Teacher's Note:
1. Turn the problem into a two-number problem by combining the known factors.
2. One negative factor is already present, so the third must be positive to keep the product negative.

Question 17. Find the value of: (−480) ÷ 12 ÷ (−4)
Answer:
Left to right:
(−480) ÷ 12 = −40.
−40 ÷ (−4) = 10.
So the value is 10.
In simple words: Two negatives give a positive result: 10.

Teacher's Note:
1. Division is not associative: always divide in order from the left.
2. 480 ÷ 12 = 40 because 12 × 40 = 480.

Section D: Long Answer Question (5 Marks)

Question 18. A service elevator operates in a hospital building. The elevator:
Descends 22 m per minute for 6 minutes
Ascends 15 m per minute for 8 minutes
Descends 10 m per minute for 5 minutes
Answer the following:
(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) Downward: (−22) × 6 and (−10) × 5; upward: (+15) × 8.
(b) (−22) × 6 = −132 m and (−10) × 5 = −50 m.
Total downward movement = −132 + (−50) = −182 m.
(c) Total upward movement = 15 × 8 = +120 m.
(d) Net movement = −182 + 120 = −62 m.
(e) The elevator is 62 m below its starting point.
In simple words: 182 m down and 120 m up leaves the elevator 62 m below the start.

Teacher's Note:
1. Add the two descents together before comparing with the ascent.
2. The sign of the net movement tells you above or below.

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

Question 19. Case Study 1: Glacier Research Expedition
A team of scientists is studying a glacier in the Himalayas. Due to melting, the glacier surface lowers at a rate of 14 meters per year for 9 years. After conservation efforts, ice accumulation causes the surface to rise at a rate of 8 meters per year for 6 years. Researchers use integer operations to determine the net change in glacier height to assess the long-term impact of climate change and conservation measures.
Answer the following questions:
(a) What integer represents the yearly decrease in height?
(b) What is the total decrease over 9 years?
(c) What is the total increase over 6 years?
(d) What is the net change in glacier height?
Answer:
(a) Yearly decrease = −14 m.
(b) Total decrease = 9 × (−14) = −126 m.
(c) Total increase = 6 × 8 = +48 m.
(d) Net change = −126 + 48 = −78 m (the surface is 78 m lower overall).
In simple words: The glacier lost 126 m and gained 48 m, so it is 78 m lower.

Teacher's Note:
1. Lowering is negative and rising is positive.
2. Rates per year are multiplied by the number of years.

Question 20. Case Study 2: Factory Production
A factory manufactures electronic components daily. Due to a power breakdown, the factory loses 1,450 units per day for 6 days. After restoring operations, production improves, and the factory gains 1,100 units per day for 9 days. The management calculates the net change in production using integer multiplication and division to evaluate operational efficiency and plan future strategies.
Answer the following questions:
(a) What integer represents the daily loss?
(b) What is the total loss over 6 days?
(c) What is the total gain over 9 days?
(d) What is the net change in production?
Answer:
(a) Daily loss = −1,450 units.
(b) Total loss = 6 × (−1,450) = −8,700 units.
(c) Total gain = 9 × 1,100 = +9,900 units.
(d) Net change = −8,700 + 9,900 = +1,200 units (production is up by 1,200 units overall).
In simple words: 8,700 units lost and 9,900 gained gives a net gain of 1,200 units.

Teacher's Note:
1. 1,450 × 6 = 1,400 × 6 + 50 × 6 = 8,400 + 300 = 8,700.
2. A positive net change means an overall gain.