CBSE Class 7 Mathematics Operations with Integers Practice Paper Worksheet Set 03 with Answers

Practice Paper 29 for Class 7 Maths Ganita Prakash Part 2 Chapter 2 Operations with Integers carries 40 marks and has 20 questions in five sections, A to E, each answered step by step below. It focuses on picking the expression with a negative or positive result, comparing negative products to find the greatest, finding unknowns such as x in x × (−15) ÷ (−5) × 6 = 108, and predicting the sign of a long product. The applied questions on a robotic elevator, a submarine, a diver and daily school attendance show how falls and rises are written as integers and combined into a net change.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. A deep-sea submarine descends at 17 m/min for 9 minutes and then ascends at 10 m/min for 6 minutes.
(i) Calculate the total descent.
(ii) Calculate the total ascent.
Answer:
Downward is negative and upward is positive.
(i) Total descent = 9 × (−17) = −153 m.
(ii) Total ascent = 6 × 10 = +60 m.
(The submarine ends −153 + 60 = −93 m, i.e. 93 m below its start.)
In simple words: 153 m down and 60 m up.

Teacher's Note:
1. 17 × 9 = 17 × 10 − 17 = 170 − 17 = 153.
2. Write the sign to show direction.

Question 12. The product of two integers is −840. One of the integers is 35.
(i) Find the other integer.
(ii) Explain why it is negative.
Answer:
(i) Other integer = (−840) ÷ 35 = −24, since 35 × 24 = 840.
Check: 35 × (−24) = −840.
(ii) The product is negative, so the two integers must have unlike signs. 35 is positive, so the other must be negative (negative ÷ positive = negative).
In simple words: −840 ÷ 35 = −24; it is negative because a positive times it gives a negative.

Teacher's Note:
1. A negative product always has one positive and one negative factor.
2. 35 × 24 = 35 × 20 + 35 × 4 = 700 + 140 = 840.

Question 13. Evaluate: (−108) × (−8) ÷ (−6)
Answer:
Left to right:
(−108) × (−8) = 864.
864 ÷ (−6) = −144.
So the value is −144.
In simple words: Three negatives give a negative result: −144.

Teacher's Note:
1. You can also do (−108) ÷ (−6) = 18 first, then 18 × (−8) = −144.
2. Check the sign by counting the negatives.

Question 14. Find x if x × (−12) ÷ 4 = −48
Answer:
Undo the division: x × (−12) = −48 × 4 = −192.
Undo the multiplication: x = −192 ÷ (−12) = 16.
Check: 16 × (−12) = −192 and −192 ÷ 4 = −48.
So x = 16.
In simple words: Multiply −48 by 4, then divide by −12: x = 16.

Teacher's Note:
1. Reverse the operations in the opposite order.
2. Dividing a negative by a negative gives a positive.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Determine the sign of: (−12) × 8 ÷ (−6) × (−3)
Answer:
Count the negative numbers: −12, −6 and −3, which is three (odd). So the result is negative.
Confirm by working left to right:
(−12) × 8 = −96.
−96 ÷ (−6) = 16.
16 × (−3) = −48.
The value is −48, which is negative.
In simple words: Three negative numbers make the answer negative (−48).

Teacher's Note:
1. The sign can be found without calculating, just by counting negatives.
2. Working out the value is a good check.

Question 16. Evaluate: (−224) ÷ (−7) × (−5) × 2
Answer:
Left to right:
(−224) ÷ (−7) = 32.
32 × (−5) = −160.
−160 × 2 = −320.
So the value is −320.
In simple words: Three negatives make it negative: −320.

Teacher's Note:
1. 224 ÷ 7 = 32 because 7 × 32 = 224.
2. Divide first because it comes first from the left.

Question 17. If x × (−15) ÷ (−5) × 6 = 108, find x
Answer:
Simplify the known numbers: (−15) ÷ (−5) = 3, and 3 × 6 = 18.
So x × 18 = 108.
x = 108 ÷ 18 = 6.
Check: 6 × (−15) = −90; −90 ÷ (−5) = 18; 18 × 6 = 108.
So x = 6.
In simple words: The numbers after x combine to 18, so x = 108 ÷ 18 = 6.

Teacher's Note:
1. Group the known numbers into one multiplier.
2. Substitute back to check.

Section D: Long Answer Question (5 Marks)

Question 18. A robotic lift descends 15 m/min for 10 minutes, ascends 9 m/min for 8 minutes, and descends again 6 m/min for 5 minutes.
(i) Represent each movement using integers
(ii) Total downward movement
(iii) Total upward movement
(iv) Net movement
(v) Final position relative to starting point
Answer:
(i) First descent −15 m/min, ascent +9 m/min, second descent −6 m/min.
(ii) First descent: 10 × (−15) = −150 m. Second descent: 5 × (−6) = −30 m.
Total downward movement = −150 + (−30) = −180 m.
(iii) Total upward movement = 8 × 9 = +72 m.
(iv) Net movement = −180 + 72 = −108 m.
(v) The lift is 108 m below its starting point.
In simple words: 180 m down and 72 m up leaves the lift 108 m below the start.

Teacher's Note:
1. List each movement with its sign before calculating.
2. Final position comes from the sign of the net movement.

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

Question 19. Case Study 1: Deep-Sea Dive
A deep-sea diver descends at a rate of 22 meters per minute for 9 minutes to reach the ocean floor. After completing the task, the diver ascends at a rate of 14 meters per minute for 6 minutes. The dive coordinator records these movements to calculate the diver’s net depth change and ensure safety standards are met.
Answer the following questions:
(a) What integer represents the diver’s downward movement per minute?
(b) What is the total depth descended?
(c) What is the total height ascended?
(d) What is the net depth change of the diver?
Answer:
(a) Downward movement per minute = −22 m.
(b) Total depth descended = 9 × (−22) = −198 m.
(c) Total height ascended = 6 × 14 = +84 m.
(d) Net depth change = −198 + 84 = −114 m (the diver ends 114 m below the starting point).
In simple words: Down 198 m and up 84 m leaves the diver 114 m below the start.

Teacher's Note:
1. 22 × 9 = 198 and 14 × 6 = 84.
2. The negative net change means the diver is still below the start.

Question 20. Case Study 2: School Attendance
In a school, a teacher records attendance changes over several days. Due to a festival, the number of students present decreases by 18 each day for 5 days. Afterward, due to a special academic drive, attendance increases by 24 students per day for 4 days. The teacher must calculate the net change in attendance using integer operations.
Answer the following questions:
(a) What integer represents the daily decrease in attendance?
(b) What is the total decrease in attendance over 5 days?
(c) What is the total increase in attendance over 4 days?
(d) What is the net change in student attendance?
Answer:
(a) Daily decrease = −18 students.
(b) Total decrease = 5 × (−18) = −90 students.
(c) Total increase = 4 × 24 = +96 students.
(d) Net change = −90 + 96 = +6 students (attendance is 6 more than at the start).
In simple words: 90 fewer then 96 more gives 6 more students overall.

Teacher's Note:
1. Decreases are negative and increases are positive.
2. A positive net change means attendance went up overall.