CBSE Class 7 Mathematics Operations with Integers Worksheet Set 01

Long Answer Questions

Question 1. The token model uses a green token for +1 and a red token for −1, and a green–red pair is a zero pair. Every multiplication starts with an empty bag: a positive multiplier means “put tokens in” and a negative multiplier means “take tokens out”.
(i) Using tokens, explain how 5 × (−3) is carried out and write its value.
(ii) Using tokens and zero pairs, explain how (−3) × 4 is carried out and write its value.
(iii) Using tokens and zero pairs, explain how (−3) × (−4) is carried out and write its value.
(iv) Write, in one sentence each, the two sign rules that your three answers suggest.
Answer:
The bag starts empty, and zero pairs may be put in freely because each is worth 0.
(i) The multiplier 5 is positive, so we put 3 red tokens in, five times. The bag holds 15 reds. So 5 × (−3) = −15.
(ii) The multiplier −3 is negative, so we must take out 4 greens, three times. Each time we first put in 4 zero pairs and then remove the 4 greens. After three rounds, 12 reds are left. So (−3) × 4 = −12.
(iii) Now we must take out 4 reds, three times. Each time we put in 4 zero pairs and remove the 4 reds. After three rounds, 12 greens are left. So (−3) × (−4) = 12.
(iv) When one integer is positive and the other negative, the product is negative. When both integers are negative, the product is positive.
In simple words: Putting in reds or taking out greens gives negatives; taking out reds gives positives.

Teacher's Note:
1. Zero pairs let us remove tokens from an empty bag.
2. The size of the product is always the product of the sizes.

Question 2. Study the multiplication of integers through a pattern.
(i) Write the products 3 × (−5), 2 × (−5), 1 × (−5) and 0 × (−5) one below the other.
(ii) Describe how the product changes each time the multiplier goes down by 1.
(iii) Continue the same pattern for the multipliers −1, −2 and −3, and write the three products you get.
(iv) State what the completed pattern tells you about the sign of (positive) × (negative) and of (negative) × (negative).
Answer:
(i) 3 × (−5) = −15
2 × (−5) = −10
1 × (−5) = −5
0 × (−5) = 0
(ii) Each time the multiplier goes down by 1, the product goes up by 5, because one fewer copy of −5 means 5 more in value.
(iii) Keeping the same step of 5: (−1) × (−5) = 5, (−2) × (−5) = 10, (−3) × (−5) = 15.
(iv) In the upper part (positive × negative) every product is negative. In the lower part (negative × negative) every product is positive. The sizes 15, 10, 5, 0, 5, 10, 15 are multiples of 5; only the sign changes.
In simple words: The products rise by 5 each step, so they become positive below zero.

Teacher's Note:
1. The pattern must continue smoothly past zero.
2. This is why negative × negative is positive.

Question 3. The number line below has four marked points.
(i) Write the integer shown by each of the points P, Q, R and S, and write the additive inverse of each of these integers.
(ii) Using the number line, find P + S and R − P. Show the jumps you make.
(iii) Which of the four integers is the smallest, and which is nearest to 0?
(iv) Two integers have a sum of 14 and a difference (first − second) of −4. Find the two integers and mark them on the same number line.
Answer:
(i) P = −7 (inverse 7); Q = −3 (inverse 3); R = 2 (inverse −2); S = 6 (inverse −6).
(ii) P + S = (−7) + 6: start at −7 and make 6 jumps to the right; we land on −1.
R − P = 2 − (−7) = 2 + 7: start at 2 and make 7 jumps to the right; we land on 9.
(iii) Numbers get smaller as we move left, so P = −7 is the smallest. R = 2 is nearest to 0 (only 2 units away; Q is 3 units away).
(iv) Pairs adding to 14: 9 and 5 give a difference of 4, so swapping them gives −4. The integers are 5 (first) and 9 (second): 5 + 9 = 14 and 5 − 9 = −4.
On the number line, 5 lies just left of S, and 9 lies three units to the right of S:

In simple words: P, Q, R, S are −7, −3, 2, 6; P + S = −1, R − P = 9; the hidden pair is 5 and 9.

Teacher's Note:
1. Subtracting a negative means moving right.
2. The further left, the smaller the integer.

Question 4. Brahmagupta, in his Brāhmasphuṭasiddhānta (628 CE), called a positive value a fortune (dhana) and a negative value a debt (ṛṇa). He wrote that the product or quotient of two fortunes is a fortune, the product or quotient of two debts is a fortune, and the product or quotient of a debt and a fortune is a debt.
(i) Rewrite each of his three statements using the words “positive” and “negative”.
(ii) Give one multiplication example and one division example for each of the three statements.
(iii) Write the three division rules in symbols for positive integers a and b, with b ≠ 0.
(iv) Why is this piece of work considered an important step in the history of arithmetic?
Answer:
(i) The product or quotient of two positive numbers is positive. The product or quotient of two negative numbers is positive. The product or quotient of a positive and a negative number (in either order) is negative.
(ii) Two fortunes: 7 × 8 = 56 and 56 ÷ 8 = 7.
Two debts: (−7) × (−8) = 56 and (−56) ÷ (−8) = 7.
A debt and a fortune: (−7) × 8 = −56 and (−56) ÷ 8 = −7 (also 7 × (−8) = −56 and 56 ÷ (−8) = −7).
(iii) (−a) ÷ b = −(a ÷ b), a ÷ (−b) = −(a ÷ b), (−a) ÷ (−b) = a ÷ b.
(iv) It was the first time anyone wrote down clear rules for multiplying and dividing positive and negative numbers. By treating a debt as a number in its own right, Brahmagupta made negative numbers part of ordinary arithmetic, which opened the way to algebra as we use it today.
In simple words: Fortune means positive and debt means negative; same kinds give a fortune, mixed kinds give a debt.

Teacher's Note:
1. Learn the words dhana and rina.
2. Give both a product and a quotient example for each rule.

Question 5. Take the three integers a = −4, b = 6 and c = −9.
(i) Verify that a × b = b × a and name the property.
(ii) Verify that a × (b × c) = (a × b) × c and name the property.
(iii) Verify that a × (b + c) = (a × b) + (a × c) and name the property.
(iv) Write the general statement of each of the three properties for any integers a, b and c.
Answer:
(i) a × b = (−4) × 6 = −24 and b × a = 6 × (−4) = −24. They are equal: the commutative property.
(ii) b × c = 6 × (−9) = −54, so a × (b × c) = (−4) × (−54) = 216. a × b = −24, so (a × b) × c = (−24) × (−9) = 216. Equal: the associative property.
(iii) b + c = −3, so a × (b + c) = (−4) × (−3) = 12. (a × b) + (a × c) = −24 + 36 = 12. Equal: the distributive property of multiplication over addition.
(iv) For any integers a, b, c: a × b = b × a; a × (b × c) = (a × b) × c; a × (b + c) = a × b + a × c.
In simple words: Both sides give −24, 216 and 12, proving the three properties.

Teacher's Note:
1. Put brackets around negative values when substituting.
2. Work out each side separately.

Question 6. Division of integers can always be checked through multiplication.
(i) Turn (−96) ÷ (−8) into a multiplication question, answer it, and write the quotient.
(ii) Write the three sign rules for division of integers in words.
(iii) Find (−144) ÷ 12, 150 ÷ (−25) and (−169) ÷ (−13).
(iv) For each answer in part (iii), write the multiplication statement that checks it.
Answer:
(i) (−96) ÷ (−8) asks: which integer multiplied by −8 gives −96? Since 8 × 12 = 96 and (−8) × 12 = −96, the quotient is 12.
(ii) Positive ÷ positive is positive. Negative ÷ negative is positive. When one is positive and the other negative, the quotient is negative. (The divisor can never be 0.)
(iii) (−144) ÷ 12 = −12; 150 ÷ (−25) = −6; (−169) ÷ (−13) = 13.
(iv) 12 × (−12) = −144; (−25) × (−6) = 150; (−13) × 13 = −169. Each check gives back the dividend.
In simple words: Turn each division into 'what times the divisor gives the dividend?'.

Teacher's Note:
1. Divisor × quotient = dividend.
2. The size comes from ordinary division.

Teacher's Note:
1. Test a guessed rule on every row.
2. Many inputs can give the same output.

Question 8. Consider the following six expressions formed from the integers 24 and 15.
(i) (−24) + (−15)
(ii) (−24) − (−15)
(iii) 24 − (−15)
(iv) (−24) × (−15)
(v) 24 × (−15)
(vi) (−360) ÷ 15
Evaluate all six expressions, showing one line of working for each. Then arrange the six values in increasing order, writing the labels (i)–(vi) in that order. Finally, state which property of integers assures you, without any fresh calculation, that (−15) × (−24) has the same value as expression (iv).
Answer:
(i) (−24) + (−15) = −39.
(ii) (−24) − (−15) = −24 + 15 = −9.
(iii) 24 − (−15) = 24 + 15 = 39.
(iv) (−24) × (−15) = 360 (both negative, so positive).
(v) 24 × (−15) = −360 (one negative, so negative).
(vi) (−360) ÷ 15 = −24 (different signs, so negative).
Increasing order: −360 < −39 < −24 < −9 < 39 < 360, so the labels are (v), (i), (vi), (ii), (iii), (iv).
Property: multiplication of integers is commutative (a × b = b × a), so (−15) × (−24) = (−24) × (−15) = 360.
In simple words: The values are −39, −9, 39, 360, −360, −24; swapping factors keeps the product.

Teacher's Note:
1. The most negative number is the smallest.
2. Subtracting a negative means adding.

Question 9. On the planet Zeta, money is counted in pibs. There are only two kinds of coins: a coin worth +11 pibs and a coin worth −7 pibs. A payment is made by handing over some coins of each kind, and the amount paid is the total value of the coins handed over. You have as many coins of each kind as you need.
(i) Show how an item costing exactly +4 pibs can be paid for.
(ii) Show how an item costing exactly +1 pib can be paid for.
(iii) Show how an amount of exactly −3 pibs can be formed.
(iv) Using your answer to part (ii), explain why every whole number of pibs can be paid exactly. Illustrate this by paying exactly +50 pibs with as few coins as you can find.
Answer:
(i) One +11 coin and one −7 coin: 11 + (−7) = 4.
(ii) Two +11 coins and three −7 coins: 2 × 11 + 3 × (−7) = 22 − 21 = 1.
(iii) One +11 coin and two −7 coins: 11 + 2 × (−7) = 11 − 14 = −3.
(iv) Part (ii) gives a bundle worth exactly 1 pib. Repeating it n times gives exactly n pibs for any positive whole number n. Five +11 coins with eight −7 coins give 55 − 56 = −1, and repeating that bundle gives any negative amount; no coins give 0. So every whole number of pibs can be paid.
For +50 pibs, fifty 1-pib bundles would need 250 coins. A much shorter way is nine +11 coins and seven −7 coins: 9 × 11 + 7 × (−7) = 99 − 49 = 50, using only 16 coins.
In simple words: A 1-pib bundle can be repeated to make any amount; 50 pibs needs only 16 coins.

Teacher's Note:
1. Once you can make 1, you can make every whole number.
2. Check each total by adding the coin values.

Teacher's Note:
1. The magic lies in how the grid is built.
2. Rearranging factors does not change a product.

Case Study Based Questions

Teacher's Note:
1. The difference of temperatures is later minus earlier.
2. A mean can be negative.

Question 12. A hospital building has floors above and below the ground. In a hospital, the ground floor is numbered 0, the floors above it are numbered +1 to +6, and the three basement levels are numbered −1, −2 and −3. Every floor, above or below the ground, is 3 m high. The lift shows the floor number on a screen, so a visitor can always read his or her position as an integer. During a rush hour the lift moves steadily, going down 2 floors every 10 seconds.
(i) The lift goes from the ground floor to the lowest basement. Which integer does the screen show, and how many metres below the ground is the lift then?
(ii) The lift is at floor +4 and goes down 7 floors. Write the addition statement and name the floor it reaches.
(iii) Starting from floor +6, where is the lift after 40 seconds of the rush-hour movement described above?
(iv) A patient at basement level −2 has to reach the operation theatre on floor +5. Through how many floors must the lift rise?
Answer:
(i) The screen shows −3. Position = (−3) × 3 = −9, that is, 9 m below the ground.
(ii) 4 + (−7) = −3. The lift reaches basement level −3.
(iii) 40 seconds = 4 spells of 10 seconds, each −2 floors: change = 4 × (−2) = −8. Position = 6 + (−8) = −2, basement level 2.
(iv) Rise = 5 − (−2) = 5 + 2 = 7 floors.
In simple words: Floors below the ground are negative, and the lift's moves are added as integers.

Teacher's Note:
1. Going down is a negative change.
2. Distance between floors = higher minus lower.

Question 13. Meera’s bank passbook records money put in and money taken out. In Meera’s passbook, money credited (put in) is written as a positive integer and money debited (taken out) as a negative integer, all in rupees. On 1 April her balance was ₹2,000. On 5 April she credited ₹1,500 and on 9 April she withdrew ₹800. Her mobile bill of ₹250 is paid automatically from the account on the 15th of every month, and this happened in four months, from April to July. No other transaction took place in these four months.
(i) Write the withdrawal of ₹800 as an integer.
(ii) What was the balance in the account just after the withdrawal on 9 April?
(iii) Write the total effect of the four automatic bill payments as a product of two integers, and find it.
(iv) Find the balance at the end of July, and write the average monthly change caused by the bill payments as a quotient of two integers.
Answer:
(i) Money taken out is negative: −800.
(ii) Balance = 2000 + 1500 + (−800) = ₹2,700.
(iii) 4 × (−250) = −1000, so ₹1,000 went out.
(iv) Balance at the end of July = 2700 + (−1000) = ₹1,700. Average monthly change = (−1000) ÷ 4 = −250, a fall of ₹250 a month.
In simple words: Credits are positive and debits negative; she ends July with ₹1,700.

Teacher's Note:
1. Write every debit with a minus sign.
2. Average change = total change ÷ number of months.

Question 14. An inter-house quiz is scored with negative marking. In an inter-house quiz there are 25 questions. Every correct answer earns +4 marks, every wrong answer earns −2 marks, and a question left unanswered earns 0 marks. Team A answered 15 questions correctly, got 8 answers wrong and left 2 questions unanswered. Team B attempted every question and got 10 of them wrong. The team with the higher total wins the trophy.
(i) How many marks did Team A earn from its correct answers?
(ii) Write the marks earned by Team A from its wrong answers as a product of two integers, and find it.
(iii) Find Team A’s total score.
(iv) Find Team B’s total score and state which team won, and by how many marks.
Answer:
(i) 15 × 4 = 60 marks.
(ii) 8 × (−2) = −16.
(iii) 60 + (−16) + 0 = 44 marks.
(iv) Team B: 25 − 10 = 15 correct. Score = 15 × 4 + 10 × (−2) = 60 − 20 = 40. Team A won by 44 − 40 = 4 marks.
In simple words: Both teams had 15 correct, but Team A lost fewer marks, 44 to 40.

Teacher's Note:
1. Unanswered questions add 0.
2. Find the correct answers of Team B first.

Question 15. A research submarine records its position with respect to sea level. A research submarine takes sea level as 0. Positions below sea level are written as negative integers, measured in metres. The submarine can dive at a steady rate of 8 m per minute and can rise at a steady rate of 5 m per minute. One morning it begins at the surface and dives for 15 minutes. It then rises quickly by 45 m to photograph a shoal of fish, and after that dives again for another 10 minutes at its usual diving rate before heading back to the surface.
(i) Write the position of the submarine after the first 15 minutes as a product of two integers, and find it.
(ii) What is its position after it rises by 45 m?
(iii) What is its position at the end of the second dive?
(iv) From this last position, how long will it take to come back to the surface?
Answer:
(i) Diving is a change of −8 m each minute: 15 × (−8) = −120 m (120 m below sea level).
(ii) −120 + 45 = −75 m.
(iii) Change in the second dive = 10 × (−8) = −80, so the position is −75 + (−80) = −155 m.
(iv) It must rise 155 m at 5 m per minute: 155 ÷ 5 = 31 minutes.
In simple words: It reaches −155 m and needs 31 minutes to come back up.

Teacher's Note:
1. Diving is negative, rising is positive.
2. Use the rising rate, not the diving rate, in part (iv).

Question 16. A cricket team compares each over with the run rate it needs. A team chasing a target needs exactly 6 runs in every over. After each over the coach writes an integer: the runs actually scored in that over minus the 6 runs required. A positive integer means the team did better than needed and a negative integer means it fell behind. In the first six overs the team scored 9, 2, 6, 0, 11 and 4 runs.
(i) Write the coach’s integer for the second over.
(ii) Write the coach’s integers for all six overs in order.
(iii) Find the sum of these six integers and say, in words, what it tells us about the team.
(iv) In each of the next four overs the team falls behind by exactly 1 run. What is the total of the coach’s integers after 10 overs?
Answer:
(i) 2 − 6 = −4.
(ii) 3, −4, 0, −6, 5, −2.
(iii) 3 + (−4) + 0 + (−6) + 5 + (−2) = −4. The team is 4 runs behind the run rate it needs.
(iv) Four more overs at −1 each: 4 × (−1) = −4. Total = −4 + (−4) = −8, so the team is 8 runs behind.
In simple words: The coach's integers add to −4 after six overs and −8 after ten.

Teacher's Note:
1. Each integer is runs scored minus 6.
2. A negative total means falling behind.

Question 17. A village pottery unit keeps a record of profit and loss on each pot. A village pottery unit sells its pots in two grades. On every perfect pot it makes a profit of ₹12, written as +12. Every pot with a crack has to be sold cheap, and on each such pot the unit makes a loss of ₹7, written as −7. At the end of a day the unit adds up all these integers to find the day’s result: a positive total means profit and a negative total means loss.
(i) On Monday the unit sold 150 perfect pots and 40 cracked pots. Find the result for Monday.
(ii) On Tuesday the kiln failed and all 95 pots sold were cracked. Find the result for Tuesday.
(iii) On Wednesday 60 pots were cracked. How many perfect pots must be sold so that the day ends with neither profit nor loss?
(iv) On Thursday only 20 perfect pots were sold and the day ended with a loss of ₹110. How many cracked pots were sold?
Answer:
(i) 150 × 12 + 40 × (−7) = 1800 − 280 = 1520: a profit of ₹1,520.
(ii) 95 × (−7) = −665: a loss of ₹665.
(iii) Cracked pots give 60 × (−7) = −420. Perfect pots must give +420, so their number is 420 ÷ 12 = 35 pots.
(iv) Perfect pots give 20 × 12 = 240. If c pots were cracked, 240 + c × (−7) = −110, so c × (−7) = −350 and c = 50. 50 cracked pots were sold. Check: 240 − 350 = −110.
In simple words: Profit is positive and loss is negative; the totals tell the result of each day.

Teacher's Note:
1. No profit, no loss means the total is 0.
2. Check part (iv) by putting 50 back.

Question 18. Children play a carrom-coin game on a number line. Four children take turns to strike a carrom coin that starts at 0 on a number line drawn on the floor. A strike that moves the coin to the right is written as a positive integer and a strike that moves it to the left as a negative integer; the size of the number is the distance moved in units. In one round the four strikes, in order, were +6, −9, +4 and −7.
(i) Write the addition statement for the round and find the final position of the coin.
(ii) What are the magnitude and the direction of the second strike?
(iii) A fifth child now strikes the coin and it comes to rest at +2. Which integer describes this fifth strike?
(iv) In a new round the coin starts at 0 and every strike moves it exactly 3 units to the left. How many strikes are needed to bring it to −45? Write this as a division of two integers.
Answer:
(i) 6 + (−9) + 4 + (−7) = −6. The coin stops 6 units to the left of 0.
(ii) The second strike is −9: magnitude 9 units, direction leftward.
(iii) From −6 to +2: 2 − (−6) = +8 (8 units to the right).
(iv) Each strike is −3, so the number of strikes is (−45) ÷ (−3) = 15.
In simple words: The coin ends at −6, needs +8 to reach 2, and 15 strikes of −3 reach −45.

Teacher's Note:
1. Magnitude is the size; direction is the sign.
2. Strike = end position − start position.

Question 19. A mine elevator moves up and down a deep shaft. An elevator in a mining shaft takes the ground level as 0. Positions above the ground are positive and positions below the ground are negative, measured in metres. The elevator descends at a steady 4 m per minute. The bottom of the shaft is at −180 m. One morning the elevator starts from a loading platform 12 m above the ground and descends without stopping. A safety board inside the cage asks the operator to work out each position and each time before the machine is started.
(i) Write an expression using integers for the position of the elevator after 20 minutes, and find that position.
(ii) How many minutes does the elevator take to go from the loading platform right down to the bottom of the shaft?
(iii) A worker has to be dropped at the −125 m level, with the elevator starting from the ground level 0. Show that the time taken is not a whole number of minutes, and state between which two whole minutes it lies.
(iv) The motor fails and the elevator’s speed becomes 0 m per minute. A trainee writes the time needed to reach −120 m as (−120) ÷ 0 minutes. Explain why this expression has no meaning.
Answer:
(i) Position = 12 + 20 × (−4) = 12 − 80 = −68 m, that is, 68 m below the ground.
(ii) Total drop from +12 to −180 = 12 − (−180) = 192 m. Time = 192 ÷ 4 = 48 minutes.
(iii) Time = 125 ÷ 4 minutes. 4 × 31 = 124 and 4 × 32 = 128, so no integer times 4 gives 125. The time is not a whole number; it lies between 31 and 32 minutes.
(iv) (−120) ÷ 0 asks which number multiplied by 0 gives −120. Every number times 0 is 0, so there is no answer: division by zero is not allowed. A cage that is not moving will never reach −120 m.
In simple words: The elevator is at −68 m after 20 minutes and needs 48 minutes to reach the bottom.

Teacher's Note:
1. Start from +12, not from 0, in parts (i) and (ii).
2. Division by zero has no meaning.

Question 20. A class plays a number game called “Halve or Flip”. In the game “Halve or Flip”, a child calls out any integer. The class then follows one rule again and again: if the number is even, take half of it; if the number is odd, multiply it by −3 and then add 1. The new number is written down and the rule is applied to it, and so on. For example, starting at 8 the class writes 8, 4, 2, 1, and then, since 1 is odd, 1 × (−3) + 1 = −2.
(i) The game starts at −6. Write the next four numbers of the chain.
(ii) Continue the same chain for four more steps after −14.
(iii) Start a fresh chain at 1 and continue until the number 1 appears again. Write the cycle you obtain and state how many numbers it contains.
(iv) Explain why a “flip” step is always followed by a “halve” step.
Answer:
(i) −6 is even: −3. −3 is odd: (−3) × (−3) + 1 = 10. 10 is even: 5. 5 is odd: 5 × (−3) + 1 = −14. The numbers are −3, 10, 5, −14.
(ii) −14 → −7; (−7) × (−3) + 1 = 22; 22 → 11; 11 × (−3) + 1 = −32. The numbers are −7, 22, 11, −32.
(iii) 1 → 1 × (−3) + 1 = −2 → −1 → (−1) × (−3) + 1 = 4 → 2 → 1. The cycle is 1, −2, −1, 4, 2, which contains 5 numbers.
(iv) A flip is used only on an odd number. An odd number times −3 is odd, and odd + 1 is even. So the result of a flip is always even, and the next step must be a halving.
In simple words: Halve even numbers, flip odd ones; a flip always gives an even number.

Teacher's Note:
1. Negative numbers can also be even or odd.
2. Odd × odd is odd; odd + 1 is even.