NCERT Solutions for Class 7th Maths Chapter 2 Figure it Out — Multiplication of Integers

Book page 42–44 Updated on2026-09-19

Q1.

Find the values of the following expressions: (a) (– 5) × (18 + (– 3)) (b) (– 7) × 4 × (– 1) (c) (– 2) × (– 1) × (– 5) × (– 3)

Answer

(a) (–5) × (18 + (–3))
= (–5) × 15
= –75

(b) (–7) × 4 × (–1)
= (–28) × (–1)
= 28

(c) (–2) × (–1) × (–5) × (–3)
= 2 × (–5) × (–3)
= (–10) × (–3)
= 30

	Negative factors	Magnitude	Answer
(a)	1 (odd)	5 × 15 = 75	–75
(b)	2 (even)	7 × 4 × 1 = 28	28
(c)	4 (even)	2 × 1 × 5 × 3 = 30	30

Why it happens: in (a) the bracket must be simplified first, and 18 + (–3) = 15 is positive, so only the –5 supplies a minus sign. In (b) and (c) simply count the minus signs — two and four, both even — so both answers are positive.

Check it yourself: part (a) also works by distributing — (–5) × 18 + (–5) × (–3) = –90 + 15 = –75. Same answer, and a useful check.

Q2.

Find the values of the following expressions: (a) (– 27) ÷ 9 (b) 84 ÷ (– 4) (c) (– 56) ÷ (– 2)

Answer

(a) (–27) ÷ 9: 27 ÷ 9 = 3, signs different → –3
(b) 84 ÷ (–4): 84 ÷ 4 = 21, signs different → –21
(c) (–56) ÷ (–2): 56 ÷ 2 = 28, signs same → 28

Turn each one into a multiplication to check it.

9 × (–3) = –27 ✓
(–4) × (–21) = 84 ✓
(–2) × 28 = –56 ✓

Why it happens: division inherits the multiplication sign rule, because a ÷ b is answering “b × ? = a”. The unknown must carry whatever sign makes the product come out right, and that is precisely the like-signs/unlike-signs rule.

Q3.

Find the integer whose product with (– 1) is: (a) 27 (b) – 31 (c) – 1 (d) 1 (e) 0

Answer

Multiplying by –1 gives the additive inverse, so the answer is always the additive inverse of the number given.

	Product wanted	Required integer	Check
(a)	27	–27	(–1) × (–27) = 27
(b)	–31	31	(–1) × 31 = –31
(c)	–1	1	(–1) × 1 = –1
(d)	1	–1	(–1) × (–1) = 1
(e)	0	0	(–1) × 0 = 0

–1 × a = –a
so if –1 × a = p, then a = –p

Why it happens: multiplying by –1 flips a number to the other side of zero without changing how far it is from zero. Doing that twice brings you home, so undoing the flip is just another flip. Part (e) is the exception that proves the rule — 0 is the only integer sitting on zero itself, so flipping it changes nothing.

Q4.

If 47 – 56 + 14 – 8 + 2 – 8 + 5 = – 4, then find the value of – 47 + 56 – 14 + 8 – 2 + 8 – 5 without calculating the full expression.

Answer

The value is 4.

Every single sign in the second expression is the opposite of the sign in the first. So the second expression is the first one multiplied by –1.

–47 + 56 – 14 + 8 – 2 + 8 – 5
= –(47 – 56 + 14 – 8 + 2 – 8 + 5)
= –(–4)
= 4

Why it happens: the first expression is really the sum 47 + (–56) + 14 + (–8) + 2 + (–8) + 5. The second is the sum of the additive inverses of those same seven numbers. Taking the inverse of every term is the same as taking the inverse of the whole sum, because multiplication by –1 distributes over addition.

Check it yourself: add the first expression step by step — 47, –9, 5, –3, –1, –9, –4. It really is –4, so the second must be 4.

Q5.

Do you remember the Collatz Conjecture from last year? Try a modified version with integers. The rule is — start with any number; if the number is even, take half of it; if the number is odd, multiply it by – 3 and add 1; repeat. An example sequence is shown below: –7, 22, 11, 32, –16, –8, –4, –2, –1, 4, 2, 1. Try this with different starting numbers: (– 21), (– 6), and so on. Describe the patterns you observe.

The example sequence drawn on page 42; the last arrow shows the chain running back into a number it has already reached.

Answer

Every start we try runs into the same short loop: 1 → –2 → –1 → 4 → 2 → 1.

First, a correction to the printed example. After 11 the rule gives 11 × (–3) + 1 = –32, not 32 — and the very next term, –16, confirms that it must be –32. The corrected chain is

–7 → 22 → 11 → –32 → –16 → –8 → –4 → –2 → –1 → 4 → 2 → 1

Now the two starts the book asks for.

Start –21: –21 is odd → (–21) × (–3) + 1 = 64
–21 → 64 → 32 → 16 → 8 → 4 → 2 → 1 → –2 → –1 → 4 → 2 → 1 → …

Start –6: –6 → –3 → 10 → 5 → –14 → –7 → 22 → 11 → –32 → –16 → –8 → –4 → –2 → –1 → 4 → 2 → 1 → …

Patterns worth describing:

An odd number always jumps to an even number, because odd × (–3) is odd and adding 1 makes it even. So an odd step is never followed by another odd step.
A negative odd number jumps to a positive number, and a positive odd number jumps to a negative one — the sign flips at every odd step.
Halving an even number over and over drives it towards ±1 quickly.
Every start we tried ends in the same 5-term cycle 1 → –2 → –1 → 4 → 2 → 1, and once in, it never gets out.

Why it happens: the two rules pull in opposite directions. Halving shrinks a number fast, while “× (–3) + 1” makes it about three times bigger but flips its sign. Since odd steps must alternate with even ones, the shrinking wins often enough to drag every chain down into the small numbers, and once you are at 1, 2, 4, –1 or –2 there is nowhere left to go but round the loop.

Try This: start at 0. It is even, so it halves to 0 and stays there forever — a one-term cycle of its own.

Q6.

In a test, (+ 4) marks are given for every correct answer and (– 2) marks are given for every incorrect answer. (a) Anita answered all the questions in the test. She scored 40 marks even though 15 of her answers were correct. How many of her answers were incorrect? How many questions are in the test? (b) Anil scored (– 10) marks even though he had 5 correct answers. How many of his answers were incorrect? Did he leave any questions unanswered?

Answer

(a) Anita got 10 answers wrong, and the test has 25 questions.

Marks from 15 correct answers = 15 × 4 = 60
She scored only 40, so the wrong answers cost her 60 – 40 = 20 marks
Each wrong answer costs 2 marks
Number of wrong answers = 20 ÷ 2 = 10

She answered every question, so
Total questions = 15 + 10 = 25

(b) Anil got 15 answers wrong, and he left 5 questions unanswered.

Marks from 5 correct answers = 5 × 4 = 20
He scored –10, so the wrong answers cost him 20 – (–10) = 30 marks
Number of wrong answers = 30 ÷ 2 = 15

Questions he attempted = 5 + 15 = 20
The test has 25 questions, so unanswered = 25 – 20 = 5

Why it happens: think of the score as starting at the “all correct so far” figure and being pulled down 2 marks at a time. For Anita the pull is 20 marks, for Anil it is 30 marks — and 30 marks of pulling drags his score right past zero into the negatives. Part (a) is what tells us the test has 25 questions, and that is the only reason part (b) can talk about unanswered questions at all.

Tip: an unanswered question is worth 0 — neither +4 nor –2. That is why Anil’s 20 attempts and 25 questions do not have to match.

Q7.

Pick the pattern — find the operations done by the machine shown below.

The machine on page 43. Each row feeds three numbers through the two hidden operations; the last star is left blank.

Answer

The machine works out a – (b × c) — the first number minus the product of the other two. The missing star is –70.

a	b	c	b × c	a – (b × c)	Result
4	8	–3	–24	4 – (–24)	28
6	9	6	54	6 – 54	–48
2	3	–2	–6	2 – (–6)	8
–9	5	–8	–40	–9 – (–40)	31
7	–4	–6	24	7 – 24	–17
–16	–6	–9	54	–16 – 54	–70

Last row:
(–6) × (–9) = 54
(–16) – 54
= –70

Why it happens: the second row is the one that gives the rule away. Its inputs are all positive and fairly small, yet the answer is –48 — far below any of them. Only a multiplication can make a number that big, and only a subtraction can turn it negative. Testing a – (b × c) on that row gives 6 – 54 = –48, and the rule then fits every other row.

Q8.

Imagine you’re in a place where the temperature drops by 5°C each hour. If the temperature is currently at 8°C, write an expression which denotes the temperature after 4 hours.

Answer

The expression is 8 + 4 × (–5), and its value is –12 °C.

Temperature after 4 hours
= 8 + 4 × (–5)
= 8 + (–20)
= –12

So the temperature will be 12 °C below zero.

Hours passed	0	1	2	3	4
Temperature (°C)	8	3	–2	–7	–12

Why it happens: a drop of 5 °C each hour is a change of –5 °C per hour, so four hours bring a change of 4 × (–5) = –20 °C. Adding that to the starting 8 °C carries the reading 12 degrees below zero. Writing the rate as a negative integer lets one expression describe the whole cooling, even as it crosses zero.

Tip: after t hours the temperature is 8 + t × (–5). Put t = 5 and you get –17 °C.

Q9.

Find 3 consecutive numbers with a product of (a) – 6, (b) 120.

Answer

(a) –3, –2, –1. (b) 4, 5, 6.

(a) (–3) × (–2) × (–1)
= 6 × (–1)
= –6

(b) 4 × 5 × 6
= 20 × 6
= 120

How to search sensibly:

Three consecutive numbers	Product
–4, –3, –2	–24
–3, –2, –1	–6
–2, –1, 0	0
1, 2, 3	6
3, 4, 5	60
4, 5, 6	120
5, 6, 7	210

Why it happens: for (a) the product is negative, so an odd number of the three must be negative. Three consecutive integers that are all negative fit, and –3, –2, –1 have magnitudes 3, 2, 1 multiplying to 6 — with three minus signs the answer is –6. For (b) the product 120 is positive and not too large, so look near the cube root of 120, which is a little under 5. Trying 4, 5, 6 works at once.

Check it yourself: could 120 come from three negative consecutive numbers? No — three negatives give a negative product. The only other option, –6, –5, –4, gives –120.

Q10.

An alien society uses a peculiar currency called ‘pibs’ with just two denominations of coins — a +13 pibs coin and a – 9 pibs coin. You have several of these coins. Is it possible to purchase an item that costs + 85 pibs? Using the two denominations, try to get the following totals: (a) + 20 (b) + 40 (c) – 50 (d) + 8 (e) + 10 (f) – 2 (g) + 1 (h) Is it possible to purchase an item that costs 1568 pibs?

Answer

Yes to all of them. Write the coins used as m coins of +13 and n coins of –9, so the total is 13m – 9n.

Total wanted	+13 coins (m)	–9 coins (n)	Check
+85 (given)	10	5	130 – 45 = 85
(a) +20	5	5	65 – 45 = 20
(b) +40	10	10	130 – 90 = 40
(c) –50	1	7	13 – 63 = –50
(d) +8	2	2	26 – 18 = 8
(e) +10	7	9	91 – 81 = 10
(f) –2	4	6	52 – 54 = –2
(g) +1	7	10	91 – 90 = 1
(h) +1568	122	2	1586 – 18 = 1568

How to find these. Follow the book’s hint and write out multiples of 13 and of 9, then look for two that differ by the amount you want.

Multiples of 13: 13, 26, 39, 52, 65, 78, 91, 104, 117, 130, …
Multiples of 9: 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, …

(d) +8: 26 – 18 = 8 → 2 coins of +13 and 2 coins of –9
(e) +10: 91 – 81 = 10 → 7 coins of +13 and 9 coins of –9
(g) +1: 91 – 90 = 1 → 7 coins of +13 and 10 coins of –9

(h) Once +1 can be made, every amount can be made. But repeating the +1 recipe 1568 times needs a huge pile of coins, so search directly instead. Look for a multiple of 13 that is 1568 plus a multiple of 9. Since 1568 = 122 × 13 – 18,

122 × 13 = 1586
1586 – 1568 = 18 = 2 × 9
So 1568 = 122 coins of +13 and 2 coins of –9

Why it happens: 13 and 9 have no common factor other than 1. Because of that, some combination 13m – 9n hits 1 exactly, and once you can build 1 you can build any integer at all by repeating. If the two coins had been, say, +12 and –9, every total would be a multiple of 3 and amounts like 85 would be impossible.

Tip: the answers above are not the only ones. Adding 9 more +13 coins and 13 more –9 coins changes the total by 9 × 13 – 13 × 9 = 0, so you can always pad a solution.

Q11.

Find the values of: (a) (32 × (– 18)) ÷ ((– 36)) (b) (32) ÷ ((– 36) × (– 18)) (c) (25 × (– 12)) ÷ ((45) × (– 27)) (d) (280 × (– 7)) ÷ ((– 8) × (– 35))

Answer

Work out the bracket on each side first, then divide.

(a) 32 × (–18) = –576, and –576 ÷ (–36) = 16

(d) 280 × (–7) = –1960, and (–8) × (–35) = 280
(–1960) ÷ 280 = –7

Parts (b) and (c) do not divide exactly, so their values are fractions, not integers.

(b) (–36) × (–18) = 648
32 ÷ 648 = 4/81 (both are divisible by 8)

(c) 25 × (–12) = –300, and 45 × (–27) = –1215
(–300) ÷ (–1215) = 300/1215 = 20/81 (both are divisible by 15)

	Numerator	Denominator	Sign	Value
(a)	–576	–36	same → +	16
(b)	32	648	same → +	4/81
(c)	–300	–1215	same → +	20/81
(d)	–1960	280	different → –	–7

Why it happens: the sign rule for division does all the sign work, and it says (b) and (c) must be positive because each has two negatives (or none). The magnitude is a separate question, and there the numbers simply do not cooperate — 32 is smaller than 648, so no integer can be the answer. Integers are closed under multiplication but not under division.

Tip: before dividing, count the negative signs across the whole expression. In (c) there are two, so you already know the answer is positive and can drop all the minus signs.

Q12.

Arrange the expressions given below in increasing order. (a) (– 348) + (– 1064) (b) (– 348) – (– 1064) (c) 348 – (– 1064) (d) (– 348) × (– 1064) (e) 348 × (– 1064) (f) 348 × 964

Answer

Work out each value first.

(a) (–348) + (–1064) = –1412
(b) (–348) – (–1064) = –348 + 1064 = 716
(c) 348 – (–1064) = 348 + 1064 = 1412
(d) (–348) × (–1064) = 370272
(e) 348 × (–1064) = –370272
(f) 348 × 964 = 335472

The two long multiplications:

348 × 1064 = 348 × 1000 + 348 × 64 = 348000 + 22272 = 370272
348 × 964 = 348 × 900 + 348 × 64 = 313200 + 22272 = 335472

Increasing order:

(e) < (a) < (b) < (c) < (f) < (d)
–370272 < –1412 < 716 < 1412 < 335472 < 370272

Why it happens: you barely need the full arithmetic to order these. The two products (d) and (e) have the same magnitude, so one is the largest and the other the smallest. The two sums and differences are much smaller in size, and among them only (a) is negative. That leaves (f), which is a product but with the smaller 964, so it sits just below (d).

Tip: with negative numbers, bigger magnitude means smaller value. –370272 is the least of all six, even though it looks like the biggest number on the page.

Q13.

Given that (– 548) × 972 = – 532656, write the values of: (a) (– 547) × 972 (b) (– 548) × 971 (c) (– 547) × 971

Answer

Use the distributive property to adjust the known product by one step at a time.

(a) (–547) × 972 = (–548 + 1) × 972
= (–548) × 972 + 972
= –532656 + 972
= –531684

(b) (–548) × 971 = (–548) × (972 – 1)
= (–548) × 972 – (–548)
= –532656 + 548
= –532108

(c) (–547) × 971 = (–547) × (972 – 1)
= (–547) × 972 + 547
= –531684 + 547
= –531137

Product	Change from the known one	Value
(–548) × 972	given	–532656
(–547) × 972	one fewer 972 of debt → add 972	–531684
(–548) × 971	one fewer –548 → add 548	–532108
(–547) × 971	from (a), add 547	–531137

Why it happens: reducing the magnitude of a negative multiplier by 1 removes one copy of the multiplicand from the debt, so the product rises by 972. Reducing the multiplicand by 1 removes one copy of –548, so the product rises by 548. In (c) both happen, but they must be applied in order — the second adjustment uses 547, not 548, because the multiplier has already changed.

Check it yourself: multiply 547 × 971 the long way. 547 × 1000 = 547000 and 547 × 29 = 15863, so 547 × 971 = 531137 — matching part (c).

Q14.

Given that 207 × (– 33 + 7) = – 5382, write the value of – 207 × (33 – 7) = _________.

Answer

The value is –5382 — exactly the same as the given product.

–33 + 7 = –26
33 – 7 = 26
so the two brackets are additive inverses of each other

–207 is the additive inverse of 207

–207 × (33 – 7)
= (–207) × 26
= –(207 × 26)
= –5382

Why it happens: going from 207 × (–26) to (–207) × 26, both factors have been replaced by their inverses. Each flip changes the sign of the product, so two flips bring the sign back to where it started. The magnitude 207 × 26 = 5382 never moved, so the answer is unchanged.

Tip: in general (–a) × (–b) = a × b, and (–a) × b = a × (–b). Spotting which of these applies saves you the whole multiplication.

Q15.

Use the numbers 3, – 2, 5, – 6 exactly once and the operations ‘+’, ‘–’, and ‘×’ exactly once and brackets as necessary to write an expression such that — (a) the result is the maximum possible (b) the result is the minimum possible

Answer

(a) Maximum = 60. One expression that reaches it:

((–2) – (3 + 5)) × (–6)
= ((–2) – 8) × (–6)
= (–10) × (–6)
= 60

(b) Minimum = –60. One expression that reaches it:

((3 + 5) – (–2)) × (–6)
= (8 + 2) × (–6)
= 10 × (–6)
= –60

Goal	Expression	Bracket value	Result
Maximum	((–2) – (3 + 5)) × (–6)	–10	60
Minimum	((3 + 5) – (–2)) × (–6)	10	–60

Why it happens: the × is used only once, so to make the result large you want the multiplication to be the last step and you want both of its parts to be as big as possible in magnitude. Putting –6 on one side of the × and gathering 3, 5 and –2 into a single bracket on the other side is the only way to build a bracket of magnitude 10. After that, the sign is chosen by which way round the subtraction goes: –10 times –6 gives +60, and 10 times –6 gives –60.

Check it yourself: try keeping the –6 inside the bracket instead, as in (3 + 5) × ((–6) – (–2)) = 8 × (–4) = –32. It uses each number and each operation once, but its magnitude is far smaller — so the arrangement above really is the extreme one.

Q16.

Fill in the blanks in at least 5 different ways with integers: (a) ___ + ___ × ___ = –36 (b) ( ___ – ___ ) × ___ = 12 (c) ( ___ – ( ___ – ___ )) = –1

Answer

Remember that in (a) the multiplication is done before the addition.

(a) ___ + ___ × ___ = –36

Filled statement	Working
0 + (–6) × 6 = –36	(–6) × 6 = –36; 0 + (–36)
4 + (–8) × 5 = –36	(–8) × 5 = –40; 4 + (–40)
(–6) + (–5) × 6 = –36	(–5) × 6 = –30; (–6) + (–30)
(–16) + 4 × (–5) = –36	4 × (–5) = –20; (–16) + (–20)
12 + (–8) × 6 = –36	(–8) × 6 = –48; 12 + (–48)
(–36) + 7 × 0 = –36	7 × 0 = 0; (–36) + 0

(b) ( ___ – ___ ) × ___ = 12

Filled statement	Working
(5 – 2) × 4 = 12	3 × 4
(2 – 5) × (–4) = 12	(–3) × (–4)
(0 – 6) × (–2) = 12	(–6) × (–2)
(–1 – 5) × (–2) = 12	(–6) × (–2)
(7 – 4) × 4 = 12	3 × 4
(–3 – 3) × (–2) = 12	(–6) × (–2)

(c) ( ___ – ( ___ – ___ )) = –1

Filled statement	Working
(2 – (5 – 2)) = –1	5 – 2 = 3; 2 – 3
(0 – (4 – 3)) = –1	4 – 3 = 1; 0 – 1
(3 – (10 – 6)) = –1	10 – 6 = 4; 3 – 4
((–5) – ((–2) – 2)) = –1	(–2) – 2 = –4; –5 + 4
((–1) – (0 – 0)) = –1	0 – 0 = 0; –1 – 0
(7 – (9 – 1)) = –1	9 – 1 = 8; 7 – 8

Why it happens: each part has a hidden equation behind it. In (a) you need first + (second × third) = –36, so choose any product you like and let the first number make up the shortfall. In (b) you need (difference) × (third) = 12, so split 12 into two factors — including the negative pairs such as (–6) × (–2) — and then build the difference you want. In (c) the expression simplifies to a – b + c, so any three integers with a – b + c = –1 will do.

Tip: once you see the hidden equation, there are infinitely many answers. Writing five is just a way of making sure you have found the pattern rather than one lucky guess.

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