NCERT Solutions for Class 7th Maths Chapter 3 End-of-chapter question set — Figure it Out
Book page 63–64 Updated on2026-09-19
Q1.
In the two rows below, colours repeat as shown. When will the blue stars meet next? [Row 1: yellow, green, orange, blue, pink, grey, repeating. Row 2: green, orange, yellow, blue, repeating.]
Answer
The blue stars meet next at the 16th star — that is, 12 places after they last met.
Row 1 repeats every 6 stars, and blue is the 4th
So row-1 blues are at 4, 10, 16, 22, 28, …
Row 2 repeats every 4 stars, and blue is the 4th
So row-2 blues are at 4, 8, 12, 16, 20, …
They already meet at position 4
After that they meet every LCM(6, 4) = 12 stars
Next meeting = 4 + 12 = position 16
Row 1 repeats every 6 stars, row 2 every 4 stars, and blue is the 4th star in both. They line up again 12 places later — at the 16th star.
Why it happens: counting from the first meeting at position 4, a row-1 blue appears after every 6 more stars and a row-2 blue after every 4 more stars. For both to appear together, the number of extra stars must be a multiple of 6 and a multiple of 4 — a common multiple. The first such number is the LCM, 12.
Check it yourself: LCM(6, 4) = 12 because 6 = 2 × 3 and 4 = 2 × 2, so the LCM needs two 2s and one 3: 2 × 2 × 3 = 12. Not 24 — the shared 2 must not be counted twice.
Q2.
(a) Is 5 × 7 × 11 × 11 a multiple of 5 × 7 × 7 × 11 × 2? (b) Is 5 × 7 × 11 × 11 a factor of 5 × 7 × 7 × 11 × 2?
Answer
(a) No.(b) No.
Prime
In 5 × 7 × 11 × 11
In 5 × 7 × 7 × 11 × 2
2
0 times
1 time
5
1 time
1 time
7
1 time
2 times
11
2 times
1 time
(a) For the first number to be a multiple of the second, it must contain the second's whole factorisation. But the second has a 2 and two 7s, and the first has no 2 and only one 7. So no.
(b) For the first to be a factor of the second, the second must contain the first. But the first has two 11s and the second has only one. So no.
5 × 7 × 11 × 11 = 4235
5 × 7 × 7 × 11 × 2 = 5390
4235 ÷ 5390 is not a whole number, and 5390 ÷ 4235 is not either ✓
Why it happens: "A is a multiple of B" and "B is a factor of A" say the same thing, so (a) and (b) are asking about the two possible directions. Neither factorisation sits inside the other: each has a prime the other is short of. When two numbers each hold something the other lacks, neither can divide the other.
Q3.
Find the HCF and LCM of the following (state your answers in the form of prime factorisations): (a) 3 × 3 × 5 × 7 × 7 and 12 × 7 × 11 (b) 45 and 36
Answer
First write both numbers fully in primes, then take minimums for the HCF and maximums for the LCM.
Check it yourself: (a) 2205 × 924 = 2,037,420 and 21 × 97020 = 2,037,420 ✓ (b) 45 × 36 = 1620 and 9 × 180 = 1620 ✓
Q4.
Find two numbers whose HCF is 1 and LCM is 66.
Answer
2 and 33 is one such pair. So are 6 and 11, 3 and 22, and 1 and 66.
HCF = 1 means the two numbers are co-prime
For co-prime numbers, LCM = product
So the product must be 66
66 = 2 × 3 × 11
Split the three primes into two groups so that no prime is shared:
Split
Pair
HCF
LCM
(none) and 2 × 3 × 11
1, 66
1
66 ✓
2 and 3 × 11
2, 33
1
66 ✓
3 and 2 × 11
3, 22
1
66 ✓
11 and 2 × 3
11, 6
1
66 ✓
Why it happens: HCF × LCM = product, so 1 × 66 = 66 tells us the product at once. Each of the three primes 2, 3 and 11 must go entirely to one number or the other — if a prime were split between them it would be shared, and the HCF would not be 1. Three primes give 2 × 2 × 2 = 8 ways to hand them out, which is the 4 pairs above counted twice.
Q5.
A cowherd took all his cows to graze in the fields. The cows came to a crossing with 3 gates. An equal number of cows passed through each gate. Later at another crossing with 5 gates again an equal number of cows passed through each gate. The same happened at the third crossing with 7 gates. If the cowherd had less than 200 cows, how many cows did he have? (Based on the folklore mathematics from Karnataka.)
Answer
He had 105 cows.
The herd splits equally into 3 → the number is a multiple of 3
It splits equally into 5 → a multiple of 5
It splits equally into 7 → a multiple of 7
So it is a common multiple of 3, 5 and 7
LCM = 3 × 5 × 7 = 105
Common multiples: 105, 210, 315, …
Less than 200 → only 105
Why it happens: 3, 5 and 7 are different primes, so the LCM is simply their product. Every common multiple is then a multiple of 105, and the next one, 210, is already past 200.
Q6.
The length, width, and height of a box are 12 cm, 18 cm, and 36 cm respectively. Which of the following sized cubes can be packed in this box without leaving gaps? (a) 9 cm (b) 6 cm (c) 4 cm (d) 3 cm (e) 2 cm
Answer
The cubes that fit are (b) 6 cm, (d) 3 cm and (e) 2 cm.
A cube of side s fills the box with no gaps only if s divides 12, 18 and 36.
Why it happens: the cube side must fit a whole number of times along each edge of the box, so it must be a common factor of all three edge lengths. The common factors are exactly the factors of the HCF, which is 6 — so 1, 2, 3 and 6 work, and nothing else does. 9 fails on the 12 cm edge and 4 fails on the 18 cm edge.
Check it yourself: a 6 cm cube fills the box with 2 × 3 × 6 = 36 cubes, and a 2 cm cube with 6 × 9 × 18 = 972 cubes.
Q7.
Among the numbers below, which is the largest number that perfectly divides both 306 and 36? (a) 36 (b) 612 (c) 18 (d) 3 (e) 2 (f) 360
Tip: 612 = 2 × 306 and 360 = 10 × 36 are multiples, not divisors. Options like these are put in to catch a reader who confuses "divides both" with "is divisible by both".
Q8.
Find the smallest number that is divisible by 3, 4, 5 and 7, but leaves a remainder of 10 when divided by 11.
Answer
The number is 2100.
Divisible by 3, 4, 5 and 7 → it is a common multiple
3 = 3, 4 = 2 × 2, 5 = 5, 7 = 7
LCM = 2 × 2 × 3 × 5 × 7 = 420
So the number is 420, 840, 1260, 1680, 2100, …
Why it happens: the remainder goes up by 2 each time you add another 420, because 420 itself leaves remainder 2 on division by 11. So the remainders run 2, 4, 6, 8, 10 — and the fifth multiple of 420 is the first to hit 10.
Q9.
Children are playing ‘Fire in the Mountain’. When the number 6 was called out, no one got out. When the number 9 was called out, no one got out. But when the number 10 was called out, some people got out. How many children could have been playing initially? (a) 72 (b) 90 (c) 45 (d) 3 (e) 36 (f) None of these
Answer
The possible numbers are (a) 72 and (e) 36.
When a number is called out, the children form groups of that size. Nobody is left out only if the total divides exactly.
No one out on 6 → the total is a multiple of 6
No one out on 9 → the total is a multiple of 9
So the total is a common multiple of 6 and 9
6 = 2 × 3, 9 = 3 × 3 → LCM = 2 × 3 × 3 = 18
Total ∈ {18, 36, 54, 72, 90, 108, …}
Some children out on 10 → the total is not a multiple of 10
Option
Multiple of 18?
Multiple of 10?
Possible?
(a) 72
yes (18 × 4)
no
yes ✓
(b) 90
yes (18 × 5)
yes — no one would be left out
no
(c) 45
no (odd, not a multiple of 6)
no
no
(d) 3
no
no
no
(e) 36
yes (18 × 2)
no
yes ✓
Check 72: 72 ÷ 6 = 12 groups ✓, 72 ÷ 9 = 8 groups ✓, 72 = 10 × 7 + 2 → 2 children left out ✓
Check 36: 36 ÷ 6 = 6 ✓, 36 ÷ 9 = 4 ✓, 36 = 10 × 3 + 6 → 6 children left out ✓
Tip: 90 is the trap. It passes the first two tests but fails the third, because 90 children make exactly 9 groups of 10 and nobody is left standing.
Q10.
Tick the correct statement(s). The LCM of two different prime numbers (m, n) can be: (a) Less than both numbers (b) In between the two numbers (c) Greater than both numbers (d) Less than m × n (e) Greater than m × n
Answer
Only (c) Greater than both numbers is correct.
Two different primes share no factor, so they are co-prime
For co-prime numbers, LCM = product LCM(m, n) = m × n
Statement
Verdict
Reason
(a) less than both
wrong
m × n is bigger than m and bigger than n
(b) in between
wrong
m × n is above both, never between them
(c) greater than both
correct ✓
m × n > m and m × n > n, since both primes are at least 2
(d) less than m × n
wrong
the LCM equals m × n, it is not less
(e) greater than m × n
wrong
the LCM can never exceed the product
Example: m = 3, n = 7 → LCM = 21, which is greater than 3 and 7, and equal to 3 × 7
Why it happens: a prime has only itself and 1 as factors, so two different primes have nothing in common except 1. With no shared prime to save, the LCM must take both primes in full, giving exactly m × n. Since n is at least 2, m × n is at least 2m, comfortably above m — and the same for n.
Q11.
A dog is chasing a rabbit that has a head start of 150 feet. It jumps 9 feet every time the rabbit jumps 7 feet. In how many leaps does the dog catch up with the rabbit?
Answer
The dog catches up in 75 leaps.
In one leap the dog covers 9 ft and the rabbit covers 7 ft
Gap closed in one leap = 9 − 7 = 2 ft
Gap to close = 150 ft
Number of leaps = 150 ÷ 2 = 75
Check after 75 leaps:
Dog has run 75 × 9 = 675 ft
Rabbit has run 75 × 7 = 525 ft, and started 150 ft ahead
Rabbit's position = 525 + 150 = 675 ft
Both are at 675 ft — the dog has caught up ✓
Why it happens: both animals leap at the same moments, so after each pair of leaps the gap shrinks by the same fixed amount, 2 ft. A gap of 150 ft therefore takes 150 ÷ 2 leaps to disappear. Notice that the answer depends only on the difference of the two jump lengths, not on how long each jump is.
Q12.
What is the smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 8, 9, 10? Do you remember the answer from Grade 6, Chapter 5?
Did you know? 360 also has 24 factors — more than any smaller number except 240 and 180 have. That is one reason a circle was divided into 360 degrees: it can be split into halves, thirds, quarters, fifths, sixths, eighths, ninths, tenths and many more, all without fractions.
Q13.
Here is a problem posed by the ancient Indian Mathematician Mahaviracharya (850 C.E.). Add together 8/15, 1/20, 7/36, 11/63 and 1/21. What do you get? How can we find this sum efficiently?
Answer
The sum is exactly 1.
The efficient way is to use the LCM of the denominators as the common denominator, instead of multiplying them all together.
15 = 3 × 5
20 = 2 × 2 × 5
36 = 2 × 2 × 3 × 3
63 = 3 × 3 × 7
21 = 3 × 7
2s: at most two · 3s: at most two · 5s: at most one · 7s: at most one
LCM = 2 × 2 × 3 × 3 × 5 × 7 = 1260
Why it is efficient: multiplying the five denominators gives 15 × 20 × 36 × 63 × 21 = 14,288,400 — a common denominator, but a needlessly huge one. The LCM 1260 is more than eleven thousand times smaller, and it works just as well, because every denominator still divides it. Small numbers mean less arithmetic and fewer mistakes, and the answer comes out already in lowest terms.
Did you know? Mahaviracharya wrote the Ganita-sara-sangraha in Karnataka around 850 CE. He enjoyed setting sums that look messy but collapse to a neat whole number — this one is a good example.