NCERT Solutions for Class 7 Maths Chapter 11 Finding Common Ground (PDF) – 2027

She should buy square tiles of side 4 ft.

The tiles must fit the room exactly, with no cutting.

Because a bigger tile covers more floor, so fewer tiles are needed.

Sameeksha wants as few tiles as possible, so she picks the 4 ft tile.

She should purchase 12 tiles.

Checking by area: 192 sq ft ÷ 16 sq ft per tile = 12 tiles ✓

No, the answer would not change. The largest square tile that fits is still 4 ft, so 12 tiles is still the fewest.

Each bag should hold 12 kg.

With 12 kg bags:

She should choose the largest common factor, 12 kg.

Jumpy lands only on multiples of his jump size, so the jump size must be a common factor of the two numbers. The longest jump is the largest common factor.

Yes. The longest jump size is exactly the HCF of the two numbers.

Jumpy lands only on multiples of his jump size. To land on both treasure numbers, the jump size must be a factor of both numbers, that is, a common factor. The longest such jump is the highest common factor — the HCF.

Yes — prime factorisation simplifies it and makes it reliable.

Yes. The prime factors always come out the same, only the order may differ.

Every route ends with one 2, two 3s and one 5.

At each step the number is divided by a prime, and the quotient is written below it. We stop when a prime is left.

Collect the primes written on the left, then the prime left at the bottom.

Check: 3 × 5 × 7 = 105 ✓ and 2 × 3 × 5 = 30 ✓

Divide by the smallest prime that works, again and again.

Check: 2 × 2 × 2 × 2 = 16, and 16 × 3 = 48, and 48 × 5 × 5 = 48 × 25 = 1200 ✓

The division method is easier.

Finishing the earlier calculation: 1200 = 40 × 30 = 5 × 8 × 5 × 6 = 5 × (2 × 2 × 2) × 5 × (2 × 3) = 2 × 2 × 2 × 2 × 3 × 5 × 5 — the same answer, but with more bookkeeping.

Yes, 28 is a factor of 840.

Reorder the prime factors so that 2 × 2 × 7 sit together (reordering does not change a product):

So 840 ÷ 28 = 30, a whole number. Hence 28 is a factor.

It should be multiplied by 30.

Yes, 14 is a factor of 840, because 2 × 7 is a subpart of the prime factorisation of 840.

Yes, 2 × 2 × 2 = 8 is a factor of 840, because the prime factorisation of 840 has three 2s.

No, 3 × 3 × 3 = 27 is not a factor of 840.

Yes. Every factor is a subpart of the prime factorisation, so listing all the subparts lists all the factors.

Take 30 = 2 × 3 × 5.

First factorise 225 by the division method.

Now form every subpart, systematically:

Factors of 225 = 1, 3, 5, 9, 15, 25, 45, 75, 225 — nine factors in all.

Yes, all the factors are in the list. Pair each factor with its partner:

The nine factors fall into four pairs plus the middle one, 15. Nothing is missing, and nothing extra has crept in.

Factorise each number into primes, then build every subpart.

Anshu’s claim is false. One counterexample is enough to settle it.

The common factors are 1, 3, 5 and 15, and the HCF is 15.

A common factor must be a subpart of both lists. Compare them:

Adding 1, the common factors are 1, 3, 5, 15. The highest is 15.

The common factors are 1, 2, 4, 7, 14 and 28, and the HCF is 28.

Common primes: 2 (112 has four, 84 has two → at most two) and 7 (one each).

The only common factor is 1, so the HCF is 1.

With no prime in common, no subpart other than the empty one is shared. So 1 is the only common factor, and it is also the HCF.

Factorise both numbers, keep the shared primes, then list every subpart built from them.

Build the HCF prime by prime: keep only the primes that occur in every number, and take each one the fewest number of times it occurs.

HCF(30, 72) = 6.

How many 2s? 30 has one 2, 72 has three 2s. The minimum is one.

How many 3s? 30 has one 3, 72 has two 3s. The minimum is one.

HCF(225, 750) = 75.

How many 3s? 225 has two, 750 has one. The minimum is one.

How many 5s? 225 has two, 750 has three. The minimum is two.

Factorise, mark the shared primes, take the minimum count of each.

No. 6, 12, 8 and 18 are not primes, so they still hide common prime factors — for example, 6 and 8 both contain 2. Breaking them into primes shows many common factors:

The shortest common length is 24 cm.

Anshu uses 24 ÷ 6 = 4 strips; Guna uses 24 ÷ 8 = 3 strips.

No. There is no largest common multiple.

After 70 days.

70 days is 10 weeks, and also Kabamai’s 7th visit.

‘Idli-vada’ is called out at the first number that is a multiple of both — the LCM.

Yes, always. ‘Idli-vada’ is called only at a number that is a multiple of both numbers — a common multiple. Counting goes up from 1, so the first such number reached is the lowest common multiple, the LCM.

Collect every prime that appears in either number, and take each one the greatest number of times it occurs in either factorisation.

Example: 12 = 2 × 2 × 3 and 18 = 2 × 3 × 3.

Yes, with every multiple.

Every multiple of 36 contains the whole prime factorisation of 36, together with some more prime factors.

Yes. It tells us exactly what a common multiple must contain.

So for each prime, keep just enough copies to cover whichever number needs more of it — the maximum of the two counts.

LCM(14, 35) = 70.

Check: 70 ÷ 14 = 5 ✓ and 70 ÷ 35 = 2 ✓

Yes. 2 × 3 × 5 × 7 = 210 is a common multiple of 14 and 35.

But it is not the lowest: 210 = 70 × 3, and the extra 3 is not needed by either number.

LCM(96, 360) = 1440.

How many 2s? 96 has five, 360 has three. Take the maximum, five.

How many 3s? 96 has one, 360 has two. Take two.

How many 5s? 96 has none, 360 has one. Take one.

Because every extra 2 doubles the number without being needed by 96 or by 360.

Take every prime that appears, each the maximum number of times.

This happens exactly when one number is a factor of the other.

The general statement: if one number is a factor of the other, their HCF is the smaller number — and the other number is a multiple of it.

The other number must be a multiple of the first.

(a) If one number is m, the other could be 2m, 3m, 4m, … — in general am, where a is any positive integer.

(b) If one number is 7k, the other could be any multiple of 7k, that is 7ka for a positive integer a — for instance 14k, 21k, 28k.

Try examples first, then state the rule.

(a) More such pairs:

(b) General statement: the LCM is one of the two numbers exactly when one number is a factor of the other — and then the LCM is the larger number.

The HCF also doubles.

HCF = 42, not 14.

Write each number as the common multiplier times the rest. The HCF is the common multiplier times whatever the two rests still share.

In (b) and (c) only.

The rule: the HCF equals the common multiplier exactly when the two other multipliers are co-prime (share no factor except 1).

Both numbers are divided by a common prime at each step, and the two quotients are written in the next row.

Multiply the numbers in the left column.

Multiply the numbers in the left column of each division.

Multiply the numbers in the left column together with the two numbers in the last row.

It contains both numbers. The left column is the common part; the last row holds what is left of each number.

It is the smallest. Removing any factor would leave out part of one number, and the two numbers in the last row share no factor, so nothing is counted twice.

It works because a step of the division never has to be a prime — any common factor will do.

Dividing by 50 is the same as dividing by 2, then 5, then 5 in one go (50 = 2 × 5 × 5). Dividing by 10 is dividing by 2 and 5 together. So the left column still multiplies to the same HCF, and the same LCM follows.

The only rule: each divisor must be a common factor of both numbers, and we must continue until the two numbers left have no common factor.

(a) 90 and 150 — spot the common factor 30 straight away.

(b) 84 and 132 — both end in an even digit and both have digit sums divisible by 3, so 12 is worth trying.

The product is greater, or the two are equal. The LCM is never bigger than the product.

They are equal exactly when the two numbers are co-prime.

Because the product is itself a common multiple, and the LCM is the smallest common multiple.

LCM(105, 95) = 1995, and it is a factor of the product — multiply it by 5 to get the product.

In every case the LCM is a factor of the product, and the multiplier is the HCF.

The pattern: product ÷ LCM = HCF, that is, HCF × LCM = product.

Yes. In every case the multiplier turned out to be the HCF.

Take 60 = 2 × 2 × 3 × 5 and 90 = 2 × 3 × 3 × 5, and count each prime.

For every prime, the HCF takes the smaller count and the LCM takes the larger count. Together they use exactly the copies found in the two numbers put together. So HCF × LCM = product of the two numbers.

It does not hold in general for three numbers.

It works only in special cases such as 2, 3, 5, where no two of the numbers share a prime factor.

The blue stars meet next at the 16th star — that is, 12 places after they last met.

(a) No.   (b) No.

(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.

(a) 3 × 3 × 5 × 7 × 7 is already prime. 12 × 7 × 11 = (2 × 2 × 3) × 7 × 11.

(b) 45 = 3 × 3 × 5 and 36 = 2 × 2 × 3 × 3.

2 and 33 is one such pair. So are 6 and 11, 3 and 22, and 1 and 66.

Split the three primes into two groups so that no prime is shared:

He had 105 cows.

Check: 105 ÷ 3 = 35 cows per gate, 105 ÷ 5 = 21, 105 ÷ 7 = 15 ✓

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.

The answer is (c) 18.

The number is 2100.

Now test each against the remainder condition.

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.

Only (c) Greater than both numbers is correct.

The dog catches up in 75 leaps.

The smallest such number is the LCM of the numbers: 360.

This is the same answer, 360, that we found in Grade 6.

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.

Writing it out by hand would take you well over a year of non-stop work.

The ring around a number is its prime factorisation, drawn as coloured arcs.

• The ring is cut into as many equal arcs as the number has prime factors, counted with repeats.
• Each arc is coloured by which prime it stands for.

Factorise each number, then draw one arc per prime factor.