Class 7 Maths Chapter 3 NCERT Solutions 2026-27

Strategic angle. Focus on HCF as the largest exact tile side. Keep the route short, but make every arithmetic jump visible.

1. The tile side must divide both 12 ft and 16 ft exactly.
2. The factors of 12 are 1,2,3,4,6,12.
3. The factors of 16 are 1,2,4,8,16.
4. The common factors are 1,2,4, so the largest possible whole-number side is 4 ft.
5. The room needs 12 ÷ 4 = 3 tiles across the breadth and 16 ÷ 4 = 4 tiles along the length.
6. Total tiles = 3 × 4 = 12.

Each tile should have side 4 ft. She needs 12 tiles. Allowing fractional feet does not change the largest possible side.

Strategic angle. Focus on HCF as the biggest equal bag weight. Keep the route short, but make every arithmetic jump visible.

1. To use as few bags as possible, each bag should be as heavy as possible.
2. The bag weight must divide both 84 and 108.
3. The common factors are 1,2,3,4,6,12.
4. The greatest common factor is 12.
5. For 84 kg, the number of bags is 84 ÷ 12 = 7.
6. For 108 kg, the number of bags is 108 ÷ 12 = 9.

Each bag should weigh 12 kg. Lekhana will use 16 bags in all.

Strategic angle. Focus on HCF as the longest common jump size. Keep the route short, but make every arithmetic jump visible.

1. For 14 and 30, factors of 14 are 1,2,7,14 and factors of 30 are 1,2,3,5,6,10,15,30. The largest common factor is 2.
2. For 7 and 11, the only common factor is 1, so the longest jump is 1.
3. For 30 and 50, common factors include 1,2,5,10, and the largest is 10.
4. For 28 and 42, common factors are 1,2,7,14, and the largest is 14.
5. A jump size that lands on both numbers must be a common factor of the two numbers.
6. The longest such jump must therefore be the highest common factor.

(a) 2, (b) 1, (c) 10, (d) 14. Yes, the longest jump size is the HCF.

Strategic angle. Focus on prime factorisation by division. Keep the route short, but make every arithmetic jump visible.

1. For 105: 105 ÷ 3=35, 35 ÷ 5=7, and 7 ÷ 7=1. Hence 105=3 × 5 × 7.
2. For 30: 30 ÷ 2=15, 15 ÷ 3=5, and 5 ÷ 5=1. Hence 30=2 × 3 × 5.
3. For 1200: 1200 → 600 → 300 → 150 → 75 → 25 → 5 → 1 by dividing by 2,2,2,2,3,5,5.
4. So 1200=2⁴ × 3 × 5².

105=3 × 5 × 7, 30=2 × 3 × 5, and 1200=2⁴ × 3 × 5².

Strategic angle. Focus on factor testing through prime powers. Keep the route short, but make every arithmetic jump visible.

1. The prime factorisation is 840=2³ × 3 × 5 × 7.
2. 14=2 × 7, and these primes occur in 840, so 14 is a factor.
3. 8=2³, and three 2s occur in 840, so 8 is a factor.
4. 27=3³, but 840 has only one factor 3, so 27 is not a factor.

14 and 8 are factors of 840. 27 is not a factor of 840.

Strategic angle. Focus on listing factors from prime factorisation. Keep the route short, but make every arithmetic jump visible.

1. 90=2 × 3² × 5, so its factors are 1,2,3,5,6,9,10,15,18,30,45,90.
2. 105=3 × 5 × 7, so its factors are 1,3,5,7,15,21,35,105.
3. 132=2² × 3 × 11, so its factors are 1,2,3,4,6,11,12,22,33,44,66,132.
4. 360=2³ × 3² × 5. Its factors are
   1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18,
   20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360.
5. 840=2³ × 3 × 5 × 7. Its factors are
   1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15,
   20, 21, 24, 28, 30, 35, 40, 42, 56, 60,
   70, 84, 105, 120, 140, 168, 210, 280, 420, 840.

90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90. 105: 1, 3, 5, 7, 15, 21, 35, 105. 132: 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132. 360: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360. 840: 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, 840.

Strategic angle. Focus on common factors and HCF. Keep the route short, but make every arithmetic jump visible.

1. gcd(50,60)=10, so common factors are factors of 10: 1,2,5,10.
2. gcd(140,275)=5, so common factors are 1,5.
3. 77=7 × 11 and 725=5² × 29, so the only common factor is 1.
4. 370=2 × 5 × 37 and 592=2⁴ × 37, so the HCF is 2 × 37=74.
5. 81 divides 243, so the common factors are 1,3,9,27,81 and the HCF is 81.

(a) common factors 1, 2, 5, 10, HCF 10; (b) 1, 5, HCF 5; (c) 1, HCF 1; (d) 1, 2, 37, 74, HCF 74; (e) 1, 3, 9, 27, 81, HCF 81.

Strategic angle. Focus on HCF through minimum prime powers. Keep the route short, but make every arithmetic jump visible.

1. 24=2³ × 3 and 180=2² × 3² × 5, so HCF is 2² × 3=12.
2. 42=2 × 3 × 7, 75=3 × 5², and 24=2³ × 3, so HCF is 3.
3. 240=2⁴ × 3 × 5 and 378=2 × 3³ × 7, so HCF is 6.
4. 400=2⁴ × 5² and 2500=2² × 5⁴, so HCF is 100.
5. 300=2² × 3 × 5² and 800=2⁵ × 5², so HCF is 100.
6. Composite factorisations may hide common prime factors. We must break the composite factors into primes before judging common factors.

(a) 12, (b) 3, (c) 6, (d) 100, (e) 100. The statement about 72 and 144 is false because 72 itself is a common factor.

Strategic angle. Focus on LCM as the first common game call. Keep the route short, but make every arithmetic jump visible.

1. For 4 and 6, 4=2² and 6=2 × 3, so the LCM is 2² × 3=12.
2. For 7 and 11, both are prime and different, so the LCM is 7 × 11=77.
3. For 14 and 30, 14=2 × 7 and 30=2 × 3 × 5, so the LCM is 2 × 3 × 5 × 7=210.
4. For 15 and 55, 15=3 × 5 and 55=5 × 11, so the LCM is 3 × 5 × 11=165.
5. The first shared call is the first common multiple. That is exactly the LCM.

(a) 12, (b) 77, (c) 210, (d) 165. Yes, the answer is always the LCM.

Strategic angle. Focus on LCM through maximum prime powers. Keep the route short, but make every arithmetic jump visible.

1. 30=2 × 3 × 5 and 72=2³ × 3², so LCM is 2³ × 3² × 5=360.
2. 36=2² × 3² and 54=2 × 3³, so LCM is 2² × 3³=108.
3. 105=3 × 5 × 7, 195=3 × 5 × 13, and 65=5 × 13, so LCM is 3 × 5 × 7 × 13=1365.
4. 222=2 × 3 × 37 and 370=2 × 5 × 37, so LCM is 2 × 3 × 5 × 37=1110.

(a) 360, (b) 108, (c) 1365, (d) 1110.

Strategic angle. Focus on HCF patterns for consecutive and co-prime numbers. Keep the route short, but make every arithmetic jump visible.

1. Consecutive even numbers can be written as 2n and 2n+2=2(n+1). Since n and n+1 are co-prime, their HCF is 2.
2. Consecutive odd numbers differ by 2 and share no factor greater than 1, so their HCF is 1.
3. Two even numbers always have 2 as a common factor, so their HCF is at least 2.
4. Two consecutive numbers always have HCF 1.
5. Co-prime numbers have HCF 1 by definition.

Consecutive even numbers have HCF 2; consecutive odd numbers have HCF 1; two even numbers have an even HCF; consecutive numbers have HCF 1; co-prime numbers have HCF 1.

Strategic angle. Focus on LCM patterns for multiple and co-prime pairs. Keep the route short, but make every arithmetic jump visible.

1. The LCM of two numbers is one of them exactly when one number is a multiple of the other.
2. The LCM of two multiples of 3 is also a multiple of 3.
3. Consecutive even numbers 2n and 2n+2 have HCF 2, so their LCM is (2n)(2n+2)2=2n(n+1).
4. Consecutive numbers are co-prime, so their LCM is their product.
5. For co-prime numbers, the LCM is the product of the numbers.

One number being a multiple of the other makes the LCM equal to the larger number. Multiples of 3 have an LCM that is a multiple of 3. Consecutive or co-prime pairs have LCM equal to the product, except consecutive even numbers share the factor 2.

Strategic angle. Focus on HCF of numbers with a shared multiplier. Keep the route short, but make every arithmetic jump visible.

1. HCF(18 × 10,18 × 15)=18 × HCF(10,15)=18 × 5=90.
2. HCF(10 × 38,10 × 21)=10 × HCF(38,21)=10 × 1=10.
3. HCF(5 × 13,5 × 20)=5 × HCF(13,20)=5 × 1=5.
4. HCF(12 × 16,12 × 20)=12 × HCF(16,20)=12 × 4=48.
5. If the remaining multipliers are co-prime, the HCF equals the common multiplier.

(a) 90, (b) 10, (c) 5, (d) 48. It is the common multiplier when the leftover factors are co-prime.

Strategic angle. Focus on efficient HCF and LCM procedures. Keep the route short, but make every arithmetic jump visible.

1. For 84 and 180, divide both by 2,2,3. The common product is 12, and the leftover quotients are 7 and 15. So HCF is 12 and LCM is 12 × 7 × 15=1260.
2. For 300 and 150, divide both by 150. The leftover quotients are 2 and 1. So HCF is 150 and LCM is 150 × 2 × 1=300.
3. For 630 and 770, divide both by 2,5,7. The common product is 70, and the leftover quotients are 9 and 11. So HCF is 70 and LCM is 70 × 9 × 11=6930.
4. For 90 and 150, common product is 30 and leftover quotients are 3 and 5. Hence HCF is 30 and LCM is 30 × 3 × 5=450.
5. For 84 and 132, common product is 12 and leftover quotients are 7 and 11. Hence HCF is 12 and LCM is 12 × 7 × 11=924.
6. Dividing by a larger common factor works because it removes several common prime factors in one step.

84 and 180: HCF 12, LCM 1260. 300 and 150: HCF 150, LCM 300. 630 and 770: HCF 70, LCM 6930. 90 and 150: HCF 30, LCM 450. 84 and 132: HCF 12, LCM 924.

Strategic angle. Focus on the HCF times LCM relation for two numbers. Keep the route short, but make every arithmetic jump visible.

1. For 45 and 105, 45=3² × 5 and 105=3 × 5 × 7. HCF is 3 × 5=15 and LCM is 3² × 5 × 7=315.
2. Then 15 × 315 = 4725 = 45 × 105.
3. For 275 and 352, 275=5² × 11 and 352=2⁵ × 11. HCF is 11 and LCM is 2⁵ × 5² × 11=8800.
4. Then 11 × 8800 = 96800 = 275 × 352.
5. For 222 and 370, 222=2 × 3 × 37 and 370=2 × 5 × 37. HCF is 2 × 37=74 and LCM is 2 × 3 × 5 × 37=1110.
6. Then 74 × 1110 = 82140 = 222 × 370.

The property holds for each pair listed. It does not generally hold for three numbers.

Strategic angle. Focus on LCM of repeating colour cycles. Keep the route short, but make every arithmetic jump visible.

1. In the top row, the blue star appears every 6 positions.
2. In the bottom row, the blue star appears every 4 positions.
3. They are blue together first at the 4th position in the shown pattern.
4. The repeat gap is lcm(6,4)=12 positions.
5. So the next meeting is at position 4+12=16.

The blue stars meet next at the 16th position.

Strategic angle. Focus on prime-factor containment for factor and multiple checks. Keep the route short, but make every arithmetic jump visible.

1. The first number is 5 × 7 × 11².
2. The second number is 2 × 5 × 7² × 11.
3. For the first number to be a multiple of the second, it would need the factors 2 and another 7. It does not have them.
4. For the first number to be a factor of the second, the second would need another 11. It does not have one.

No to both parts.

Strategic angle. Focus on HCF and LCM from prime-power comparison. Keep the route short, but make every arithmetic jump visible.

1. The first pair is 3² × 5 × 7² and 2² × 3 × 7 × 11.
2. Their HCF is 3 × 7.
3. Their LCM is 2² × 3² × 5 × 7² × 11.
4. For 45 and 36, we have 45=3² × 5 and 36=2² × 3².
5. The HCF is 3² and the LCM is 2² × 3² × 5.

(a) HCF =3 × 7, LCM =2² × 3² × 5 × 7² × 11. (b) HCF =3², LCM =2² × 3² × 5.

Strategic angle. Focus on choosing a co-prime factor pair for a given LCM. Keep the route short, but make every arithmetic jump visible.

1. Factorise 66=2 × 3 × 11.
2. We need two co-prime numbers whose product is 66.
3. Take 6=2 × 3 and 11.
4. They share no common factor other than 1.
5. Their LCM is 6 × 11=66.

One valid pair is 6 and 11.

Strategic angle. Focus on LCM of 3, 5 and 7 under 200. Keep the route short, but make every arithmetic jump visible.

1. The number of cows must be divisible by 3, 5, and 7.
2. The LCM of 3, 5, and 7 is 3 × 5 × 7=105.
3. Multiples of 105 below 200 are only 105.
4. So the cowherd had 105 cows.

105 cows.

Strategic angle. Focus on common divisors of box dimensions. Keep the route short, but make every arithmetic jump visible.

1. A cube side length must divide all three dimensions: 12, 18, and 36.
2. 9 does not divide 12, so 9 cm is not possible.
3. 6 divides 12, 18, and 36, so 6 cm is possible.
4. 4 does not divide 18, so 4 cm is not possible.
5. 3 divides 12, 18, and 36, so 3 cm is possible.
6. 2 divides 12, 18, and 36, so 2 cm is possible.

The possible cube sizes are 6 cm, 3 cm, and 2 cm.

Strategic angle. Focus on HCF using the Euclidean algorithm. Keep the route short, but make every arithmetic jump visible.

1. Find the HCF of 306 and 36.
2. 306 = 36 × 8 + 18.
3. 36 = 18 × 2 + 0.
4. So the HCF is 18.
5. Among the given options, 18 is the largest divisor of both numbers.

18.

Strategic angle. Focus on LCM plus a remainder check. Keep the route short, but make every arithmetic jump visible.

1. The number must be a multiple of lcm(3,4,5,7)=420.
2. Check the multiples 420, 840, 1260, 1680, 2100, ….
3. On dividing by 11, their remainders are 2,4,6,8,10,….
4. The first remainder 10 occurs at 2100.

2100.

Strategic angle. Focus on divisibility by 6 and 9 but not by 10. Keep the route short, but make every arithmetic jump visible.

1. The number of children must be divisible by 6 and 9.
2. So it must be divisible by lcm(6,9)=18.
3. It must not be divisible by 10, because some children got out when 10 was called.
4. 72 is divisible by 18 and not by 10.
5. 36 is divisible by 18 and not by 10.
6. 90 is divisible by 10, and 45, 3 are not divisible by 6.

72 and 36 could be the initial numbers of children.

Strategic angle. Focus on LCM of two different primes. Keep the route short, but make every arithmetic jump visible.

1. Two different prime numbers have no common factor other than 1.
2. So their LCM is their product m × n.
3. This product is greater than both primes.
4. It is not less than the product and not greater than the product.

Only statement (c), greater than both numbers, is correct.

Strategic angle. Focus on relative gain per leap. Keep the route short, but make every arithmetic jump visible.

1. Use relative gain per leap, because both animals jump once in each comparison.
2. In one leap, the dog gains 9 ft while the rabbit gains 7 ft.
3. The gain per leap is 9-7=2 ft.
4. The dog must close a gap of 150 ft.
5. Number of leaps =150 ÷ 2=75.

75 leaps.

Strategic angle. Focus on LCM of a list of numbers. Keep the route short, but make every arithmetic jump visible.

1. Find the LCM of the listed numbers.
2. The highest power of 2 needed is 2³, from 8.
3. The highest power of 3 needed is 3², from 9.
4. The factor 5 is needed from 5 or 10.
5. So the LCM is 2³ × 3² × 5=8 × 9 × 5=360.

360.

Strategic angle. Focus on LCM as a common denominator for fractions. Keep the route short, but make every arithmetic jump visible.

1. The LCM of 15,20,36,63,21 is 1260.
2. 815=6721260.
3. 120=631260.
4. 736=2451260.
5. 1163=2201260.
6. 121=601260.

1.