Question 1

Find the H.C.F. of 9⁄10, 12⁄25, 18⁄35, and 21⁄40.

Accepted Answer

Given data Fractions: 9⁄10, 12⁄25, 18⁄35, 21⁄40 Concept / Approach H.C.F. of fractions = HCF of numerators divided by LCM of denominators. Step-by-step HCF of numerators HCF(9, 12, 18, 21) = 3LCM of denominators LCM(10, 25, 35, 40)Prime factors: 10 = 2 × 5; 25 = 5^2; 35 = 5 × 7; 40 = 2^3 × 5LCM = 2^3 × 5^2 × 7 = 8 × 25 × 7 = 1400Therefore H.C.F. = 3⁄1400 = 3⁄1400 Common pitfalls Using LCM of numerators by mistake, which is for LCM of fractions, not HCF. Final Answer 3⁄1400

Question 2

Find the greatest number that divides 43, 91, and 183 leaving the same remainder in each case.

Accepted Answer

Given data Same remainder when dividing 43, 91, 183 by the required number. Concept / Approach Greatest such number equals GCD of pairwise differences. Step-by-step Differences: 91 − 43 = 48; 183 − 91 = 92; 183 − 43 = 140GCD(48, 92) = 4GCD(4, 140) = 4 Verification 43 mod 4 = 3; 91 mod 4 = 3; 183 mod 4 = 3. Same remainder. Common pitfalls Taking GCD of the numbers themselves instead of their pairwise differences. Final Answer Greatest required number = 4.

Question 3

Express 252 as a product of prime factors.

Accepted Answer

Given data Number to factorize into primes: 252. Concept / Approach Use successive division by the smallest possible prime at each step (prime factor tree). Stop when all factors are prime; then group equal primes into powers. Step-by-step calculation 252 ÷ 2 = 126126 ÷ 2 = 6363 ÷ 3 = 2121 ÷ 3 = 77 is prime ⇒ stop.Thus 252 = 2 × 2 × 3 × 3 × 7 = 2^2 × 3^2 × 7. Verification / Alternative Multiply back: 2^2 × 3^2 × 7 = 4 × 9 × 7 = 36 × 7 = 252 ✔ Common pitfalls Missing one of the 3's (since 63 = 3 × 21 = 3 × 3 × 7). Stopping at a composite factor (e.g., 21) instead of fully factoring to primes. Final Answer Prime factorization: 2^2 × 3^2 × 7.

Question 4

Find the G.C.D. (H.C.F.) of 1.08, 0.36 and 0.90.

Accepted Answer

Given data Decimals: 1.08, 0.36, 0.90 Concept / Approach To compute G.C.D. of decimals, clear decimals by multiplying all numbers by 10^k, where k is the maximum number of decimal places. After finding the G.C.D. of the integers, divide back by the same 10^k. Step-by-step calculation Max decimal places = 2 ⇒ multiply by 100.Numbers become: 108, 36, 90.GCD(108, 36) = 36.GCD(36, 90) = 18.Hence integer G.C.D. = 18 ⇒ decimal G.C.D. = 18 ÷ 100 = 0.18. Verification Check: 1.08 ÷ 0.18 = 6; 0.36 ÷ 0.18 = 2; 0.90 ÷ 0.18 = 5 ⇒ all integers. Common pitfalls Using average or minimum instead of G.C.D. Forgetting to scale back after finding the integer G.C.D. Final Answer G.C.D. = 0.18.

Question 5

Find the greatest 4-digit number divisible by 15, 25, 40, and 75.

Accepted Answer

Given data Divisibility by all of: 15, 25, 40, 75. Concept / Approach The required numbers are multiples of L.C.M.(15, 25, 40, 75). Find the largest multiple ≤ 9999. Step-by-step calculation Prime forms: 15 = 3 × 5; 25 = 5^2; 40 = 2^3 × 5; 75 = 3 × 5^2.LCM = 2^3 × 3 × 5^2 = 8 × 3 × 25 = 600.Largest 4-digit multiple of 600: ⌊9999 ÷ 600⌋ = 16 ⇒ 16 × 600 = 9600. Verification 9600÷15, 9600÷25, 9600÷40, 9600÷75 are all integers. Common pitfalls Treating it as G.C.D. problem instead of L.C.M. Rounding 9999/600 incorrectly (it must be the floor). Final Answer Greatest such number: 9600.

Question 6

Find the greatest possible length that exactly measures 7 m, 3 m 85 cm, and 12 m 95 cm.

Accepted Answer

Given data Lengths: 7 m, 3 m 85 cm, 12 m 95 cm. Concept / Approach Convert to a common unit (centimetres), then compute the H.C.F. of the lengths. Step-by-step conversion & H.C.F. 7 m = 700 cm3 m 85 cm = 385 cm12 m 95 cm = 1295 cmHCF(700, 385) ⇒ 700 − 385 = 315; 385 − 315 = 70; 315 ÷ 70 = remainder 35; 70 ÷ 35 = 0 ⇒ HCF = 35.HCF(35, 1295): 1295 ÷ 35 = 37 remainder 0 ⇒ HCF remains 35.Hence greatest exact measuring length = 35 cm (i.e., 0.35 m). Verification 700/35 = 20; 385/35 = 11; 1295/35 = 37 ⇒ all integers. Common pitfalls Mixing metres and centimetres during H.C.F. computation. Accidentally taking L.C.M. instead of H.C.F. Final Answer Greatest length: 35 cm.

Question 7

Two numbers are in the ratio 2:3 and their L.C.M. is 48. Find the sum of the numbers.

Accepted Answer

Given data Numbers in ratio 2:3 ⇒ let them be 2k and 3k. LCM(2k, 3k) = 48. Concept / Approach For coprime 2 and 3, LCM(2k, 3k) = 6k. Solve for k, then compute the numbers and their sum. Step-by-step calculation 6k = 48 ⇒ k = 8.Numbers: 2k = 16 and 3k = 24.Sum = 16 + 24 = 40. Verification LCM(16, 24) = 48 ✔ Common pitfalls Using G.C.D. instead of L.C.M. relation; or forgetting that 2 and 3 are coprime. Final Answer Sum of the numbers = 40.

Question 8

Find the highest common factor (H.C.F.) of 36 and 84.

Accepted Answer

Given data Numbers: 36 and 84. Concept / Approach Use prime factorization or Euclidean algorithm to find H.C.F. Step-by-step (prime factorization) 36 = 2^2 × 3^284 = 2^2 × 3 × 7Common prime powers: 2^2 × 3 = 4 × 3 = 12. Alternative (Euclidean) 84 mod 36 = 12, 36 mod 12 = 0 ⇒ H.C.F. = 12. Final Answer H.C.F. = 12.

Question 9

Find the smallest number to add to 2497 so the sum is divisible by 5, 6, 4, and 3.

Accepted Answer

Given data We need k ≥ 0 such that 2497 + k is divisible by each of 5, 6, 4, 3. Concept / Approach Let L = L.C.M.(5, 6, 4, 3). Then 2497 + k ≡ 0 (mod L). Step-by-step calculation Prime forms: 5 = 5; 6 = 2 × 3; 4 = 2^2; 3 = 3.L = 2^2 × 3 × 5 = 60.2497 mod 60: 60 × 41 = 2460 ⇒ remainder 37.Need k ≡ −37 ≡ 23 (mod 60).Minimum such k = 23. Verification (2497 + 23) = 2520, which is divisible by 60 (and by 5, 6, 4, 3 individually). Common pitfalls Checking divisibility by each number separately without using L.C.M., increasing error risk. Final Answer Required minimal addition = 23.

Question 10

Two numbers are in the ratio 3:4 and their H.C.F. is 4. Find their L.C.M.

Accepted Answer

Given data Numbers = 3h and 4h with H.C.F. = h = 4. Concept / Approach When ratio terms are coprime (3 and 4), LCM(3h, 4h) = 12h. Alternatively: product = (H.C.F.) × (L.C.M.). Step-by-step computation h = 4 ⇒ numbers are 12 and 16.LCM(12, 16) = 2^2 × 3 × 2^4/2^2? Or directly = (12 × 16) / HCF(12,16) = (192)/4 = 48. Verification Multiples: 12 → 12,24,36,48; 16 → 16,32,48 ⇒ first common is 48. Common pitfalls Assuming H.C.F. is 1 for 3:4; remember the actual numbers share factor h. Final Answer L.C.M. = 48.

Question 11

Find the L.C.M. of 24, 36, and 40.

Accepted Answer

Given data Numbers: 24, 36, 40. Concept / Approach Use prime powers: take the highest power of each prime appearing in any number. Prime factorization 24 = 2^3 × 336 = 2^2 × 3^240 = 2^3 × 5LCM = 2^3 × 3^2 × 5 = 8 × 9 × 5 = 360. Verification 360 ÷ 24 = 15; 360 ÷ 36 = 10; 360 ÷ 40 = 9 ⇒ all integers. Final Answer L.C.M. = 360.

Question 12

Find the smallest number which, when diminished by 7, is divisible by 12, 16, 18, 21, and 28.

Accepted Answer

Given data We seek the least N such that N − 7 is divisible by 12, 16, 18, 21, and 28 simultaneously. Concept / Approach Let L = L.C.M.(12, 16, 18, 21, 28). Then N − 7 must be a multiple of L ⇒ N = Lk + 7. Minimal N occurs at k = 1. Compute L.C.M. Prime forms: 12 = 2^2 × 3; 16 = 2^4; 18 = 2 × 3^2; 21 = 3 × 7; 28 = 2^2 × 7.Take highest powers: 2^4, 3^2, 7^1 ⇒ L = 16 × 9 × 7 = 1008. Smallest N N = 1008 × 1 + 7 = 1015. Verification N − 7 = 1008, clearly divisible by 12, 16, 18, 21, and 28. Common pitfalls Answering 1008 (forgetting to add 7 back). Missing 3^2 in L.C.M. due to 18. Final Answer Smallest such number = 1015.

Question 13

A, B, and C start together around a circular track. A takes 252 seconds per lap, B takes 308 seconds, and C takes 198 seconds. After how much time will they all be back together at the starting point?

Accepted Answer

Given data A's lap time = 252 s B's lap time = 308 s C's lap time = 198 s Concept / Approach They meet together at the start after the least common multiple (LCM) of their lap times. Use prime factorization to compute LCM. Step-by-step calculation (prime factors)252 = 2^2 × 3^2 × 7308 = 2^2 × 7 × 11198 = 2 × 3^2 × 11LCM = 2^2 × 3^2 × 7 × 11 = 4 × 9 × 7 × 11 = 2772 s Verification2772 ÷ 252 = 11; 2772 ÷ 308 = 9; 2772 ÷ 198 = 14 — all integers, so 2772 s is a common multiple and the least by construction.Convert to minutes: 2772 s = 46 min 12 s. Common pitfalls Adding or averaging lap times (not valid). The meeting time is the LCM, not the sum/average. Missing a prime power when forming the LCM. Final Answer2772 seconds (i.e., 46 minutes 12 seconds)

Question 14

Reduce the fraction 128352/238368 to lowest terms.

Accepted Answer

Concept / Approach Divide numerator and denominator by their greatest common divisor (GCD). Step-by-step calculationGCD(128352, 238368) = 18336128352 ÷ 18336 = 7238368 ÷ 18336 = 13So, 128352/238368 = 7/13 Quick check7/13 is already in lowest terms (13 is prime and does not divide 7). Final Answer7/13

Question 15

Find the least number which leaves a remainder of 3 when divided by 5, 6, 7, and 8, but is divisible by 9.

Accepted Answer

Concept / Approach If a number leaves the same remainder 3 on division by 5, 6, 7, and 8, then it is of the form N = LCM(5,6,7,8) × k + 3. Additionally, N must be divisible by 9. Step-by-step calculationLCM(5,6,7,8) = 840N = 840k + 3Divisibility by 9: 840k + 3 ≡ 0 (mod 9)840 ≡ 3 (mod 9) ⇒ 3k + 3 ≡ 0 ⇒ 3(k + 1) ≡ 0 (mod 9)k + 1 ≡ 0 (mod 3) ⇒ k ≡ 2 (mod 3)Smallest k = 2 ⇒ N = 840×2 + 3 = 1683 Verification1683 ÷ 9 = 187 (exact); 1683 leaves remainder 3 when divided by 5, 6, 7, and 8 since 1680 is a common multiple. Final Answer1683

Question 16

Three numbers are in the ratio 3 : 4 : 5, and their LCM is 2400. What is their HCF?

Accepted Answer

Concept / Approach If numbers are in ratio 3:4:5 and their HCF is h, the numbers are 3h, 4h, and 5h. Then LCM = h × LCM(3,4,5). Step-by-step calculationLCM(3,4,5) = 60Given LCM = 2400 = h × 60h = 2400 ÷ 60 = 40 Final Answer40

Question 17

What is the least number which, when doubled, is exactly divisible by 12, 18, 21, and 30?

Accepted Answer

Concept / Approach Let the required number be n. Then 2n must be a common multiple of 12, 18, 21, and 30. The least such multiple is the LCM; hence take n = LCM/2. Step-by-step calculationLCM(12,18,21,30) = 12602n = 1260 ⇒ n = 1260/2 = 630 Verification2×630 = 1260 which is divisible by 12, 18, 21, and 30. Final Answer630

Question 18

Find the least multiple of 7 that leaves a remainder of 4 when divided by 6, 9, 15, and 18.

Accepted Answer

Concept / Approach If N leaves remainder 4 upon division by 6, 9, 15, and 18, then N ≡ 4 (mod LCM(6,9,15,18)). Simultaneously, N must be a multiple of 7. Step-by-step calculationLCM(6,9,15,18) = 90N ≡ 4 (mod 90) and N ≡ 0 (mod 7)Let N = 4 + 90k. Require 4 + 90k ≡ 0 (mod 7)90 ≡ 6 (mod 7) ⇒ 4 + 6k ≡ 0 ⇒ 6k ≡ 3 (mod 7)Inverse of 6 (mod 7) is 6 ⇒ k ≡ 3×6 ≡ 18 ≡ 4 (mod 7)Smallest k = 4 ⇒ N = 4 + 90×4 = 364 Verification364 ÷ 7 = 52 (multiple), and 364 leaves remainder 4 upon division by 6, 9, 15, and 18. Final Answer364

Question 19

The product of two numbers is 2028 and their HCF is 13. How many such pairs of numbers are possible?

Accepted Answer

Concept / Approach Let the numbers be 13a and 13b with gcd(a, b) = 1. Then (13a)(13b) = 2028 ⇒ 169ab = 2028 ⇒ ab = 12. Each prime power in 12 must be assigned wholly to either a or b to keep gcd(a, b) = 1. Step-by-step reasoning12 = 2^2 × 3Possible coprime unordered factor pairs (a, b): (1, 12) and (3, 4)Corresponding number pairs: (13×1, 13×12) and (13×3, 13×4) NoteIf ordered pairs were asked, the count would be 4. For unordered pairs (usual convention), the count is 2. Final Answer2

Question 20

Six bells start together and toll at intervals of 2, 4, 6, 8, 10, and 12 seconds respectively. In 30 minutes, how many times do they toll together?

Accepted Answer

Concept / Approach They toll together every LCM of the given intervals. Step-by-step calculationLCM(2,4,6,8,10,12) = 120 seconds30 minutes = 1800 secondsNumber of meetings = (1800 ÷ 120) + 1 = 15 + 1 = 16 Why +1?They also toll together at time t = 0 (the very start). Since 1800 is a multiple of 120, the final instant is also included. Final Answer16

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