Question 1

Let N be the greatest number that divides 1305, 4665, and 6905 leaving the same remainder in each case. What is the sum of the digits of N?

Accepted Answer

Concept / Approach If N leaves the same remainder on dividing several numbers, N must divide all pairwise differences. Step-by-step calculation4665 − 1305 = 33606905 − 4665 = 22406905 − 1305 = 5600N = GCD(3360, 2240, 5600) = 1120Sum of digits of 1120 = 1 + 1 + 2 + 0 = 4 Final Answer4

Question 2

The sum of two numbers is 55. Their HCF is 5 and their LCM is 120. What is the sum of the reciprocals of the two numbers?

Accepted Answer

Concept / Approach For positive integers a and b, ab = HCF × LCM. The sum of reciprocals is (a + b)/ab. Step-by-step calculationa + b = 55ab = 5 × 120 = 600Sum of reciprocals = (a + b)/ab = 55/600 = 11/120 Final Answer11/120

Question 3

The product of two numbers is 4107. If their H.C.F. is 37, find the greater number.

Accepted Answer

Given data Product of two numbers = 4107 H.C.F. (gcd) of the two numbers = 37 Concept / Approach If HCF = h, then numbers can be written as h·x and h·y with gcd(x, y) = 1. Then product = h^2 · x · y ⇒ x · y = product / h^2. Step-by-step calculation h = 37 ⇒ h^2 = 37^2 = 1369x · y = 4107 ÷ 1369 = 3Coprime factor pairs of 3 are (1, 3)Numbers are 37·1 = 37 and 37·3 = 111 Verification 37 × 111 = 4107 and gcd(37, 111) = 37. Common pitfalls Forgetting to divide the product by h^2 (not just by h). Final Answer Greater number = 111.

Question 4

Three pairwise coprime numbers have ab = 551 and bc = 1073. Find a + b + c.

Accepted Answer

Given data a, b, c are pairwise coprime a·b = 551 b·c = 1073 Concept / Approach If a, b, c are pairwise coprime, then b = gcd(a·b, b·c). Once b is found, a = 551 / b and c = 1073 / b. Step-by-step calculation gcd(551, 1073):1073 − 551 = 522551 − 522 = 29522 ÷ 29 = 18 remainder 0 ⇒ gcd = 29Thus b = 29a = 551 ÷ 29 = 19c = 1073 ÷ 29 = 37Sum = 19 + 29 + 37 = 85 Verification Pairwise gcds: gcd(19,29)=1, gcd(29,37)=1, gcd(19,37)=1. Common pitfalls Assuming b is a common prime factor without checking gcd. Final Answer a + b + c = 85.

Question 5

Which of the following fractions is the largest?

Accepted Answer

7⁄8

Question 6

Find the least number that leaves a remainder of 8 when divided by 12, 15, 20, and 54.

Accepted Answer

Given data N leaves remainder 8 upon division by 12, 15, 20, and 54. Concept / Approach N − 8 must be divisible by all four numbers ⇒ N − 8 is a multiple of their L.C.M. Step-by-step calculation Prime factors: 12 = 2^2×3, 15 = 3×5, 20 = 2^2×5, 54 = 2×3^3LCM = 2^2 × 3^3 × 5 = 4 × 27 × 5 = 540Smallest N = 540 + 8 = 548 Verification 548 − 8 = 540 is divisible by 12, 15, 20 and 54. Common pitfalls Answering 540 (forgetting to add back the remainder 8). Final Answer Required number = 548.

Question 7

Which of the following numbers has the most divisors?

Accepted Answer

48

Question 8

The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. Find the larger number.

Accepted Answer

Given data H.C.F. = 23 The co-prime parts contributing to L.C.M. are 13 and 14. Concept / Approach If HCF = h and the co-prime parts are p and q, the numbers can be taken as h·p and h·q (since gcd(p, q) = 1). Step-by-step calculation Numbers: 23×13 = 299 and 23×14 = 322L.C.M. = 23×13×14 and H.C.F. = 23, as required. Verification gcd(299, 322) = 23 (since 299 = 23×13 and 322 = 23×14). Common pitfalls Multiplying 13 and 14 again into a single number, which is not the intent. Final Answer Larger number = 322.

Question 9

Find the greatest number that divides 1657 and 2037 leaving remainders 6 and 5 respectively.

Accepted Answer

Given data 1657 leaves remainder 6; 2037 leaves remainder 5 when divided by the required number. Concept / Approach If numbers leave (possibly different) remainders r1, r2 on division by d, then d divides the difference of the adjusted numbers: gcd(1657 − 6, 2037 − 5). Step-by-step gcd 1657 − 6 = 16512037 − 5 = 2032gcd(1651, 2032):2032 − 1651 = 3811651 ÷ 381 = 4 remainder 127381 ÷ 127 = 3 remainder 0 ⇒ gcd = 127 Verification 1657 mod 127 = 6 and 2037 mod 127 = 5. Common pitfalls Taking gcd of 1657 and 2037 directly without adjusting for remainders. Final Answer Greatest number = 127.

Question 10

The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one number is 275, find the other.

Accepted Answer

Given data H.C.F. = 11 L.C.M. = 7700 One number = 275 Concept / Approach For any two numbers a and b: a × b = HCF × LCM. Step-by-step calculation a × b = 11 × 7700 = 84700Other number = 84700 ÷ 275 = 308 Verification gcd(275, 308) = 11 and lcm(275, 308) = 7700 (since 275×308 = 84700 = 11×7700). Common pitfalls Dividing 7700 by 275 (incorrect method) instead of using product = HCF × LCM. Final Answer Other number = 308.

Question 11

HCF with common remainders: Find the greatest number that divides 263, 935, and 1383 leaving a remainder of 7 in each case.

Accepted Answer

Introduction / Context:When a number D divides several numbers leaving the same remainder r, D must divide the pairwise differences after subtracting r. This converts a “common remainder” problem into a Highest Common Factor (HCF) computation. Given Data / Assumptions: Numbers: 263, 935, 1383. Common remainder r = 7. We seek the largest divisor D. Concept / Approach:If each number N leaves remainder 7 upon division by D, then D | (N − 7). So compute 263 − 7, 935 − 7, and 1383 − 7, and then take their HCF. Step-by-Step Solution: Reduce by remainder: 263 − 7 = 256, 935 − 7 = 928, 1383 − 7 = 1376Compute HCF(256, 928): 928 mod 256 = 160; 256 mod 160 = 96; 160 mod 96 = 64; 96 mod 64 = 32; 64 mod 32 = 0 ⇒ HCF = 32Check with 1376: 1376 / 32 = 43 (exact) ⇒ HCF remains 32Therefore, the greatest such divisor is 32 Verification / Alternative check:Confirm all original numbers give remainder 7 when divided by 32: 263 = 32*8 + 7, 935 = 32*29 + 7, 1383 = 32*43 + 7. Why Other Options Are Wrong:31 and 35 do not divide all adjusted values (256, 928, 1376). 30 is not a factor of 256 or 928. Common Pitfalls:Forgetting to subtract the common remainder before finding the HCF; directly taking HCF of the raw numbers would be incorrect. Final Answer:32

Question 12

Polynomial HCF condition: If (x + 2) is the HCF of x^2 + ax + b and x^2 + cx + d, which relation among a, b, c, d must hold?

Accepted Answer

Introduction / Context:When (x + 2) is a common factor (and actually the HCF) of two polynomials, both polynomials must vanish at x = −2. Converting that fact to coefficient relations is a standard method to determine constraints among a, b, c, d. Given Data / Assumptions: Polynomials: f(x) = x^2 + ax + b and g(x) = x^2 + cx + d. (x + 2) is a common factor ⇒ f(−2) = 0 and g(−2) = 0. Concept / Approach:Evaluate each polynomial at x = −2 to get two linear equations: 4 − 2a + b = 0 and 4 − 2c + d = 0. Rearranging yields equalities that can be combined into a single relation matching one of the answer choices. Step-by-Step Solution: f(−2): 4 − 2a + b = 0 ⇒ b = 2a − 4g(−2): 4 − 2c + d = 0 ⇒ d = 2c − 4Eliminate the constant: b − 2a = −4 and d − 2c = −4 ⇒ b − 2a = d − 2cRearrange: b + 2c = 2a + d Verification / Alternative check:Substitute b = 2a − 4 and d = 2c − 4 into b + 2c − (2a + d); it simplifies to 0 identically, confirming the relation. Why Other Options Are Wrong:They do not follow from f(−2) = g(−2) = 0 simultaneously; testing with sample coefficients that satisfy the correct relation will make other proposed relations fail. Common Pitfalls:Evaluating at x = +2 instead of −2, or mixing the signs when rearranging the equalities. Final Answer:b + 2c = 2a + d

Question 13

Find the greatest measuring length: Determine the greatest possible length (in centimeters) that measures exactly 4 m 3 cm, 4 m 34 cm, and 4 m 65 cm.

Accepted Answer

Introduction / Context:“Measuring exactly” multiple lengths using the greatest possible length is a Highest Common Factor (HCF) problem. Converting all measures to the same unit (centimeters) and computing the HCF yields the required maximum measuring length. Given Data / Assumptions: Lengths: 4 m 3 cm, 4 m 34 cm, 4 m 65 cm. 1 m = 100 cm. Concept / Approach:Convert each to centimeters and compute their HCF. The HCF is the largest length that divides all three exactly with no remainder. Step-by-Step Solution: Convert: 4 m 3 cm = 403 cm; 4 m 34 cm = 434 cm; 4 m 65 cm = 465 cmHCF(403, 434): 434 − 403 = 31; and 403 = 31 × 13 ⇒ HCF = 31Now HCF(31, 465): 465 = 31 × 15 ⇒ HCF remains 31Therefore, the greatest measuring length is 31 cm Verification / Alternative check:Divide each: 403/31 = 13, 434/31 = 14, 465/31 = 15 — all integers. Why Other Options Are Wrong:28, 29, 32, and 27 do not divide all three lengths exactly (at least one will leave a remainder). Common Pitfalls:Forgetting to convert meters to centimeters consistently; attempting to take the HCF in mixed units leads to errors. Final Answer:31 cm

Question 14

Compare simple fractions: Which of the following is the greatest fraction among 7/8, 6/7, 4/5, and 5/6?

Accepted Answer

Introduction / Context:To compare fractions with different denominators quickly, one can convert them to decimals, use cross-multiplication, or place them over a common denominator. Here the values are close to 1, so careful comparison is needed. Given Data / Assumptions: Fractions: 7/8, 6/7, 4/5, 5/6. All are positive proper fractions less than 1. Concept / Approach:Use either decimal approximations or cross-multiplication pairwise. Since all denominators are small, decimals are fast and reliable to three decimal places. Step-by-Step Solution: 7/8 = 0.8756/7 ≈ 0.8575/6 ≈ 0.8334/5 = 0.800Hence, 7/8 is the greatest. Verification / Alternative check:Cross-multiplication: compare 7/8 vs 6/7 ⇒ 7*7 = 49 vs 6*8 = 48 ⇒ 7/8 larger; similarly 7/8 vs 5/6 ⇒ 7*6 = 42 vs 5*8 = 40 ⇒ 7/8 larger; and 7/8 vs 4/5 ⇒ 7*5 = 35 vs 4*8 = 32 ⇒ 7/8 larger. Why Other Options Are Wrong:Each has a smaller decimal value than 0.875, so they cannot be the greatest among the four. Common Pitfalls:Rounding too aggressively (e.g., 0.86 vs 0.88) can reverse close comparisons. Use exact fractions or enough decimal places. Final Answer:7/8

Question 15

Find the LCM of fractions clearly What is the least common multiple (LCM) of the fractions 2/3, 4/9, 5/6, and 7/12? Use the standard rule: for positive fractions, LCM = LCM of numerators / HCF of denominators.

Accepted Answer

Introduction / Context:This problem asks for the least common multiple (LCM) of several fractions. For fractions, there is a well-known formula that converts the multi-step process into two simpler integer tasks: take the LCM of all numerators and divide it by the highest common factor (HCF or GCD) of all denominators. Given Data / Assumptions: Fractions: 2/3, 4/9, 5/6, 7/12. All numbers are positive. Rule: LCM of fractions = LCM(numerators) / HCF(denominators). Concept / Approach:Compute two values independently: (1) LCM of numerators and (2) HCF of denominators. Then divide the first by the second. This avoids converting each fraction to many equivalent forms. Step-by-Step Solution:Numerators: 2, 4, 5, 7 ⇒ LCM = 140.Denominators: 3, 9, 6, 12 ⇒ HCF = gcd(3, 9, 6, 12) = 3.Therefore LCM of the fractions = 140 / 3. Verification / Alternative check:Confirm 140/3 is a common multiple: (140/3) ÷ (2/3) = 70, (140/3) ÷ (4/9) = 105, (140/3) ÷ (5/6) = 56, (140/3) ÷ (7/12) = 80. All are integers; trying any smaller candidate fails for at least one fraction. Why Other Options Are Wrong:1/18 and 1/36 are far too small. 35/9 is not a common multiple for all given fractions. 28/3 is a common multiple for some but not the least for all. Common Pitfalls:Mixing up the rule (using HCF of numerators / LCM of denominators, which is actually for the HCF of fractions), or computing LCM and HCF pairwise rather than across the full set. Final Answer:140/3

Question 16

Compute the LCM of three fractions step by step Find the least common multiple (LCM) of the fractions 3/4, 6/7, and 9/8 using LCM(numerators) / HCF(denominators).

Accepted Answer

Introduction / Context:LCM for fractions reduces to one LCM on integers and one HCF on integers. This avoids unwieldy conversions to common denominators for each fraction. Given Data / Assumptions: Fractions: 3/4, 6/7, 9/8. Rule: LCM(fractions) = LCM of numerators / HCF of denominators. Concept / Approach:Find LCM(3, 6, 9) and HCF(4, 7, 8) separately, then divide the former by the latter. Step-by-Step Solution:LCM of numerators: LCM(3, 6, 9) = 18.HCF of denominators: gcd(4, 7, 8) = 1 (since 7 is coprime to both 4 and 8).Therefore LCM of fractions = 18 / 1 = 18. Verification / Alternative check:(18) ÷ (3/4) = 24; (18) ÷ (6/7) = 21; (18) ÷ (9/8) = 16. All quotients are integers, confirming 18 works for all three. Why Other Options Are Wrong:3, 6 are too small to be common multiples for all three. 3/56 and 9/28 misapply the fraction-HCF rule or invert the intended formula. Common Pitfalls:Interchanging the two rules (HCF vs LCM), or assuming you must first find a common denominator across all fractions. Final Answer:18

Question 17

Find the HCF of several fractions Determine the highest common factor (HCF) of 1/2, 3/4, 5/6, 7/8, and 9/10 using the standard fraction-HCF rule.

Accepted Answer

Introduction / Context:For fractions, the HCF (also called GCD) uses a reversed version of the LCM rule: take HCF of numerators and divide by LCM of denominators. This yields the greatest fraction that divides each given fraction exactly. Given Data / Assumptions: Fractions: 1/2, 3/4, 5/6, 7/8, 9/10. Rule: HCF(fractions) = HCF(numerators) / LCM(denominators). Concept / Approach:Compute HCF(1, 3, 5, 7, 9) and LCM(2, 4, 6, 8, 10), then form the fraction HCF/LCM in lowest terms. Step-by-Step Solution:HCF of numerators: gcd(1, 3, 5, 7, 9) = 1.LCM of denominators: LCM(2, 4, 6, 8, 10) = 120.Therefore HCF of fractions = 1 / 120. Verification / Alternative check:Check that 1/120 divides each given fraction: e.g., (1/2) ÷ (1/120) = 60, (3/4) ÷ (1/120) = 90, etc. All quotients are integers, confirming correctness. Why Other Options Are Wrong:1/2 and 1/10 are too large to divide all given fractions exactly; 9/120 is not the greatest common factor by rule; 3/20 does not divide 1/2 an integer number of times when considered with all others. Common Pitfalls:Taking LCM of numerators and HCF of denominators (reversing the rule), or reducing early and losing track of the largest common divisor. Final Answer:1/120

Question 18

Repair and compute the LCM with powers Interpret the numbers as 3^3, 4, 4^2, and 3 (standard notation compressed in the source). Find the least common multiple (LCM) of 3^3, 4, 4^2, and 3.

Accepted Answer

Introduction / Context:The raw stem showed compressed spacing: “3 3, 4, 4 2 and 3”. Applying the Recovery-First policy, we interpret this in standard exam style as powers: 3^3, 4, 4^2, and 3. We now compute the LCM of these integers. Given Data / Assumptions: Numbers: 3^3 = 27, 4 = 2^2, 4^2 = 16 = 2^4, and 3 = 3^1. All are positive integers. Concept / Approach:LCM takes the highest power of each prime that appears in the factorisation across all numbers. Step-by-Step Solution:Prime 2: highest power among {2^2, 2^4} is 2^4 = 16.Prime 3: highest power among {3^3, 3^1} is 3^3 = 27.Therefore LCM = 2^4 * 3^3 = 16 * 27 = 432. Verification / Alternative check:432 ÷ 27 = 16, ÷ 16 = 27, ÷ 4 = 108, ÷ 3 = 144, all integers; any smaller candidate would miss at least one highest prime power. Why Other Options Are Wrong:12 and 48 do not contain sufficient powers (miss 3^3 or 2^4). “None of these” is invalid since 432 fits perfectly. 96 is still short of 3^3. Common Pitfalls:Adding exponents instead of taking maximum exponents, or misreading the compressed stem and losing the powers. Final Answer:432

Question 19

Find the LCM of two numbers using gcd-lcm relation Compute the least common multiple (LCM) of 87 and 145 efficiently.

Accepted Answer

Introduction / Context:The fastest way to find the LCM of two numbers is to use the identity LCM(a, b) * gcd(a, b) = a * b. This avoids full prime factorisation if you can quickly see a common factor. Given Data / Assumptions: a = 87, b = 145. gcd(87, 145) is needed. Concept / Approach:Factor lightly: 87 = 3 * 29; 145 = 5 * 29. Their gcd is 29. Then LCM = (87 * 145) / 29. Step-by-Step Solution:gcd(87, 145) = 29.LCM = (87 * 145) / 29 = 87 * 5 = 435. Verification / Alternative check:435 ÷ 87 = 5 and 435 ÷ 145 = 3, both integers. No smaller positive number is a multiple of both 87 and 145. Why Other Options Are Wrong:870 and 1740 are multiples but not the least; 1305 is 435 * 3; 2610 is 435 * 6. Only 435 is the minimal common multiple. Common Pitfalls:Missing the shared factor 29; multiplying the numbers without dividing by the gcd; mixing up HCF and LCM definitions. Final Answer:435

Question 20

Identify a pair of co-primes (gcd equals 1) From the given pairs, select the pair whose greatest common divisor is exactly 1.

Accepted Answer

Introduction / Context:Two integers are co-prime if their only common positive divisor is 1. This is equivalent to saying their gcd (HCF) is 1. We check each pair quickly using small prime factors. Given Data / Assumptions: Pairs: (14, 35), (18, 25), (31, 93), (32, 62), (21, 28). Co-prime ⇔ gcd = 1. Concept / Approach:Factor each pair lightly: look for obvious shared primes such as 2, 3, 5, 7, 11, 31. Step-by-Step Solution:(14, 35): 14 = 2 * 7, 35 = 5 * 7 ⇒ gcd ≥ 7 ⇒ not co-prime.(18, 25): 18 = 2 * 3^2, 25 = 5^2 ⇒ no common primes ⇒ gcd = 1 ⇒ co-prime.(31, 93): 93 = 3 * 31 ⇒ gcd ≥ 31 ⇒ not co-prime.(32, 62): 32 = 2^5, 62 = 2 * 31 ⇒ gcd ≥ 2 ⇒ not co-prime.(21, 28): 21 = 3 * 7, 28 = 4 * 7 ⇒ gcd ≥ 7 ⇒ not co-prime. Verification / Alternative check:Euclidean algorithm on (18, 25): 25 mod 18 = 7, 18 mod 7 = 4, 7 mod 4 = 3, 4 mod 3 = 1 ⇒ gcd = 1. Why Other Options Are Wrong:Each has a clear shared factor (7, 31, 2, or 7 respectively), so gcd is greater than 1. Common Pitfalls:Assuming numbers with different parity must be co-prime (not always), or overlooking large shared primes like 31. Final Answer:(18, 25)

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