Question 1

Product relation between HCF and LCM: For the two integers 18 and 15, compute the product HCF × LCM.

Accepted Answer

Introduction / Context: For any two positive integers a and b, the fundamental identity holds: HCF(a, b) * LCM(a, b) = a * b. This allows quick computation of one product given the numbers themselves. Given Data / Assumptions: a = 18, b = 15 We want HCF * LCM for this pair. Concept / Approach: Use the identity directly: the product of HCF and LCM equals the product of the two numbers. There is no need to compute HCF and LCM separately in this case. Step-by-Step Solution: Compute a * b = 18 * 15 = 270.Therefore HCF(18,15) * LCM(18,15) = 270. Verification / Alternative check: Indeed, HCF(18,15) = 3 and LCM(18,15) = 90, and 3 * 90 = 270, confirming the identity. Why Other Options Are Wrong: 120, 150, 175 do not equal 18 * 15 and thus contradict the fundamental relation. Common Pitfalls: Mistakenly adding HCF and LCM or computing them separately with errors. Final Answer: 270

Question 2

Find the missing number from HCF and LCM: Two numbers have HCF = 15 and LCM = 225. If one number is 75, find the other number.

Accepted Answer

Introduction / Context: For two numbers m and n, the identity m * n = HCF(m, n) * LCM(m, n) lets us find the unknown when one number and both HCF and LCM are known. This avoids trial-and-error completely. Given Data / Assumptions: HCF = 15 LCM = 225 One number = 75 Concept / Approach: Compute the product of HCF and LCM to get m * n, then divide by the known number to obtain the unknown partner. Ensure the quotient is an integer to validate consistency. Step-by-Step Solution: m * n = 15 * 225 = 3375.Given one number is 75, the other is 3375 / 75 = 45. Verification / Alternative check: Check HCF(75, 45) = 15 and LCM(75, 45) = 225; their product is 3375, matching the identity. Why Other Options Are Wrong: 105, 90, 60 do not jointly yield HCF 15 and LCM 225 with 75 under the product identity. Common Pitfalls: Forgetting the product identity or miscalculating 3375/75. Final Answer: 45

Question 3

Numbers from HCF and ratio: The HCF of three numbers is 23. If they are in the ratio 1 : 2 : 3, find the actual numbers.

Accepted Answer

Introduction / Context: When numbers are in a given ratio and share a known HCF, each number can be expressed as HCF times the respective ratio part (provided the ratio parts are coprime in aggregate). This is a standard reconstruction technique in HCF–ratio problems. Given Data / Assumptions: HCF = 23 Ratio = 1 : 2 : 3 Concept / Approach: Let the numbers be 23x, 23y, 23z with x : y : z = 1 : 2 : 3. Because the common factor extracted is the HCF, the simplest choice is x = 1, y = 2, z = 3 to obtain the minimal triple satisfying the ratio and HCF conditions. Step-by-Step Solution: Number 1 = 23 * 1 = 23.Number 2 = 23 * 2 = 46.Number 3 = 23 * 3 = 69. Verification / Alternative check: HCF(23, 46, 69) = 23, and 23 : 46 : 69 reduces to 1 : 2 : 3 as required. Why Other Options Are Wrong: All other triplets either do not have HCF equal to 23 or do not preserve the exact 1 : 2 : 3 ratio. Common Pitfalls: Multiplying the ratio parts by another common factor, which would increase the HCF beyond 23. Final Answer: 23, 46, 69

Question 4

Using HCF–LCM product with a sum constraint: The HCF and LCM of two numbers m and n are 6 and 210, respectively. If m + n = 72, find the value of 1/m + 1/n.

Accepted Answer

Introduction / Context: This question combines the product identity for two numbers with a sum constraint to compute the sum of reciprocals directly. It illustrates how HCF and LCM can determine mn, and with m + n given, one can form (m + n)/(mn) without solving for m and n individually. Given Data / Assumptions: HCF(m, n) = 6 LCM(m, n) = 210 m + n = 72 Find 1/m + 1/n. Concept / Approach: Use the identity m * n = HCF * LCM. Then 1/m + 1/n = (m + n) / (m * n). This bypasses the need to find m and n separately and leverages given aggregate information efficiently. Step-by-Step Solution: Compute mn = 6 * 210 = 1260.Compute (m + n) / (mn) = 72 / 1260.Reduce the fraction: divide numerator and denominator by 18 ⇒ 72/1260 = 4/70 = 2/35. Verification / Alternative check: There exist integer pairs with HCF 6 and LCM 210 satisfying m + n = 72 (e.g., solving t^2 − 72t + 1260 = 0). Regardless of the specific pair, (m + n)/(mn) remains 2/35. Why Other Options Are Wrong: 1/35, 3/35, 5/37 do not equal 72/1260 after reduction and thus contradict the computed ratio. Common Pitfalls: Miscalculating mn or simplifying 72/1260 incorrectly. Final Answer: 2/35

Question 5

Two numbers are in the ratio 5 : 6 and their least common multiple (LCM) is 480. Determine the highest common factor (HCF) of the two numbers, showing clear reasoning.

Accepted Answer

Introduction / Context: This problem connects the ratio of two numbers with their least common multiple (LCM) to find the highest common factor (HCF). In number theory, the product of two numbers equals the product of their HCF and LCM. When a ratio is given, representing the numbers as multiples of a common factor streamlines the calculation. Given Data / Assumptions: Ratio of numbers = 5 : 6 LCM = 480 Both numbers are positive integers Concept / Approach: If two numbers are 5k and 6k for some positive integer k, then HCF(5k, 6k) = k (since 5 and 6 are coprime) and LCM(5k, 6k) = 30k. We are given the LCM, so we can solve for k directly. The HCF is then k itself. Step-by-Step Solution: Let the numbers be 5k and 6k.LCM(5k, 6k) = 30k.Given LCM = 480 ⇒ 30k = 480.Solve for k: k = 480 / 30 = 16.Therefore, HCF = k = 16. Verification / Alternative check: Compute numbers explicitly: 5k = 80 and 6k = 96. Then HCF(80, 96) = 16 and LCM(80, 96) = (80*96)/16 = 480, confirming consistency. Why Other Options Are Wrong: 20 and 24 arise from misusing the product-HCF-LCM relation; 6 and 5 incorrectly treat the ratio terms themselves as the HCF; they ignore scaling by k. Common Pitfalls: Confusing the ratio numbers with the actual numbers; forgetting that LCM scales linearly with k when the ratio terms are coprime; or mixing up HCF and LCM in the identity number1 * number2 = HCF * LCM. Final Answer: 16

Question 6

A number is exactly divisible by both 11 and 13. Identify which statement must be true about this number.

Accepted Answer

Introduction / Context: This is a fundamental divisibility question about combined factors. When a number is divisible by two integers, we investigate what stronger conditions follow, particularly relating to their product and least common multiple (LCM). Given Data / Assumptions: The number is divisible by 11. The same number is divisible by 13. 11 and 13 are distinct primes. Concept / Approach: If a number N is divisible by prime p and prime q, then N is divisible by p*q, because primes have no nontrivial common factors. In general, N is divisible by LCM(11, 13). With primes, LCM equals their product 11*13 = 143. Step-by-Step Solution: Since 11 | N and 13 | N, and gcd(11, 13) = 1, LCM(11, 13) = 11*13.Therefore, N is divisible by 11*13 = 143. Verification / Alternative check: Take N = 143 as the smallest example. It is divisible by both 11 and 13. Any integer multiple of 143 also shares this property, confirming the rule. Why Other Options Are Wrong: Divisible by (11 + 13) = 24 is not guaranteed. Divisible by (13 - 11) = 2 is not guaranteed. Divisible by (13 ÷ 11) is meaningless for integers. “Divisible by 22 only” ignores the 13 condition. Common Pitfalls: Confusing sum or difference with necessary divisibility; forgetting that for primes, the LCM equals the product; and misreading divisibility statements involving division. Final Answer: Divisible by (11 x 13)

Question 7

The least common multiple (LCM) of two numbers is 48. The two numbers are in the ratio 2 : 3. Determine the sum of the numbers with full working.

Accepted Answer

Introduction / Context: Reiterating the standard method: when numbers are in a simple ratio and the LCM is known, we scale the ratio terms to meet the LCM and then compute the actual numbers and their sum. Given Data / Assumptions: LCM = 48 Ratio = 2 : 3 Concept / Approach: Let the numbers be 2k and 3k. For coprime 2 and 3, LCM(2k, 3k) = 6k. Match 6k to 48 to find k, then sum the two values. Step-by-Step Solution: 6k = 48 ⇒ k = 8.Numbers: 16 and 24.Sum = 16 + 24 = 40. Verification / Alternative check: LCM(16, 24) = 48; ratio 16:24 simplifies to 2:3. Why Other Options Are Wrong: 28, 32, 64, and 36 do not correspond to the correct scaled values that achieve LCM 48. Common Pitfalls: Forgetting that 2 and 3 are coprime so the LCM with scaling is 6k; arithmetic slips when solving for k. Final Answer: 40

Question 8

The product of two whole numbers is 1500 and their highest common factor (HCF) is 10. Find the least common multiple (LCM).

Accepted Answer

Introduction / Context: This question directly applies the foundational identity for two positive integers: product = HCF * LCM. Given the product and HCF, we can solve immediately for the LCM. Given Data / Assumptions: Product of the two numbers = 1500 HCF = 10 Numbers are whole (nonnegative integers) Concept / Approach: Use the identity: number1 * number2 = HCF * LCM. Rearranging gives LCM = (number1 * number2) / HCF. Substitute the given values. Step-by-Step Solution: LCM = (product) / HCF = 1500 / 10 = 150. Verification / Alternative check: If the numbers were, for instance, 30 and 50, their product is 1500 and HCF is 10, giving LCM = 150. This aligns with the computed value and the identity. Why Other Options Are Wrong: 15000 misplaces the division; 1500 and 15 confuse product versus factors; 300 doubles the correct result without basis. Common Pitfalls: Inverting the formula, or assuming HCF and LCM are themselves the numbers. Always apply product = HCF * LCM carefully. Final Answer: 150

Question 9

Find the least number that must be added to 2497 so the resulting sum is exactly divisible by 5, 6, 4, and 3.

Accepted Answer

Introduction / Context: We are asked for the smallest nonnegative addition that makes a given number divisible by several integers at once. This is a classic application of least common multiple (LCM) and modular arithmetic. Given Data / Assumptions: Starting number: 2497 Divisors: 5, 6, 4, and 3 We want 2497 + x to be divisible by each divisor. Concept / Approach: Compute LCM(5, 6, 4, 3). The sum must be a multiple of this LCM. Then find the smallest x such that 2497 + x ≡ 0 (mod LCM). Step-by-Step Solution: LCM(5, 6, 4, 3) = LCM(5, 6) = 30; LCM(30, 4) = 60; LCM(60, 3) = 60.Compute remainder of 2497 modulo 60: 2497 = 60*41 + 37, so 2497 ≡ 37 (mod 60).We need x such that 37 + x ≡ 0 (mod 60) ⇒ x ≡ 23 (mod 60).Smallest nonnegative x is 23. Verification / Alternative check: 2497 + 23 = 2520. Dividing: 2520/5 = 504, /6 = 420, /4 = 630, /3 = 840. All are integers. Why Other Options Are Wrong: 3, 13, 33, and 17 do not make 2497 + x a multiple of 60; they leave nonzero remainders modulo 60. Common Pitfalls: Taking LCM incorrectly (e.g., multiplying all divisors without reduction) or using the wrong remainder when computing x modulo the LCM. Final Answer: 23

Question 10

Find the least positive integer which, when divided by each of 12, 16, and 18, leaves a remainder of 5 in every case.

Accepted Answer

Introduction / Context: This is a remainder (congruence) problem. A number leaving the same remainder when divided by several moduli can be written as that common remainder plus a multiple of the least common multiple (LCM) of those moduli. Given Data / Assumptions: Divisors: 12, 16, 18 Common remainder: 5 We seek the least positive such number. Concept / Approach: Let N be the number. Then N ≡ 5 (mod 12), N ≡ 5 (mod 16), and N ≡ 5 (mod 18). Therefore, N − 5 is a common multiple of 12, 16, and 18; hence N − 5 is a multiple of LCM(12, 16, 18). Step-by-Step Solution: Compute LCM: 12 = 2^2*3, 16 = 2^4, 18 = 2*3^2.LCM = 2^4 * 3^2 = 16 * 9 = 144.Thus N = 5 + 144t for integer t ≥ 0.Least positive solution corresponds to t = 1 ⇒ N = 149. Verification / Alternative check: 149 ÷ 12 leaves remainder 5; 149 ÷ 16 leaves 5; 149 ÷ 18 leaves 5. All conditions hold. Why Other Options Are Wrong: 139, 144, 154, and 161 do not equal 5 plus a multiple of 144, so they fail at least one of the remainder checks. Common Pitfalls: Using gcd instead of LCM; choosing t = 0 which gives 5 (too small compared to the options); or miscomputing the LCM exponents. Final Answer: 149

Question 11

What is the least number that must be subtracted from 1294 so that the remainder, when divided by 9, 11, and 13, leaves the same remainder 6 in each case?

Accepted Answer

Introduction / Context: We want to adjust a given number downward so that the resulting remainder is congruent to 6 modulo 9, 11, and 13 simultaneously. This is a modular arithmetic problem relying on least common multiples (LCMs). Given Data / Assumptions: Original number: 1294 Desired property: R ≡ 6 (mod 9, 11, 13) We seek the least nonnegative subtraction x where R = 1294 − x. Concept / Approach: If R ≡ 6 modulo each of 9, 11, and 13, then R − 6 is divisible by LCM(9, 11, 13). Compute 1294 − 6 and reduce modulo that LCM to find the smallest x so that 1294 − x − 6 is exactly that multiple. Step-by-Step Solution: LCM(9, 11, 13) = 9*11*13 = 1287 (since they are pairwise coprime).Compute 1294 − 6 = 1288.We need x such that 1288 − x is a multiple of 1287 ⇒ choose x = 1 to get 1287.Thus the least number to subtract is 1. Verification / Alternative check: R = 1294 − 1 = 1293. Then 1293 − 6 = 1287 which is divisible by 9, 11, and 13. Hence 1293 leaves remainder 6 when divided by each of 9, 11, and 13. Why Other Options Are Wrong: 0, 2, 3, and 5 do not reduce 1294 sufficiently to land exactly one LCM step down from 1294 − 6. They fail at least one modulus check. Common Pitfalls: Forgetting to subtract the target remainder first; computing gcd instead of LCM; arithmetic slips with 1287. Final Answer: 1

Question 12

Find the unique number between 900 and 1000 that leaves a remainder of 23 when divided by both 38 and 57.

Accepted Answer

Introduction / Context: This problem asks for a number within a range satisfying two simultaneous remainder conditions. It is a straightforward application of the Chinese Remainder Theorem when moduli are not coprime but share a common factor appropriately, or by using the least common multiple (LCM) if the same remainder is required. Given Data / Assumptions: N ≡ 23 (mod 38) N ≡ 23 (mod 57) 900 ≤ N ≤ 1000 Concept / Approach: If the desired remainder is the same for both moduli, then N − 23 must be a multiple of LCM(38, 57). Compute that LCM and then list candidates within the range by adding multiples of the LCM to 23. Step-by-Step Solution: 38 = 2*19, 57 = 3*19 ⇒ LCM(38, 57) = 2*3*19 = 114.General solution: N = 23 + 114t for integer t.Find t with 900 ≤ 23 + 114t ≤ 1000. Try t = 8 ⇒ N = 23 + 912 = 935 (in range). t = 9 ⇒ N = 1049 (exceeds 1000).Therefore the required number is 935. Verification / Alternative check: 935 ÷ 38 leaves 23 (since 38*24 = 912; 935 − 912 = 23). 935 ÷ 57 leaves 23 (since 57*16 = 912; 935 − 912 = 23). Both conditions hold. Why Other Options Are Wrong: 945, 925, 955, and 965 are not of the form 23 + 114t and fail at least one modulus. Common Pitfalls: Using gcd instead of LCM; not ensuring the same remainder approach; or missing the range restriction when enumerating solutions. Final Answer: 935

Question 13

Three positive integers are pairwise coprime. The product of the first two is 437, and the product of the last two is 551. Find the sum of all three integers.

Accepted Answer

Introduction / Context: This problem blends factorization with the concept of pairwise coprimality. By comparing two products that share one variable, we can isolate each integer using greatest common divisors (gcds) and prime factorizations. Given Data / Assumptions: a, b, c are pairwise coprime (any two have gcd 1). a*b = 437 b*c = 551 All are positive integers. Concept / Approach: Since a and c are coprime (pairwise condition), the only common factor between ab and bc is b itself. Therefore gcd(437, 551) equals b. Factor the numbers to compute the gcd and then find a and c by division. Step-by-Step Solution: Prime factorization: 437 = 19*23; 551 = 19*29.gcd(437, 551) = 19 ⇒ b = 19.Then a = 437 / 19 = 23; c = 551 / 19 = 29.Sum = a + b + c = 23 + 19 + 29 = 71. Verification / Alternative check: Check pairwise coprime: gcd(23, 19) = 1, gcd(19, 29) = 1, gcd(23, 29) = 1. Products match the given values. Everything is consistent. Why Other Options Are Wrong: 91, 81, 70, and 73 result from incorrect gcd identification or misfactorization, leading to wrong values for a, b, or c. Common Pitfalls: Overlooking pairwise coprime condition; assuming common factors between a and c; arithmetic slips in factorization or division. Final Answer: 71

Question 14

What least number must be subtracted from 1936 so that the remainder, when divided by 9, 10, and 15, leaves the same remainder 7 in each case?

Accepted Answer

Introduction / Context: We need to reduce 1936 to a number that is congruent to 7 modulo 9, 10, and 15. This again uses the idea that if a number leaves the same remainder with several moduli, then subtracting that remainder yields a multiple of their LCM. Given Data / Assumptions: Original number: 1936 Desired remainder: 7 for each divisor 9, 10, and 15 Find the least nonnegative subtraction x so that R = 1936 − x meets the condition. Concept / Approach: If R ≡ 7 (mod 9, 10, 15), then R − 7 is divisible by LCM(9, 10, 15). Compute that LCM, reduce 1936 − 7 modulo it, and choose the smallest subtraction to hit an exact multiple. Step-by-Step Solution: LCM(9, 10, 15) = 90 (since 9 = 3^2, 10 = 2*5, 15 = 3*5 ⇒ LCM = 2*3^2*5 = 90).Compute 1936 − 7 = 1929.Find x such that 1929 − x is a multiple of 90. 1929 mod 90 = 39.Therefore x = 39 gives 1929 − 39 = 1890 which is 90*21.Hence the least number to subtract is 39. Verification / Alternative check: R = 1936 − 39 = 1897. Then R − 7 = 1890 which is divisible by 9, 10, and 15. Thus R leaves remainder 7 in each division. Why Other Options Are Wrong: 46, 53, and 44 correspond to incorrect remainders modulo 90; 36 misses the exact multiple by 3. Common Pitfalls: Not subtracting the target remainder first; computing LCM improperly; or miscalculating remainders modulo 90. Final Answer: 39

Question 15

Distribution using H.C.F: An inspector of schools wants to distribute 84 balls and 180 bats equally among as many boys as possible, with each boy receiving the same number of balls and the same number of bats. What is the greatest possible number of…

Accepted Answer

Introduction / Context: This is a classic highest common factor (H.C.F) application. To distribute two different items equally among the maximum number of recipients, the number of recipients must divide each item count exactly. Therefore, we seek the largest number that divides both 84 (balls) and 180 (bats). Given Data / Assumptions: Balls = 84 Bats = 180 Each boy must get the same number of balls and the same number of bats. We need the greatest possible number of boys. Concept / Approach: The greatest possible number of boys equals HCF(84, 180). Factorization or Euclidean Algorithm may be used. If HCF = k, then both 84 and 180 are multiples of k, allowing equal distribution without remainder. Step-by-Step Solution: Prime-factorize 84 = 2^2 * 3 * 7Prime-factorize 180 = 2^2 * 3^2 * 5Common prime powers: 2^2 and 3^1 ⇒ HCF = 2^2 * 3 = 12Thus, the maximum number of boys is 12. Verification / Alternative check: Using the Euclidean Algorithm: HCF(180, 84) ⇒ 180 mod 84 = 12; 84 mod 12 = 0 ⇒ HCF = 12. With 12 boys, each gets 84/12 = 7 balls and 180/12 = 15 bats (both integers). Why Other Options Are Wrong: 16, 15, 14, and 18 do not divide both 84 and 180 exactly, so equal distribution would fail for at least one item. Common Pitfalls: Using LCM instead of HCF, or checking divisibility against only one of the item counts. Both counts must be divisible. Final Answer: 12

Question 16

Greatest measuring length (in cm): Find the greatest possible length that can be used to measure exactly 7 m, 3 m 85 cm, and 12 m 95 cm. Give your answer in centimeters.

Accepted Answer

Introduction / Context: The “greatest measuring length” that measures given lengths exactly is an HCF problem. Convert all quantities to a common unit and compute the HCF to ensure exact measurement without remainder. Given Data / Assumptions: Lengths: 7 m, 3 m 85 cm, 12 m 95 cm. 1 m = 100 cm. Answer required in centimeters. Concept / Approach: Convert to centimeters, then find HCF. If L is the measuring unit, each length should be a multiple of L. The largest such L is HCF of the converted values. Step-by-Step Solution: 7 m = 700 cm3 m 85 cm = 385 cm12 m 95 cm = 1295 cmHCF(700, 385) ⇒ 700 − 385 = 315; HCF(385, 315) = 70; thus HCF so far = 70HCF(70, 1295) ⇒ 1295 mod 70 = 35 ⇒ HCF = 35Therefore, the greatest measuring length is 35 cm. Verification / Alternative check: 700/35 = 20, 385/35 = 11, 1295/35 = 37. All are integers, so 35 cm works. Why Other Options Are Wrong: 15, 25, 42, and 21 cm are not the highest common measure; some may divide some lengths but not all, or are smaller than the true HCF. Common Pitfalls: Forgetting to convert meters to centimeters consistently, or stopping at an intermediate common factor (like 70) without checking against all lengths. Final Answer: 35 cm

Question 17

Two numbers with given sum and H.C.F: The sum of two numbers is 216 and their H.C.F is 27. Which pair are the numbers?

Accepted Answer

Introduction / Context: When the HCF is known, express the two numbers as multiples of the HCF. The sum condition then constrains the multipliers. Co-primeness of the multipliers is essential when the extracted common factor is the full HCF. Given Data / Assumptions: a + b = 216 HCF(a, b) = 27 Concept / Approach: If HCF is 27, write a = 27x and b = 27y with HCF(x, y) = 1. Then 27(x + y) = 216 ⇒ x + y = 8. Find co-prime positive integer pairs that sum to 8: (1,7) and (3,5). Both yield valid numbers; we must see which are present among options. Step-by-Step Solution: Possible pairs for (x, y): (1, 7) ⇒ numbers: 27 and 189Another pair: (3, 5) ⇒ numbers: 81 and 135Check options: 27 and 189 is listed; 81 and 135 is not listed (but appears as a distractor in option_e for completeness here). Hence the correct option among the provided choices is 27 and 189. Verification / Alternative check: HCF(27, 189) = 27; sum = 216. Conditions satisfied. Why Other Options Are Wrong: 54 and 162 have HCF 54, not 27; 108 and 108 have HCF 108; “None of these” is unnecessary since a valid option exists. Common Pitfalls: Choosing numbers that sum to 216 but have a larger HCF than required, or forgetting the co-primeness condition on (x, y). Final Answer: 27 and 189

Question 18

Product and G.C.D (interpreted as H.C.F): The product of two two-digit numbers is 2160 and their greatest common divisor is 12. Identify the numbers.

Accepted Answer

Introduction / Context: For any two positive integers a and b, the relation a * b = GCD(a, b) * LCM(a, b) holds. When product and GCD are given, you can confirm candidate pairs by checking both the product and the GCD requirement. Given Data / Assumptions: a * b = 2160 GCD(a, b) = 12 a and b are two-digit numbers. Concept / Approach: Test options: both numbers must be multiples of 12 to have GCD 12 (not strictly necessary, but typical); more importantly, confirm GCD and product exactly. Dismissing pairs that fail the GCD check is fastest. Step-by-Step Solution: Option (36, 60): GCD(36, 60) = 12, product = 2160 ⇒ fits both conditions.Option (72, 30): GCD = 6 ⇒ fails GCD requirement.Option (96, 25): product 2400 ⇒ fails product; also GCD ≠ 12.Therefore, the valid pair is 36 and 60. Verification / Alternative check: Compute LCM using a * b / GCD = 2160 / 12 = 180. Indeed, LCM(36, 60) = 180, confirming consistency. Why Other Options Are Wrong: They fail either the stated GCD or the product condition. “None of these” is incorrect because a valid pair exists. Common Pitfalls: Assuming any factor pair of 2160 works without checking the GCD; or miscomputing the GCD due to incomplete factorization. Final Answer: 36 and 60

Question 19

H.C.F–L.C.M relationship: The H.C.F of two numbers is 16 and their L.C.M is 160. If one number is 32, find the other number.

Accepted Answer

Introduction / Context: For two integers a and b, the fundamental relation is a * b = HCF(a, b) * LCM(a, b). With one number known, you can immediately find the other by rearranging this relation. Given Data / Assumptions: HCF = 16 LCM = 160 One number = 32 Concept / Approach: Compute product a * b = 16 * 160 = 2560. If a = 32, then b = 2560 / 32. Ensure the result is an integer and matches the HCF/LCM structure (i.e., multiples align correctly). Step-by-Step Solution: a * b = HCF * LCM = 16 * 160 = 2560Given a = 32 ⇒ b = 2560 / 32 = 80Hence, the other number is 80. Verification / Alternative check: HCF(32, 80) = 16 and LCM(32, 80) = (32 * 80) / 16 = 160. Both conditions satisfy the given information. Why Other Options Are Wrong: 48, 96, 112, or 64 do not simultaneously satisfy the HCF 16 and LCM 160 conditions with 32. Common Pitfalls: Forgetting the product relation or dividing the wrong way around. Always compute other = (HCF * LCM) / known. Final Answer: 80

Question 20

Find L.C.M from product and H.C.F: The product of two numbers is 1600 and their H.C.F is 5. What is the L.C.M of the two numbers?

Accepted Answer

Introduction / Context: The relation a * b = HCF(a, b) * LCM(a, b) allows quick computation of LCM when product and HCF are given. This is a common test of fundamental number theory identities used in aptitude exams. Given Data / Assumptions: a * b = 1600 HCF(a, b) = 5 Concept / Approach: Rearrange to LCM = (a * b) / HCF. Because both a and b are integers, the division must be exact and yield an integer LCM. Step-by-Step Solution: LCM = (a * b) / HCF = 1600 / 5 = 320Therefore, LCM = 320. Verification / Alternative check: Pick a factor pair of 1600, e.g., 20 and 80, whose HCF is 20 (not 5); but if numbers were 40 and 40, HCF is 40. This shows the specific pair is not needed; the identity already ensures the correct LCM value, independent of the particular pair. Why Other Options Are Wrong: 1605 and 1595 are near the product and included to trap careless reading; 8000 is product * 5, not product / HCF; 400 does not satisfy the identity with HCF 5. Common Pitfalls: Multiplying instead of dividing by HCF; or assuming specific numbers are needed when the identity suffices. Final Answer: 320

Problems on H.C.F and L.C.M Questions