Question 1

Two numbers are coprime and their product is 117. What is their L.C.M. (least common multiple)?

Accepted Answer

Introduction / Context: The relationship between the GCD (HCF) and LCM of two integers a and b is a*b = gcd(a, b) * lcm(a, b). If a and b are coprime, their gcd is 1, which simplifies the relationship greatly. Given Data / Assumptions: a and b are coprime ⇒ gcd(a, b) = 1. Product a*b = 117. Concept / Approach: Using a*b = gcd(a, b) * lcm(a, b), substitute gcd(a, b) = 1 and a*b = 117 to obtain lcm(a, b) directly. No factorization of 117 is required for the conclusion, although 117 = 9*13 shows a possible coprime pair is not necessary here. Step-by-Step Solution: Given gcd(a, b) = 1 and a*b = 117.Then lcm(a, b) = (a*b) / gcd(a, b) = 117 / 1 = 117. Verification / Alternative check: If a = 9 and b = 13 (which are coprime), lcm(9, 13) = 117 indeed, matching the formula. Any coprime pair with product 117 yields the same LCM value 117. Why Other Options Are Wrong: 9, 13, 39, 81: None equals the product 117 for coprime numbers; only 117 satisfies the identity for all coprime factorizations of 117. Common Pitfalls: Assuming the LCM is one of the factors instead of using the product–gcd–lcm identity. Final Answer: 117

Question 2

The L.C.M. of two numbers is 2376 and their H.C.F. is 33. If one number is 297, find the other number.

Accepted Answer

Introduction / Context: The fundamental identity for two positive integers a and b is a*b = gcd(a, b) * lcm(a, b). Knowing the LCM, HCF, and one number allows us to compute the other directly by rearrangement. Given Data / Assumptions: LCM = 2376. HCF = 33. One number a = 297. Find the other number b. Concept / Approach: Use b = (HCF * LCM) / a. Ensure a divides the product exactly, since the result must be an integer. Perform simplification before multiplication if convenient. Step-by-Step Solution: Compute b = (33 * 2376) / 297.Observe 2376 / 297 = 8 (since 297 * 8 = 2376).Therefore b = 33 * 8 = 264. Verification / Alternative check: Check gcd(297, 264) = 33 and lcm(297, 264) = (297*264)/33 = 2376. Both match the given data, confirming correctness. Why Other Options Are Wrong: 216, 642, 792, 198: These do not satisfy both the specified HCF and LCM simultaneously with 297 under the product identity. Common Pitfalls: Miscalculating the quotient 2376/297. Forgetting to divide by the known number after multiplying HCF and LCM. Final Answer: 264

Question 3

The H.C.F. and L.C.M. of two numbers are 13 and 1989, respectively. If one number is 117, determine the other number.

Accepted Answer

Introduction / Context: We use the product identity for two integers a and b: a*b = gcd(a, b) * lcm(a, b). Given gcd and lcm along with one number, the other number follows immediately from algebraic manipulation. Given Data / Assumptions: gcd = 13. lcm = 1989. One number a = 117. Find b. Concept / Approach: Compute b = (gcd * lcm) / a. Simplify before multiplying if helpful, to avoid large intermediate values and reduce arithmetic errors. Step-by-Step Solution: b = (13 * 1989) / 117.Note 117 = 9 * 13; thus (13 * 1989) / 117 = 1989 / 9.Compute 1989 / 9 = 221.Hence the other number is 221. Verification / Alternative check: Check gcd(117, 221) = 13 and lcm(117, 221) = (117*221)/13 = 1989. Both conditions are satisfied, verifying the result. Why Other Options Are Wrong: 121, 131, 231, 143: These candidates do not simultaneously satisfy the given gcd and lcm with 117 under the product identity. Common Pitfalls: Cancelling incorrectly or multiplying before simplifying, which can lead to arithmetic slips. Final Answer: 221

Question 4

Three numbers are in the ratio 3 : 4 : 5 and their L.C.M. is 1200. Determine the H.C.F. (greatest common divisor) of the three numbers.

Accepted Answer

Introduction / Context: When numbers are in a given ratio, they can be represented as multiples of a common factor (their HCF). The LCM of such numbers relates directly to this common factor and the LCM of the ratio parts. We are to find the HCF given the ratio and the LCM of the actual numbers. Given Data / Assumptions: Numbers are 3h, 4h, and 5h for some positive integer h (the HCF). LCM of the actual numbers is 1200. Concept / Approach: LCM(3h, 4h, 5h) = h * LCM(3, 4, 5) because h is a common factor across all. Since 3, 4, and 5 are pairwise coprime, LCM(3, 4, 5) = 60. Set h * 60 = 1200 to solve for h, which is the HCF of the actual numbers. Step-by-Step Solution: Compute LCM of ratio parts: LCM(3, 4, 5) = 60.Set h * 60 = 1200 ⇒ h = 1200 / 60 = 20.Therefore, HCF of the numbers is h = 20. Verification / Alternative check: The numbers would be 60, 80, and 100 (multiplying 3, 4, 5 by h = 20). Their LCM is indeed 1200 and their HCF is 20, confirming all conditions. Why Other Options Are Wrong: 40, 30, 80, 60: These do not satisfy h * 60 = 1200, or they imply numbers whose LCM would differ from 1200. Common Pitfalls: Confusing HCF with LCM of the ratio parts; remember HCF is the common multiplier h. Final Answer: 20

Question 5

Four numbers are proportional to 10 : 12 : 15 : 18. Given that their highest common factor (HCF) is 3, determine the least common multiple (LCM) of the actual four numbers.

Accepted Answer

Introduction / Context: This problem tests the relationship between a set of numbers given in a ratio and their actual values when a highest common factor (HCF) is specified. Once the actual numbers are reconstructed from the ratio using the given HCF, the least common multiple (LCM) can be computed by prime factorization or by reasoning about maximum prime powers. Given Data / Assumptions: Ratio: 10 : 12 : 15 : 18. HCF of the four actual numbers is 3. All quantities are positive integers. Concept / Approach: If numbers are in the ratio k1 : k2 : k3 : k4 and the HCF of the actual numbers is h, then the actual numbers can be taken as h * k1, h * k2, h * k3, h * k4 provided that HCF(k1, k2, k3, k4) = 1. Then compute the LCM of these four actual integers using the highest power of each prime that appears across them. Step-by-Step Solution: Base ratio parts: 10, 12, 15, 18. Their overall HCF is 1, so actual numbers = 3*10, 3*12, 3*15, 3*18 = 30, 36, 45, 54.Prime factors: 30 = 2*3*5; 36 = 2^2*3^2; 45 = 3^2*5; 54 = 2*3^3.LCM takes maximum powers: 2^2 * 3^3 * 5 = 4 * 27 * 5 = 540. Verification / Alternative check: Confirm each divides 540: 540/30=18, 540/36=15, 540/45=12, 540/54=10; all integers, so 540 is a common multiple. No smaller candidate has the required prime powers. Why Other Options Are Wrong: 420, 620, 680 miss at least one required prime power (e.g., 3^3) and will not be divisible by all four numbers simultaneously. Common Pitfalls: Using the LCM of the ratio parts and forgetting to multiply by the HCF. Here, because the HCF is incorporated in the actual numbers already, the prime-power method is safest. Final Answer: 540

Question 6

Let the highest common factor (HCF) of two positive integers a and b be 12, with a > b > 12. Which ordered pair (a, b) satisfies these conditions?

Accepted Answer

Introduction / Context: This item checks understanding of the definition of HCF and the implications for the possible forms of a and b. If HCF(a, b) = 12, both a and b must be multiples of 12, and no larger common factor may divide both simultaneously. Given Data / Assumptions: a and b are positive integers. HCF(a, b) = 12. a > b > 12. Concept / Approach: Write a = 12x and b = 12y with HCF(x, y) = 1. The inequality a > b > 12 implies x > y ≥ 2 (since b > 12 must be at least 24). Check each option against these conditions and verify the HCF directly if needed. Step-by-Step Solution: Option D: a = 36, b = 24 → both multiples of 12.Compute HCF(36, 24): 36 = 12*3, 24 = 12*2, HCF = 12.The inequality holds: 36 > 24 > 12. Verification / Alternative check: Other options fail one or more conditions: (12, 24) or (24, 12) violate a > b > 12; (24, 36) reverses order and b is not > 12 for the larger-first order; hence only (36, 24) works. Why Other Options Are Wrong: (12, 24) and (24, 12) do not satisfy a > b > 12 as required. (24, 36) does not have a > b; it is reversed and also b = 36 is not less than a. Common Pitfalls: Forgetting that HCF imposes co-primeness on the scaled-down pair (x, y) and missing the strict order a > b > 12. Final Answer: 36, 24

Question 7

Two numbers have HCF + LCM = 403 and also LCM = 12 times HCF. If one number is 93, find the other number.

Accepted Answer

Introduction / Context: This problem blends two key identities: (i) for two integers m and n, HCF * LCM = m * n, and (ii) here an extra relation LCM = 12 * HCF plus their sum. Using these, we can compute HCF and LCM first, then use the product identity to find the unknown integer directly. Given Data / Assumptions: H + L = 403. L = 12H. Known number = 93. Concept / Approach: From L = 12H and H + L = 403, solve for H, L. Then use m * n = H * L to find the partner of 93: other = (H * L) / 93. Ensure the result is an integer to be consistent with integer inputs. Step-by-Step Solution: H + 12H = 403 ⇒ 13H = 403 ⇒ H = 31.Thus L = 12 * 31 = 372.Product m * n = H * L = 31 * 372 = 11532.Given one number is 93, the other = 11532 / 93 = 124. Verification / Alternative check: Check: HCF(93,124) = 31 and LCM = (93*124)/31 = 372. Sum 31 + 372 = 403 and 372 = 12 * 31, all satisfied. Why Other Options Are Wrong: 115, 122, 138 do not yield HCF 31 and LCM 372 with 93 under the product relation. Common Pitfalls: Mistakes in dividing 11532 by 93 or in solving 13H = 403. Final Answer: 124

Question 8

Two numbers have LCM equal to 20 times their HCF, and (LCM + HCF) = 2520. If one number is 480, what is three times (the triple of) the other number?

Accepted Answer

Introduction / Context: This is a structured HCF–LCM question with an additional sum constraint. Once HCF is determined, the product HCF * LCM gives the product of the two numbers, enabling computation of the unknown. Finally, the question asks for the triple of the other number, a simple scaling step at the end. Given Data / Assumptions: LCM = 20 * HCF. LCM + HCF = 2520. One number = 480. Find 3 * (the other number). Concept / Approach: Let HCF = h, then LCM = 20h and h + 20h = 2520 ⇒ 21h = 2520 ⇒ h = 120. Product of the numbers mn = h * LCM = 120 * 2400 = 288000. Hence the other number is 288000 / 480. Multiply the result by 3 to produce the required triple. Step-by-Step Solution: h = 2520 / 21 = 120.LCM = 20h = 2400.mn = 120 * 2400 = 288000.Other number = 288000 / 480 = 600.Triple of the other number = 3 * 600 = 1800. Verification / Alternative check: Check HCF(480, 600) = 120 and LCM = (480*600)/120 = 2400. Sum = 120 + 2400 = 2520 and LCM = 20 * HCF, all consistent. Why Other Options Are Wrong: 1200, 1500, 2100 do not match the computed triple 1800 obtained from the identities and arithmetic. Common Pitfalls: Miscalculating the product mn or the final multiplication by 3. Final Answer: 1800

Question 9

Greatest divisor with given remainders: Find the largest integer that divides 130, 305, and 245 leaving remainders 6, 9, and 17 respectively.

Accepted Answer

Introduction / Context: When a divisor leaves different remainders with different numbers, subtracting the respective remainders from those numbers produces a set of integers all divisible by the same unknown divisor. The problem then reduces to computing the HCF of these adjusted numbers. Given Data / Assumptions: N divides 130 with remainder 6 ⇒ N | (130 − 6) = 124. N divides 305 with remainder 9 ⇒ N | (305 − 9) = 296. N divides 245 with remainder 17 ⇒ N | (245 − 17) = 228. Concept / Approach: The required divisor N is the HCF of 124, 296, and 228. Apply the Euclidean algorithm sequentially to find the HCF efficiently. Step-by-Step Solution: HCF(124, 296): 296 − 2*124 = 48; HCF(124,48).HCF(124,48): 124 mod 48 = 28; 48 mod 28 = 20; 28 mod 20 = 8; 20 mod 8 = 4; 8 mod 4 = 0 ⇒ HCF = 4.Now HCF(4, 228): since 228 mod 4 = 0, overall HCF remains 4. Verification / Alternative check: Check directly: 130 ≡ 6 (mod 4); 305 ≡ 1 (mod 4) but given remainder 9 reduces to 1 mod 4; 245 ≡ 1 (mod 4) while remainder 17 reduces to 1 mod 4. Consistent. Why Other Options Are Wrong: 5, 14, 24 are not common divisors of 124, 296, and 228; they fail at least one divisibility check. Common Pitfalls: Mistaking the LCM approach; here we must use HCF after subtracting remainders. Final Answer: 4

Question 10

Find the smallest positive number N such that: • N leaves remainder 4 when divided by each of 16, 18, and 20; and • N is exactly divisible by 7.

Accepted Answer

Introduction / Context: This is a linear congruence with a divisibility constraint. A number that leaves the same remainder r with several moduli is congruent to r modulo the LCM of those moduli. Then, impose the divisibility requirement to select the least positive solution. Given Data / Assumptions: N ≡ 4 (mod 16), N ≡ 4 (mod 18), N ≡ 4 (mod 20). N is divisible by 7. Concept / Approach: First, compute L = LCM(16, 18, 20). All solutions to the congruences are N = L*k + 4 for integer k ≥ 0. Then choose the smallest k for which N is a multiple of 7. Use modular arithmetic to solve (L*k + 4) ≡ 0 (mod 7). Step-by-Step Solution: Prime factors: 16 = 2^4; 18 = 2*3^2; 20 = 2^2*5.LCM L = 2^4 * 3^2 * 5 = 16 * 9 * 5 = 720.General solution: N = 720k + 4.Impose N ≡ 0 (mod 7): 720k + 4 ≡ 0 (mod 7). Since 720 ≡ 6 (mod 7), we need 6k + 4 ≡ 0 ⇒ 6k ≡ 3 (mod 7).Multiply both sides by 6 (inverse of 6 mod 7): k ≡ 18 ≡ 4 (mod 7).Least nonnegative k = 4 ⇒ N = 720*4 + 4 = 2884. Verification / Alternative check: Check: 2884 ÷ 16 leaves 4, ÷ 18 leaves 4, ÷ 20 leaves 4; and 2884 ÷ 7 = 412 exactly. Why Other Options Are Wrong: 2256, 865, 3332 fail either the remainder condition or the divisibility-by-7 condition (or both). Common Pitfalls: Losing track of the LCM or using some but not all moduli in forming L. Final Answer: 2884

Question 11

Three runners complete one lap of a circular track in 2 h, 4 h, and 5.5 h respectively. After how many hours will they all be together again at the starting point?

Accepted Answer

Introduction / Context: When multiple periodic events start together and repeat at different periods, the first time they coincide again is given by the least common multiple (LCM) of their periods. Here, periods are times to complete one lap. Because one time is fractional in hours, converting to minutes simplifies the LCM computation. Given Data / Assumptions: Times per lap: 2 h, 4 h, 5.5 h. All start together at time 0. Concept / Approach: Convert hours to minutes to avoid decimals, find the LCM in minutes, then convert back to hours. The LCM ensures each runner has completed an integer number of laps at the meeting instant. Step-by-Step Solution: Convert: 2 h = 120 min; 4 h = 240 min; 5.5 h = 330 min.Prime factors: 120 = 2^3*3*5; 240 = 2^4*3*5; 330 = 2*3*5*11.LCM = 2^4 * 3 * 5 * 11 = 16 * 165 = 2640 minutes.Convert to hours: 2640 / 60 = 44 h. Verification / Alternative check: Check multiples: 44/2 = 22 laps, 44/4 = 11 laps, 44/5.5 = 8 laps; all integers, so they meet together at 44 h. Why Other Options Are Wrong: 40 h, 20 h, 22 h are not common multiples of all three periods in hours; some runner would not be at the start at those times. Common Pitfalls: Taking the greatest common divisor instead of the least common multiple, or mishandling the 5.5 h decimal. Final Answer: 44 h

Question 12

Find two integers whose LCM is 1188 and HCF is 9. Choose the correct pair from the options.

Accepted Answer

Introduction / Context: For two integers m and n, the relation m * n = HCF(m, n) * LCM(m, n) is fundamental. Any candidate pair must satisfy both the product identity and the given HCF. This allows quick validation of each option without exhaustive search. Given Data / Assumptions: LCM = 1188. HCF = 9. We must select a valid pair (order does not matter). Concept / Approach: Check the candidate pair’s HCF and ensure product / HCF equals the stated LCM. If both conditions are satisfied, the pair is valid. If at least one valid pair exists, “Data inadequate” is inappropriate. Step-by-Step Solution: Option A: (27, 396): HCF(27, 396) = 9 (since 396 mod 27 = 18, 27 mod 18 = 9, 18 mod 9 = 0).LCM = (27 * 396) / 9 = 10692 / 9 = 1188. This matches the requirement.Option C: (36, 99): HCF = 9 but LCM = (36*99)/9 = 396, not 1188; invalid. Verification / Alternative check: The identity HCF * LCM = product holds perfectly for (27, 396): 9 * 1188 = 10692 = 27 * 396. Why Other Options Are Wrong: (9, 27) has HCF 9 but LCM 27, not 1188. (36, 99) produces the wrong LCM. “Data inadequate” is wrong because a valid pair exists and is identified. Common Pitfalls: Assuming uniqueness; multiple pairs can exist, but you only need one correct option. Final Answer: 27, 396

Question 13

A school has 391 boys and 323 girls. They are to be divided into the largest possible number of identical classes so that each class has the same number of boys and the same number of girls. How many classes are formed, and how many boys and girls a…

Accepted Answer

Introduction / Context: This item focuses on using the highest common factor (HCF) to form the maximum number of identical classes from two groups. The number of classes must divide each of the two populations so that every class has the same count of boys and the same count of girls. Given Data / Assumptions: Boys = 391. Girls = 323. All classes must be identical with equal composition. Concept / Approach: If the number of classes is k, then k must divide both 391 and 323. The largest possible number of classes is thus k = HCF(391, 323). Then the number of boys per class is 391 / k and the number of girls per class is 323 / k. Step-by-Step Solution: Compute HCF(391, 323): 391 − 323 = 68.323 mod 68 = 323 − 4*68 = 323 − 272 = 51.68 mod 51 = 17; 51 mod 17 = 0 ⇒ HCF = 17.Number of classes = 17.Boys per class = 391 / 17 = 23; Girls per class = 323 / 17 = 19. Verification / Alternative check: Total check: 17 classes * (23 boys) = 391 boys; 17 * (19 girls) = 323 girls; composition identical for each class. Why Other Options Are Wrong: Only option B correctly states both the number of classes (17) and the per-class composition (23 boys, 19 girls). The others swap labels or give incorrect class counts. Common Pitfalls: Mistaking “23 classes” for “23 per class”, or mixing the counts of boys and girls per class. Final Answer: 17 classes; 23 boys and 19 girls in each

Question 14

Two numbers have product 7168 and highest common factor (HCF) 16. Considering all possible such pairs, find the sum of all the distinct numbers involved.

Accepted Answer

Introduction / Context: When product and HCF are known, we can parameterize the two numbers as multiples of the HCF with co-prime multipliers. This approach lets us enumerate all valid unordered pairs and then compute the requested sum over distinct values appearing in those pairs. Given Data / Assumptions: a * b = 7168. HCF(a, b) = 16. a, b are positive integers. Concept / Approach: Let a = 16x and b = 16y with HCF(x, y) = 1. Then 256xy = 7168 ⇒ xy = 7168 / 256. Factor the right-hand side, and list the coprime factor pairs (x, y). Each gives a valid unordered pair (a, b). Finally, add all distinct numbers obtained from these pairs. Step-by-Step Solution: Compute xy: 7168 / 256 = 28.Coprime factor pairs of 28: (1, 28) and (4, 7). (2, 14) is not coprime.Corresponding (a, b) pairs: (16*1, 16*28) = (16, 448) and (16*4, 16*7) = (64, 112). Orders reversed are the same unordered pairs.Distinct numbers appearing: 16, 448, 64, 112. Sum = 16 + 448 + 64 + 112 = 640. Verification / Alternative check: Check HCFs: HCF(16, 448) = 16; HCF(64, 112) = 16. Products: 16*448 = 7168 and 64*112 = 7168, both correct. Why Other Options Are Wrong: 860, 460 do not match the computed sum of distinct elements across all valid pairs. “Data inadequate” is incorrect because the parameterization produces a finite, explicit set of pairs. Common Pitfalls: Including non-coprime pairs like (2, 14) which would make the HCF greater than 16. Final Answer: 640

Question 15

Two positive integers have a sum of 1215 and their highest common factor (HCF) is 81. How many unordered pairs of such numbers exist, and what are all the pairs?

Accepted Answer

Introduction / Context: This problem uses the relationship between a pair of integers, their highest common factor (HCF), and their sum. By factoring out the HCF, we reduce the question to counting coprime pairs that add to a fixed total. We then lift those back to the original numbers. Given Data / Assumptions: Sum of the numbers = 1215 HCF (greatest common divisor) = 81 Numbers are positive integers We count unordered pairs (order does not matter) Concept / Approach: Let the numbers be 81a and 81b where gcd(a, b) = 1. Then 81a + 81b = 1215 ⇒ a + b = 1215 / 81. Once we find a + b, we list all positive coprime pairs (a, b) with that sum and convert to the original numbers (81a, 81b). Each unordered coprime pair yields one valid pair of original numbers. Step-by-Step Solution: 1215 / 81 = 15 ⇒ a + b = 15, with gcd(a, b) = 1.List positive unordered pairs summing to 15: (1,14), (2,13), (3,12), (4,11), (5,10), (6,9), (7,8).Keep only coprime pairs: (1,14), (2,13), (4,11), (7,8). (Pairs with 3,12 and 6,9 and 5,10 are not coprime.)Multiply each by 81: (81,1134), (162,1053), (324,891), (567,648).Hence, there are 4 unordered pairs. Verification / Alternative check: All listed pairs sum to 1215 and each pair has HCF 81, since gcd(81a, 81b) = 81*gcd(a, b) = 81 when gcd(a, b) = 1. Why Other Options Are Wrong: Answers 1, 2, and 3 underestimate the number of coprime decompositions of 15; 5 overestimates by including at least one non-coprime pair. Common Pitfalls: Forgetting to enforce gcd(a, b) = 1; double-counting ordered pairs as two; or missing one of the valid coprime pairs that sum to 15. Final Answer: 4 (pairs: (81,1134), (162,1053), (324,891), (567,648))

Question 16

There are 21 mango trees, 42 apple trees, and 56 orange trees to be planted in rows. Each row must contain the same number of trees and only one variety per row. What is the minimum total number of rows required?

Accepted Answer

Introduction / Context: This is a greatest common divisor (GCD) application. To minimize the number of rows, we maximize how many trees go in each row, which must be the same for all varieties. That maximum per-row count is the GCD of the three totals. Given Data / Assumptions: Mango trees = 21 Apple trees = 42 Orange trees = 56 Each row: same number of trees; single variety only Concept / Approach: If x is the number of trees per row, then x must divide 21, 42, and 56. The largest such x is gcd(21, 42, 56). The minimum number of rows then equals 21/x + 42/x + 56/x. Step-by-Step Solution: gcd(21, 42) = 21; gcd(21, 56) = 7 ⇒ gcd(21, 42, 56) = 7.Trees per row = 7.Rows for each variety: 21/7 = 3 (mango), 42/7 = 6 (apple), 56/7 = 8 (orange).Total minimum rows = 3 + 6 + 8 = 17. Verification / Alternative check: Any larger per-row count would not divide at least one total; any smaller per-row count increases rows, so 17 is minimal. Why Other Options Are Wrong: 15, 18, and 20 correspond to using a non-maximal divisor; 3 counts only one variety’s rows, not the total. Common Pitfalls: Using LCM instead of GCD; mixing varieties in a single row; or summing counts incorrectly. Final Answer: 17

Question 17

Five bells start together and then ring at intervals of 6 s, 7 s, 8 s, 9 s, and 12 s respectively. In one hour, how many additional times (excluding the initial start) will they ring together?

Accepted Answer

Introduction / Context: When events repeat at fixed intervals, the times they coincide are multiples of the least common multiple (LCM) of their intervals. Here we find the LCM of the five bell intervals and count how many multiples lie within one hour, excluding the initial time zero. Given Data / Assumptions: Intervals: 6 s, 7 s, 8 s, 9 s, 12 s Total duration: 1 hour = 3600 s Exclude the start at t = 0 Concept / Approach: They ring together every LCM(6, 7, 8, 9, 12) seconds. The count of together-rings after the start equals floor(3600 / LCM). Step-by-Step Solution: Prime factors: 6 = 2*3; 7 = 7; 8 = 2^3; 9 = 3^2; 12 = 2^2*3.LCM = 2^3 * 3^2 * 7 = 8 * 9 * 7 = 504 seconds.Number of coincidences after start = floor(3600 / 504) = 7. Verification / Alternative check: They meet at t = 504, 1008, 1512, 2016, 2520, 3024, and 3528 seconds (seven times) within 3600 s. Why Other Options Are Wrong: 3, 5, 6, 9 result from incorrect LCM or duration division, or from including/excluding the start/end inconsistently. Common Pitfalls: Using gcd instead of LCM; adding one for the start when the question says to exclude it; rounding errors in the division. Final Answer: 7

Question 18

Three containers hold mixtures measured as 403 kg, 434 kg, and 465 kg. Find the greatest measuring capacity (in kg) that can exactly measure each quantity.

Accepted Answer

Introduction / Context: This is a highest common factor (HCF) or greatest common divisor (GCD) question. The largest measure that fits exactly into each total is the GCD of the three weights. Given Data / Assumptions: Quantities: 403 kg, 434 kg, 465 kg We seek the greatest capacity dividing all three exactly Concept / Approach: Compute the GCD via prime factorization. The common prime factors with the smallest exponents across all numbers define the GCD. Step-by-Step Solution: 403 = 13 * 31434 = 2 * 217 = 2 * 7 * 31465 = 5 * 93 = 5 * 3 * 31Common factor across all three = 31 ⇒ GCD = 31 kg. Verification / Alternative check: 403/31 = 13, 434/31 = 14, 465/31 = 15, all integers; hence 31 kg measures each exactly. Why Other Options Are Wrong: 1 kg is trivial but not the greatest. 7 kg and 13 kg do not divide all three. 41 kg divides none of them exactly. Common Pitfalls: Arithmetic mistakes in factorization; stopping at a smaller common factor and missing the larger prime 31. Final Answer: 31 kg

Question 19

Timber lengths 42 m, 49 m, and 63 m are to be cut into planks of equal length with no waste. What is the greatest possible length of each plank?

Accepted Answer

Introduction / Context: To maximize the uniform plank length that exactly divides all given lengths, we take the greatest common divisor (GCD) of the three lengths. This ensures no waste and equal-size planks. Given Data / Assumptions: Lengths: 42 m, 49 m, 63 m Planks must have equal maximum possible length Concept / Approach: The largest length that divides all three numbers evenly is gcd(42, 49, 63). Use prime factorization or Euclidean algorithm to compute it. Step-by-Step Solution: gcd(42, 49) = 7 (since 42 = 2 * 3 * 7, 49 = 7^2)gcd(7, 63) = 7 (since 63 = 3^2 * 7)Therefore, the greatest plank length = 7 m. Verification / Alternative check: 42/7 = 6, 49/7 = 7, 63/7 = 9; all integral. Why Other Options Are Wrong: 14 m, 21 m, 42 m, and 63 m do not divide all three lengths without remainder. Common Pitfalls: Mistaking LCM for GCD or not checking divisibility across all lengths. Final Answer: 7 m

Question 20

Six bells start together and ring at intervals of 2 s, 4 s, 6 s, 8 s, 10 s, and 12 s. In 30 minutes, how many times will they ring together (including the start)?

Accepted Answer

Introduction / Context: When multiple periodic events begin together, they will next coincide at multiples of the least common multiple (LCM) of their intervals. Counting coincidences over a duration includes the initial coincidence at t = 0 if asked to include the start. Given Data / Assumptions: Intervals: 2 s, 4 s, 6 s, 8 s, 10 s, 12 s Duration: 30 minutes = 1800 s Include the initial start together Concept / Approach: Find LCM of the intervals to get the repeat time for simultaneous ringing. Then the count of coincidences including the start is floor(1800 / LCM) + 1. Step-by-Step Solution: Prime factors: 2, 4 = 2^2, 6 = 2*3, 8 = 2^3, 10 = 2*5, 12 = 2^2*3.LCM = 2^3 * 3 * 5 = 8 * 3 * 5 = 120 s.Number of coincidences = floor(1800 / 120) + 1 = 15 + 1 = 16. Verification / Alternative check: They coincide at t = 0, 120, 240, …, 1800. That’s 16 equally spaced times including t = 0 and t = 1800. Why Other Options Are Wrong: 4, 10, 14, 15 arise from excluding the start or miscomputing the LCM or number of intervals within 1800 s. Common Pitfalls: Forgetting to include the start as instructed; using gcd instead of LCM; arithmetic slips in 1800/120. Final Answer: 16

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