Question 1

Find the greatest possible length that can be used as a measuring rod to measure each of the following lengths exactly (with no remainder):  • 7 m • 3 m 85 cm • 12 m 95 cm  What is the greatest such length?

Accepted Answer

Introduction / Context: This question is an application of HCF to measurement. The largest length that can measure several distances exactly is the HCF of those distances, provided they are all expressed in the same units. We convert meters and centimeters into only centimeters, then find the gcd. Given Data / Assumptions: First length: 7 m = 700 cm Second length: 3 m 85 cm = 385 cm Third length: 12 m 95 cm = 1295 cm We need the greatest length that divides all three lengths exactly Concept / Approach: If the measuring rod has length L (in centimeters), then L must be a common divisor of 700, 385, and 1295. The largest such L is the HCF of these three numbers: L = HCF(700, 385, 1295) Step-by-Step Solution: Step 1: Compute gcd(700, 385).700 mod 385 = 315.385 mod 315 = 70.315 mod 70 = 35.70 mod 35 = 0, so gcd(700, 385) = 35.Step 2: Compute gcd(35, 1295).1295 ÷ 35 = 37 exactly, so 1295 mod 35 = 0.Therefore gcd(35, 1295) = 35.Step 3: Hence HCF(700, 385, 1295) = 35.So the greatest possible length of the measuring rod is 35 cm. Verification / Alternative check: Check divisibility: 700 ÷ 35 = 20, 385 ÷ 35 = 11, and 1295 ÷ 35 = 37. All are integers, so 35 cm measures each length exactly. Any longer rod length, such as 42 cm or 70 cm, would fail to divide one of the lengths perfectly. Why Other Options Are Wrong: 15 cm: Does not divide 385 cm or 1295 cm exactly.25 cm: 385 cm is not divisible by 25.42 cm: 385 cm and 1295 cm are not divisible by 42.70 cm: 385 cm is not divisible by 70. Common Pitfalls: Not converting all measurements to the same unit (centimeters) before finding the HCF.Using LCM instead of HCF, which would give a larger and incorrect length.Stopping after the first gcd computation without including the third number. Final Answer: 35 cm

Question 2

If the sum of two numbers is 55, and the HCF (GCD) and LCM of these numbers are 5 and 120 respectively, then find the sum of the reciprocals of the two numbers.  (Express your answer as a simplified fraction.)

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Introduction / Context: This question links the properties of HCF and LCM with reciprocals. Once we determine the two numbers from their sum, HCF, and LCM, we can easily compute the sum of their reciprocals. The key is to use the relationship between HCF, LCM, and the product of the numbers, and to factor out the HCF to obtain a simpler pair of coprime numbers. Given Data / Assumptions: Sum of the two numbers: x + y = 55 HCF(x, y) = 5 LCM(x, y) = 120 We need (1/x) + (1/y) Concept / Approach: Let x = 5a and y = 5b because 5 is the HCF. Then gcd(a, b) = 1. The LCM of x and y is 5ab, and we are told: 5ab = 120 ⇒ ab = 24 Also the sum gives: x + y = 5a + 5b = 55 ⇒ a + b = 11 We need to find coprime integers a and b such that a + b = 11 and ab = 24. Then we can reconstruct x and y and find the sum of reciprocals. Step-by-Step Solution: Step 1: From LCM relation: ab = 120 / 5 = 24.Step 2: From sum relation: a + b = 55 / 5 = 11.Step 3: Find integer pairs (a, b) with product 24 and sum 11.Possible factor pairs of 24 are (1, 24), (2, 12), (3, 8), (4, 6).Among these, 3 + 8 = 11 and gcd(3, 8) = 1, so (a, b) = (3, 8) works.Step 4: Therefore, x = 5a = 15 and y = 5b = 40.Step 5: Sum of reciprocals:1/x + 1/y = 1/15 + 1/40 = (40 + 15) / 600 = 55 / 600 = 11 / 120. Verification / Alternative check: Check that the numbers 15 and 40 match the original conditions: HCF(15, 40) = 5, LCM(15, 40) = (15 * 40) / 5 = 120, and sum = 15 + 40 = 55. The reciprocal sum 1/15 + 1/40 simplifies to 11/120, confirming the answer. Why Other Options Are Wrong: 13/125, 7/60, 14/57, 1/5: These fractions do not equal 1/15 + 1/40, and each would correspond to different number pairs that do not satisfy the given HCF, LCM, and sum conditions. Common Pitfalls: Attempting to guess x and y directly without using the structured approach with a and b.Forgetting to divide the sum and product by the HCF to move into the coprime pair (a, b).Not simplifying the final fraction when computing the sum of reciprocals. Final Answer: 11/120

Question 3

A drink vendor has 80 liters of Mazza, 144 liters of Pepsi, and 368 liters of Sprite. He wants to pack them in cans such that:  • Each can has the same capacity (in liters), • No can contains a mix of two different drinks, and • All the drink is pac…

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Introduction / Context: This question is very similar in structure to another HCF-based packing problem. To minimize the number of cans, the capacity of each can should be the largest possible value that divides each of the three drink quantities exactly. Once that capacity is known, we count how many cans are needed for each drink and add them together. Given Data / Assumptions: Mazza = 80 liters Pepsi = 144 liters Sprite = 368 liters All cans have equal capacity No mixing of drinks in the same can, and no leftovers Concept / Approach: If the can capacity is x liters, then x must divide 80, 144, and 368. The maximum such x is the HCF of the three volumes: x = HCF(80, 144, 368) The least number of cans is then computed by dividing each volume by x and summing the results. Step-by-Step Solution: Step 1: Compute gcd(80, 144).144 mod 80 = 64.80 mod 64 = 16.64 mod 16 = 0, so gcd(80, 144) = 16.Step 2: Compute gcd(16, 368).368 ÷ 16 = 23 exactly, so 368 mod 16 = 0.Therefore HCF(80, 144, 368) = 16, so can capacity x = 16 liters.Step 3: Number of cans of each drink:Mazza: 80 / 16 = 5 cans.Pepsi: 144 / 16 = 9 cans.Sprite: 368 / 16 = 23 cans.Step 4: Total cans = 5 + 9 + 23 = 37. Verification / Alternative check: Using a smaller can size that also divides all three volumes (for example, 8 liters) yields more cans: 80/8 + 144/8 + 368/8 = 10 + 18 + 46 = 74, which is larger than 37. Since 16 is the greatest common divisor, it gives the minimum possible number of cans. Why Other Options Are Wrong: 35, 36, 38, 46: These totals do not correspond to the distribution obtained when using the maximum possible can capacity of 16 liters. Any different total implies a smaller can size or an invalid capacity that does not divide all three volumes exactly. Common Pitfalls: Using LCM instead of HCF, which is not appropriate for this type of packing problem.Choosing a can size that divides some but not all of the drink volumes.Forgetting to add the cans needed for each drink after determining the capacity. Final Answer: 37

Question 4

The HCF (GCD) and LCM of two numbers are given as 21 and 84 respectively, and the ratio of the two numbers is 1 : 4.  (Note: Since HCF cannot be greater than LCM, interpret the data as HCF = 21 and LCM = 84.)  What is the larger of the two numbers?

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Introduction / Context: This question involves HCF, LCM, and the ratio of two numbers. A key fact is that for two positive integers, the HCF cannot exceed the LCM. The statement in the question lists HCF and LCM in reverse order, but the only meaningful interpretation is that HCF = 21 and LCM = 84. Using the ratio 1 : 4, we can reconstruct the two numbers and then identify the larger one. Given Data / Assumptions: Interpreted HCF = 21 Interpreted LCM = 84 Ratio of numbers = 1 : 4 Numbers are positive integers Concept / Approach: If the ratio is 1 : 4, we can let the two numbers be x and 4x. The HCF of x and 4x is x, because x divides both and there is no larger common divisor than x itself. Using the given HCF, we set x = 21. Then the numbers become 21 and 84, and we can confirm the LCM from this pair. Step-by-Step Solution: Step 1: Let the two numbers be x and 4x based on the ratio 1 : 4.Step 2: HCF(x, 4x) = x.Step 3: The HCF is given as 21, so x = 21.Step 4: Therefore, the numbers are 21 and 84.Step 5: The larger number is 84. Verification / Alternative check: Check the LCM of 21 and 84: since 84 is a multiple of 21, LCM(21, 84) = 84, which matches the interpreted LCM. The ratio 21 : 84 simplifies to 1 : 4, also matching the given ratio. This confirms that the interpretation and resulting numbers are consistent. Why Other Options Are Wrong: 48: If the larger number were 48 with ratio 1 : 4, the smaller number would be 12. HCF(12, 48) = 12, not 21.108: With ratio 1 : 4, the smaller number would be 27. HCF(27, 108) = 27, not 21.12: Too small to be the larger number in a ratio of 1 : 4 and does not match the HCF condition.96: This would imply the smaller number is 24, and HCF(24, 96) = 24, not 21. Common Pitfalls: Not recognizing that HCF cannot be greater than LCM and accepting the original order without correction.Failing to apply the ratio correctly when constructing the pair of numbers.Overlooking the fact that for numbers x and 4x, the gcd is always x. Final Answer: 84

Question 5

Find the HCF (GCD) of the following three numbers:  1) 4 × 27 × 3125 2) 8 × 9 × 25 × 7 3) 16 × 81 × 5 × 11 × 49  What is the HCF of these three values?

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Introduction / Context: This question involves computing the HCF of three composite numbers that are each given as products. The most systematic approach is to factor each component into prime factors, combine them, and then identify the common primes with the minimum exponents. This is exactly how the HCF is defined in terms of prime powers. Given Data / Assumptions: A = 4 × 27 × 3125 B = 8 × 9 × 25 × 7 C = 16 × 81 × 5 × 11 × 49 We need HCF(A, B, C) Concept / Approach: Write each of A, B, and C in prime factor form. Then the HCF is the product of all primes that appear in every number, each taken to the smallest exponent with which it appears in any of the three factorizations. Step-by-Step Solution: Step 1: Factorize A.4 = 2^2, 27 = 3^3, 3125 = 5^5.So A = 2^2 * 3^3 * 5^5.Step 2: Factorize B.8 = 2^3, 9 = 3^2, 25 = 5^2, 7 = 7^1.So B = 2^3 * 3^2 * 5^2 * 7.Step 3: Factorize C.16 = 2^4, 81 = 3^4, 5 = 5^1, 11 = 11^1, 49 = 7^2.So C = 2^4 * 3^4 * 5^1 * 11 * 7^2.Step 4: Identify primes common to A, B, and C: these are 2, 3, and 5.Prime 2: minimum exponent is min(2, 3, 4) = 2.Prime 3: minimum exponent is min(3, 2, 4) = 2.Prime 5: minimum exponent is min(5, 2, 1) = 1.Step 5: HCF = 2^2 * 3^2 * 5^1 = 4 * 9 * 5 = 180. Verification / Alternative check: Since 180 = 2^2 * 3^2 * 5, and each of A, B, and C contains at least 2^2, 3^2, and 5^1, 180 divides all three numbers. Any attempt to increase the exponent of 2, 3, or 5 in the HCF would break divisibility for at least one of the numbers, so 180 is indeed the greatest common divisor. Why Other Options Are Wrong: 90: Contains only 2^1, but A has 2^2 as the minimum common exponent, so 90 is not the HCF.120: Has 2^3, which exceeds the exponent of 2 in A, so it does not divide A.360: Has higher exponents and does not divide B or A exactly.60: Too small and does not reflect the full common prime power structure. Common Pitfalls: Including primes like 7 or 11 in the HCF even though they are not common to all three numbers.Using maximum exponents instead of minimum ones, which gives the LCM instead of the HCF.Making mistakes when factoring large components such as 3125 or 81. Final Answer: 180

Question 6

A rectangular room is 6 meters 24 centimeters long and 4 meters 32 centimeters wide. The floor must be covered completely using square tiles of equal size (no cutting and no gaps).  What is the least number of such square tiles required to cover the…

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Introduction / Context: This is a standard tiling problem that uses HCF. To cover a rectangular floor with square tiles of equal size without cutting, the side length of each tile must divide both the length and width of the room exactly. To minimize the number of tiles, we choose the largest such tile, which corresponds to the HCF of the two dimensions. Given Data / Assumptions: Length of room = 6 m 24 cm = 624 cm Width of room = 4 m 32 cm = 432 cm Tiles are square and all identical No cutting or gaps are allowed Concept / Approach: Let the side of each square tile be s cm. Then s must be a common divisor of 624 and 432. To minimize the number of tiles, s should be equal to the HCF of 624 and 432. Once s is known, the number of tiles is: Number of tiles = (624 / s) * (432 / s) Step-by-Step Solution: Step 1: Compute gcd(624, 432).624 mod 432 = 192.432 mod 192 = 48.192 mod 48 = 0, so gcd(624, 432) = 48.Step 2: Therefore, the side of each square tile is s = 48 cm.Step 3: Number of tiles along the length = 624 / 48 = 13.Step 4: Number of tiles along the width = 432 / 48 = 9.Step 5: Total number of tiles = 13 * 9 = 117. Verification / Alternative check: If a larger tile size than 48 cm were chosen, it would not divide both 624 and 432 exactly, making tiling without cutting impossible. Any smaller tile size that still divides both dimensions, such as 24 cm, would lead to more tiles (26 by 18 = 468 tiles), which is not minimal. Thus, using tiles of side 48 cm and a total of 117 tiles is optimal. Why Other Options Are Wrong: 107, 127, 137, 156: None of these counts correspond to a valid square tile size that divides both 624 cm and 432 cm and also yields a minimal number of tiles. They represent either smaller tile sizes or tilings that are not consistent with the HCF-based calculation. Common Pitfalls: Not converting meters and centimeters into a single unit before finding the HCF.Using LCM instead of HCF for the tile side, which would give a very large and impractical result.Forgetting to multiply the number of tiles along the length and width to obtain the total. Final Answer: 117

Question 7

Find the least common multiple (LCM) of the three positive integers 24, 36, and 40. Show clearly how you use prime factorization to obtain the smallest number that is exactly divisible by all three.

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Introduction / Context: The question is about finding the least common multiple (LCM) of three integers: 24, 36, and 40. LCM problems test understanding of prime factorization, common multiples, and the idea of the smallest positive number that each given integer can divide without leaving a remainder. Given Data / Assumptions: First number = 24 Second number = 36 Third number = 40 All numbers are positive integers Concept / Approach: To find the LCM of several integers, we use prime factorization. Each number is written as a product of primes raised to powers. For LCM, we take each prime that appears in any of the factorizations and use the highest power of that prime. Multiplying these highest powers gives the required LCM. This ensures the result is divisible by all the original numbers and is the smallest such common multiple. Step-by-Step Solution: 24 = 2^3 * 3 36 = 2^2 * 3^2 40 = 2^3 * 5 For prime 2, highest power is 2^3. For prime 3, highest power is 3^2. For prime 5, highest power is 5^1. LCM = 2^3 * 3^2 * 5 = 8 * 9 * 5 = 360. Verification / Alternative check: Check divisibility of 360: 360 / 24 = 15, which is an integer. 360 / 36 = 10, which is an integer. 360 / 40 = 9, which is an integer. Since 360 is divisible by all three numbers, it is a common multiple. Any smaller candidate would miss at least one required prime power, so 360 is the least such multiple. Why Other Options Are Wrong: 120 and 240 are not divisible by 36 because they lack the factor 3^2. 480 and 720 are multiples of 360 but not the least common multiple, they are larger than necessary. Common Pitfalls: Students often mistakenly add or average the numbers, or they forget to use the highest power of each prime when computing the LCM. Another common error is to find any common multiple but fail to confirm that it is the smallest one. Working systematically with prime factorization prevents these mistakes. Final Answer: 360

Question 8

The highest common factor (HCF) of the numbers 3240, 3600, and a third unknown number is 36, and the least common multiple (LCM) of all three is 2^4 * 3^5 * 5^2 * 7^2. Using the relationship between HCF, LCM, and prime factorization, determine the v…

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Introduction / Context: This question tests a deeper understanding of how the highest common factor (HCF) and least common multiple (LCM) relate to the prime factorizations of three numbers. Instead of only computing HCF or LCM, we are asked to reconstruct an unknown number from the given combined HCF and LCM information, which is a more advanced application frequently seen in aptitude exams. Given Data / Assumptions: HCF of 3240, 3600, and the third number = 36 LCM of 3240, 3600, and the third number = 2^4 * 3^5 * 5^2 * 7^2 First known number = 3240 Second known number = 3600 Third number is positive and must be consistent with the given HCF and LCM. Concept / Approach: For each prime, the exponent in the HCF is the minimum exponent among the three numbers, and the exponent in the LCM is the maximum exponent among them. By factoring the known numbers and comparing with the required HCF and LCM prime powers, we can deduce the prime factorization of the unknown third number and then compute it directly. Step-by-Step Solution: 3240 = 2^3 * 3^4 * 5^1 3600 = 2^4 * 3^2 * 5^2 HCF = 36 = 2^2 * 3^2 LCM = 2^4 * 3^5 * 5^2 * 7^2 Let the third number be N = 2^a * 3^b * 5^c * 7^d. From HCF for prime 2: min(3, 4, a) = 2 so a = 2. From HCF for prime 3: min(4, 2, b) = 2 so b ≥ 2. From HCF for prime 5: min(1, 2, c) = 0 so c = 0. From LCM for prime 3: max(4, 2, b) = 5 so b = 5. From LCM for prime 7: max(0, 0, d) = 2 so d = 2. Thus N = 2^2 * 3^5 * 7^2 = 4 * 243 * 49 = 47628. Verification / Alternative check: We can check that N shares HCF 36 with the other numbers and that the LCM using all three numbers matches the given expression. The exponents for 2 and 3 satisfy both minimum and maximum constraints, while 5 and 7 appear with required exponents in the overall LCM. This confirms that 47628 is consistent with the given information. Why Other Options Are Wrong: 35280 and 42146 either introduce incorrect prime factors or fail to match the HCF condition. 49874 and 24157 do not generate the specified LCM when combined with 3240 and 3600. Common Pitfalls: A frequent mistake is to assume that the product of numbers is always simply HCF multiplied by LCM, which is only guaranteed for two numbers, not for three. Another common error is ignoring the prime exponent rules for minimum (HCF) and maximum (LCM). Working prime by prime prevents such confusion. Final Answer: 47628

Question 9

Six bells start tolling together at the same instant and then toll repeatedly at fixed intervals of 2, 4, 6, 8, 10, and 12 seconds respectively. Over a total duration of 30 minutes, how many times in all will all six bells toll together at exactly t…

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Introduction / Context: This problem models a classic synchronization situation. Multiple bells ring at fixed time intervals. We are asked how many times they all ring together within a given time frame. Such questions are practical applications of the least common multiple (LCM) concept in time and periodic processes. Given Data / Assumptions: Bell intervals: 2 seconds, 4 seconds, 6 seconds, 8 seconds, 10 seconds, 12 seconds Total time considered: 30 minutes 30 minutes = 30 * 60 = 1800 seconds All bells toll together first at time 0 seconds. Concept / Approach: All bells will toll together every time a moment occurs that is a common multiple of all their individual intervals. The smallest such time gap is the least common multiple of the intervals. Once that LCM is known, the number of simultaneous tolls is the count of multiples of this LCM that lie within the total time, including the starting moment at 0 seconds. Step-by-Step Solution: Find LCM of 2, 4, 6, 8, 10, 12. Prime factorizations: 2 = 2, 4 = 2^2, 6 = 2 * 3, 8 = 2^3, 10 = 2 * 5, 12 = 2^2 * 3. Take highest powers of each prime: 2^3, 3^1, 5^1. LCM = 2^3 * 3 * 5 = 8 * 3 * 5 = 120 seconds. Number of intervals inside 1800 seconds = 1800 / 120 = 15. Include the initial toll at time 0 seconds, so total together tolls = 15 + 1 = 16. Verification / Alternative check: The bells toll together at times 0, 120, 240, 360, and so on up to 1800 seconds. Counting these gives 16 moments: this matches the calculation based on division and confirms the answer. Why Other Options Are Wrong: 10, 15, 18, and 20 either ignore the initial together toll at time zero or assume a wrong LCM, or they miscount the number of multiples up to 1800 seconds. Common Pitfalls: Two common mistakes are forgetting to include the initial toll at time zero and confusing the sum of intervals with the LCM. Always remember that the bells are already together at the starting instant and that the LCM, not simple addition, determines when repeated alignment occurs. Final Answer: 16

Question 10

Three tanks have capacities of 98 litres, 182 litres, and 266 litres respectively. Find the capacity of the largest measuring cylinder that can be used to measure out and exactly fill each tank an integer number of times, without leaving any remaind…

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Introduction / Context: This question is a real life application of the highest common factor (HCF). When we want one measuring container to exactly measure several different quantities, the largest possible size of that container is the HCF of the given capacities. It ensures that each capacity is a whole number multiple of the measuring cylinder without any leftover liquid. Given Data / Assumptions: Tank 1 capacity = 98 litres Tank 2 capacity = 182 litres Tank 3 capacity = 266 litres We want the largest possible measuring cylinder that measures each capacity exactly. Concept / Approach: The highest common factor of the three capacities is the largest integer that divides each of them exactly. We find the HCF by prime factorization of each capacity and then taking the product of all common primes with their smallest exponent across the three numbers. Step-by-Step Solution: 98 = 2 * 7^2 182 = 2 * 7 * 13 266 = 2 * 7 * 19 Common prime factors in all three: 2^1 and 7^1. HCF = 2 * 7 = 14 litres. Thus, a 14 litre cylinder will exactly measure each tank: 98 / 14 = 7 times, 182 / 14 = 13 times, and 266 / 14 = 19 times. Verification / Alternative check: Checking divisibility: 98 mod 14 = 0, so 14 divides 98 exactly. 182 mod 14 = 0, so 14 divides 182 exactly. 266 mod 14 = 0, so 14 divides 266 exactly. No larger option among the choices divides all three capacities exactly, so 14 litres is indeed the largest possible measure. Why Other Options Are Wrong: 7 litres divides all capacities, but it is not the largest possible measure. 21, 42, and 98 litres do not divide each of 98, 182, and 266 without leaving a remainder, so they cannot be used as the required measuring cylinder capacity. Common Pitfalls: Common errors include choosing a common divisor that is not the greatest one, or confusing HCF with LCM. Here we must maximize the cylinder size, which is why we focus on the HCF and not on the least common multiple of the capacities. Final Answer: 14 litres

Question 11

Determine the highest common factor (HCF) of the three positive integers 1, 2, and 3, and explain why this value is the greatest integer that divides all three numbers exactly.

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Introduction / Context: This question checks a basic but very important concept in number theory, the highest common factor (HCF) of a small set of integers. Even though the numbers are simple, understanding why the HCF is what it is lays the foundation for more complex problems involving divisibility, fractions, and simplifying ratios. Given Data / Assumptions: Numbers involved: 1, 2, and 3 All are positive integers We are looking for the greatest integer that divides each of them exactly. Concept / Approach: The HCF of a set of integers is the largest positive integer that divides every number in the set without leaving any remainder. For small numbers, we can list the divisors and find their intersection. For larger numbers, we often use prime factorization or the Euclidean algorithm. For this very small set, simple listing is enough. Step-by-Step Solution: Divisors of 1: 1 Divisors of 2: 1, 2 Divisors of 3: 1, 3 Common divisors of 1, 2, and 3: only 1. Therefore, the greatest common divisor or HCF is 1. Verification / Alternative check: Confirm divisibility: 1 divided by 1 gives 1. 2 divided by 1 gives 2. 3 divided by 1 gives 3. All results are integers. No integer greater than 1 divides 1, so there cannot be any higher common factor than 1. Why Other Options Are Wrong: 2 and 3 do not divide 1, so they cannot be common factors of the entire set. 0 is not considered as an HCF in standard number theory, and 6 is larger than any of the given numbers and does not divide them all. Common Pitfalls: A typical misunderstanding is to think that the HCF must be one of the larger given numbers or that 0 can be a highest common factor. Remember that the HCF must divide every number exactly and must be positive. With 1 in the list, the HCF can never exceed 1. Final Answer: 1

Question 12

If N is the greatest number that will divide 1305, 4665, and 6905 leaving the same remainder in each case, find the sum of the digits of N using the idea of equal remainders and highest common factor.

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Introduction / Context: This question uses a standard trick in number theory. When a number divides several integers leaving the same remainder each time, the differences between those integers are all exactly divisible by that number. This converts a remainder based question into a highest common factor (HCF) problem, which can then be solved systematically. Given Data / Assumptions: N is the greatest number that divides 1305, 4665, and 6905 leaving the same remainder in each case. We are asked to find the sum of the digits of N. All given numbers are positive integers. Concept / Approach: If a number N divides several integers leaving the same remainder r, then: (4665 - 1305), (6905 - 4665), and (6905 - 1305) are all multiples of N. Therefore, N must be a common divisor of these differences, and the greatest such N is the HCF of these differences. Once N is found, we just add its digits for the final answer. Step-by-Step Solution: Compute differences: 4665 - 1305 = 3360 6905 - 4665 = 2240 6905 - 1305 = 5600 Find HCF of 3360, 2240, and 5600. First, HCF(3360, 2240) = 1120. Then HCF(1120, 5600) = 1120. So N = 1120. Sum of digits of N = 1 + 1 + 2 + 0 = 4. Verification / Alternative check: We can check that 1120 divides all differences: 3360 / 1120 = 3, 2240 / 1120 = 2, 5600 / 1120 = 5. All results are integers, confirming that 1120 is a common divisor. Since we used the HCF, it is the greatest such number, and the digit sum 4 is therefore correct. Why Other Options Are Wrong: 6, 7, 8, and 9 do not represent the sum of digits of the correct greatest number 1120. They might correspond to digit sums of smaller divisors, but the question specifically asks for the greatest one. Common Pitfalls: A common error is to try to find N directly from the original numbers without using differences, or to confuse the idea with LCM. Another mistake is to take the difference of only one pair and not compute the HCF of all three differences, which might result in a smaller divisor rather than the greatest one. Final Answer: 4

Question 13

Find the greatest common divisor (GCD) or highest common factor (HCF) of the decimal numbers 1.08, 0.36, and 0.90 by converting them into equivalent integers and then simplifying your result back to decimal form.

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Introduction / Context: This question explores the concept of the greatest common divisor (GCD) for decimal numbers. Although GCD is usually taught with integers, decimal values can be handled by scaling them up to integers, performing the GCD calculation, and then scaling the result back down to a decimal. This technique is very useful in quantitative aptitude and data interpretation problems. Given Data / Assumptions: Numbers: 1.08, 0.36, and 0.90 We want the greatest common divisor of these three decimal values. All numbers have at most two decimal places. Concept / Approach: To find the GCD of decimal numbers, we eliminate the decimals by multiplying each number by a suitable power of 10 so that all become integers. We then compute the GCD of the integers using standard methods. Finally, we divide the result by the same power of 10 to obtain the GCD in decimal form. Step-by-Step Solution: Multiply each number by 100 to remove decimal places: 1.08 * 100 = 108 0.36 * 100 = 36 0.90 * 100 = 90 Now find GCD of 108, 36, and 90. GCD(108, 36) = 36. GCD(36, 90) = 18. So GCD of 108, 36, and 90 is 18. Scale back by dividing by 100: 18 / 100 = 0.18. Verification / Alternative check: Check divisibility of the original decimals by 0.18: 1.08 / 0.18 = 6 0.36 / 0.18 = 2 0.90 / 0.18 = 5 All quotients are integers, showing that 0.18 divides all the numbers exactly and is a common divisor. Since it comes from the integer GCD, it is the greatest common divisor in decimal form. Why Other Options Are Wrong: 0.90, 0.36, and 1.08 are larger than 0.18 and do not divide all three values exactly. 0.06 divides all three numbers but is not the greatest such divisor, because 0.18 is a larger common divisor. Common Pitfalls: Students sometimes try to work directly with decimals without scaling, which can be confusing. Another mistake is not using the same power of 10 for all numbers or forgetting to scale back at the end. Always convert decimals to integers in a consistent way, find the GCD, and then convert back to decimals for the final answer. Final Answer: 0.18

Question 14

Find the greatest number that will divide 964, 1238, and 1400 leaving remainders of 41, 31, and 51 respectively in each case. Use the idea of equal remainders and highest common factor to determine this largest divisor.

Accepted Answer

Introduction / Context: This is another example of a remainder based highest common factor (HCF) question. When a number divides several integers and leaves potentially different remainders, we can transform the problem by subtracting those remainders and turning it into a pure divisibility question. Then the largest such divisor is the HCF of the adjusted values. Given Data / Assumptions: A number N divides 964 leaving remainder 41. The same N divides 1238 leaving remainder 31. The same N divides 1400 leaving remainder 51. We want the greatest such N. Concept / Approach: If N divides several numbers and leaves remainders r1, r2, r3, the quantities (number - remainder) for each case are exact multiples of N. Thus, N must be a common divisor of these adjusted values. The greatest such N is the highest common factor of those adjusted numbers. So we compute the three adjusted numbers and then find their HCF. Step-by-Step Solution: Compute adjusted values: 964 - 41 = 923 1238 - 31 = 1207 1400 - 51 = 1349 Now find HCF(923, 1207, 1349). First compute HCF of 923 and 1207. Using Euclidean method, HCF(923, 1207) = 71. Now HCF(71, 1349) is also 71. Therefore, N = 71. Verification / Alternative check: Check each original number with N = 71: 964 / 71 = 13 remainder 41. 1238 / 71 = 17 remainder 31. 1400 / 71 = 19 remainder 51. All remainders match the given values, confirming that 71 is a valid divisor. Since it comes from the HCF of the adjusted values, it is the greatest such divisor. Why Other Options Are Wrong: 64, 58, 69, and 82 either do not divide all the adjusted values exactly or produce different remainders than those specified in the problem, so they cannot be the required greatest number. Common Pitfalls: A typical error is to try to work directly with the original numbers and remainders without forming the adjusted numbers by subtraction. Another mistake is to compute HCF using only two of the adjusted numbers and to forget to include the third, which can lead to a divisor that is not common to all three cases. Final Answer: 71

Question 15

Determine the least common multiple (LCM) of the two integers 15 and 12, showing clearly how prime factorization or another standard method leads to the smallest number divisible by both.

Accepted Answer

Introduction / Context: This question focuses on finding the least common multiple (LCM) of two commonly used integers, 15 and 12. LCM is an essential concept for working with fractions, common denominators, and problems involving synchronized events in arithmetic and algebra. Given Data / Assumptions: First integer = 15 Second integer = 12 We want the smallest positive integer that is divisible by both 15 and 12. Concept / Approach: The LCM of two numbers can be found via prime factorization. We factor each number into primes and then take all distinct primes with their highest exponent appearing in either factorization. Multiplying these together yields the LCM. Another method is to use the relationship LCM * HCF = product of the two numbers, but prime factorization is very clear here. Step-by-Step Solution: Prime factorization of 15: 15 = 3 * 5. Prime factorization of 12: 12 = 2^2 * 3. Collect highest powers of each prime: 2^2, 3, 5. LCM = 2^2 * 3 * 5 = 4 * 3 * 5. LCM = 12 * 5 = 60. Verification / Alternative check: Check that 60 is a multiple of both numbers: 60 / 15 = 4, which is an integer. 60 / 12 = 5, which is an integer. There is no smaller positive number than 60 that is divisible by both, because any multiple of 15 that is also a multiple of 12 must include the full prime components 2^2, 3, and 5, giving at least 60. Why Other Options Are Wrong: 12 and 30 are multiples of only one of the numbers or are too small to accommodate both factorizations fully. 45 and 90 are multiples of 15, but 45 is not divisible by 12, and 90 is larger than necessary, so it is not the least common multiple. Common Pitfalls: Common issues include simply multiplying the numbers without reducing by common factors, which often produces a common multiple but not the least one. Another mistake is forgetting to include the highest power of each prime when using prime factorization. Carefully listing prime factors avoids these errors. Final Answer: 60

Question 16

Three numbers are in the ratio 3 : 4 : 5 and their least common multiple (LCM) is 3600. Using the properties of ratios, LCM, and highest common factor (HCF), determine the HCF of these three numbers.

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Introduction / Context: This question connects ratios with LCM and HCF. When numbers are given in a simple ratio, we can treat the actual numbers as multiples of a common factor. Using this structure, along with the given LCM, we can determine the HCF. This is a standard pattern seen in many aptitude tests involving number properties. Given Data / Assumptions: The three numbers are in the ratio 3 : 4 : 5. The LCM of the three numbers is 3600. All numbers are positive integers. Concept / Approach: If the numbers are in the ratio 3 : 4 : 5, we can write them as 3k, 4k, and 5k, where k is a positive integer that represents the HCF. Since 3, 4, and 5 are pairwise co prime, the LCM of 3k, 4k, and 5k is 3 * 4 * 5 * k = 60k. The given LCM value allows us to solve for k, which is then the HCF of the three numbers. Step-by-Step Solution: Let the numbers be 3k, 4k, and 5k. Since 3, 4, and 5 are co prime pairwise, LCM(3, 4, 5) = 3 * 4 * 5 = 60. Therefore, LCM(3k, 4k, 5k) = 60k. Given LCM = 3600, so 60k = 3600. Solve for k: k = 3600 / 60 = 60. Thus, HCF of the three numbers is k = 60. Verification / Alternative check: The actual numbers are 3 * 60 = 180, 4 * 60 = 240, and 5 * 60 = 300. The HCF of 180, 240, and 300 is clearly 60. Their LCM should be 60 multiplied by LCM of 3, 4, 5 which is 60 * 60 = 3600, matching the given LCM, so the reasoning is consistent. Why Other Options Are Wrong: 40, 100, 120, and 20 do not satisfy the relationship LCM = 60 * HCF in this setup. Using any of these values as the HCF would produce an LCM that is different from 3600, contradicting the given condition. Common Pitfalls: A frequent mistake is to misunderstand the role of k and treat 3, 4, and 5 as the actual numbers rather than as a ratio pattern. Another error is to compute HCF and LCM independently without using the powerful structure that the ratio provides. Remember that for numbers in a co prime ratio, the LCM of the actual numbers is the product of the ratio terms multiplied by the HCF. Final Answer: 60

Question 17

A palace has gold, silver, and bronze coins in quantities of 18000, 9600, and 3600 respectively. All coins must be packed into rooms so that each room contains the same total number of coins, and each room holds coins of only one type. What is the m…

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Introduction / Context: This problem is a practical application of the highest common factor (HCF) in the context of distributing objects evenly. The palace has three types of coins in large quantities, and the goal is to minimize the number of rooms while keeping the number of coins per room the same and using only one type of coin in each room. Such questions blend HCF with logical interpretation of distribution conditions. Given Data / Assumptions: Gold coins = 18000 Silver coins = 9600 Bronze coins = 3600 Each room contains coins of only one type. Every room has the same total number of coins. We want the minimum number of rooms, which means the maximum possible coins per room. Concept / Approach: Since each room must have the same number of coins and all three totals must be split into equal sized groups, the total number of coins per room must be a common divisor of 18000, 9600, and 3600. To minimize the number of rooms, we maximize the coins per room, so we choose the highest common factor of the three quantities. Then we compute how many rooms are needed for each type and add them together. Step-by-Step Solution: Find HCF of 18000, 9600, and 3600. 18000 = 2^4 * 3^2 * 5^3 9600 = 2^7 * 3^1 * 5^2 3600 = 2^4 * 3^2 * 5^2 Minimum powers for common primes: 2^4, 3^1, 5^2. HCF = 2^4 * 3 * 5^2 = 16 * 3 * 25 = 1200 coins per room. Gold rooms = 18000 / 1200 = 15. Silver rooms = 9600 / 1200 = 8. Bronze rooms = 3600 / 1200 = 3. Total rooms required = 15 + 8 + 3 = 26. Verification / Alternative check: Each room has 1200 coins. This fits exactly into each type total and respects the rule that a room cannot mix coin types. Any larger number of coins per room would not divide all three totals simultaneously, so would violate the equal room size condition. Hence 26 rooms is minimal. Why Other Options Are Wrong: 24, 18, 12, and 30 correspond either to smaller common divisors, different interpretations, or arrangements with unequal numbers of coins per room. They do not satisfy all the given constraints while also minimizing the number of rooms. Common Pitfalls: Some learners misinterpret the condition and try to make the number of rooms the HCF directly, or they allow each type of coin to have a different coin count per room, which breaks the given requirement. The key is that every room must have the same number of coins and that no room mixes coin types, so the group size must be the HCF of all three totals. Final Answer: 26

Question 18

Three containers hold mixtures in quantities of 1365 litres, 1560 litres, and 1755 litres respectively. Find the capacity of the largest measuring vessel that can be used to measure each of these quantities exactly an integer number of times.

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Introduction / Context: This problem is a direct application of the highest common factor (HCF) in a measurement context. We want one measuring vessel that will exactly measure three given quantities of liquid without leaving any remainder. In such cases, the largest possible capacity of the vessel is equal to the HCF of the three quantities. Given Data / Assumptions: First container: 1365 litres Second container: 1560 litres Third container: 1755 litres We want the largest measure that exactly divides all three capacities. Concept / Approach: The largest measure that can be used with all three containers is their HCF. This is because each capacity must be an exact multiple of the measure. To find the HCF, we perform prime factorization or use the Euclidean algorithm. Since the numbers are moderately large, prime factorization combined with divisibility checks is very helpful. Step-by-Step Solution: First, find the HCF of 1365 and 1560. 1365 factors: 1365 = 3 * 5 * 7 * 13. 1560 factors: 1560 = 2^3 * 3 * 5 * 13. Common factors so far: 3 * 5 * 13 = 195. Now check 1755. 1755 = 3 * 5 * 3 * 3 * 13 = 3^3 * 5 * 13. 1755 is divisible by 195 because 1755 / 195 = 9. Therefore, HCF(1365, 1560, 1755) = 195 litres. Verification / Alternative check: We can verify: 1365 / 195 = 7, an integer. 1560 / 195 = 8, an integer. 1755 / 195 = 9, an integer. No option larger than 195 divides all three values exactly, so 195 litres is the largest possible measure that satisfies the requirement. Why Other Options Are Wrong: 37, 97, 129, and 65 either do not divide all three capacities exactly or are smaller than the maximum common factor 195, so they cannot be the largest possible common measure. Common Pitfalls: One mistake is to confuse HCF with LCM, which would give a very large number that is not a measure but a combined multiple. Another error is to check divisibility only for one or two containers and overlook the third. It is essential to ensure that the chosen measure divides all three quantities without remainder. Final Answer: 195 litres

Question 19

Find the highest common factor (HCF) of the two integers 865 and 2595, and explain why this value is the greatest integer that divides both numbers without leaving a remainder.

Accepted Answer

Introduction / Context: This question asks for the highest common factor (HCF) of two numbers, 865 and 2595. The HCF is widely used in simplifying fractions, solving ratio problems, and understanding divisibility properties. When one number is a factor of another, the HCF turns out to be the smaller number, which is an important observation in such problems. Given Data / Assumptions: First number = 865 Second number = 2595 We are looking for the largest integer that divides both numbers exactly. Concept / Approach: If one number divides the other exactly, the smaller number is the HCF. To check this, we can perform division or use factorization. Alternatively, we can apply the Euclidean algorithm, which repeatedly uses the remainder to reduce the problem until the remainder becomes zero. The last non zero remainder is the HCF. Step-by-Step Solution: Observe that 2595 = 3 * 865. Compute: 865 * 3 = 2595. This means 865 divides 2595 exactly with no remainder. Also, of course, 865 divides itself exactly. Therefore, the HCF of 865 and 2595 is 865. Verification / Alternative check: Using the Euclidean algorithm: 2595 divided by 865 gives quotient 3 and remainder 0. When the remainder becomes 0, the divisor at that step is the HCF. Thus, HCF(2595, 865) = 865. This confirms our earlier reasoning that 865 is the highest common factor. Why Other Options Are Wrong: 5, 25, 35, and 95 are all factors of 865 and 2595, but they are smaller than 865. Since 865 itself divides both numbers exactly, it is clearly the greatest such factor, and the others cannot be the HCF. Common Pitfalls: Students sometimes stop after finding a small common factor and forget to check if one number is a factor of the other. Another error is to confuse HCF with LCM and attempt unnecessary multiplications instead of looking for the largest common divisor. Always check whether the smaller number divides the larger one before moving to more complicated methods. Final Answer: 865

Question 20

Find the least common multiple (LCM) of the two integers 72 and 84, using prime factorization or another standard method, and identify the correct value from the options given.

Accepted Answer

Introduction / Context: The question asks for the least common multiple (LCM) of 72 and 84. LCM problems are very common in aptitude tests and are used in questions involving common cycles, meeting times, and common denominators in fractions. Prime factorization provides a clear and reliable way to compute the LCM. Given Data / Assumptions: First number = 72 Second number = 84 We wish to find the smallest positive integer that both 72 and 84 divide exactly. Concept / Approach: To find the LCM, we express each number as a product of prime factors. For each prime that appears in either factorization, we take the highest power that appears in any of the numbers. The LCM is the product of these highest powers. This guarantees that the resulting number is divisible by each original number and is the smallest such common multiple. Step-by-Step Solution: Prime factorization of 72: 72 = 2^3 * 3^2. Prime factorization of 84: 84 = 2^2 * 3 * 7. Take highest powers for each prime: For 2: max(3, 2) = 3, so use 2^3. For 3: max(2, 1) = 2, so use 3^2. For 7: presence only in 84, so use 7^1. LCM = 2^3 * 3^2 * 7 = 8 * 9 * 7. Compute: 8 * 9 = 72, then 72 * 7 = 504. Verification / Alternative check: Check divisibility: 504 / 72 = 7, which is an integer. 504 / 84 = 6, which is an integer. No smaller candidate in the options is divisible by both 72 and 84, confirming that 504 is indeed the least common multiple. Why Other Options Are Wrong: 144, 252, 420, and 756 either fail to be divisible by both 72 and 84 or are not the smallest such multiple. For example, 252 is divisible by 84 but not by 72, and 756 is a larger multiple of 504 but not the least common multiple. Common Pitfalls: A common error is to multiply the numbers directly without reducing by common factors, producing a multiple but not the least common multiple. Another mistake is to forget the prime 7 from 84 or use too small a power of 2 or 3. Careful and complete prime factorization avoids these problems. Final Answer: 504

HCF and LCM Questions