Question 1

A rectangular courtyard is 3.78 meters long and 5.25 meters wide. It is to be paved exactly with square tiles, all of the same size. What is the largest possible side length of each square tile in centimeters?

Accepted Answer

Introduction / Context: This is a classic highest common factor (HCF) or greatest common divisor problem disguised as a geometry and tiling question. The idea is to cover a rectangular courtyard fully with square tiles of equal size, without cutting any tile. The largest possible tile side corresponds directly to the greatest common divisor of the courtyard dimensions expressed in the same unit. Given Data / Assumptions: Length of the courtyard = 3.78 meters. Width of the courtyard = 5.25 meters. Tiles are square and identical. We want the largest possible side length of the tile in centimeters. Concept / Approach: To find the largest square tile that can exactly pave the rectangle, we: Convert all dimensions to the same unit (here, centimeters). Find the greatest common divisor of the length and the width in centimeters. The HCF will be the largest side length of the square tile that divides both dimensions exactly. Step-by-Step Solution: Step 1: Convert 3.78 meters to centimeters: 3.78 * 100 = 378 cm. Step 2: Convert 5.25 meters to centimeters: 5.25 * 100 = 525 cm. Step 3: We need the HCF of 378 and 525. Step 4: Factor 378: 378 = 2 * 3 * 3 * 3 * 7 (or 2 * 3^3 * 7). Step 5: Factor 525: 525 = 3 * 5 * 5 * 7 (or 3 * 5^2 * 7). Step 6: Identify common prime factors: both have one 3 and one 7. Step 7: Multiply the common factors: HCF = 3 * 7 = 21. Step 8: Therefore, the largest possible tile side is 21 centimeters. Verification / Alternative Check: Check that 21 cm divides both dimensions exactly: 378 ÷ 21 = 18 tiles along the length, and 525 ÷ 21 = 25 tiles along the width. Both results are whole numbers, so 21 cm tiles will cover the courtyard exactly without cutting. If we try a larger size like 42 cm, 378 ÷ 42 = 9 is an integer but 525 ÷ 42 is not an integer, confirming that 42 cm is too large. Why Other Options Are Wrong: 14 cm divides 378 and 525, but it is not the greatest common divisor; a larger size, 21 cm, still works. 42 cm does not divide 525 exactly, so the width cannot be tiled perfectly with 42 cm squares. Thus, the correct largest size is 21 cm, so the option “None of these” is also incorrect. Common Pitfalls: One common mistake is failing to convert dimensions to the same units before finding the HCF. Another is guessing tile sizes from the options without verifying exact divisibility. Always compute the highest common factor in such questions to ensure you have the largest possible tile size that fits both dimensions exactly. Final Answer: The largest possible side length of each square tile is 21 cms.

Question 2

What is the least common multiple (L.C.M.) of the numbers 22, 54, 108, 135 and 198?

Accepted Answer

Introduction / Context: Finding the least common multiple (L.C.M.) of a set of numbers is important in many arithmetic and algebra problems, especially those involving synchronization of cycles, finding common periods, or adding and subtracting fractions. In this question, we must determine the L.C.M. of five numbers: 22, 54, 108, 135 and 198. Given Data / Assumptions: Numbers: 22, 54, 108, 135 and 198. We need the smallest positive number that is divisible by each of these. All numbers are positive integers. Concept / Approach: To find the L.C.M.: Write the prime factorization of each number. For each distinct prime, take the highest power that appears in any factorization. Multiply these highest powers together to get the L.C.M. This approach guarantees the smallest number that contains all required prime factors in sufficient powers. Step-by-Step Solution: Step 1: Factor each number into primes. 22 = 2 * 11. 54 = 2 * 3^3 (since 54 = 2 * 27 = 2 * 3^3). 108 = 2^2 * 3^3 (since 108 = 4 * 27). 135 = 3^3 * 5 (since 135 = 27 * 5). 198 = 2 * 3^2 * 11 (since 198 = 2 * 99 = 2 * 9 * 11). Step 2: Identify all distinct primes: 2, 3, 5 and 11. Step 3: For each prime, find the highest power appearing in any number. For 2: highest power is 2^2 (from 108). For 3: highest power is 3^3 (from 54, 108 and 135). For 5: highest power is 5^1 (from 135). For 11: highest power is 11^1 (from 22 and 198). Step 4: Multiply these highest powers: L.C.M. = 2^2 * 3^3 * 5 * 11. Step 5: Compute step by step: 2^2 = 4, 3^3 = 27, so 4 * 27 = 108. Step 6: Multiply 108 by 5: 108 * 5 = 540. Step 7: Multiply 540 by 11: 540 * 11 = 5940. Verification / Alternative Check: Check divisibility: 5940 ÷ 22 = 270 exactly. 5940 ÷ 54 = 110 exactly. 5940 ÷ 108 = 55 exactly. 5940 ÷ 135 = 44 exactly. 5940 ÷ 198 = 30 exactly. Since 5940 is divisible by each number without remainder, it is a common multiple. Because we used the highest prime powers method, it is also the least such common multiple. Why Other Options Are Wrong: 330 and 1980 are too small; they are not divisible by all the given numbers, especially 108 and 135. 11880 is a multiple of 5940, but it is larger and therefore not the least common multiple. Thus, 5940 is the smallest positive integer divisible by all five numbers. Common Pitfalls: A common mistake is to forget one of the primes or to choose a lower power than required, which leads to a number that is not a common multiple. Others may create a multiple that works but is not the least one. Sticking to prime factorization and highest powers prevents these errors. Final Answer: The L.C.M. of 22, 54, 108, 135 and 198 is 5940.

Question 3

Two co-prime numbers have a product equal to 117. For such co-prime numbers, what will be their least common multiple (L.C.M.)?

Accepted Answer

Introduction: This aptitude question checks your understanding of basic number theory concepts, especially highest common factor (H.C.F.) and least common multiple (L.C.M.) for co-prime numbers. Knowing how L.C.M. and H.C.F. behave for such numbers is frequently tested in competitive exams and placements. Given Data / Assumptions: The product of two co-prime numbers is 117. The two numbers are co-prime, meaning their H.C.F. is 1. We are asked to find the L.C.M. of these two numbers. Concept / Approach: For any two positive integers a and b, the fundamental relationship is: a * b = HCF(a, b) * LCM(a, b) If the numbers are co-prime, then HCF(a, b) = 1. Therefore, the product of the numbers becomes equal to their L.C.M. Understanding and applying this identity directly leads to the answer without needing to know the individual numbers. Step-by-Step Solution: Step 1: Let the two co-prime numbers be a and b. Step 2: Given that a * b = 117. Step 3: Since a and b are co-prime, HCF(a, b) = 1. Step 4: Use the identity a * b = HCF(a, b) * LCM(a, b). Step 5: Substitute values: 117 = 1 * LCM(a, b). Step 6: Therefore, LCM(a, b) = 117. Verification / Alternative check: We can also factorise 117 as 117 = 9 * 13 = 3^2 * 13. Numbers such as 9 and 13 are co-prime, and their product is 117. The L.C.M. of 9 and 13, since they are co-prime, is 9 * 13 = 117, confirming the same result. Why Other Options Are Wrong: Equal to their H.C.F.: For co-prime numbers, H.C.F. = 1, but L.C.M. is generally much larger. 1: This would be the H.C.F., not the L.C.M. Cannot be calculated from the given data: The product and co-prime condition are enough to determine the L.C.M. 39: This is a factor of 117, but not equal to the required L.C.M. Common Pitfalls: Students often confuse H.C.F. and L.C.M., or forget the relationship between product, H.C.F. and L.C.M. Another mistake is thinking that without knowing the individual numbers, L.C.M. cannot be found, which is not true here because of the co-prime condition. Final Answer: Thus, the least common multiple of the two co-prime numbers whose product is 117 is 117.

Question 4

The L.C.M. of two numbers is 495 and their H.C.F. is 5. If the sum of the two numbers is 100, what is the difference between these two numbers?

Accepted Answer

Introduction: This problem tests your understanding of the relationship between the highest common factor (H.C.F.), least common multiple (L.C.M.), and the product of two numbers. It also involves forming and solving a quadratic equation using the sum and product of the numbers, a common pattern in aptitude exams. Given Data / Assumptions: L.C.M. of the two numbers = 495. H.C.F. of the two numbers = 5. Sum of the two numbers = 100. We are required to find the difference between the two numbers. Concept / Approach: For any two positive integers a and b, we use the identity: a * b = HCF(a, b) * LCM(a, b) Once the product and sum of the numbers are known, we can take a and b as roots of a quadratic equation: x^2 - (sum)x + (product) = 0 Solving this equation gives the two numbers, and their difference can then be computed easily. Step-by-Step Solution: Step 1: Let the two numbers be a and b. Step 2: Product a * b = HCF * LCM = 5 * 495 = 2475. Step 3: Sum a + b = 100 (given). Step 4: Form the quadratic: x^2 - (a + b)x + a * b = 0. Step 5: So, x^2 - 100x + 2475 = 0. Step 6: Compute the discriminant: D = 100^2 - 4 * 2475 = 10000 - 9900 = 100. Step 7: Roots are x = (100 ± 10) / 2 → x = 55 and x = 45. Step 8: The difference between the numbers = 55 - 45 = 10. Verification / Alternative check: Check the conditions: Sum = 55 + 45 = 100, product = 55 * 45 = 2475. Also, H.C.F.(55,45) = 5 and L.C.M.(55,45) = 2475 / 5 = 495. All conditions are satisfied, so the solution is consistent. Why Other Options Are Wrong: 90: This would require the numbers to be 5 and 95, which do not have L.C.M. 495. 70: Would correspond to numbers like 15 and 85, again not matching the given L.C.M. and H.C.F. 46: No pair of integers satisfying the product and L.C.M. conditions gives this difference. 55: This would imply one of the numbers is zero or negative in this context, which is not valid. Common Pitfalls: A common mistake is to forget the product relation with L.C.M. and H.C.F., or to incorrectly compute the quadratic discriminant. Some students also try trial and error with random pairs instead of using a systematic algebraic approach. Final Answer: The difference between the two numbers is 10.

Question 5

Two prime numbers x and y (with x greater than y) have a least common multiple (L.C.M.) equal to 161. What is the value of the expression 3y - x?

Accepted Answer

Introduction: This question assesses your understanding of prime numbers and how their least common multiple (L.C.M.) is determined. It also checks your ability to factorise a composite number and then substitute values into a simple algebraic expression involving these primes. Given Data / Assumptions: x and y are prime numbers. x is greater than y. L.C.M. of x and y is 161. We need to find the value of 3y - x. Concept / Approach: For two distinct prime numbers, the L.C.M. is simply the product of the two primes. Hence, if the L.C.M. is known, factorising it into prime factors will directly give the two primes. Once x and y are identified, we can easily evaluate the expression 3y - x. Step-by-Step Solution: Step 1: Given LCM(x, y) = 161. Step 2: Factorise 161 to identify its prime factors. Step 3: 161 ÷ 7 = 23, so 161 = 7 * 23. Step 4: Both 7 and 23 are prime numbers, so x and y must be 23 and 7 (since x > y). Step 5: Thus, x = 23 and y = 7. Step 6: Compute the required expression: 3y - x = 3 * 7 - 23 = 21 - 23 = -2. Verification / Alternative check: Check that L.C.M.(7, 23) = 7 * 23 = 161, which matches the given condition. Therefore, the assignment of x and y is correct, and the expression has been evaluated accurately. Why Other Options Are Wrong: 1: This would require 3y - x = 1, which does not hold for the prime pair 7 and 23. -1: No arrangement of these primes with the given x > y condition yields -1 for 3y - x. 2 and 0: These values would arise only for different pairs or different ordering, which would then violate the given L.C.M. or x > y condition. Common Pitfalls: Students may mistakenly assume that 161 is divisible by smaller primes like 11 or 13 and waste time checking. Another common error is to swap x and y incorrectly after factorisation, forgetting that x must be greater than y, which changes the value of 3y - x. Final Answer: The value of the expression 3y - x is -2.

Question 6

The sum of two numbers is 55. Their H.C.F. is 5 and their L.C.M. is 120. What is the sum of the reciprocals of these two numbers?

Accepted Answer

Introduction: This question combines number theory with fractions. It tests your understanding of the relationship between two numbers, their H.C.F., their L.C.M., and how to find the sum of their reciprocals. Such questions commonly appear in competitive exams to check conceptual clarity. Given Data / Assumptions: Sum of the two numbers = 55. H.C.F. of the two numbers = 5. L.C.M. of the two numbers = 120. We need to find the sum of the reciprocals of the numbers. Concept / Approach: Let the numbers be a and b. We know the identity: a * b = HCF(a, b) * LCM(a, b) Once we find the product a * b and know a + b, we can compute the sum of reciprocals: 1/a + 1/b = (a + b) / (a * b) This removes the need to explicitly determine a and b, making the calculation straightforward. Step-by-Step Solution: Step 1: Let the two numbers be a and b. Step 2: Given HCF(a, b) = 5 and LCM(a, b) = 120. Step 3: Compute the product: a * b = 5 * 120 = 600. Step 4: The sum a + b is given as 55. Step 5: Sum of reciprocals is 1/a + 1/b = (a + b) / (a * b). Step 6: Substitute values: 1/a + 1/b = 55 / 600. Step 7: Simplify the fraction: 55 / 600 = 11 / 120. Verification / Alternative check: We can also find the actual numbers. Since a * b = 600 and a + b = 55, the numbers are roots of x^2 - 55x + 600 = 0. Solving, we get x = 40 and x = 15. Then 1/40 + 1/15 = (3 + 8) / 120 = 11 / 120, which matches our earlier result. Why Other Options Are Wrong: 55/601: This uses an incorrect denominator and does not follow from the product relation. 120/11: This is the reciprocal of the correct answer, not the required value. 601/55: This is a large number with no basis in the calculations. 5/120: This is a simplified incorrect fraction and does not match the computed sum of reciprocals. Common Pitfalls: A typical error is to try to compute the two numbers directly without using the identity for product. Another mistake is inverting the final fraction or mixing up numerator and denominator in the sum of reciprocals formula. Final Answer: The sum of the reciprocals of the two numbers is 11/120.

Question 7

The H.C.F. of two numbers is 11 and their L.C.M. is 385. If one of the numbers lies between 75 and 125, what is that number?

Accepted Answer

Introduction: This problem checks how well you understand the relationship between the highest common factor (H.C.F.), least common multiple (L.C.M.), and the actual numbers. It also tests your ability to use factorisation and simple inequalities to select the correct number from a given range. Given Data / Assumptions: H.C.F. of the two numbers = 11. L.C.M. of the two numbers = 385. One of the numbers lies between 75 and 125. We need to find that number. Concept / Approach: Let the two numbers be 11m and 11n, since their H.C.F. is 11. Then, if m and n are co-prime, the L.C.M. of the two numbers is: LCM = 11 * m * n We are given that LCM = 385, so we can solve for m * n. We then find all pairs of co-prime factors (m, n) of this product and check which resulting number lies between 75 and 125. Step-by-Step Solution: Step 1: Let the numbers be 11m and 11n with gcd(m, n) = 1. Step 2: LCM(11m, 11n) = 11 * m * n. Step 3: Given LCM = 385, so 11 * m * n = 385. Step 4: Therefore, m * n = 385 / 11 = 35. Step 5: Factor pairs of 35 are (1,35) and (5,7), both co-prime pairs. Step 6: Possible numbers are 11 * 1 = 11, 11 * 35 = 385, and 11 * 5 = 55, 11 * 7 = 77. Step 7: Among these, the numbers between 75 and 125 are only 77. Verification / Alternative check: Check the pair (55, 77): H.C.F.(55,77) = 11 and L.C.M.(55,77) = (55 * 77) / 11 = 385. This pair satisfies the given conditions, and one of the numbers, 77, falls in the required range. Therefore, 77 is the correct choice. Why Other Options Are Wrong: 99: With 99 as one number, it is not possible to pair another integer so that H.C.F. = 11 and L.C.M. = 385. 110: This number exceeds the structure implied by the factors of 385 and 11. 88 and 121: Neither fits as a member of any pair with H.C.F. 11 and L.C.M. 385. Any number not derived from the factor pairs of 35 will violate the H.C.F. and L.C.M. conditions. Common Pitfalls: Students sometimes forget that after factoring out the H.C.F., the remaining co-factors must be co-prime. Others may guess numbers in the range without checking both H.C.F. and L.C.M. conditions, leading to incorrect answers. Final Answer: The number that lies between 75 and 125 is 77.

Question 8

Find the H.C.F. (greatest common divisor) of the following three numbers: 4 × 27 × 3125, 8 × 9 × 25 × 7, and 16 × 81 × 5 × 11 × 49.

Accepted Answer

Introduction: This question tests your ability to find the highest common factor (H.C.F.) of large numbers given in factorised product form. It reinforces prime factorisation skills and understanding of how to extract common factors across multiple composite numbers. Given Data / Assumptions: We are given three numbers in product form: First number: 4 × 27 × 3125 Second number: 8 × 9 × 25 × 7 Third number: 16 × 81 × 5 × 11 × 49 We need to find the H.C.F. of these three numbers. Concept / Approach: To find the H.C.F. of several numbers, we: Express each number as a product of prime powers. For each prime, take the smallest exponent that appears in all the numbers. Multiply these common prime factors together to get the H.C.F. Step-by-Step Solution: Step 1: Factorise each component. 4 = 2^2, 27 = 3^3, 3125 = 5^5. First number = 2^2 * 3^3 * 5^5. 8 = 2^3, 9 = 3^2, 25 = 5^2, 7 = 7^1. Second number = 2^3 * 3^2 * 5^2 * 7^1. 16 = 2^4, 81 = 3^4, 5 = 5^1, 11 = 11^1, 49 = 7^2. Third number = 2^4 * 3^4 * 5^1 * 7^2 * 11^1. Step 2: Identify common primes: 2, 3, and 5 appear in all three numbers. Step 3: Take minimum exponents across the three numbers. For 2: min(2, 3, 4) = 2. For 3: min(3, 2, 4) = 2. For 5: min(5, 2, 1) = 1. Step 4: H.C.F. = 2^2 * 3^2 * 5^1 = 4 * 9 * 5 = 180. Verification / Alternative check: We can divide all three original numbers by 180 and confirm that the quotients are integers, and that no higher common factor exists. Since we already used minimum prime exponents, 180 is guaranteed to be the greatest common divisor. Why Other Options Are Wrong: 360 and 540: These include higher powers of 2 or 3 than are common to all three numbers. 1260: This introduces extra factors like 7 and 11 that are not common to all three numbers. 90: This is a common factor but not the greatest, since 180 is larger and still divides all three numbers. Common Pitfalls: Errors often arise from miscalculating prime powers, forgetting that H.C.F. uses minimum exponents, or mixing H.C.F. with L.C.M. (which uses maximum exponents). Careful prime factorisation avoids these issues. Final Answer: The highest common factor of the three given numbers is 180.

Question 9

The H.C.F. of two numbers is 8. Which of the following values can never be the L.C.M. of these two numbers?

Accepted Answer

Introduction: This question examines your understanding of the basic relationship between the highest common factor (H.C.F.) and the least common multiple (L.C.M.) of two integers. It focuses on necessary divisibility conditions that any valid L.C.M. must satisfy when the H.C.F. is given. Given Data / Assumptions: H.C.F. of two numbers = 8. We have four candidate values for the L.C.M.: 32, 48, 60, 152, and an extra option 88. We must determine which value cannot possibly be the L.C.M. Concept / Approach: Key facts: The H.C.F. of two integers always divides their L.C.M. Therefore, if H.C.F. = 8, then any possible L.C.M. must be a multiple of 8. If a number is not divisible by 8, it cannot be the L.C.M. of two numbers whose H.C.F. is 8. Step-by-Step Solution: Step 1: Check each option for divisibility by 8. Step 2: 32 ÷ 8 = 4 (integer) → possible L.C.M. Step 3: 48 ÷ 8 = 6 (integer) → possible L.C.M. Step 4: 60 ÷ 8 = 7.5 (not an integer) → cannot be L.C.M. Step 5: 152 ÷ 8 = 19 (integer) → possible L.C.M. Step 6: 88 ÷ 8 = 11 (integer) → possible L.C.M. Step 7: Hence, 60 is the only option that is not divisible by 8. Verification / Alternative check: For an L.C.M. candidate L and H.C.F. H, the product of the two numbers a and b is H * L. If H does not divide L, there is no way to get integer values for a and b with H as their H.C.F. Therefore, any candidate not divisible by 8 is automatically invalid as an L.C.M. Why Other Options Are Wrong: 32: It is a multiple of 8, so it can be an L.C.M. for suitable numbers such as 8 and 32. 48: Also a multiple of 8, so valid for some pair of numbers with H.C.F. 8. 152: Divisible by 8, hence it can be an L.C.M. for appropriate numbers. 88: Another multiple of 8 and thus can serve as an L.C.M. Common Pitfalls: Students often forget the simple rule that H.C.F. must divide L.C.M. and may attempt unnecessarily complex calculations. Others confuse H.C.F. and L.C.M. relationships or only check whether the L.C.M. is larger than the H.C.F., which is not sufficient. Final Answer: The value that can never be the L.C.M. of two numbers with H.C.F. 8 is 60.

Question 10

A heap of coconuts is divided into groups of 2, of 3, and of 5. Each time, exactly one coconut is left over. What is the least possible number of coconuts in the heap?

Accepted Answer

Introduction: This is a classic remainder and least common multiple type question, testing your understanding of congruences and modular arithmetic. It asks for the smallest number that leaves the same remainder when divided by several different divisors. Given Data / Assumptions: A heap of coconuts is divided into groups of 2: remainder 1. The same heap is divided into groups of 3: remainder 1. The heap is also divided into groups of 5: remainder 1. We need the least positive number of coconuts satisfying all these conditions. Concept / Approach: If a number N leaves remainder 1 when divided by 2, 3, and 5, that means: N ≡ 1 (mod 2) N ≡ 1 (mod 3) N ≡ 1 (mod 5) Equivalently, N - 1 is divisible by 2, 3, and 5. Therefore, N - 1 must be a multiple of the L.C.M. of 2, 3, and 5. The smallest such N is obtained by adding 1 to this L.C.M. Step-by-Step Solution: Step 1: Compute LCM(2, 3, 5). Step 2: LCM(2, 3) = 6. Step 3: LCM(6, 5) = 30. Step 4: So N - 1 must be a multiple of 30. Step 5: The smallest positive multiple is 30 itself. Step 6: Therefore, N = 30 + 1 = 31. Verification / Alternative check: Check N = 31: 31 ÷ 2 = 15 remainder 1. 31 ÷ 3 = 10 remainder 1. 31 ÷ 5 = 6 remainder 1. All three conditions are satisfied, and since we used the least non-zero multiple of 30, 31 is the least positive solution. Why Other Options Are Wrong: 63: 63 ÷ 5 leaves remainder 3, not 1. 16: 16 ÷ 3 leaves remainder 1, but 16 ÷ 5 leaves remainder 1, and 16 ÷ 2 leaves 0 remainder, so one condition fails. 27: 27 ÷ 5 leaves remainder 2, not 1. 46: 46 ÷ 3 leaves remainder 1, but 46 ÷ 5 leaves remainder 1 and 46 ÷ 2 leaves 0 remainder, failing the requirement for division by 2. Common Pitfalls: Some students test each option directly without recognising the underlying L.C.M. and modular arithmetic shortcut. Others may mistakenly use H.C.F. instead of L.C.M. or forget that N - 1, not N, must be a multiple of 2, 3, and 5. Final Answer: The least possible number of coconuts in the heap is 31.

Question 11

Find the least common multiple (L.C.M.) of the three numbers 80, 85, and 90.

Accepted Answer

Introduction: Finding the least common multiple (L.C.M.) of several numbers is a standard aptitude skill. This question uses moderately sized numbers to test your comfort with prime factorisation and the rule for computing L.C.M. from prime powers. Given Data / Assumptions: The three numbers are 80, 85, and 90. We are required to find their L.C.M. Concept / Approach: To find the L.C.M. of several numbers: Express each number as a product of prime powers. For each prime that appears in any factorisation, take the highest exponent. Multiply all these prime powers together to obtain the L.C.M. Step-by-Step Solution: Step 1: Factorise each number. 80 = 8 * 10 = 2^4 * 5. 85 = 5 * 17 = 5 * 17. 90 = 9 * 10 = 2 * 3^2 * 5. Step 2: Collect all distinct primes: 2, 3, 5, and 17. Step 3: Choose the highest exponent of each prime across the three numbers. For 2: max exponent is 4 (from 80). For 3: max exponent is 2 (from 90). For 5: max exponent is 1 (all contain at most 5^1). For 17: exponent is 1 (from 85). Step 4: L.C.M. = 2^4 * 3^2 * 5 * 17. Step 5: Compute stepwise: 2^4 = 16; 3^2 = 9. Step 6: 16 * 9 = 144. Step 7: 144 * 5 = 720. Step 8: 720 * 17 = 12240. Verification / Alternative check: Confirm that 12240 is divisible by each number: 12240 ÷ 80 = 153 (integer). 12240 ÷ 85 = 144 (integer). 12240 ÷ 90 = 136 (integer). Since 12240 is divisible by all three numbers and is obtained using maximum prime powers, it is indeed their least common multiple. Why Other Options Are Wrong: 11440, 11998, 12880, 8160: None of these numbers are obtained using the correct prime power combination, and some are not even divisible by all three numbers. For example, 11440 is not divisible by 85, and 8160 is not divisible by 85 either. Common Pitfalls: A common mistake is to multiply the numbers directly instead of using prime factors, which leads to unnecessarily large and incorrect results. Another error is to use minimum exponents of primes (which corresponds to H.C.F., not L.C.M.). Final Answer: The least common multiple of 80, 85, and 90 is 12240.

Question 12

A drink vendor has 368 litres of Maaza, 80 litres of Pepsi, and 144 litres of Sprite. He wants to pack these drinks in cans so that: • each can contains the same whole number of litres of a single drink, and • no can contains a mixture of two drinks…

Accepted Answer

Introduction: This question uses the concepts of highest common factor (H.C.F.) and divisibility in a real-life context. It checks whether you can identify a common capacity for containers such that the total number of containers used is minimised while volumes remain whole numbers. Given Data / Assumptions: Maaza: 368 litres. Pepsi: 80 litres. Sprite: 144 litres. All cans must have the same capacity in litres. Each can is filled with only one kind of drink. We want the least possible number of cans. Concept / Approach: To minimise the total number of cans, we need each can to have the greatest possible capacity that divides evenly into each of the three drink quantities. This capacity is the H.C.F. of 368, 80, and 144. Once we know the capacity, the number of cans for each drink is volume divided by capacity, and the total is the sum of these counts. Step-by-Step Solution: Step 1: Find HCF(368, 80, 144). Step 2: HCF(368, 80) = 16. Step 3: HCF(16, 144) = 16. Step 4: Therefore, maximum can capacity = 16 litres. Step 5: Number of cans for Maaza = 368 / 16 = 23. Step 6: Number of cans for Pepsi = 80 / 16 = 5. Step 7: Number of cans for Sprite = 144 / 16 = 9. Step 8: Total cans required = 23 + 5 + 9 = 37. Verification / Alternative check: If we used a smaller common divisor, such as 8 litres per can, the number of cans would be: 368 / 8 = 46, 80 / 8 = 10, 144 / 8 = 18; total = 74 cans. This is clearly more than 37, so 16 litres per can is better. Because 16 is the H.C.F., no larger capacity works for all three volumes, so 37 is indeed the minimum number of cans. Why Other Options Are Wrong: 47, 46, 41, 35: These totals would correspond to either smaller can sizes or non-integer volumes per can, violating the conditions. They do not arise from using the true H.C.F. as the capacity. Common Pitfalls: Learners sometimes choose the least common multiple or mistakenly use any common divisor instead of the greatest one. Others forget that the can capacity must divide all three quantities exactly, leading to fractional cans, which are not allowed. Final Answer: The least number of cans required is 37.

Question 13

There are 1001 pens and 910 pencils to be distributed among students. Each student must receive the same number of pens and the same number of pencils. What is the maximum possible number of students?

Accepted Answer

Introduction: This question is a direct application of the highest common factor (H.C.F.) concept in a distribution context. It tests whether you can use H.C.F. to maximise the number of groups (students) while ensuring equal distribution of items (pens and pencils) to each student. Given Data / Assumptions: Total pens = 1001. Total pencils = 910. Each student must receive the same number of pens. Each student must also receive the same number of pencils. No pen or pencil should be left undistributed. Concept / Approach: If each student gets the same number of pens and pencils, then the number of students must divide both 1001 and 910 exactly. To maximise the number of students, we must find the greatest number that divides both totals, which is the H.C.F. of 1001 and 910. Step-by-Step Solution: Step 1: Compute HCF(1001, 910). Step 2: Factorise 1001 = 7 * 11 * 13. Step 3: Factorise 910 = 7 * 13 * 10 = 2 * 5 * 7 * 13. Step 4: Common prime factors are 7 and 13. Step 5: Therefore, H.C.F. = 7 * 13 = 91. Step 6: Hence, the maximum number of students possible is 91. Verification / Alternative check: Check divisibility: 1001 ÷ 91 = 11 pens per student. 910 ÷ 91 = 10 pencils per student. Both are integers, meaning all pens and pencils are distributed equally among 91 students with no remainder left. Why Other Options Are Wrong: 910 or 1001: These would require each student to receive fewer than one item of the other type, which is impossible with whole items. 1911: This is greater than both 1001 and 910, so it cannot divide either total. 70: Although 70 divides 910, it does not divide 1001 exactly. Common Pitfalls: Some students mistakenly compute the least common multiple or simply take a random common divisor instead of the greatest. Others may not factorise properly and miss the correct H.C.F. Using prime factorisation or the Euclidean algorithm systematically avoids such errors. Final Answer: The maximum number of students among whom the pens and pencils can be distributed equally is 91.

Question 14

The H.C.F. of 3240, 3600, and a third number is 36. The L.C.M. of these three numbers is 2^4 × 3^5 × 5^2 × 7^2. What is the value of the third number?

Accepted Answer

Introduction: This is a higher-level H.C.F. and L.C.M. question involving three numbers. It requires careful use of prime factorisation and understanding how H.C.F. and L.C.M. relate to the prime exponents of all the numbers involved. Such questions are common in advanced aptitude and number theory practice. Given Data / Assumptions: Numbers: 3240, 3600, and an unknown third number N. H.C.F.(3240, 3600, N) = 36. L.C.M.(3240, 3600, N) = 2^4 × 3^5 × 5^2 × 7^2. We need to determine the value of N. Concept / Approach: We use prime factorisation of the known numbers: For H.C.F. across all numbers, take the minimum exponent of each prime among the three numbers. For L.C.M., take the maximum exponent of each prime among the three numbers. Using the known H.C.F. and L.C.M., we deduce the prime exponents of the third number N so that both conditions are satisfied. Step-by-Step Solution: Step 1: Prime factorise 3240 and 3600. 3240 = 2^3 * 3^4 * 5^1. 3600 = 2^4 * 3^2 * 5^2. Step 2: HCF(3240, 3600) = 2^3 * 3^2 * 5^1 = 360. Step 3: HCF(3240, 3600, N) = 36 = 2^2 * 3^2. → So N must have at least 2^2 and 3^2 and must not have factor 5 (otherwise HCF would include 5). Step 4: LCM of all three is given as 2^4 * 3^5 * 5^2 * 7^2. From 3240 and 3600, maximum exponents so far: 2^4, 3^4, 5^2. Step 5: To reach the LCM exponents, N must contribute exponents: For 3: need 3^5, so N has 3^5. For 7: need 7^2, so N has 7^2. For 2: max exponent given is already 4 from 3600, so N can have 2^a with a ≤ 4 but at least 2 (for HCF). Step 6: To keep LCM exponent of 2 at 4, we choose smallest valid a = 2. Step 7: Therefore, N = 2^2 * 3^5 * 7^2. Step 8: Compute N: 2^2 = 4, 3^5 = 243, 7^2 = 49. Step 9: First 4 * 243 = 972. Step 10: Then 972 * 49 = 47628. Verification / Alternative check: Check H.C.F. condition: gcd(3240, 3600, 47628) is 36, and the L.C.M. computed from these three numbers matches 2^4 * 3^5 * 5^2 * 7^2. This confirms that N = 47628 is consistent with the given information. Why Other Options Are Wrong: 49874, 24157, 42146, 95256: These numbers do not have the exact prime factor structure required to produce both the specified H.C.F. and L.C.M. with 3240 and 3600. In particular, they either introduce unwanted factors, omit needed powers of 3 or 7, or disturb the H.C.F. from 36. Common Pitfalls: Many learners confuse how to use exponents for H.C.F. and L.C.M., sometimes taking maximum exponents for H.C.F. or minimum exponents for L.C.M., which is reversed. Others may not fully factorise the given numbers, leading to mistakes in deducing the third number. Careful factorisation and systematic reasoning are crucial here. Final Answer: The value of the third number is 47628.

Question 15

Find the least common multiple (L.C.M.) of the two numbers 54 and 60.

Accepted Answer

Introduction: This problem reinforces the method of finding the least common multiple (L.C.M.) using prime factorisation. Working with two composite numbers like 54 and 60 helps you practise identifying prime powers and using them correctly to compute the L.C.M. Given Data / Assumptions: The numbers are 54 and 60. We need to find their L.C.M. Concept / Approach: For two numbers a and b: Prime factorise each number. For each prime that appears in either factorisation, take the highest exponent. Multiply those prime powers together to get the L.C.M. This ensures the L.C.M. is the smallest number divisible by both a and b. Step-by-Step Solution: Step 1: Factorise 54. 54 = 6 * 9 = 2 * 3^3. Step 2: Factorise 60. 60 = 6 * 10 = 2^2 * 3 * 5. Step 3: Gather all primes: 2, 3, and 5. Step 4: Take maximum exponents. For 2: max exponent is 2 (from 60). For 3: max exponent is 3 (from 54). For 5: max exponent is 1 (from 60). Step 5: L.C.M. = 2^2 * 3^3 * 5. Step 6: Compute: 2^2 = 4; 3^3 = 27. Step 7: 4 * 27 = 108. Step 8: 108 * 5 = 540. Verification / Alternative check: Check divisibility: 540 ÷ 54 = 10 (integer). 540 ÷ 60 = 9 (integer). Since 540 is divisible by both numbers and is generated from maximum prime exponents, it is indeed their least common multiple. Why Other Options Are Wrong: 600: Divisible by 60 but not by 54, since 600 ÷ 54 is not an integer. 60 and 54: Each is one of the numbers, not a common multiple of both. 108: Divisible by 54 but not by 60, so it cannot be the L.C.M. Common Pitfalls: Many students mistakenly multiply the numbers directly instead of using prime factorisation, getting a much larger multiple. Others may forget to include a prime (like 5 here) or use the smaller exponent instead of the larger one when computing the L.C.M. Final Answer: The least common multiple of 54 and 60 is 540.

Question 16

The H.C.F. of 210 and 55 is expressible in the form 210 × 5 + 55P. What is the value of P?

Accepted Answer

Introduction: This question tests your understanding of the Euclidean algorithm and the idea that the highest common factor (H.C.F.) of two integers can be expressed as a linear combination of those integers. It is a fundamental concept in number theory and is often used to solve Diophantine equations. Given Data / Assumptions: Two numbers: 210 and 55. Their H.C.F. can be written as 210 × 5 + 55P. We need to find the integer value of P. Concept / Approach: First, we compute the H.C.F. of 210 and 55 using either prime factorisation or the Euclidean algorithm. Then we equate this H.C.F. to the given expression 210 × 5 + 55P and solve for P. Note that the linear combination may involve a negative coefficient. Step-by-Step Solution: Step 1: Compute HCF(210, 55). 210 ÷ 55 = 3 remainder 45 (since 55 * 3 = 165 and 210 - 165 = 45). 55 ÷ 45 = 1 remainder 10. 45 ÷ 10 = 4 remainder 5. 10 ÷ 5 = 2 remainder 0. Therefore, H.C.F. = 5. Step 2: Given that HCF = 210 × 5 + 55P. Step 3: Substitute HCF = 5 into the expression: 5 = 210 × 5 + 55P. Step 4: Compute 210 × 5 = 1050. Step 5: So 5 = 1050 + 55P. Step 6: Rearrange: 55P = 5 - 1050 = -1045. Step 7: Therefore, P = -1045 / 55 = -19. Verification / Alternative check: Verify that 210 × 5 + 55 × (-19) = 1050 - 1045 = 5, which matches the H.C.F. This confirms that P = -19 is correct and that the linear combination indeed yields the greatest common divisor. Why Other Options Are Wrong: -23: Would give 210 × 5 + 55 × (-23) = 1050 - 1265 = -215, not equal to 5. 16 and 27: These positive values of P make 210 × 5 + 55P far larger than 5 and not equal to the H.C.F. 5: Substituting gives 210 × 5 + 55 × 5 = 1050 + 275 = 1325, again not the H.C.F. Common Pitfalls: Students sometimes assume that both coefficients in the linear combination must be positive, which is not true. Others may miscalculate the H.C.F. or make algebraic errors while solving for P. Carefully following the steps of the Euclidean algorithm and algebraic manipulation avoids these mistakes. Final Answer: The correct value of P is -19.

Question 17

The difference between two numbers is 14. Their L.C.M. is 441 and their H.C.F. is 7. What are the two numbers?

Accepted Answer

Introduction: This question tests multiple number theory concepts at once: the relationship between H.C.F., L.C.M., and the numbers themselves, along with the given difference between the two numbers. Solving it requires both algebraic reasoning and understanding of prime factorisation. Given Data / Assumptions: Difference between the two numbers = 14. L.C.M. of the numbers = 441. H.C.F. of the numbers = 7. We need to find the two numbers. Concept / Approach: Let the two numbers be a and b. Since their H.C.F. is 7, we can write: a = 7x and b = 7y where gcd(x, y) = 1. Also: LCM(a, b) = 7xy = 441 and a - b = 14 ⇒ 7|x - y| = 14 ⇒ |x - y| = 2. We must find co-prime integers x and y that satisfy these conditions. Step-by-Step Solution: Step 1: From LCM, 7xy = 441. Step 2: So xy = 441 / 7 = 63. Step 3: From the difference, |a - b| = 14 ⇒ |7x - 7y| = 14. Step 4: Therefore, 7|x - y| = 14 ⇒ |x - y| = 2. Step 5: We need co-prime positive integers x and y with product 63 and difference 2. Step 6: Factor pairs of 63: (1,63), (3,21), (7,9). Step 7: Only the pair (7,9) has difference 2 and gcd(7,9) = 1. Step 8: Therefore, x = 7 and y = 9 (or vice versa). Step 9: The numbers are a = 7x = 49 and b = 7y = 63. Step 10: So the two numbers are 49 and 63. Verification / Alternative check: Check the conditions: Difference: 63 - 49 = 14 (matches). H.C.F.(49, 63) = 7. LCM(49, 63) = (49 * 63) / 7 = 441. All given conditions are satisfied, so the answer is correct. Why Other Options Are Wrong: 64 and 48: Difference is 16, and H.C.F. is not 7. 62 and 46: Difference 16 again, and their H.C.F. is not 7. 64 and 49: Difference 15, not 14. 56 and 42: Difference 14, but H.C.F. is 14 and L.C.M. is not 441. Common Pitfalls: Some learners forget to factor out the H.C.F. into the form 7x and 7y, or do not enforce that x and y must be co-prime. Others may not systematically check factor pairs of 63 for the required difference and coprimality, leading to incorrect guesses. Final Answer: The two numbers are 63 and 49.

Question 18

The H.C.F. of two numbers is 21 and their L.C.M. is 84. If the ratio of the two numbers is 1 : 4, what is the larger of the two numbers?

Accepted Answer

Introduction: This question checks your understanding of how H.C.F., L.C.M., and the ratio of two numbers are related. It is a typical aptitude problem where you must reconstruct the original numbers from their H.C.F., L.C.M., and ratio. Given Data / Assumptions: H.C.F. of the two numbers = 21. L.C.M. of the two numbers = 84. The ratio of the numbers is 1 : 4. We must find the larger of the two numbers. Concept / Approach: If two numbers are in the ratio m : n and have H.C.F. h, then the numbers can be written as: h * m and h * n Their L.C.M. is then: h * m * n (when m and n are co-prime) We use the ratio and H.C.F. to get the numbers, and check consistency with the given L.C.M. Step-by-Step Solution: Step 1: Let the numbers be 21 * 1 and 21 * 4, since the ratio is 1 : 4. Step 2: So the numbers are 21 and 84. Step 3: Check the H.C.F. of 21 and 84. H.C.F.(21, 84) = 21 (correct). Step 4: Compute their L.C.M. using the product relation. Product = 21 * 84 = 1764. Using HCF * LCM = product: 21 * LCM = 1764 ⇒ LCM = 1764 / 21 = 84. Step 5: The calculated L.C.M. matches the one given. Step 6: Therefore, the larger number is 84. Verification / Alternative check: Besides the H.C.F. and L.C.M. check, the ratio condition is satisfied: 21 : 84 = 1 : 4 (dividing both by 21). Thus, the numbers 21 and 84 correctly meet all given constraints and 84 is indeed the larger number. Why Other Options Are Wrong: 48 and 12: Neither pair with any other number in the ratio 1 : 4 gives H.C.F. 21 and L.C.M. 84. 108: With the smaller number in 1 : 4 ratio, L.C.M. and H.C.F. would not match the given values. 21: This is the smaller number, not the larger one. Common Pitfalls: A common mistake is to ignore the ratio and try to guess numbers randomly from the options. Others may misapply the relation between H.C.F., L.C.M., and the product of numbers. Correctly interpreting the ratio and using the H.C.F. as a common factor leads quickly to the solution. Final Answer: The larger of the two numbers is 84.

Question 19

In number theory, find the highest common factor (HCF) of the three numbers 4 × 27 × 3125, 8 × 9 × 25 × 7, and 16 × 81 × 5 × 11 × 49.

Accepted Answer

Introduction: This aptitude question tests your understanding of how to find the highest common factor (HCF) of large numbers that are expressed as products of smaller factors. Such questions become very simple when you use prime factorisation instead of trying to divide directly. Given Data / Assumptions: First number: 4 × 27 × 3125 Second number: 8 × 9 × 25 × 7 Third number: 16 × 81 × 5 × 11 × 49 We need to calculate the HCF of these three numbers. Concept / Approach: The HCF of a set of numbers is found by taking the common prime factors with their smallest powers. So we first express each number as a product of prime numbers. Then we identify which primes are common to all three numbers and select the minimum exponent of each common prime. Finally, we multiply these to get the HCF. Step-by-Step Solution: First number: 4 × 27 × 3125 4 = 2^2, 27 = 3^3, 3125 = 5^5, so first number = 2^2 × 3^3 × 5^5 Second number: 8 × 9 × 25 × 7 8 = 2^3, 9 = 3^2, 25 = 5^2, 7 = 7, so second number = 2^3 × 3^2 × 5^2 × 7 Third number: 16 × 81 × 5 × 11 × 49 16 = 2^4, 81 = 3^4, 5 = 5, 11 = 11, 49 = 7^2, so third number = 2^4 × 3^4 × 5 × 7^2 × 11 Common primes to all three: 2, 3, and 5 Smallest exponent of 2 is 2, of 3 is 2, of 5 is 1 Therefore, HCF = 2^2 × 3^2 × 5 = 4 × 9 × 5 = 180 Verification / Alternative check: We can check quickly: 180 divides each number because each number contains at least 2^2, 3^2, and 5 as factors. Also, no larger number with these primes will divide all three, since at least one number would not contain the higher power. So 180 is confirmed as the highest common factor. Why Other Options Are Wrong: 360 is 2^3 × 3^2 × 5 and requires 2^3 in every number, which the first number does not have. 540 includes 3^3, which is not present in all three. 1260 contains extra prime factors (7) or higher powers, so it cannot be a common factor of all three numbers. Common Pitfalls: A common mistake is to multiply all visible numbers directly and then guess the HCF without doing factorisation. Another error is to include primes like 7 or 11 in the HCF just because they appear in one or two numbers rather than all three. Always work with prime factors systematically. Final Answer: The highest common factor of the given three numbers is 180.

Question 20

A rectangular room has dimensions 6 metres 24 centimetres by 4 metres 32 centimetres. What is the least number of equal square tiles required to completely cover the floor of the room?

Accepted Answer

Introduction: This question checks your understanding of using the highest common factor (HCF) to solve geometry and tiling problems. We want to cover a floor with identical square tiles without cutting any tile and without leaving gaps. Given Data / Assumptions: Length of the room = 6 metres 24 centimetres Width of the room = 4 metres 32 centimetres Tiles are square and all have the same size We need the least possible number of tiles Concept / Approach: To minimise the number of square tiles, each tile should be as large as possible. The side of the largest possible square that can exactly tile the floor must be the HCF of the two dimensions (expressed in the same unit). Once we know the side of each tile, we divide the area of the floor by the area of one tile to get the number of tiles. Step-by-Step Solution: Convert dimensions to centimetres: Length = 6 m 24 cm = 6 × 100 + 24 = 624 cm Width = 4 m 32 cm = 4 × 100 + 32 = 432 cm Find HCF of 624 and 432: 624 = 2^4 × 3 × 13 432 = 2^4 × 3^3 Common prime factors with smallest powers: 2^4 × 3 = 16 × 3 = 48 So side of each largest square tile = 48 cm Number of tiles along length = 624 / 48 = 13 Number of tiles along width = 432 / 48 = 9 Total tiles required = 13 × 9 = 117 Verification / Alternative check: We can check by multiplication: 117 × (48 × 48) must equal 624 × 432. Since the ratios along length and width are integers (13 and 9), the tiling is exact and no cutting is needed. This confirms the correctness of the HCF based approach. Why Other Options Are Wrong: 107, 127, and 137 are just random numbers around the correct value and do not come from the factorisation. They would correspond to non integer tile dimensions, which would not fit the room exactly when using square tiles of equal size. Common Pitfalls: Many learners forget to convert metres into centimetres before finding the HCF. Another mistake is using the least common multiple (LCM) instead of HCF, which would give incorrect dimensions and far too many tiles. Always remember that maximum tile size corresponds to the HCF of the two side lengths. Final Answer: The minimum number of equal square tiles required to cover the room exactly is 117.

HCF and LCM Questions