Question 1

The product of two positive integers is 4941 and their least common multiple (LCM) is 81. What is the highest common factor (HCF) of these two numbers?

Accepted Answer

Introduction:This question evaluates your grasp of the standard relationship between the highest common factor (HCF), least common multiple (LCM) and the product of two numbers. It is a direct formula-based problem that can be solved very quickly once you know the rule. Given Data / Assumptions: Product of the two numbers = 4941. LCM of the two numbers = 81. Both numbers are positive integers. We need to find the HCF. Concept / Approach:For any two positive integers a and b, the following well-known relationship holds:a * b = HCF(a, b) * LCM(a, b)Rearranging this formula allows us to compute the HCF when the product and LCM are known:HCF(a, b) = (a * b) / LCM(a, b) Step-by-Step Solution:Step 1: Start from the formula.a * b = HCF * LCMStep 2: Substitute given values.4941 = HCF * 81Step 3: Solve for HCF.HCF = 4941 / 81Step 4: Do the division.81 * 60 = 4860, remainder = 8181 * 61 = 4860 + 81 = 4941Therefore, HCF = 61 Verification / Alternative check:Once we get HCF = 61, we can imagine two numbers whose product is 4941 and which share this factor. Writing one number as 61 * p and the other as 61 * q, their product becomes 61 * 61 * p * q = 4941. Dividing 4941 by 61 gives 81, so p * q = 81, and LCM(61p, 61q) must be 81 * 61 / 61 = 81, matching the given LCM. Why Other Options Are Wrong:60: 60 * 81 = 4860, which is less than 4941.59: 59 * 81 = 4779, not equal to 4941.35: 35 * 81 = 2835, far from the required product.27: 27 * 81 = 2187, also incorrect. Common Pitfalls:Typical mistakes include attempting to factorize 4941 manually without using the formula, or randomly trying divisors. Always remember that product = HCF * LCM simplifies such questions greatly. Final Answer:The highest common factor (HCF) of the two numbers is 61.

Question 2

The highest common factor (HCF) and least common multiple (LCM) of two numbers are 11 and 825 respectively. If one of the numbers is 275, what is the other number?

Accepted Answer

Introduction:This question again uses the fundamental relationship between HCF, LCM and the product of two numbers. Such questions are designed to check how quickly you can convert verbal data into a mathematical formula and solve it. Given Data / Assumptions: HCF of the two numbers = 11. LCM of the two numbers = 825. One of the numbers = 275. Find the other number. Concept / Approach:Use the identity for two positive integers a and b:a * b = HCF(a, b) * LCM(a, b)With HCF, LCM and one number known, the other number can be obtained directly by division. Step-by-Step Solution:Step 1: Apply the formula.a * b = HCF * LCMStep 2: Substitute values.275 * (other number) = 11 * 825Step 3: Calculate the right-hand side.11 * 825 = 9075Step 4: Solve for the other number.Other number = 9075 / 275Step 5: Perform the division.275 * 33 = 9075, so the other number is 33 Verification / Alternative check:Check HCF and LCM for 33 and 275. Factorize: 33 = 3 * 11, 275 = 5^2 * 11. Common factor is 11, so HCF = 11. LCM is (33 * 275) / 11 = 9075 / 11 = 825. Both values match the given information, confirming the answer. Why Other Options Are Wrong:45: 275 * 45 = 12375, not equal to 11 * 825 = 9075.53: 275 * 53 = 14575, also incorrect.75: 275 * 75 = 20625, does not match required product.99: 275 * 99 = 27225, far greater than 9075. Common Pitfalls:Some candidates attempt to guess factors of 825 or 275 and get confused. Always rely on the direct formula and perform clean arithmetic. Factorization is only needed for verification, not for the main computation here. Final Answer:The other number is 33.

Question 3

Two numbers are in the ratio 15 : 11. If their highest common factor (HCF) is 13, what is the greater of the two numbers?

Accepted Answer

Introduction:This question checks whether you understand how the highest common factor (HCF) relates to numbers given in a specific ratio. This is a common pattern in ratio and HCF questions asked in competitive examinations. Given Data / Assumptions: The ratio of two numbers is 15 : 11. Their HCF is 13. We must find the greater number. Concept / Approach:If two numbers are in the ratio m : n and their HCF is h, then the actual numbers can be written as:First number = m * hSecond number = n * hThis works because the HCF is the common factor that multiplies the ratio terms to give the actual numbers. Step-by-Step Solution:Step 1: Identify ratio terms.m = 15, n = 11, and HCF = 13Step 2: Compute the actual numbers.First number = 15 * 13 = 195Second number = 11 * 13 = 143Step 3: Identify the greater number.Between 195 and 143, the greater number is 195 Verification / Alternative check:Check that the HCF of 195 and 143 is indeed 13. Factorization: 195 = 3 * 5 * 13, 143 = 11 * 13. The common factor is 13, so HCF = 13. Now, the ratio 195 : 143 simplifies by dividing both sides by 13, giving 15 : 11, which matches the given ratio. Thus, our computation is fully consistent. Why Other Options Are Wrong:125, 175, 143, 169: None of these, when paired with an appropriate second number using HCF = 13, will retain the exact ratio 15 : 11. For example, 143 is actually the smaller number, not the greater one. The other values do not follow from multiplying 15 or 11 by 13. Common Pitfalls:A common mistake is to forget that HCF must be a factor of both numbers, or to treat the ratio 15 : 11 as actual numbers rather than scaled versions. Always multiply the ratio terms by the HCF to recover the real numbers. Final Answer:The greater number is 195.

Question 4

What is the highest common factor (HCF) of the two integers 6345 and 2160?

Accepted Answer

Introduction:Finding the highest common factor (HCF), also known as greatest common divisor (GCD), of two numbers is a core arithmetic skill used in many aptitude questions. This problem asks you to determine the HCF of 6345 and 2160 using systematic factorization or the Euclidean algorithm. Given Data / Assumptions: First number = 6345. Second number = 2160. Both are positive integers. We have to find their HCF. Concept / Approach:The most efficient method is the Euclidean algorithm, which uses repeated division to find the greatest common divisor. The idea is:HCF(a, b) = HCF(b, a mod b)and this is repeated until the remainder becomes zero. The last non-zero remainder is the HCF. Step-by-Step Solution:Step 1: Arrange numbers.Take a = 6345 and b = 2160.Step 2: Apply Euclidean algorithm.6345 ÷ 2160 gives quotient 2 and remainder r1.2 * 2160 = 4320, so r1 = 6345 − 4320 = 2025.Now HCF(6345, 2160) = HCF(2160, 2025).Step 3: Continue with 2160 and 2025.2160 − 2025 = 135, so remainder r2 = 135.Now HCF(2160, 2025) = HCF(2025, 135).Step 4: Divide 2025 by 135.2025 ÷ 135 = 15 exactly, remainder 0.Therefore, HCF = 135 Verification / Alternative check:Check that 135 divides both numbers exactly. 6345 ÷ 135 = 47, which is an integer. 2160 ÷ 135 = 16, also an integer. Thus 135 is a common factor. Any greater common divisor would need to be larger than 135, but 135 already embodies all common prime factors, so it is indeed the highest common factor. Why Other Options Are Wrong:45: Although 45 divides both numbers, it is not the largest such factor since 135 is bigger and also divides both.270: 2160 ÷ 270 = 8, but 6345 ÷ 270 is not an integer.15 and 90: Both divide the numbers but are smaller than 135, so they cannot be the HCF. Common Pitfalls:Students may stop at a smaller common factor like 45 or 90, forgetting to check if there is a larger one. Using the Euclidean algorithm avoids guessing and ensures you reach the greatest common factor systematically. Final Answer:The highest common factor of 6345 and 2160 is 135.

Question 5

What is the least common multiple (LCM) of the integers 16, 24, 36, 52 and 54?

Accepted Answer

Introduction:This question tests your ability to find the least common multiple (LCM) of several integers. LCM problems are widely used in aptitude exams to examine how well you work with prime factorization and multiples. Given Data / Assumptions: The numbers are 16, 24, 36, 52 and 54. All numbers are positive integers. We need the smallest positive integer that is exactly divisible by each of these numbers. Concept / Approach:The LCM of a list of numbers is obtained by taking the highest power of each prime that appears in the prime factorization of any of the numbers. We then multiply these highest powers together:LCM = product of all primes raised to their maximum exponent appearing in any number Step-by-Step Solution:Step 1: Prime factorization of each number.16 = 2^424 = 2^3 * 336 = 2^2 * 3^252 = 2^2 * 1354 = 2 * 3^3Step 2: Take highest powers of all primes involved.For prime 2: maximum exponent = 4 (from 16).For prime 3: maximum exponent = 3 (from 54).For prime 13: maximum exponent = 1 (from 52).Step 3: Form the LCM.LCM = 2^4 * 3^3 * 13Step 4: Compute the value.2^4 = 163^3 = 2716 * 27 = 432432 * 13 = 5616 Verification / Alternative check:Check divisibility: 5616 ÷ 16, 24, 36, 52 and 54 should all be integers. For example, 5616 ÷ 16 = 351, 5616 ÷ 24 = 234, 5616 ÷ 36 = 156, 5616 ÷ 52 = 108, and 5616 ÷ 54 = 104, all integers. Hence 5616 is a common multiple, and because we have used maximum prime powers, it is the least such multiple. Why Other Options Are Wrong:5618 and 5216: These are not constructed from the proper prime powers and will not be divisible by all the given numbers.432: Although it is multiple of 16, 24 and 36, it is not divisible by 52 and 54.10368: This is a multiple of 5616 and thus a common multiple but not the least one, so it cannot be the LCM. Common Pitfalls:Errors often occur when students miss one of the numbers or take the wrong highest power of a prime factor. Writing the factorization clearly and picking the maximum exponents systematically prevents such mistakes. Final Answer:The least common multiple of 16, 24, 36, 52 and 54 is 5616.

Question 6

What is the least common multiple (LCM) of the two integers 64 and 56?

Accepted Answer

Introduction:This is a straightforward LCM calculation involving two numbers. It tests your understanding of prime factorization and how to use it to calculate the least common multiple quickly and accurately. Given Data / Assumptions: First number = 64. Second number = 56. We are asked to find their least common multiple (LCM). Concept / Approach:The LCM of two integers can be found by using prime factorization. We express each number as a product of prime powers and then take the highest power of each prime that appears in any factorization.LCM = product of primes raised to maximum exponents appearing in either number Step-by-Step Solution:Step 1: Prime factorize each number.64 = 2^656 = 2^3 * 7Step 2: Take the highest exponent for each prime.For prime 2: max exponent is 6 (from 64).For prime 7: max exponent is 1 (from 56).Step 3: Form the LCM.LCM = 2^6 * 7Step 4: Compute the value.2^6 = 6464 * 7 = 448 Verification / Alternative check:Check that 448 is divisible by both numbers. 448 ÷ 64 = 7, and 448 ÷ 56 = 8. Both quotients are integers, so 448 is a common multiple. Because it is constructed from prime factors with minimal exponents that cover both numbers, it is the least such multiple. Why Other Options Are Wrong:488, 484, 408, 512: None of these are the true LCM. For example, 512 is a multiple of 64 but not of 56. The others either fail divisibility by 64 or 56 or both. Only 448 satisfies divisibility by both numbers and is the smallest such number. Common Pitfalls:Learners sometimes mistakenly multiply the two numbers directly (64 * 56) instead of factoring them, which gives 3584, a common multiple but not the least. Another error is miscalculating the prime exponents, such as using 2^3 instead of 2^6. Careful factorization prevents these errors. Final Answer:The least common multiple of 64 and 56 is 448.

Question 7

What is the least common multiple (LCM) of the two integers 57 and 93?

Accepted Answer

Introduction:This problem asks you to find the least common multiple (LCM) of two numbers, 57 and 93. Such LCM questions are simple when you apply prime factorization or use the relationship between product, HCF and LCM. Given Data / Assumptions: First number = 57. Second number = 93. We need their least common multiple. Concept / Approach:We can use prime factorization:57 = 3 * 1993 = 3 * 31The LCM is obtained by taking each prime that appears in either factorization with the highest power that appears. Alternatively, for two numbers, we can also use:LCM(a, b) = (a * b) / HCF(a, b) Step-by-Step Solution:Step 1: Prime factorize.57 = 3 * 1993 = 3 * 31Step 2: Identify distinct primes.Primes involved are 3, 19 and 31.Step 3: LCM is product of these primes with exponent 1.LCM = 3 * 19 * 31Step 4: Compute stepwise.3 * 19 = 5757 * 31 = 1767 Verification / Alternative check:Check divisibility: 1767 ÷ 57 = 31, an integer, and 1767 ÷ 93 = 19, also an integer. Thus 1767 is a common multiple of 57 and 93. There is no smaller positive integer with this property because we used the minimal prime factors needed, so 1767 is indeed the least common multiple. Why Other Options Are Wrong:1567, 1576, 1919: These are not multiples of both 57 and 93. Dividing them by 57 or 93 does not give an integer.114: While 114 is a common multiple of 57 and 93 (114 ÷ 57 = 2, 114 ÷ 93 is not an integer), so even this is not a common multiple of both. Hence only 1767 qualifies. Common Pitfalls:One common mistake is to assume that LCM must be close to or smaller than the product a * b, without actually doing prime factorization. Another is to miscalculate the multiplication 57 * 31. Taking it step by step avoids such arithmetic slips. Final Answer:The least common multiple of 57 and 93 is 1767.

Question 8

What is the highest common factor (HCF) of the integers 133 and 112?

Accepted Answer

Introduction:This question is about finding the highest common factor (HCF) or greatest common divisor (GCD) of two numbers, 133 and 112. It helps reinforce mental factorization skills that are important in many quantitative aptitude topics such as fractions, ratios and simplification. Given Data / Assumptions: First number = 133. Second number = 112. We need to determine their HCF. Concept / Approach:The HCF of two numbers is the largest positive integer that divides both of them exactly. We can find it either by prime factorization or by the Euclidean algorithm. Here, prime factorization is quite simple. Step-by-Step Solution:Step 1: Factorize 133.133 = 7 * 19Step 2: Factorize 112.112 = 16 * 7 = 2^4 * 7Step 3: Identify common prime factors.Common prime factor is 7.Step 4: Compute the HCF.Since 7 is the only common prime factor and appears to the power 1 in both, HCF = 7 Verification / Alternative check:We can verify by direct division. 133 ÷ 7 = 19, an integer, and 112 ÷ 7 = 16, also an integer. Try any larger candidate such as 14 or 19: 133 is not divisible by 14, and 112 is not divisible by 19. Hence no number larger than 7 divides both numbers, confirming that 7 is indeed the highest common factor. Why Other Options Are Wrong:15 and 6: Neither of these divides 133 exactly, so they cannot be the HCF.19: While 19 divides 133, it does not divide 112, so it cannot be a common factor.14: 14 divides 112 but not 133, so it is also not a common factor of both. Common Pitfalls:Some students check small divisors randomly and may miss the correct one if they do not factorize systematically. Others think a larger factor like 19 or 14 may work without verifying divisibility for both numbers. Always factor or apply the Euclidean algorithm to be sure. Final Answer:The highest common factor of 133 and 112 is 7.

Question 9

What is the least common multiple (LCM) of the integers 120 and 450?

Accepted Answer

Introduction:This problem focuses on finding the least common multiple (LCM) of two numbers, 120 and 450. Such questions are directly useful in solving problems related to repeating events, scheduling and synchronization in aptitude tests. Given Data / Assumptions: First number = 120. Second number = 450. We need their least common multiple. Concept / Approach:We use prime factorization to determine the LCM by taking each prime factor with the highest power that appears in either number.120 = 2^3 * 3 * 5450 = 2 * 3^2 * 5^2Then we select the maximum power of each prime: 2^3, 3^2 and 5^2, and multiply them to get the LCM. Step-by-Step Solution:Step 1: Factorize 120.120 = 2^3 * 3 * 5Step 2: Factorize 450.450 = 2 * 3^2 * 5^2Step 3: Choose highest powers of each prime.For 2: max exponent = 3 ⇒ 2^3For 3: max exponent = 2 ⇒ 3^2For 5: max exponent = 2 ⇒ 5^2Step 4: Form the LCM.LCM = 2^3 * 3^2 * 5^2Step 5: Compute numerically.2^3 = 83^2 = 95^2 = 258 * 9 = 7272 * 25 = 1800 Verification / Alternative check:Check that 1800 is divisible by both numbers. 1800 ÷ 120 = 15 and 1800 ÷ 450 = 4. Both are integers, so 1800 is a common multiple. Because we used the largest exponents of all relevant primes minimally, no smaller positive number can be a common multiple, confirming 1800 as the LCM. Why Other Options Are Wrong:2400, 3600, 4800: These are multiples of 1800 or other larger numbers, so although they may be common multiples, they are not the least.900: 900 is divisible by 450, but 900 ÷ 120 = 7.5, which is not an integer, so it is not even a common multiple. Common Pitfalls:Candidates may directly multiply 120 and 450 to get 54000 and mistakenly think that is the LCM. Prime factorization or using LCM = (a * b) / HCF ensures much smaller and correct answers for such problems. Final Answer:The least common multiple of 120 and 450 is 1800.

Question 10

Three numbers are in the ratio 3 : 4 : 5 and their L.C.M. is 2400. What is the H.C.F. of these three numbers?

Accepted Answer

Introduction / Context:This question tests your understanding of the relationship between numbers given in a ratio and their highest common factor (H.C.F.) and least common multiple (L.C.M.). Problems like this appear frequently in aptitude exams to check your grasp of basic number theory concepts involving ratios, H.C.F., and L.C.M. Given Data / Assumptions: The three numbers are in the ratio 3 : 4 : 5. Their L.C.M. is 2400. We are asked to find the H.C.F. of the three numbers. Concept / Approach:If three numbers are in the ratio 3 : 4 : 5, they can be written as 3k, 4k, and 5k for some positive integer k. The L.C.M. of these numbers will be k multiplied by the L.C.M. of 3, 4, and 5. Once we use the given L.C.M. to find k, we can reconstruct the actual numbers and then compute their H.C.F. by inspection, since the H.C.F. will be k itself when the ratio numbers are co-prime to each other. Step-by-Step Solution:Step 1: Let the three numbers be 3k, 4k, and 5k.Step 2: Since 3, 4, and 5 are pairwise co-prime, their L.C.M. is 3 * 4 * 5 = 60.Step 3: The L.C.M. of 3k, 4k, and 5k is therefore 60k.Step 4: We are told that the L.C.M. is 2400, so 60k = 2400.Step 5: Solve for k: k = 2400 / 60 = 40.Step 6: So the actual numbers are 3 * 40 = 120, 4 * 40 = 160, and 5 * 40 = 200.Step 7: The H.C.F. of 120, 160, and 200 is clearly 40, since all three are multiples of 40 and no larger common factor exists. Verification / Alternative check:Check with prime factors: 120 = 2^3 * 3 * 5, 160 = 2^5 * 5, 200 = 2^3 * 5^2.The common part is 2^3 * 5 = 8 * 5 = 40, which matches our H.C.F. This confirms that 40 is the correct highest common factor. Why Other Options Are Wrong:Option b (80) is too large because 80 does not divide 120 exactly. Option c (120) also fails because 120 does not divide 160 or 200 without leaving remainders. Option d (200) is larger than one of the numbers (160), so it cannot be a common factor. Only 40 divides all three numbers exactly. Common Pitfalls:A common mistake is to confuse ratio numbers with the actual numbers, taking 3, 4, and 5 themselves as the numbers. Another error is to miscalculate the L.C.M. of 3, 4, and 5 or to forget that the L.C.M. of the scaled numbers is the L.C.M. of the ratio multiplied by k. Some students also try to guess the H.C.F. directly from the L.C.M. without using the ratio properly, which leads to wrong answers. Final Answer:The H.C.F. of the three numbers is 40.

Question 11

The greatest number that leaves remainders 6 and 5 when dividing 1657 and 2037 respectively is equal to what value?

Accepted Answer

Introduction / Context:This is a classic highest common factor (H.C.F.) question where a divisor leaves different remainders when dividing two numbers. It tests whether you understand how to translate remainder information into an equivalent problem involving differences and greatest common divisors, a concept that appears very often in aptitude and number theory questions. Given Data / Assumptions: When the required number divides 1657, the remainder is 6. When the same number divides 2037, the remainder is 5. The number we seek is the greatest possible integer that satisfies both conditions. Concept / Approach:If a number d divides 1657 and leaves remainder 6, then 1657 - 6 is exactly divisible by d. Similarly, if the same number leaves remainder 5 on dividing 2037, then 2037 - 5 is divisible by d. Therefore, d must divide the difference between these adjusted values as well. So we first compute 1651 and 2032, then find the H.C.F. of these two numbers to get the greatest possible divisor satisfying both remainder conditions. Step-by-Step Solution:Step 1: Let the required greatest number be d.Step 2: From 1657 with remainder 6, we get 1657 - 6 = 1651, which is divisible by d.Step 3: From 2037 with remainder 5, we get 2037 - 5 = 2032, which is also divisible by d.Step 4: Therefore, d is a common divisor of 1651 and 2032.Step 5: The greatest such divisor is the H.C.F. of 1651 and 2032.Step 6: Find H.C.F.: 2032 - 1651 = 381.Step 7: Next, 1651 ÷ 381 gives a remainder: 381 * 4 = 1524, 1651 - 1524 = 127.Step 8: Now compute 381 ÷ 127. This is exact because 127 * 3 = 381.Step 9: Hence, the H.C.F. of 1651 and 2032 is 127, so the required greatest number is 127. Verification / Alternative check:Check 1657 ÷ 127: 127 * 13 = 1651 and 1657 - 1651 = 6, correct remainder.Check 2037 ÷ 127: 127 * 16 = 2032 and 2037 - 2032 = 5, correct remainder. As both conditions are satisfied, 127 is confirmed as the correct greatest number. Why Other Options Are Wrong:Option a (123), option c (235), and option d (305) do not simultaneously satisfy the remainder conditions for both 1657 and 2037. If you perform the division and check the remainders, you will see that they either fail for one of the numbers or do not divide the adjusted values 1651 and 2032 exactly. Hence, they cannot be the required greatest number. Common Pitfalls:Students sometimes attempt to find a number that directly divides the original numbers with the given remainders without converting to the equivalent H.C.F. problem. Another frequent error is to subtract the remainders incorrectly or to compute the H.C.F. carelessly. Always remember that you must work with 1657 - 6 and 2037 - 5, not with 1657 and 2037 directly, when the remainders are known. Final Answer:The greatest number that satisfies the given conditions is 127.

Question 12

The product of two numbers is 2028 and their H.C.F. is 13. How many distinct pairs of such positive integers are possible?

Accepted Answer

Introduction / Context:This problem combines the ideas of highest common factor (H.C.F.), factorization, and counting possible pairs of numbers. It checks whether you can use the relationship between H.C.F., co-prime factors, and the product of two numbers to count how many pairs of numbers satisfy the given conditions. This is a popular style of question in quantitative aptitude sections. Given Data / Assumptions: The product of the two positive integers is 2028. The H.C.F. (greatest common divisor) of the two numbers is 13. We must find how many such distinct pairs of numbers exist. Concept / Approach:If two numbers have H.C.F. 13, we can write them as 13a and 13b, where a and b are co-prime (their H.C.F. is 1). The product of the numbers is then 13a * 13b = 13^2 * ab. This product is given as 2028, which allows us to find ab. Once ab is known, we look for co-prime factor pairs (a, b) whose product equals that value. Each such pair gives one possible pair of original numbers 13a and 13b. Step-by-Step Solution:Step 1: Let the numbers be 13a and 13b with H.C.F. 13, so H.C.F.(a, b) = 1.Step 2: Product = 13a * 13b = 169ab.Step 3: Given product is 2028, so 169ab = 2028.Step 4: Compute ab = 2028 / 169 = 12.Step 5: We now need all co-prime positive integer pairs (a, b) such that ab = 12.Step 6: Factor pairs of 12 are (1, 12), (2, 6), (3, 4) and their reverses.Step 7: Check co-primeness: gcd(1, 12) = 1 (valid), gcd(2, 6) = 2 (invalid), gcd(3, 4) = 1 (valid).Step 8: So valid co-prime pairs are (1, 12) and (3, 4). Each pair corresponds to numbers (13, 156) and (39, 52).Step 9: As we count pairs without ordering, {13, 156} and {39, 52} give 2 distinct pairs. Verification / Alternative check:For pair (13, 156): product = 13 * 156 = 2028; H.C.F. is 13.For pair (39, 52): product = 39 * 52 = 2028; H.C.F. is also 13. No other co-prime factor pairs of 12 exist, so there are exactly 2 valid pairs. Why Other Options Are Wrong:Option a (1) undercounts the pairs by missing one valid combination. Option c (3) and option d (4) overcount, usually due to treating ordered pairs (a, b) and (b, a) as different or including non co-prime pairs like (2, 6). The question asks for the number of such pairs of numbers, which naturally refers to distinct unordered pairs of positive integers. Common Pitfalls:A common error is forgetting that a and b must be co-prime. Some candidates take all factor pairs of 12 without checking gcd, and others treat (a, b) and (b, a) as separate pairs. Always carefully check co-prime conditions and interpret whether the question refers to ordered or unordered pairs. Most exam questions consider pairs like (13, 156) and (156, 13) as the same pair. Final Answer:The number of such pairs is 2.

Question 13

A rectangular courtyard 3.78 m long and 5.25 m wide is to be paved completely with square tiles of equal size. What is the largest possible side (in cm) of each square tile?

Accepted Answer

Introduction / Context:This question is a practical application of the highest common factor (H.C.F.) concept in the context of tiling a rectangular floor with square tiles. It is a standard type of aptitude problem that tests your ability to convert measurements to consistent units and then use H.C.F. to find the largest possible tile size that divides both length and breadth exactly. Given Data / Assumptions: Length of the courtyard = 3.78 meters. Breadth (width) of the courtyard = 5.25 meters. The courtyard is to be paved exactly with square tiles of equal side length. We are asked to find the largest possible side of a tile in centimetres. Concept / Approach:To cover the rectangle exactly with square tiles, the side of each square must be a common divisor of both the length and the breadth. The largest possible tile corresponds to the greatest common divisor of the two dimensions. Since the dimensions are given in meters with decimals, we convert them to centimetres and compute the H.C.F. of the resulting integers. That H.C.F. gives the side of the largest square tile in centimetres. Step-by-Step Solution:Step 1: Convert lengths to centimetres: 3.78 m = 378 cm and 5.25 m = 525 cm.Step 2: We need the largest integer that divides both 378 and 525 exactly; this is their H.C.F.Step 3: Find H.C.F. using prime factorization or Euclid's algorithm.Step 4: Factorize: 378 = 2 * 3^3 * 7 and 525 = 3 * 5^2 * 7.Step 5: Common prime factors are 3^1 and 7^1, so H.C.F. = 3 * 7 = 21.Step 6: Therefore, the largest possible side of the square tile is 21 cm. Verification / Alternative check:Check divisibility: 378 ÷ 21 = 18, an integer.Also, 525 ÷ 21 = 25, an integer. Since both dimensions are exact multiples of 21 cm and no larger common factor exists, 21 cm is indeed the maximum side length for such square tiles. Why Other Options Are Wrong:Option a (14 cms) divides 378 but does not divide 525 exactly with a larger greatest common factor already found. Option c (42 cms) is too large because 378 is not an exact multiple of 42. Option d (None of these) is incorrect because we have found a valid tile size of 21 cms that satisfies all conditions. Common Pitfalls:Many students forget to convert metres to centimetres first and directly attempt to compute the H.C.F. of the decimals, which can cause confusion. Others miscalculate the prime factorization or choose a common factor but not the greatest one. Always ensure you work in consistent units and verify that your final tile size divides both dimensions exactly with no remainder. Final Answer:The largest possible side of each square tile is 21 cms.

Question 14

Find the L.C.M. of the numbers 22, 54, 108, 135 and 198.

Accepted Answer

Introduction / Context:This numerical aptitude question tests your ability to find the least common multiple (L.C.M.) of several integers. L.C.M. problems are frequently asked in exams because they are fundamental to understanding multiples, divisibility, and synchronizing events or cycles based on different periods. Given Data / Assumptions: The numbers given are 22, 54, 108, 135, and 198. We need to compute the least common multiple of these numbers. All numbers are positive integers. Concept / Approach:The L.C.M. of a set of numbers is the smallest positive integer that is divisible by each of the numbers. The most systematic way to find the L.C.M. is to use prime factorization. We factor each number into primes, then take the highest power of every prime that appears in any factorization. Multiplying these highest powers gives the L.C.M. Step-by-Step Solution:Step 1: Prime factorize each number.22 = 2 * 11.54 = 2 * 3^3.108 = 2^2 * 3^3.135 = 3^3 * 5.198 = 2 * 3^2 * 11.Step 2: Identify all distinct prime factors: 2, 3, 5, and 11.Step 3: Take the highest power of each prime that appears: 2^2 (from 108), 3^3 (from 54, 108, 135), 5^1 (from 135), and 11^1 (from 22 or 198).Step 4: Multiply these highest powers: L.C.M. = 2^2 * 3^3 * 5 * 11.Step 5: Compute: 2^2 = 4; 3^3 = 27. So L.C.M. = 4 * 27 * 5 * 11.Step 6: 4 * 27 = 108; 108 * 5 = 540; 540 * 11 = 5940.Step 7: Therefore, the L.C.M. is 5940. Verification / Alternative check:Check divisibility: 5940 ÷ 22, 5940 ÷ 54, 5940 ÷ 108, 5940 ÷ 135, and 5940 ÷ 198 all give integer results. Since all the original numbers divide 5940 and no smaller number containing all required prime powers exists, 5940 is indeed the least common multiple. Why Other Options Are Wrong:Option a (330) and option b (1980) are too small; they fail to be divisible by all of the given numbers, especially 108 and 135. Option d (11880) is a common multiple but not the least one; it is simply 2 * 5940. Only 5940 is the smallest integer divisible by each of the five numbers. Common Pitfalls:Many students incorrectly combine factors or forget to take the highest power of each prime, which leads to a number that is not divisible by all given values. Another common error is to try to find the L.C.M. pairwise and to accumulate mistakes. Prime factorization keeps the process systematic and reduces the chance of missing any prime power. Final Answer:The L.C.M. of 22, 54, 108, 135, and 198 is 5940.

Question 15

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 / Context:This question combines three important number theory ideas: sum of two numbers, their highest common factor (H.C.F.), and their least common multiple (L.C.M.). Such problems test your understanding of the relationship between product, H.C.F., and L.C.M., as well as your ability to solve simple quadratic equations arising from these relationships. 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 must find the difference between the two numbers. Concept / Approach:For any two positive integers a and b, the product a * b is equal to H.C.F. * L.C.M. when H.C.F. and L.C.M. are known. Using this, we can find the product of the numbers. Together with the sum, we can set up a quadratic equation whose roots are the two numbers. Once the numbers are found, subtract the smaller from the larger to get the required difference. Step-by-Step Solution:Step 1: Let the two numbers be x and y.Step 2: Given H.C.F. = 5 and L.C.M. = 495.Step 3: Use the relation x * y = H.C.F. * L.C.M. = 5 * 495 = 2475.Step 4: Also given x + y = 100.Step 5: Let x and y be roots of the quadratic t^2 - (sum)t + product = 0.Step 6: So t^2 - 100t + 2475 = 0.Step 7: Compute the discriminant: D = 100^2 - 4 * 2475 = 10000 - 9900 = 100.Step 8: Roots are t = (100 ± √100) / 2 = (100 ± 10) / 2.Step 9: Thus, t = 55 or t = 45, so the two numbers are 55 and 45.Step 10: The difference between the two numbers is 55 - 45 = 10. Verification / Alternative check:Check H.C.F. of 55 and 45: gcd(55, 45) = 5.Check L.C.M.: (55 * 45) / 5 = 2475 / 5 = 495.Check sum: 55 + 45 = 100. All conditions are satisfied, confirming that the solution is consistent and the difference is 10. Why Other Options Are Wrong:Option b (46), option c (70), and option d (90) correspond to wrong pairs of numbers that do not satisfy all three conditions simultaneously. If you try to reconstruct possible pairs with those differences, you will see that they either fail the product condition or the H.C.F. and L.C.M. relation. Common Pitfalls:Candidates sometimes forget the key formula relating product, H.C.F., and L.C.M., or they miscompute the product. Another common mistake is solving the quadratic incorrectly or mixing up the roots. Always double-check the discriminant and verify that both the sum and product match the original data for the found numbers. Final Answer:The difference between the two numbers is 10.

Question 16

The sum of two numbers is 528 and their H.C.F. is 33. How many distinct pairs of positive integers satisfy these conditions?

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Introduction / Context:This question checks whether you can use the concept of H.C.F. to express numbers in a common form and then count the number of valid pairs. It combines number theory with basic counting and factorization and is a common type in competitive exams where you must find how many pairs of numbers fit certain conditions. Given Data / Assumptions: The sum of two positive integers is 528. Their H.C.F. is 33. We must find the number of pairs of such numbers. Concept / Approach:If two numbers have H.C.F. 33, we can write them as 33a and 33b, where a and b are co-prime positive integers. Their sum is 33a + 33b = 33(a + b). Since the sum is 528, we can find a + b. Every pair of co-prime positive integers (a, b) with that sum corresponds to a valid pair of numbers. So our task reduces to counting co-prime pairs whose sum is a fixed number. Step-by-Step Solution:Step 1: Let the two numbers be 33a and 33b with H.C.F.(a, b) = 1.Step 2: Sum of the numbers is 33a + 33b = 33(a + b).Step 3: Given sum = 528, so 33(a + b) = 528.Step 4: Divide to get a + b = 528 / 33 = 16.Step 5: We now need pairs of positive integers (a, b) such that a + b = 16 and gcd(a, b) = 1.Step 6: Possible positive integer pairs (a, b) with sum 16 are (1, 15), (2, 14), (3, 13), (4, 12), (5, 11), (6, 10), (7, 9), and (8, 8).Step 7: Check gcd for each: gcd(1, 15) = 1, gcd(2, 14) = 2, gcd(3, 13) = 1, gcd(4, 12) = 4, gcd(5, 11) = 1, gcd(6, 10) = 2, gcd(7, 9) = 1, gcd(8, 8) = 8.Step 8: Valid co-prime pairs are (1, 15), (3, 13), (5, 11), and (7, 9).Step 9: Therefore there are 4 valid pairs of (a, b), giving 4 pairs of numbers (33a, 33b). Verification / Alternative check:Example pair: a = 1, b = 15 gives numbers 33 and 495 with H.C.F. 33 and sum 528.Similarly for (3, 13), (5, 11), and (7, 9) you get other valid pairs with H.C.F. 33 and sum 528. No other co-prime pairs exist, so there are exactly 4 possible pairs. Why Other Options Are Wrong:Options b (6), c (8), and d (12) overcount because they treat pairs with common factors as valid or fail to enforce the co-prime condition required by the H.C.F. representation. Only the four pairs listed respect both the co-prime condition and the sum requirement. Common Pitfalls:Students often count all pairs with the correct sum without checking if a and b are co-prime. Others may mistakenly think that (a, b) and (b, a) are separate pairs, but the question is about unordered pairs of numbers. A disciplined approach of listing and checking gcd values prevents these mistakes. Final Answer:The number of such pairs is 4.

Question 17

The product of two co-prime numbers is 117. What should be the L.C.M. of these two numbers?

Accepted Answer

Introduction / Context:This is a concept-based question on co-prime numbers and the relationship between highest common factor (H.C.F.) and least common multiple (L.C.M.). It tests your theoretical understanding rather than long calculation, making it a quick scoring problem if you know the basic rule. Given Data / Assumptions: The two numbers are co-prime, meaning their H.C.F. is 1. The product of the two numbers is 117. We are asked to find the L.C.M. of the two numbers. Concept / Approach:For any two positive integers a and b, the relation between their product, H.C.F. (denoted h), and L.C.M. (denoted L) is: a * b = h * L. When a and b are co-prime, their H.C.F. is 1. That means a * b = 1 * L, or L = a * b. So for co-prime numbers, their L.C.M. is simply equal to the product of the numbers. Step-by-Step Solution:Step 1: Let the two co-prime numbers be a and b.Step 2: Given that a and b are co-prime, H.C.F.(a, b) = 1.Step 3: The product of the numbers is a * b = 117.Step 4: Use the formula a * b = H.C.F. * L.C.M.Step 5: Substitute H.C.F. = 1, so 117 = 1 * L.C.M.Step 6: Therefore L.C.M. = 117. Verification / Alternative check:Since a and b are co-prime, any factorization of 117 into primes can represent the two numbers by grouping the prime factors in any way. For example, 117 = 9 * 13, where gcd(9, 13) = 1. The L.C.M. of 9 and 13 is (9 * 13) / 1 = 117, which matches our result. Any other co-prime factor pair will still produce an L.C.M. equal to 117. Why Other Options Are Wrong:Option a (1) is incorrect because it confuses H.C.F. with L.C.M.; co-prime numbers have H.C.F. 1, not L.C.M. 1. Option c (Equal to their H.C.F) is also wrong because their H.C.F. is 1, while the L.C.M. here is clearly larger. Option d (cannot be calculated) is incorrect, as there is enough information to find the L.C.M. using the standard formula. Common Pitfalls:Students sometimes mix up H.C.F. and L.C.M., assuming that if numbers are co-prime, both must be small. Others forget the formula a * b = H.C.F. * L.C.M. Knowing this relationship is crucial to quickly answering many exam questions without trial and error. Final Answer:The L.C.M. of the two co-prime numbers is 117.

Question 18

Find the greatest common divisor (G.C.D.) of the decimal numbers 1.08, 0.36 and 0.9.

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Introduction / Context:This question tests your understanding of finding the greatest common divisor (G.C.D., also known as H.C.F.) when the numbers are given in decimal form. Converting decimals into whole numbers by using a common factor of 10 is a standard strategy. Such problems appear in aptitude tests to examine your comfort with decimals and factors together. Given Data / Assumptions: The numbers are 1.08, 0.36, and 0.9. We are asked to determine their greatest common divisor. All numbers are positive decimal numbers. Concept / Approach:To find the G.C.D. of decimal numbers, we can scale them by a power of 10 so that all numbers become integers. We then compute the G.C.D. of these integers and finally scale back by dividing by the same power of 10. This makes the calculation much easier and avoids confusion with decimal arithmetic. Step-by-Step Solution:Step 1: Express each decimal with two decimal places: 1.08, 0.36, and 0.90.Step 2: Multiply each by 100 to remove decimals: 1.08 * 100 = 108, 0.36 * 100 = 36, 0.9 * 100 = 90.Step 3: Now find the G.C.D. of the integers 108, 36, and 90.Step 4: First, gcd(108, 36) = 36, because 36 divides 108 exactly.Step 5: Next, gcd(36, 90) = 18.Step 6: So the G.C.D. of 108, 36, and 90 is 18.Step 7: Since we multiplied by 100 earlier, we now divide the G.C.D. by 100: 18 / 100 = 0.18.Step 8: Therefore, the greatest common divisor of 1.08, 0.36, and 0.9 is 0.18. Verification / Alternative check:Check divisibility: 1.08 ÷ 0.18 = 6, 0.36 ÷ 0.18 = 2, and 0.9 ÷ 0.18 = 5. All quotients are integers, confirming that 0.18 is a common divisor. No larger decimal (greater than 0.18) divides all three numbers exactly, so this is indeed the greatest common divisor. Why Other Options Are Wrong:Option a (0.03) is a common divisor but not the greatest, since 0.18 is larger and still divides all numbers. Option b (0.9) fails because 0.9 does not divide 1.08 or 0.36 exactly. Option d (0.108) does not divide 0.9 exactly into an integer quotient. Therefore these options cannot represent the G.C.D. Common Pitfalls:Some students attempt to work directly with decimals and make arithmetic mistakes. Others forget to apply the same scaling factor to all numbers. A reliable technique is to convert all decimals to integers using a common power of 10, compute the G.C.D. of the integers, and then scale back to obtain the final answer in decimal form. Final Answer:The greatest common divisor of the three decimals is 0.18.

Question 19

Find the smallest positive integer which, when decreased by 7, is exactly divisible by 12, 16, 18, 21 and 28.

Accepted Answer

Introduction / Context:This problem is a standard application of least common multiple (L.C.M.) in which a number, after being decreased by a certain constant, becomes divisible by several integers. Such questions test whether you can recognise that the adjusted number must be a common multiple and then find the smallest original number that satisfies the condition. Given Data / Assumptions: We seek a smallest number N. N, when diminished by 7 (N - 7), is divisible by 12, 16, 18, 21, and 28. We must find this smallest positive integer N. Concept / Approach:If N - 7 is divisible by several numbers, then N - 7 must be a common multiple of those numbers. To get the smallest such N, we take N - 7 as the least common multiple of 12, 16, 18, 21, and 28. Then we add 7 back to obtain N. Therefore the key step is computing the L.C.M. of the given set of integers. Step-by-Step Solution:Step 1: Let N be the required number.Step 2: Given that N - 7 is divisible by 12, 16, 18, 21, and 28.Step 3: Therefore, N - 7 must be a common multiple of these five numbers.Step 4: To get the smallest possible N, take N - 7 equal to the L.C.M. of 12, 16, 18, 21, and 28.Step 5: Compute prime factors: 12 = 2^2 * 3, 16 = 2^4, 18 = 2 * 3^2, 21 = 3 * 7, 28 = 2^2 * 7.Step 6: Take highest powers: 2^4, 3^2, and 7^1.Step 7: L.C.M. = 2^4 * 3^2 * 7 = 16 * 9 * 7 = 144 * 7 = 1008.Step 8: So N - 7 = 1008, giving N = 1008 + 7 = 1015. Verification / Alternative check:Check N - 7 = 1015 - 7 = 1008.Now verify divisibility: 1008 ÷ 12 = 84, 1008 ÷ 16 = 63, 1008 ÷ 18 = 56, 1008 ÷ 21 = 48, 1008 ÷ 28 = 36. All quotients are integers, so N = 1015 satisfies all the conditions, and since 1008 is the L.C.M., no smaller N will work. Why Other Options Are Wrong:Option a (1008) corresponds to N - 7, not N itself. Option c (1022) and option d (1032) do not yield N - 7 values that are common multiples of all five numbers. Testing them will show at least one divisor fails in each case. Common Pitfalls:A common mistake is to stop at the L.C.M. and forget to add the 7 back, leading to the wrong answer 1008. Others miscalculate the L.C.M. because they do not consistently take the highest power of each prime across all numbers. Being systematic in prime factorization and remembering to undo the subtraction is crucial. Final Answer:The smallest required number is 1015.

Question 20

The L.C.M. of two prime numbers x and y (with x > y) is 161. What is the value of the expression 3y - x?

Accepted Answer

Introduction / Context:This is a short conceptual question involving prime numbers and least common multiple (L.C.M.). It checks whether you can factor a composite number into prime factors and then use the definition of L.C.M. for prime numbers to deduce the values of x and y before evaluating a simple expression in terms of them. Given Data / Assumptions: x and y are prime numbers. x is greater than y. Their L.C.M. is 161. We must compute 3y - x. Concept / Approach:When two numbers are prime and distinct, their L.C.M. is simply their product. Therefore, if the L.C.M. is 161, and both numbers are prime, 161 must factor into exactly two prime factors, which are x and y. After factoring 161 into primes, we identify which is larger and which is smaller, then substitute into the expression 3y - x. Step-by-Step Solution:Step 1: Factor 161 into primes.Step 2: Check divisibility: 161 ÷ 7 = 23, so 161 = 7 * 23.Step 3: Both 7 and 23 are prime numbers.Step 4: Since x > y, set x = 23 and y = 7.Step 5: Compute 3y - x = 3 * 7 - 23.Step 6: 3 * 7 = 21, so 3y - x = 21 - 23 = -2. Verification / Alternative check:Check L.C.M. of 7 and 23: because these primes are distinct, L.C.M. = 7 * 23 = 161. This matches the given L.C.M., confirming that x and y are correctly assigned as 23 and 7 respectively, and therefore the expression 3y - x evaluates to -2. Why Other Options Are Wrong:Options b (-1), c (1), and d (2) correspond to incorrect combinations or incorrect identification of x and y. Any alternative assignment of primes, or miscalculation of 3y - x, would not match the factorization of 161 and would fail to satisfy the condition x > y while still giving L.C.M. 161. Common Pitfalls:Some students forget that for distinct primes, the L.C.M. equals the product, and they may look for other combinations of factors or mis-factorise 161. Others might accidentally swap x and y or plug them into the expression incorrectly. A quick prime factorization and careful substitution into 3y - x avoids these errors. Final Answer:The value of the expression 3y - x is -2.

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