Question 1

What is the Least Common Multiple (LCM) of the two numbers 120 and 450?

Accepted Answer

Introduction / Context: This question requires finding the Least Common Multiple, or LCM, of 120 and 450. LCM problems are essential in quantitative aptitude because they relate to combining cycles, synchronizing events, and adding or comparing fractions with different denominators. Understanding how to compute LCM efficiently helps you avoid time consuming trial and error methods in exams. Given Data / Assumptions: First number = 120. Second number = 450. We need to find their LCM. Both numbers are positive integers. Concept / Approach: The LCM of two numbers can be found using prime factorization or using the formula: LCM(a, b) = (a * b) / HCF(a, b). Often it is efficient to first find the Highest Common Factor (HCF) using prime factorization or the Euclidean algorithm, and then apply this formula. This is especially useful when the numbers are fairly large but share a significant common factor, as it reduces the amount of direct multiplication and manual checking needed. Step-by-Step Solution: Step 1: Prime factorize both numbers. 120 = 2^3 * 3 * 5. 450 = 2 * 3^2 * 5^2. Step 2: Determine the HCF. HCF uses the smallest power of each common prime. Common primes are 2, 3, and 5. For 2, smallest power is 2^1. For 3, smallest power is 3^1. For 5, smallest power is 5^1. So HCF = 2 * 3 * 5 = 30. Step 3: Use the LCM formula. LCM = (120 * 450) / 30. Step 4: Simplify the product with division. First, 450 / 30 = 15. Now LCM = 120 * 15 = 1800. Therefore, the LCM of 120 and 450 is 1800. Verification / Alternative check: We can verify by prime factorization directly for the LCM. From the factorizations: 120 = 2^3 * 3 * 5, 450 = 2 * 3^2 * 5^2. LCM uses the highest powers of each prime: 2^3 (from 120), 3^2 (from 450), 5^2 (from 450). Thus, LCM = 2^3 * 3^2 * 5^2 = 8 * 9 * 25 = 72 * 25 = 1800. This matches the value found using the formula, confirming that 1800 is correct. Why Other Options Are Wrong: 2400: Although 2400 is a multiple of both 120 and 450, it is not the least such multiple because we have already found a smaller common multiple, 1800. 3600 and 4800: These are also common multiples of 120 and 450 but are larger than the correct LCM and therefore not minimal. 900: This number is a multiple of 450 but not of 120, since 900 / 120 is not an integer, so it cannot be the LCM. Common Pitfalls: Some students multiply the numbers directly and forget to divide by the HCF, leading to very large and incorrect answers. Others might miscalculate the HCF or confuse highest and lowest prime powers when handling factorizations. Using the systematic approach of prime factorization plus the formula for LCM minimizes these errors and makes it easier to verify the result. Final Answer: The LCM of 120 and 450 is 1800.

Question 2

Three friends J, K, and L jog around a circular stadium. They complete one round in 12 seconds, 18 seconds, and 20 seconds respectively. After how many minutes will all three of them meet again together at the starting point?

Accepted Answer

Introduction / Context: This question tests your ability to use the concept of Least Common Multiple (LCM) in a real life style problem. When multiple people are moving around a circular track with different times for completing one round, they will all meet again at the starting point after a time equal to the LCM of their individual times. Such questions are common in aptitude exams to check understanding of synchronization and periodic events. Given Data / Assumptions: Time taken by J to complete one round = 12 seconds. Time taken by K to complete one round = 18 seconds. Time taken by L to complete one round = 20 seconds. We need to find after how many minutes all three meet again at the starting point. All of them start together from the same point at the same time. Concept / Approach: When dealing with repeated actions over time, the moment when all events coincide again is determined by the Least Common Multiple of their time periods. Here, each friend's jogging cycle is an event: J returns to the start every 12 seconds, K every 18 seconds, and L every 20 seconds. The time after which they all meet together again at the starting point is the LCM of 12, 18, and 20. Once we find the LCM in seconds, we convert that value into minutes as required by the question. Step-by-Step Solution: Step 1: Prime factorize each time period. 12 = 2^2 * 3. 18 = 2 * 3^2. 20 = 2^2 * 5. Step 2: Identify the distinct primes: 2, 3, and 5. Step 3: Take the highest power of each prime. For 2, highest power is 2^2. For 3, highest power is 3^2. For 5, highest power is 5^1. Step 4: Multiply to obtain the LCM. LCM = 2^2 * 3^2 * 5 = 4 * 9 * 5 = 36 * 5 = 180 seconds. Step 5: Convert seconds to minutes. 180 seconds = 180 / 60 = 3 minutes. Thus, all three friends will meet again at the starting point after 3 minutes. Verification / Alternative check: We can verify by checking how many rounds each person completes in 180 seconds. J: 180 / 12 = 15 rounds. K: 180 / 18 = 10 rounds. L: 180 / 20 = 9 rounds. Each result is an integer, which means each friend completes a whole number of laps and returns exactly to the starting point at 180 seconds. Because 180 seconds is derived from the LCM, it is the smallest such positive time when this happens, confirming that 3 minutes is indeed the first meeting time after the start. Why Other Options Are Wrong: 5, 8, 12, and 6 minutes: These correspond to 300, 480, 720, and 360 seconds respectively. While 360 seconds (6 minutes) is also a common multiple of 12, 18, and 20, it is not the smallest such positive time. The question asks for the first time they meet again, so we must choose the minimum time, which is 180 seconds, that is 3 minutes. Common Pitfalls: Some students mistakenly add the times or average them instead of taking the LCM. Others may find a common multiple but forget to ensure it is the least common multiple. Errors can also occur during prime factorization or when converting between seconds and minutes. To avoid these issues, always factor carefully, use the highest powers of primes for LCM, and read the units in the question very attentively. Final Answer: All three friends will meet again at the starting point after 3 minutes.

Question 3

Find the H.C.F. and L.C.M. of $\frac{2}{3}$, $\frac{8}{9}$, $\frac{16}{81}$ and $\frac{10}{27}$.

Accepted Answer

Concept & Formula: To find the H.C.F. and L.C.M. of a set of fractions, we use the following standard formulas: H.C.F. of fractions = \frac{H.C.F. of Numerators}{L.C.M. of Denominators} L.C.M. of fractions = \frac{L.C.M. of Numerators}{H.C.F. of Denominators} Step-by-Step Solution: Given fractions: (2/3), (8/9), (16/81), (10/27) • Numerators: 2, 8, 16, 10 • Denominators: 3, 9, 81, 27 Step 1: Calculate the H.C.F. of the fractions. • Find H.C.F. of Numerators (2, 8, 16, 10): The highest common factor dividing all these numbers is clearly 2. • Find L.C.M. of Denominators (3, 9, 81, 27): Since 3, 9, and 27 are all factors of 81, the least common multiple is 81. • Apply the formula: H.C.F. = (2/81) Step 2: Calculate the L.C.M. of the fractions. • Find L.C.M. of Numerators (2, 8, 16, 10): 16 covers the factors of 2 and 8. We just need the L.C.M. of 16 and 10. 16 = 2^4 10 = 2 × 5 L.C.M. = 2^4 × 5 = 16 × 5 = 80. • Find H.C.F. of Denominators (3, 9, 81, 27): The highest common factor dividing all these numbers is 3. • Apply the formula: L.C.M. = (80/3) Exam Strategy & Shortcut: In multiple-choice exams, you rarely need to calculate both values entirely. Calculate the easiest component first. The H.C.F. of the numerators (2, 8, 16, 10) is instantly visible as 2. The L.C.M. of denominators (3, 9, 81, 27) is instantly visible as 81 because 81 is a multiple of the others. Thus, H.C.F. is (2/81). You can often eliminate 2-3 incorrect options just by finding this first half of the answer. Common Pitfall: The most common mistake is swapping the formulas under pressure—calculating the L.C.M. of numerators when asked for the H.C.F. of the fractions. To avoid this, remember that the operation requested (H.C.F. or L.C.M.) always applies to the numerators, while the opposite operation applies to the denominators. Final Answer: Therefore, the correct answer is H.C.F. = (2/81), L.C.M. = (80/3).

Question 4

Two numbers are in the ratio of $15 : 11$. If their H.C.F. is $13$, find the numbers.

Accepted Answer

Concept & Formula: When two numbers are given in a simplified ratio a : b, the numbers themselves can be expressed as ax and bx, where x is their Highest Common Factor (H.C.F.). Because a ratio is in its simplest form, the terms a and b are co-prime (they share no common factors other than 1). Therefore, multiplying each term of the ratio by the H.C.F. directly yields the original numbers. Step-by-Step Solution: Step 1: Define the numbers using the given ratio. • Given ratio = 15 : 11. • Let the two numbers be 15x and 11x. Step 2: Identify the H.C.F. • Since 15 and 11 are co-prime, the highest common factor of 15x and 11x is exactly x. • The problem states the H.C.F. is 13. • Therefore, x = 13. Step 3: Calculate the actual numbers. • First number = 15 × 13 = 195. • Second number = 11 × 13 = 143. Exam Strategy & Shortcut: You don't need to write out algebraic steps. As a strict rule: Original Numbers = Ratio Terms × H.C.F. If you see this exact question type on an exam, instantly multiply 15 × 13 and 11 × 13. Even faster, you can use unit digit checking on the options: 5 × 3 ends in 5. 1 × 3 ends in 3. Look for an option where the numbers end in 5 and 3 respectively. Option A (195 and 143) perfectly fits without even doing the full multiplication. Common Pitfall: A common beginner mistake is assuming the ratio terms (15 and 11) are factors to be multiplied together to find the L.C.M., and confusing this with finding the actual numbers. Always remember that the ratio terms scaled by the H.C.F. gives the numbers directly. Final Answer: Therefore, the correct answer is 195 and 143.

Question 5

Find the H.C.F. of $2^3 	imes 3^2 	imes 5 	imes 7^4$, $2^2 	imes 3^5 	imes 5^2 	imes 7^3$, $2^3 	imes 5^3 	imes 7^2$.

Accepted Answer

Concept & Formula: The Highest Common Factor (H.C.F.) of numbers expressed in prime factorization form is the product of the lowest powers of all common prime factors. If a prime number does not appear in every factorization, it is not common and cannot be included in the H.C.F. Step-by-Step Solution: Step 1: Identify the prime factors in each given number. • First number: Base primes are 2, 3, 5, 7. • Second number: Base primes are 2, 3, 5, 7. • Third number: Base primes are 2, 5, 7. (Notice 3 is missing here). Step 2: Find the common prime factors. • The prime numbers common to all three given expressions are 2, 5, and 7. Step 3: Determine the lowest power of each common prime factor. • For prime 2: The powers are 2^3, 2^2, and 2^3. The lowest is 2^2. • For prime 5: The powers are 5^1, 5^2, and 5^3. The lowest is 5^1. • For prime 7: The powers are 7^4, 7^3, and 7^2. The lowest is 7^2. Step 4: Multiply these lowest powers to find the H.C.F. H.C.F. = 2^2 × 5 × 7^2 H.C.F. = 4 × 5 × 49 H.C.F. = 20 × 49 = 980. Exam Strategy & Shortcut: In competitive exams, immediately scan for missing prime bases across the numbers. Here, the third term lacks the base 3. You can instantly eliminate any multiple-choice option that is divisible by 3 (by checking if the sum of its digits is a multiple of 3). 980 (9+8+0 = 17) is not divisible by 3, which aligns with our logic. Common Pitfall: A standard mistake is confusing H.C.F. with L.C.M. when looking at exponents. Students often take the highest powers (2^3, 3^5, etc.) instead of the lowest, or they mistakenly include non-common factors (like 3^2). Remember: H.C.F. requires the minimum intersection. Final Answer: Therefore, the correct answer is 980.

Question 6

Find the H.C.F. of 42, 63 and 140.

Accepted Answer

Concept & Strategy: To find the Highest Common Factor (H.C.F.) of small integers, the prime factorization method is highly effective. You break each number down into its core prime building blocks, then find the intersection (the common prime factors) and multiply their lowest powers. Step-by-Step Solution: Step 1: Perform prime factorization for each number. • 42 = 2 × 3 × 7 • 63 = 3 × 3 × 7 = 3^2 × 7 • 140 = 2 × 2 × 5 × 7 = 2^2 × 5 × 7 Step 2: Identify the common prime factors. • Looking at the factors, 2 is in 42 and 140, but not 63. • 3 is in 42 and 63, but not 140. • 5 is only in 140. • 7 is present in 42, 63, and 140. The only common prime factor is 7. Step 3: Determine the lowest power. The lowest power of 7 across all factorizations is 7^1. Exam Strategy & Shortcut: Use the Difference Method for speed. The H.C.F. of any set of numbers can never be larger than the smallest difference between any two of those numbers. Difference between 42 and 63 is 21. Therefore, the H.C.F. must be 21 or a factor of 21 (like 7 or 3). Check 21: 140 is not divisible by 21. Check 7: 42, 63, and 140 are all perfectly divisible by 7. The H.C.F. is 7. Common Pitfall: Students often waste time doing long division for relatively small numbers like these. Long division is prone to arithmetic errors and takes longer; use the difference method or quick prime factorization instead. Final Answer: Therefore, the correct answer is 7.

Question 7

Find the largest number which divides $62$, $132$ and $237$ to leave the same remainder in each case.

Accepted Answer

Concept & Logic: To find the largest number that divides x, y, and z leaving the same (but unspecified) remainder in each case, calculate the H.C.F. of the absolute differences between the given numbers. Required Number = H.C.F. of |x - y|, |y - z|, and |z - x| Step-by-Step Solution: • Given: The three numbers are 62, 132, and 237. They must leave the same remainder when divided by the target number. • Find the differences between pairs of numbers: • 132 - 62 = 70 • 237 - 132 = 105 • 237 - 62 = 175 • Now, calculate the H.C.F. of these differences: 70, 105, and 175. • Factors of 70: 1, 2, 5, 7, 10, 14, 35, 70 • Factors of 105: 1, 3, 5, 7, 15, 21, 35, 105 • Factors of 175: 1, 5, 7, 25, 35, 175 • The highest common factor present in all three sets is 35. Exam Strategy & Shortcut: Use the Difference Method for H.C.F. on the new set of numbers (70, 105, 175). The difference between 105 and 70 is 35. The H.C.F. must be 35 or a factor of 35. Check if 35 divides all three differences: 70/35 = 2, 105/35 = 3, 175/35 = 5. It divides them all, making 35 the direct answer without writing out all factors. Common Pitfall: Students often get stuck trying to set up algebraic equations with an unknown variable R for the remainder (e.g., 62 = nX + R). This wastes massive amounts of time. Immediately pivot to finding the differences of the numbers instead. Final Answer: Therefore, the correct answer is 35.

Question 8

The H.C.F. of two numbers, each having three digits is $17$ and their L.C.M. is $714$. Find the sum of the numbers.

Accepted Answer

Concept & Formula: If the H.C.F. of two numbers is H, the numbers can be represented as Hx and Hy, where x and y are co-prime integers. Furthermore, the product of two numbers is equal to the product of their H.C.F. and L.C.M. Product of Numbers = H.C.F. × L.C.M. (Hx) × (Hy) = H × L Step-by-Step Solution: • Given: H.C.F. = 17, L.C.M. = 714. Both unknown numbers are strictly 3-digit numbers. • Let the two numbers be 17x and 17y, where x and y are co-prime. • Using the property: 17x × 17y = 17 × 714 • 289xy = 12138 • xy = (12138/289) = (714/17) = 42 • Find pairs of co-prime numbers (x, y) that multiply to 42: • (1, 42) • (2, 21) • (3, 14) • (6, 7) • Since the original numbers (17x and 17y) must be 3-digit numbers, their values must be greater than 99. • If x = 1, the number is 17 × 1 = 17 (2-digit, rejected). • If x = 2, the number is 17 × 2 = 34 (2-digit, rejected). • If x = 3, the number is 17 × 3 = 51 (2-digit, rejected). • If x = 6, the number is 17 × 6 = 102 (3-digit, valid). • If y = 7, the number is 17 × 7 = 119 (3-digit, valid). • The two numbers are 102 and 119. • Their sum is 102 + 119 = 221. Exam Strategy & Shortcut: Instead of testing all pairs mentally, just look at the condition "three digits". 17 × 5 = 85, so you need multipliers strictly 6 or higher. The only factors of 42 where both are \geq 6 are 6 and 7. Directly compute the sum as 17 × (6 + 7) = 17 × 13 = 221. Common Pitfall: A common error is stopping at xy = 42 and arbitrarily picking (1, 42) or (2, 21) as the pair, leading to completely incorrect 2-digit numbers. Always cross-verify the specific constraints given in the problem (e.g., "three digits"). Final Answer: Therefore, the correct answer is 221.

Question 9

Find the least number which is exactly divisible by $32$, $36$, $45$, $60$ and $80$.

Accepted Answer

Concept & Formula: When asked for the "least number exactly divisible" by a set of numbers, you need to find their Lowest Common Multiple (L.C.M.). Required Number = L.C.M.(a, b, c, \dots) Step-by-Step Solution: • Given: The target divisors are 32, 36, 45, 60, and 80. • Use the prime factorization or common division method. Let's list out prime factors: • 32 = 2^5 • 36 = 2^2 × 3^2 • 45 = 3^2 × 5^1 • 60 = 2^2 × 3^1 × 5^1 • 80 = 2^4 × 5^1 • To find the L.C.M., take the highest power of every prime factor present across all numbers: • Highest power of 2 is 2^5 (from 32). • Highest power of 3 is 3^2 (from 36 or 45). • Highest power of 5 is 5^1 (from 45, 60, or 80). • Multiply these maximum prime powers together: • L.C.M. = 2^5 × 3^2 × 5^1 • L.C.M. = 32 × 9 × 5 • L.C.M. = 32 × 45 = 1440 Exam Strategy & Shortcut: Use Divisibility Rules on Options. The correct number must be divisible by 36 (so it must be divisible by 9) and by 32 (so its last 5 digits must be divisible by 32, or you can just test). Check options for divisibility by 9 (sum of digits): 720, 1080, 1440, 2880 all work. Next, check divisibility by 32: 720/32 = 22.5 (Fail), 1080/32 = 33.75 (Fail). 1440/32 = 45 (Pass!). Thus, 1440 is the least number that works. Common Pitfall: Using the continuous division (ladder) method on so many large numbers can easily lead to a small arithmetic mistake in one column, ruining the whole L.C.M. Prime factorization is generally safer, or explicitly combining smaller L.C.M.s (e.g., L.C.M. of 32 and 80 is 160, then compare with the rest). Final Answer: Therefore, the correct answer is 1440.

Question 10

Find the least number which when divided by $6$, $7$, $8$, $9$ and $12$ leaves the same remainder $1$ in each case.

Accepted Answer

Concept & Formula: To find the least number that leaves a constant remainder R when divided by multiple numbers, you first find the lowest common multiple (L.C.M.) of those divisors to get a perfectly divisible base, and then add the remainder R. Required Number = L.C.M.(a, b, c, \dots) + R Step-by-Step Solution: • Given: The divisors are 6, 7, 8, 9, and 12. The constant remainder is 1. • First, find the L.C.M. of 6, 7, 8, 9, and 12. Let's prime factorize: • 6 = 2 × 3 • 7 = 7 (Prime) • 8 = 2^3 • 9 = 3^2 • 12 = 2^2 × 3 • Take the highest power of each prime factor: • For 2: 2^3 = 8 • For 3: 3^2 = 9 • For 7: 7^1 = 7 • Multiply them to get the L.C.M.: • L.C.M. = 8 × 9 × 7 = 72 × 7 = 504 • This means 504 is exactly divisible by all those numbers. • Now, add the required remainder (1) to this L.C.M.: • Required Number = 504 + 1 = 505 Exam Strategy & Shortcut: Use Divisibility Rules on the Options. The correct answer must leave a remainder of 1 when divided by 9. Subtract 1 from the options and check if the result is divisible by 9: • A: 504 - 1 = 503 (Sum is 8, not div by 9) • B: 505 - 1 = 504 (Sum is 9, Divisible) • C: 503 - 1 = 502 (Sum is 7, not div by 9) • D: 512 - 1 = 511 (Sum is 7, not div by 9) Option B is the only viable candidate in seconds! Common Pitfall: A very frequent mistake is adding the remainder 1 to the numbers *before* taking the L.C.M., or forgetting to add the remainder entirely at the end (choosing 504). Always calculate the pure L.C.M. first, and only append the remainder at the final step. Final Answer: Therefore, the correct answer is 505.

Question 11

Find the greatest number of five digits which is divisible by $15$, $21$ and $36$.

Accepted Answer

Concept & Strategy: To find the greatest N-digit number divisible by a set of numbers, first find their L.C.M. Then, divide the largest possible N-digit number (e.g., 99999) by this L.C.M. Subtract the resulting remainder from the N-digit number to get a perfect multiple. Required Number = (Largest N-digit number) - Remainder Step-by-Step Solution: • Given: Target divisors are 15, 21, and 36. We need a 5-digit max number. • The largest 5-digit number is 99999. • Find the L.C.M. of 15, 21, and 36: • 15 = 3 × 5 • 21 = 3 × 7 • 36 = 2^2 × 3^2 • L.C.M. = 2^2 × 3^2 × 5 × 7 = 4 × 9 × 35 = 1260 • Divide 99999 by the L.C.M. (1260): • 1260 × 70 = 88200 • 1260 × 79 = 99540 • Division yields a quotient of 79 and a remainder of 459. • Subtract the remainder from the original largest 5-digit number: • Required Number = 99999 - 459 = 99540 Exam Strategy & Shortcut: Don't calculate long division if you don't have to. The correct number must be divisible by 36, which means it must be divisible by 9. Check the options for divisibility by 9 (sum of digits): • 99999 (sum 45, divisible, but leaves rem 9 when div by 10/36 fail) • 99540 (sum 27, divisible by 9) • 98260 (sum 25, fail) • 99640 (sum 28, fail) Option B is clearly the only mathematically viable distractor that perfectly handles the 9 and 20 constraints. Common Pitfall: Students often confuse "greatest number" phrasing with finding the Highest Common Factor (H.C.F.). Here, "greatest number" acts as a boundary condition for the magnitude of the answer, not the factor operation. You must still use L.C.M. Final Answer: Therefore, the correct answer is 99540.

Question 12

Find the largest number which when subtracted from $10000$, the remainder is divisible by $32$, $36$, $48$ and $54$.

Accepted Answer

Concept & Logic: Let the required number be X. The problem states that when X is subtracted from 10000, the resulting difference is completely divisible by a given set of numbers. This means the difference must be the L.C.M. of those divisors (to ensure X is as large as possible, we subtract the smallest possible multiple, which is the L.C.M.). 10000 - X = L.C.M.(a, b, c, d) X = 10000 - L.C.M.(a, b, c, d) Step-by-Step Solution: • Given: Base number 10000. Divisors are 32, 36, 48, and 54. • First, find the L.C.M. of the divisors using prime factorization: • 32 = 2^5 • 36 = 2^2 × 3^2 • 48 = 2^4 × 3 • 54 = 2 × 3^3 • Extract the highest powers of the prime factors: • For 2: 2^5 = 32 • For 3: 3^3 = 27 • Multiply to find the L.C.M.: • L.C.M. = 32 × 27 = 864 • Now apply the core logic equation. We know 10000 - X = 864. • X = 10000 - 864 = 9136 Exam Strategy & Shortcut: You can work directly from the options using divisibility rules. The question asks for an option X such that (10000 - X) is a multiple of 54 (and thus a multiple of 9). • 10000 - 9136 = 864 (Digit sum is 18 \rightarrow Divisible by 9) • 10000 - 864 = 9136 (Digit sum is 19 \rightarrow Fail) Option B successfully creates a difference that satisfies the divisibility condition. Common Pitfall: A very common trap is doing all the hard work to calculate the L.C.M. (864) and then immediately choosing 864 as the final answer. You must re-read the prompt carefully: it asks for the number *subtracted from* 10000 to get the L.C.M., not the L.C.M. itself. Final Answer: Therefore, the correct answer is 9136.

Question 13

Find the least number which when divided by 20, 25, 35 and 40 leaves remainders 14, 19, 29 and 34 respectively.

Accepted Answer

Concept & Logic: When a number divided by divisors a, b, c leaves remainders p, q, r respectively, and the difference between each divisor and its remainder is a constant K (i.e., a-p = b-q = c-r = K), the required number is the L.C.M. of the divisors minus that constant difference. Required Number = L.C.M.(a, b, c) - K Step-by-Step Solution: • Given: Divisors are 20, 25, 35, and 40. Remainders are 14, 19, 29, and 34. • Find the common difference (K) between each divisor and its respective remainder: • 20 - 14 = 6 • 25 - 19 = 6 • 35 - 29 = 6 • 40 - 34 = 6 • Since the difference is constant (K = 6), calculate the L.C.M. of the divisors (20, 25, 35, 40). • Using prime factorization or division method, the L.C.M. of 20, 25, 35, 40 is 1400. • Subtract the common difference from the L.C.M.: • Required Number = 1400 - 6 = 1394. Exam Strategy & Shortcut: Use Option Elimination. The problem states that dividing the number by 40 leaves a remainder of 34. This means if you add 6 to the correct option, it must be perfectly divisible by 40 (and 20, 25, 35). Let's test the options by adding 6: • 1394 + 6 = 1400 (Divisible by 40, ends in 00 so divisible by 25. This works perfectly). • 1400 + 6 = 1406 (Not divisible by 40). This identifies the answer in seconds without finding the LCM. Common Pitfall: Students often intuitively *add* the constant difference (6) to the L.C.M. instead of subtracting it. Adding 6 gives 1406, which would leave a constant remainder of 6 in every case, completely violating the staggered remainders required by the question. Final Answer: Therefore, the correct answer is 1394.

Question 14

What is the least number which when divided by the numbers 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but when divided by 13 leaves no remainder?

Accepted Answer

Concept & Logic: To find a number that leaves a constant remainder R when divided by a set of numbers, it must be of the form (L.C.M. × k) + R, where k is an integer (1, 2, 3 \dots). To satisfy an additional divisibility condition (like being perfectly divisible by 13), we must find the smallest value of k that makes the entire expression divisible by that number. Required Number = (L.C.M. × k) + R Step-by-Step Solution: • Given: Divisors are 3, 5, 6, 8, 10, and 12. Constant remainder is 2. The final number must be a multiple of 13. • First, find the L.C.M. of 3, 5, 6, 8, 10, and 12. • The L.C.M. is 120. • The required number must be of the form: 120k + 2. • We need to find the least value of k such that (120k + 2) is exactly divisible by 13. To make testing easier, break 120 down into a multiple of 13 plus a remainder: • 120k = (117k + 3k) *(since 117 is exactly divisible by 13)* • So, (120k + 2) = 117k + (3k + 2) • Since 117k is always divisible by 13, we only need to find a k where (3k + 2) is divisible by 13. Test small values for k: • If k = 1: 3(1) + 2 = 5 (No) • If k = 2: 3(2) + 2 = 8 (No) • If k = 3: 3(3) + 2 = 11 (No) • If k = 8: 3(8) + 2 = 26 (Yes! 26 is divisible by 13). • Substitute k = 8 back into the original form: • Required number = 120(8) + 2 = 960 + 2 = 962. Exam Strategy & Shortcut: Use Divisibility Rules on Options. The question demands the answer is exactly divisible by 13. Test the options directly: • 962 / 13 = 74 (Perfectly divisible). • 122 / 13 = 9.38 (Fail). • 242 / 13 = 18.61 (Fail). If multiple options are divisible by 13, subtract 2 from them and check if the result is divisible by 10 (ends in 0) and 12. 962 - 2 = 960, which works perfectly. Common Pitfall: A standard trap is ignoring the k multiplier and simply calculating L.C.M. + Remainder = 120 + 2 = 122. While 122 satisfies the first condition, it completely fails the secondary condition of being divisible by 13. Always apply the multiplier form for compound conditions. Final Answer: Therefore, the correct answer is 962.

Question 15

The traffic lights at three different road crossings change after every 48 sec., 72 sec. and 108 sec. respectively. If they all change simultaneously at 8:20:00 hours, then at what time will they again change simultaneously?

Accepted Answer

Concept & Strategy: When periodic events (like bells ringing, lights changing, or people running a track) occur at different intervals, they will all happen together again at a time interval equal to the Lowest Common Multiple (L.C.M.) of their individual periods. Simultaneous Interval = L.C.M.(T_1, T_2, T_3, \dots) Step-by-Step Solution: • Given: Intervals of change are 48 sec, 72 sec, and 108 sec. Initial simultaneous change was at 8:20:00. • Find the L.C.M. of 48, 72, and 108 using prime factorization: • 48 = 2^4 × 3 • 72 = 2^3 × 3^2 • 108 = 2^2 × 3^3 • Extract the highest powers of the prime factors: • Highest power of 2 is 2^4 = 16. • Highest power of 3 is 3^3 = 27. • Multiply them to get the L.C.M.: • L.C.M. = 16 × 27 = 432 seconds. • This means the lights will change together again after 432 seconds. • Convert 432 seconds into minutes and seconds: • 432 / 60 = 7 minutes with a remainder of 12 seconds. • So, 432 seconds = 7 minutes 12 seconds. • Add this time interval to the initial time (8:20:00): • 8 hours : 20 minutes : 00 seconds • + 0 hours : 7 minutes : 12 seconds • = 8:27:12 hours. Exam Strategy & Shortcut: Instead of heavy prime factorization, look for common large factors to speed up L.C.M. calculation. All numbers are multiples of 12. • 48 = 12 × 4 • 72 = 12 × 6 • 108 = 12 × 9 The L.C.M. of (4, 6, 9) is simply 36. Multiply the common factor back: 12 × 36 = 432 seconds. Then convert to minutes rapidly. Common Pitfall: A careless mistake in time addition. Students calculate 432 seconds correctly but mess up the base-60 conversion (e.g., assuming 100 seconds in a minute), leading to an incorrect final timestamp. Always meticulously divide by 60 for time. Final Answer: Therefore, the correct answer is 8:27:12 hrs.

Question 16

Seema, Meena and Reema begin to jog around a circular stadium and they complete their revolutions in 54 seconds, 42 seconds and 63 seconds respectively. After how much time will they come together at the starting point?

Accepted Answer

Concept & Strategy: When multiple entities move in a circular path starting at the same time and place, they will all meet again at the starting point after a time equal to the Lowest Common Multiple (L.C.M.) of the times taken by each to complete one full revolution. Meeting Time = L.C.M.(T_1, T_2, T_3, \dots) Step-by-Step Solution: • Given: Revolution times are 54 seconds, 42 seconds, and 63 seconds. • Calculate the L.C.M. of 54, 42, and 63. Let's use prime factorization: • 54 = 2 × 3^3 • 42 = 2 × 3 × 7 • 63 = 3^2 × 7 • Take the highest power of each prime factor present: • For 2: 2^1 = 2 • For 3: 3^3 = 27 • For 7: 7^1 = 7 • Multiply to find the L.C.M.: • L.C.M. = 2 × 27 × 7 = 54 × 7 = 378 seconds. • The three girls will meet again at the starting point exactly 378 seconds later. • Convert 378 seconds into minutes and seconds: • 378 / 60 gives a quotient of 6 (minutes) and a remainder of 18 (seconds). • So, 378 seconds = 6 minutes 18 seconds. Exam Strategy & Shortcut: Use the options backwards. The correct total seconds must be a multiple of 63 (which means it must be a multiple of 9, so its digit sum is 9). Convert the options to seconds: • Option A: (6 × 60) + 18 = 378 (Digit sum = 18, div by 9. Works). • Option B: (6 × 60) + 36 = 396 (Digit sum = 18. Does it divide by 7? 396 / 7 = 56.5. Fails). • Option C: (5 × 60) + 48 = 348 (Fails digit sum of 9). Option A is confirmed in seconds without calculating the full LCM from scratch. Common Pitfall: Students sometimes confuse "meeting at the starting point" with "meeting anywhere on the track." If the question asks when they meet *anywhere*, relative speeds are required. Because they specify meeting *at the starting point*, pure L.C.M. of individual lap times is the only correct method. Final Answer: Therefore, the correct answer is 6 min 18 sec.

Question 17

Arrange the fractions $\frac{17}{18}$, $\frac{31}{36}$, $\frac{43}{45}$, $\frac{59}{60}$ in the ascending order.

Accepted Answer

Concept & Strategy: To reliably compare and order multiple fractions, convert them into equivalent fractions with a common denominator. The common denominator will be the L.C.M. of the original denominators. Once the denominators are identical, simply sort the fractions based on the values of their numerators. Step-by-Step Solution: • Given: The fractions are (17/18), (31/36), (43/45), and (59/60). • Find the L.C.M. of the denominators (18, 36, 45, and 60). • L.C.M. of 18, 36, 45, 60 = 180. • Convert each fraction to an equivalent fraction with a denominator of 180: • (17/18) = (17 × 10/18 × 10) = (170/180) • (31/36) = (31 × 5/36 × 5) = (155/180) • (43/45) = (43 × 4/45 × 4) = (172/180) • (59/60) = (59 × 3/60 × 3) = (177/180) • Compare the new numerators: • 155 < 170 < 172 < 177 • Therefore, their corresponding equivalent fractions follow the same ascending order: • (155/180) < (170/180) < (172/180) < (177/180) • Substitute back the original fractions: • (31/36) < (17/18) < (43/45) < (59/60) Exam Strategy & Shortcut: Use the Numerator-Denominator Difference Trick. Look at the gap between the numerator and denominator for each fraction: • (17/18) (Gap = 1) • (31/36) (Gap = 5) • (43/45) (Gap = 2) • (59/60) (Gap = 1) For proper fractions where values are close to 1, the fraction with the largest gap is generally the smallest. The gap of 5 implies (31/36) is the smallest. The gap of 2 implies (43/45) is next. The gap of 1 is shared by two fractions, so compare them directly: (17/18) vs (59/60). Since 59/60 is closer to 60/60, it is larger. Result: (31/36) < (17/18) < (43/45) < (59/60). Common Pitfall: Attempting to convert all fractions into exact decimals by long division under time pressure is a major mistake. It takes far too much time and increases the risk of calculation errors. Stick to L.C.M. equivalence or gap observation methods. Final Answer: Therefore, the correct answer is (31/36) < (17/18) < (43/45) < (59/60).

Question 18

A number $n$ is said to be perfect if the sum of all its divisors (excluding $n$ itself) is equal to $n$. An example of perfect number is

Accepted Answer

Concept & Strategy: A "Perfect Number" is a positive integer that is equal to the sum of its proper divisors. Proper divisors are all the positive integers that divide the number evenly, excluding the number itself. The strategy here is straightforward: extract the proper divisors for each option and calculate their sum to see which one equals the original number. Step-by-Step Solution: We will test each option to find the perfect number: • Test Option (a) 6: Divisors of 6 (excluding 6): 1, 2, 3 Sum of divisors = 1 + 2 + 3 = 6 Since 6 = 6, this is a perfect number! • Test Option (b) 9: Divisors of 9 (excluding 9): 1, 3 Sum of divisors = 1 + 3 = 4 Since 4 
eq 9, it is not a perfect number. • Test Option (c) 15: Divisors of 15 (excluding 15): 1, 3, 5 Sum of divisors = 1 + 3 + 5 = 9 Since 9 
eq 15, it is not a perfect number. • Test Option (d) 21: Divisors of 21 (excluding 21): 1, 3, 7 Sum of divisors = 1 + 3 + 7 = 11 Since 11 
eq 21, it is not a perfect number. Exam Strategy & Shortcut: Memorization Benchmark: Perfect numbers are rare in early mathematics. The first few perfect numbers are mathematically famous: 6, 28, 496, and 8128. Memorizing at least the first two (6 and 28) is highly recommended for aptitude exams as they frequently appear as trivia or direct questions, allowing you to bypass calculation entirely. Common Pitfall: A common error is forgetting to exclude the number itself (n) when calculating the sum of divisors. If you include the number itself, the sum for a perfect number will always equal 2n instead of n. Final Answer: Therefore, the correct answer is 6.

Question 19

$\frac{1095}{1168}$ when expressed in simplest form is

Accepted Answer

Concept & Strategy: To reduce a fraction to its simplest form, we need to find the Highest Common Factor (HCF) of its numerator and denominator, and then divide both by this HCF. Alternatively, we can use the difference method: the HCF of two numbers divides their difference. This helps quickly narrow down potential factors without doing long division. Step-by-Step Solution: • Given: The fraction is (1095/1168). • Calculation / Deduction: 1. Let's find the difference between the denominator and the numerator: 1168 - 1095 = 73 2. The HCF must be a factor of 73. Since 73 is a prime number, the HCF is highly likely to be 73. 3. Test if both numbers are divisible by 73: 1095 \div 73 = 15 1168 \div 73 = 16 4. Since both divide perfectly, we reduce the fraction by dividing the top and bottom by 73: (1095 \div 73/1168 \div 73) = (15/16) Exam Strategy & Shortcut: Unit Digit Estimation: Look at the options. The fraction ends in 5 on top and 8 on the bottom. We need a multiplier k such that k × Numerator Option ends in 5, and k × Denominator Option ends in 8. Let's test option (b) (15/16). If 15 × k ends in 5, k must be odd (like 3, 5, 7, \dots). If 16 × k ends in 8, k could be 3 (16 × 3 = 48) or 8 (16 × 8 = 128). Let's test k ending in 3: 15 × 3 = 45 (ends in 5), 16 × 3 = 48 (ends in 8). Scaling this up, let k = 73. 15 × 73 = 1095. 16 × 73 = 1168. The match is instantaneous. Common Pitfall: Trying to systematically divide 1095 and 1168 by small primes (2, 3, 5, 7, 11\dots). Since their only common factor is the large prime 73, you will waste immense time and likely assume the fraction cannot be reduced. Always use the difference method for large numbers. Final Answer: Therefore, the correct answer is (15/16).

Question 20

Reduce $\frac{128352}{238368}$ to its lowest terms.

Accepted Answer

Concept & Strategy: Reducing massive fractions like (128352/238368) using standard division or finding the HCF is too time-consuming for an aptitude exam. Instead, we reverse-engineer the problem using the given multiple-choice options. We assume one of the options is correct, which means its denominator multiplied by some factor k equals 238368, and its numerator multiplied by the same k equals 128352. Step-by-Step Solution: • Given: The fraction (128352/238368). • Calculation / Deduction: 1. First, quickly estimate the value of the fraction to eliminate wild options. (128/238) is roughly (12/24) or (1/2) (0.5). 2. Check the options based on this estimate: (3/4) = 0.75 (Too big) (5/13) \approx 0.38 (Too small) (7/13) \approx 0.53 (Very close) (9/13) \approx 0.69 (Too big) 3. Option (c) is the only logical candidate. Let's verify it strictly. 4. If (7/13) is the exact ratio, then 238368 \div 13 should give us our exact multiplier k. 238368 \div 13 = 18336 5. Now multiply this multiplier k by the numerator 7: 18336 × 7 = 128352 6. This perfectly matches the original numerator, confirming our ratio. Exam Strategy & Shortcut: Cross-Multiplication Unit Digit Check: If (128352/238368) = (7/13), then 128352 × 13 = 238368 × 7. Check just the unit digits: Left side: 2 × 3 = 6. Right side: 8 × 7 = 56 (ends in 6). The unit digits match perfectly. Doing this mentally takes 5 seconds and solidly confirms your estimation. Common Pitfall: Students often attempt long division to find the HCF of 128352 and 238368. For six-digit numbers, this is a trap designed to drain your exam time. Never calculate the HCF directly if simple fractional options are provided. Final Answer: Therefore, the correct answer is (7/13).

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