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 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. $$ 	ext{Required Number} = 	ext{H.C.F. of } |x - y|, |y - z|, 	ext{ and } |z - x| $$ ### 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 4

$\frac{1095}{1168}$ 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 $\frac{1095}{1168}$. * **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 $73$: $$\frac{1095 \div 73}{1168 \div 73} = \frac{15}{16}$$ ### 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 $k 	imes 	ext{Numerator Option}$ ends in $5$, and $k 	imes 	ext{Denominator Option}$ ends in $8$. Let's test option (b) $\frac{15}{16}$. If $15 	imes k$ ends in $5$, $k$ must be odd (like $3, 5, 7, \dots$). If $16 	imes k$ ends in $8$, $k$ could be $3$ ($16 	imes 3 = 48$) or $8$ ($16 	imes 8 = 128$). Let's test $k$ ending in $3$: $15 	imes 3 = 45$ (ends in $5$), $16 	imes 3 = 48$ (ends in $8$). Scaling this up, let $k = 73$. $15 	imes 73 = 1095$. $16 	imes 73 = 1168$. 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 $\frac{15}{16}$.**

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