Question 1

The H.C.F. of $\frac{9}{10}, \frac{12}{25}, \frac{18}{35}$ and $\frac{21}{40}$ is

Accepted Answer

Concept & Formula: When calculating the Highest Common Factor (HCF) of several fractions, we apply the foundational fraction rule: H.C.F. = \frac{H.C.F. of Numerators}{L.C.M. of Denominators} Step-by-Step Solution: • Given Fractions: (9/10), (12/25), (18/35), and (21/40). • Step 1: Calculate the HCF of the numerators (9, 12, 18, 21). All of these numbers are clearly multiples of 3. 9 = 3 × 3 12 = 3 × 4 18 = 3 × 6 21 = 3 × 7 The highest number that divides all of them is 3. So, the numerator of our answer is 3. • Step 2: Calculate the LCM of the denominators (10, 25, 35, 40). Prime factorize each: 10 = 2 × 5 25 = 5^2 35 = 5 × 7 40 = 2^3 × 5 Take the highest power of each prime factor (2, 5, 7): LCM = 2^3 × 5^2 × 7 LCM = 8 × 25 × 7 LCM = 200 × 7 = 1400 • Step 3: Assemble the fraction. Result = (3/1400) Exam Strategy & Shortcut: Calculate the HCF of the numerators first, which is 3. Instantly eliminate option (b) and (d). You are left with (3/5) and (3/1400). Now look at the denominators. For the denominator of the answer to be 5, the LCM of 10, 25, 35, 40 would have to be 5, which is logically impossible (the LCM must be greater than or equal to the largest number, 40). Thus, without doing the heavy math, option (c) is the only mathematically viable choice. Common Pitfall: Calculating the full LCM for the denominators using the ladder method wastes a lot of time. In multiple-choice exams, determining just the numerator and doing a quick logical check on the remaining denominator options is the key to finishing on time. Final Answer: Therefore, the correct answer is (3/1400).

Question 2

The L.C.M. of $\frac{1}{3}, \frac{5}{6}, \frac{2}{9}, \frac{4}{27}$ is

Accepted Answer

Concept & Formula: To find the Least Common Multiple (LCM) of a set of fractions, we use the standard formula: L.C.M. of fractions = \frac{L.C.M. of Numerators}{H.C.F. of Denominators} Step-by-Step Solution: • Given Fractions: (1/3), (5/6), (2/9), and (4/27). • Step 1: Calculate the L.C.M. of the numerators (1, 5, 2, 4). The multiples of the highest number (5) that are also divisible by 4 and 2 is 20. L.C.M. (1, 5, 2, 4) = 20 • Step 2: Calculate the H.C.F. of the denominators (3, 6, 9, 27). The factors are 3, 2 × 3, 3^2, and 3^3. The highest common factor shared by all is clearly 3. H.C.F. (3, 6, 9, 27) = 3 • Step 3: Assemble the fraction. Result = (20/3) Exam Strategy & Shortcut: First, scan the numerators: 1, 5, 2, 4. Their L.C.M. is obviously 20. Look at the options—only option (c) has 20 in the numerator. You can confidently select (c) without even calculating the H.C.F. for the denominators. Common Pitfall: Students often mix up the formula, calculating the L.C.M. of the denominators and the H.C.F. of the numerators (which would give (1/54), option a). Remember: "Whatever they ask for goes on top." If they ask for L.C.M., put L.C.M. in the numerator. Final Answer: Therefore, the correct answer is (20/3).

Question 3

The L.C.M. of $\frac{3}{4}, \frac{6}{7}, \frac{9}{8}$ is

Accepted Answer

Concept & Formula: To evaluate the Least Common Multiple (LCM) of given fractions, the corresponding rule is: L.C.M. of fractions = \frac{L.C.M. of Numerators}{H.C.F. of Denominators} Step-by-Step Solution: • Given Fractions: (3/4), (6/7), and (9/8). • Step 1: Determine the L.C.M. of numerators (3, 6, 9). The numbers are 3, 2 × 3, and 3^2. The L.C.M. is 2 × 3^2 = 18. • Step 2: Determine the H.C.F. of denominators (4, 7, 8). The factors are 2^2, 7, and 2^3. Since 7 is a prime number and doesn't divide 4 or 8, the only common factor is 1. • Step 3: Final Calculation. Result = (18/1) = 18 Exam Strategy & Shortcut: Recognize immediately that the denominator 7 is prime, while 4 and 8 are powers of 2. Their H.C.F. must be 1. Thus, the answer is just the L.C.M. of 3, 6, and 9. Since 18 is the smallest number divisible by all three, the answer is 18. Takes 5 seconds visually. Common Pitfall: Students might mistakenly calculate the L.C.M. of both numerator and denominator, or just focus on the largest numbers. Another mistake is choosing 9 (option c) by just looking at the largest numerator, forgetting that 9 is not divisible by 6. Final Answer: Therefore, the correct answer is 18.

Question 4

The H.C.F. of $1.75$, $5.6$ and $7$ is

Accepted Answer

Concept & Strategy: To find the Highest Common Factor (HCF) of decimals, avoid working with them directly as it can cause alignment errors. Strategy: Convert all decimals to whole numbers by multiplying them by a power of 10 (based on the maximum decimal places). Find the HCF of the new whole numbers, then divide the result by that same power of 10. Step-by-Step Solution: • Given: 1.75, 5.6, and 7. • Step 1: Equalize decimal places. The maximum number of decimal places is two (in 1.75). Rewrite all numbers with two decimal places: 1.75, 5.60, 7.00 • Step 2: Convert to whole numbers. Multiply by 100: 175, 560, 700 • Step 3: Find the H.C.F. of 175, 560, 700. Prime factorization: 175 = 5^2 × 7 560 = 2^4 × 5 × 7 700 = 2^2 × 5^2 × 7 The common prime factors are 5 and 7. The lowest powers are 5^1 and 7^1. H.C.F. = 5 × 7 = 35 • Step 4: Convert back to decimal. Divide the H.C.F. by 100: 35 \div 100 = 0.35 Exam Strategy & Shortcut: Look at 1.75 and 7. You know 1.75 × 4 = 7. So, 1.75 perfectly divides 7. The H.C.F. must be a factor of 1.75. Option (c) 3.5 is larger than 1.75, so it's eliminated. Now check if 0.35 works for 5.6. Since 35 × 16 = 560, it works perfectly. Common Pitfall: The most fatal mistake is multiplying by different powers of 10 for each number (e.g., 1.75 × 100, 5.6 × 10, 7 × 1) resulting in 175, 56, 7, which gives a completely incorrect H.C.F. You must equalize the decimal places first before removing the decimal. Final Answer: Therefore, the correct answer is 0.35.

Question 5

The G.C.D. of $1.08$, $0.36$ and $0.9$ is

Accepted Answer

Concept & Strategy: The Greatest Common Divisor (GCD) is the exact same concept as the Highest Common Factor (HCF). To evaluate decimals, first equalize the number of decimal places across all terms, multiply by a base of 10 to make them integers, find the GCD of those integers, and finally restore the decimal point. Step-by-Step Solution: • Given: 1.08, 0.36, and 0.9. • Step 1: Equalize decimal places. The maximum decimal places is two. Rewrite them as: 1.08, 0.36, 0.90 • Step 2: Remove decimals (Multiply by 100). The numbers become 108, 36, and 90. • Step 3: Calculate GCD of (108, 36, 90). Notice that 36 × 3 = 108. So the GCD of 108 and 36 is just 36. Now, find the GCD of 36 and 90. 36 = 18 × 2 90 = 18 × 5 The highest common factor is clearly 18. • Step 4: Restore the decimal point. Divide the result by the multiplier (100): 18 \div 100 = 0.18 Exam Strategy & Shortcut: Convert in your head: 108, 36, 90. You know 18 is a massive common factor here (18 × 6, 18 × 2, 18 × 5). Since 2, 6, 5 share no common factors, 18 is the GCD. Divide by 100 to get 0.18. This takes less than 15 seconds if you are well-versed in your 18 times table. Common Pitfall: A frequent error is omitting the zero in 0.90. Students often treat 0.9 as just 9 when converting, finding the GCD of 108, 36, 9 (which is 9), leading to an incorrect answer of 0.09. Always pad with zeros to match the longest decimal before converting to integers. Final Answer: Therefore, the correct answer is 0.18.

Question 6

The L.C.M. of $3$, $2.7$ and $0.09$ is

Accepted Answer

Concept & Strategy: Just like with H.C.F., the strategy for calculating the L.C.M. of decimals is to convert them into like decimals (same number of decimal places) by adding trailing zeros. Then, treat them as integers, find their L.C.M., and place the decimal point back in the result. Step-by-Step Solution: • Given: The numbers are 3, 2.7, and 0.09. • Step 1: The maximum number of decimal places is two (in 0.09). Rewrite all numbers with two decimal places: 3.00, 2.70, and 0.09. • Step 2: Remove the decimal points by multiplying each by 100. We now need the L.C.M. of 300, 270, and 9. • Step 3: Perform prime factorization for each number: • 300 = 2^2 × 3 × 5^2 • 270 = 2 × 3^3 × 5 • 9 = 3^2 • Step 4: The L.C.M. is found by multiplying the highest power of every prime factor present. L.C.M. = 2^2 × 3^3 × 5^2 L.C.M. = 4 × 27 × 25 = 100 × 27 = 2700 • Step 5: Adjust the decimal point by dividing by 100 (since we multiplied by 100 initially). 2700 / 100 = 27 Exam Strategy & Shortcut: Multiple Check: The L.C.M. of a set of numbers must be a multiple of the largest number in that set (or equal to it). The largest number here is 3. The L.C.M. must therefore be \ge 3. Looking at the options, (a) 2.7, (b) 0.27, and (c) 0.027 are all strictly less than 3 and thus physically impossible to be multiples of 3. By pure elimination, only 27 can be the L.C.M. Common Pitfall: A very common mistake is finding the L.C.M. of 3, 27, and 9 (ignoring place value alignment) which gives 27, and then guessing where to put the decimal point back (often guessing 0.27). Always pad the numbers to equal length (3.00, 2.70, 0.09) before dropping decimals to maintain mathematical integrity. Final Answer: Therefore, the correct answer is 27.

Question 7

H.C.F. of $3240$, $3600$ and a third number is $36$ and their L.C.M. is $2^4 	imes 3^5 	imes 5^2 	imes 7^2$. The third number is

Accepted Answer

Concept & Formula: When analyzing the H.C.F. and L.C.M. of three or more numbers via prime factorization: • H.C.F. is formed by taking the lowest power of each prime factor common to all numbers. • L.C.M. is formed by taking the highest power of every prime factor present across any of the numbers. By comparing the prime factorizations of the known numbers against the given H.C.F. and L.C.M., we can deduce the exact prime factor makeup of the unknown third number. Step-by-Step Solution: • Given: • Number 1 (N_1) = 3240 • Number 2 (N_2) = 3600 • Unknown Number 3 (N_3) = ? • H.C.F. = 36 • L.C.M. = 2^4 × 3^5 × 5^2 × 7^2 • Step 1: Prime factorize N_1, N_2, and the H.C.F. • 3240 = 324 × 10 = 18^2 × (2 × 5) = (2 × 3^2)^2 × (2 × 5) = 2^3 × 3^4 × 5^1 • 3600 = 36 × 100 = (2^2 × 3^2) × (2^2 × 5^2) = 2^4 × 3^2 × 5^2 • H.C.F. (36) = 2^2 × 3^2 • Step 2: Let N_3 = 2^a × 3^b × 5^c × 7^d. We will determine a, b, c, d by applying H.C.F. and L.C.M. rules. • Step 3 (Analyze base 2): • H.C.F. has 2^2. This means the *minimum* power of 2 among N_1(2^3), N_2(2^4), and N_3(2^a) must be 2. Thus, a = 2. (This eliminates options c and d). • Step 4 (Analyze base 3): • L.C.M. has 3^5. This means the *maximum* power of 3 among N_1(3^4), N_2(3^2), and N_3(3^b) must be 5. Neither N_1 nor N_2 has 3^5, so N_3 must provide it. Thus, b = 5. • Step 5 (Analyze base 5): • H.C.F. has no 5 (or 5^0). N_1 has 5^1 and N_2 has 5^2. To force the minimum power of 5 to be 0, N_3 must not contain 5 as a factor. Thus, c = 0. (This eliminates option b). • Step 6 (Analyze base 7): • L.C.M. has 7^2. Neither N_1 nor N_2 has any 7s. Thus, N_3 must provide the 7^2. Thus, d = 2. • Conclusion: Bringing it all together, N_3 = 2^2 × 3^5 × 7^2. Exam Strategy & Shortcut: Surgical Elimination: You do not need to calculate everything to find the answer. Look at the L.C.M.: it contains 7^2 and 3^5. By quick mental prime factorization, 3240 maxes out at 3^4, and 3600 maxes out at 3^2. Neither provides 3^5, and neither provides 7^2. Therefore, the third number must contain exactly 3^5 × 7^2. Only options (a) and (d) have this. Next, look at the H.C.F.: 36 means 2^2. Since 3240 has 2^3 and 3600 has 2^4, the third number must be the one providing the 2^2 limitation. Therefore, it must be 2^2 × 3^5 × 7^2. Common Pitfall: A common error is confusing the minimum and maximum power rules, leading students to select an option like (d) 2^3 × 3^5 × 7^2. If the third number had 2^3, then the lowest power of 2 across all three numbers would be 2^3 = 8. But the H.C.F. is 36, which only contains 2^2 = 4. Thus, the third number strictly anchors the H.C.F. to 2^2. Final Answer: Therefore, the correct answer is 2^2 × 3^5 × 7^2.

Question 8

The product of two numbers is $4107$. If the H.C.F. of these numbers is $37$, then the greater number is

Accepted Answer

Concept & Formula: When the Highest Common Factor (H.C.F.) of two numbers is given, the numbers can be algebraically represented as H × a and H × b, where H is the H.C.F., and a and b are strictly co-prime integers (meaning they share no common factors other than 1). The product of the two numbers is given by the formula: (H × a) × (H × b) = Product Step-by-Step Solution: • Given: The H.C.F. (H) is 37. The product of the two numbers is 4107. • Step 1: Represent the numbers using the H.C.F. Let the two numbers be 37a and 37b, where a and b are co-prime. • Step 2: Set up the equation using the given product. 37a × 37b = 4107 1369 × ab = 4107 • Step 3: Solve for the product of the co-prime factors, ab. ab = (4107/1369) ab = 3 • Step 4: Determine the possible co-prime pairs for (a, b) that multiply to 3. Since 3 is a prime number, the only possible pair is (1, 3). So, let a = 1 and b = 3. • Step 5: Calculate the actual numbers. First number: 37 × 1 = 37 Second number: 37 × 3 = 111 • Step 6: Identify the greater number. Between 37 and 111, the greater number is 111. Exam Strategy & Shortcut: Divisibility & Logical Deduction: You know the numbers must be multiples of 37. Looking at the options, you are searching for a number that, when multiplied by another multiple of 37, yields 4107. Since the last digit of the product is 7, and the H.C.F. ends in 7, you can quickly test the unit digits. If the greater number is 111 (ends in 1), the smaller number must end in 7 to make the product end in 7 (1 × 7 = 7). Since 37 ends in 7, 37 × 111 is a perfect candidate. 37 × 111 = 4107. The math checks out instantly. Common Pitfall: A frequent mistake is forgetting to divide the product by the square of the H.C.F. (H^2). Students often divide 4107 by just 37 once, obtaining 111, and mistakenly assume this is the product ab. While 111 happens to be the final answer here by coincidence, this flawed method will yield completely incorrect results on almost any other exam question of this type. Final Answer: Therefore, the correct answer is 111.

Question 9

Three numbers which are co-prime to each other are such that the product of the first two is $551$ and that of the last two is $1073$. The sum of the three numbers is

Accepted Answer

Concept & Logic: Let the three numbers be A, B, and C. If they are all pairwise co-prime, it means no two numbers share a common factor greater than 1. We are given two overlapping products: Product 1: A × B Product 2: B × C The middle number, B, is common to both products. Therefore, B must be exactly equal to the Highest Common Factor (H.C.F.) of the two given product values. Step-by-Step Solution: • Given: Let the three numbers be x, y, and z. x × y = 551 y × z = 1073 • Step 1: Identify the common variable. The number y is a factor of both 551 and 1073. Since the numbers are co-prime to each other, y is the *only* common factor between these two products. Thus, y = H.C.F.(551, 1073). • Step 2: Calculate the H.C.F. of 551 and 1073. Using the long division method or difference method: Difference = 1073 - 551 = 522. The H.C.F. must be a factor of 522. Let's test divisibility. 551 is not divisible by 2 or 3. Let's use Euclidean division: 1073 = 551 × 1 + 522 551 = 522 × 1 + 29 522 = 29 × 18 + 0 The remainder is 0, so the H.C.F. is 29. Therefore, the middle number y = 29. • Step 3: Find the first number x. x × 29 = 551 x = (551/29) = 19 • Step 4: Find the third number z. 29 × z = 1073 z = (1073/29) = 37 • Step 5: Calculate the sum of the three numbers. Sum = x + y + z = 19 + 29 + 37 = 85 Exam Strategy & Shortcut: Difference Method for Fast H.C.F.: Finding the H.C.F. of large odd numbers can be tedious. Instead, find their difference: 1073 - 551 = 522. The H.C.F. must divide this difference. 522 is even, so divide it by 2 \rightarrow 261. Divide by 3 \rightarrow 87. Divide by 3 again \rightarrow 29. 29 is prime. Try dividing 551 by 29; it works perfectly (19). You now have x=19, y=29. Then find z = 1073 / 29 = 37. Add them up for 85. Common Pitfall: Students often waste immense amounts of time trying to brute-force prime factorize 551 and 1073 independently without realizing they share a massive common factor. Always look for the conceptual overlap (the middle number) and use the H.C.F. to unlock both factorizations simultaneously. Final Answer: Therefore, the correct answer is 85.

Question 10

Three numbers are in the ratio of $3: 4: 5$ and their L.C.M. is $2400$. Their H.C.F. is

Accepted Answer

Concept & Formula: When a set of numbers is given in a ratio a:b:c, they can be represented as ax, bx, and cx, where the variable x represents the Highest Common Factor (H.C.F.) of the numbers. If the base ratio integers (a, b, c) are mutually co-prime (meaning no common factor exists across all of them), the Least Common Multiple (L.C.M.) is calculated by multiplying the ratio integers together with the H.C.F. L.C.M. = a × b × c × H.C.F. Step-by-Step Solution: • Given: Ratio of numbers = 3: 4: 5 L.C.M. = 2400 • Step 1: Assign algebraic variables. Let the three numbers be 3x, 4x, and 5x. Here, x represents the H.C.F. we need to find. • Step 2: Formulate the L.C.M. expression. The integers 3, 4, and 5 are pairwise co-prime (they share no common factors). The L.C.M. of 3x, 4x, and 5x is simply the L.C.M. of (3, 4, 5) multiplied by the common variable x. L.C.M. = (3 × 4 × 5) × x L.C.M. = 60x • Step 3: Solve for x using the given L.C.M. value. 60x = 2400 x = (2400/60) x = 40 • Conclusion: Since x represents the common factor, the H.C.F. is exactly 40. Exam Strategy & Shortcut: Direct Formula Application: If you recognize that 3, 4, and 5 are co-prime, you can skip writing equations entirely. You just multiply the ratio numbers and divide the L.C.M. by that product. 3 × 4 × 5 = 60 H.C.F. = (2400/60) = 40 This is a 5-second mental math problem. Common Pitfall: The main danger occurs if the ratio integers are *not* co-prime (for example, if the ratio was 2:4:5). In that case, you cannot simply multiply them together (2 × 4 × 5 × x). You would have to find the actual L.C.M. of the ratio integers first (L.C.M. of 2, 4, 5 is 20, so it would be 20x). Always verify that the ratio components share no common factors before applying the direct multiplication shortcut. Here, 3, 4, 5 are clear, so the shortcut works flawlessly. Final Answer: Therefore, the correct answer is 40.

Question 11

The sum of two numbers is $528$ and their H.C.F. is $33$. The number of pairs of numbers satisfying the above conditions is

Accepted Answer

Concept & Formula: Let the two numbers be 33a and 33b, where a and b are strictly co-prime integers (H.C.F.(a, b) = 1). Using the sum condition: 33a + 33b = Sum The number of valid co-prime pairs (a, b) determines the total number of pairs of numbers that satisfy the given conditions. Step-by-Step Solution: • Given: Sum = 528, H.C.F. = 33. • Step 1: Represent the numbers as 33a and 33b. • Step 2: Formulate the equation: 33a + 33b = 528 33(a + b) = 528 • Step 3: Isolate (a + b): a + b = (528/33) = 16 • Step 4: List all positive integer pairs (a, b) such that a + b = 16, and filter for co-prime pairs: • (1, 15) \rightarrow Co-prime (H.C.F. = 1). (Valid) • (2, 14) \rightarrow Both even (H.C.F. = 2). (Invalid) • (3, 13) \rightarrow Co-prime (H.C.F. = 1). (Valid) • (4, 12) \rightarrow Share factor 4. (Invalid) • (5, 11) \rightarrow Co-prime (H.C.F. = 1). (Valid) • (6, 10) \rightarrow Both even (H.C.F. = 2). (Invalid) • (7, 9) \rightarrow Co-prime (H.C.F. = 1). (Valid) • (8, 8) \rightarrow Share factor 8. (Invalid) • Step 5: Count the valid pairs: (1, 15), (3, 13), (5, 11), and (7, 9). There are exactly 4 pairs. Exam Strategy & Shortcut: Coprimality Count: Once you find a + b = 16, you can find the number of valid pairs by calculating how many numbers less than 16 are co-prime to 16 (Euler's totient function \phi(16)) and dividing by 2 (since order doesn't matter for pairs). Since 16 = 2^4, the numbers not co-prime to it are all even numbers. The odd numbers less than 16 are 1, 3, 5, 7, 9, 11, 13, 15 (8 numbers). Pairs = 8 / 2 = 4. This avoids listing and checking every single pair manually. Common Pitfall: A very common mistake is counting all possible additive combinations that sum to 16 (which would give 8 pairs, leading to option c). Always remember to filter out any pairs that share a common factor, as their inclusion would change the H.C.F. from 33 to a higher multiple. Final Answer: Therefore, the correct answer is 4.

Question 12

The number of number-pairs lying between $40$ and $100$ with their H.C.F. as $15$ is

Accepted Answer

Concept & Strategy: If the H.C.F. of a pair of numbers is 15, then both numbers must be distinct multiples of 15. The strategy is to find all multiples of 15 that fall strictly within the range [40, 100], and then determine how many unique pairs can be formed from these multiples such that their H.C.F. remains exactly 15. Step-by-Step Solution: • Given: Range = 40 to 100, H.C.F. = 15. • Step 1: List all multiples of 15 between 40 and 100: • 15 × 3 = 45 • 15 × 4 = 60 • 15 × 5 = 75 • 15 × 6 = 90 So, the available numbers are: \{45, 60, 75, 90\}. • Step 2: Form all possible unique pairs from this set and verify their H.C.F.: • Pair 1: (45, 60) \rightarrow H.C.F.(15×3, 15×4) = 15 × H.C.F.(3,4) = 15. (Valid) • Pair 2: (45, 75) \rightarrow H.C.F.(15×3, 15×5) = 15 × H.C.F.(3,5) = 15. (Valid) • Pair 3: (45, 90) \rightarrow H.C.F.(45, 90) = 45. (Invalid, H.C.F. is not 15) • Pair 4: (60, 75) \rightarrow H.C.F.(15×4, 15×5) = 15 × H.C.F.(4,5) = 15. (Valid) • Pair 5: (60, 90) \rightarrow H.C.F.(60, 90) = 30. (Invalid, H.C.F. is not 15) • Pair 6: (75, 90) \rightarrow H.C.F.(15×5, 15×6) = 15 × H.C.F.(5,6) = 15. (Valid) • Step 3: Count the valid pairs: (45, 60), (45, 75), (60, 75), and (75, 90). There are 4 valid pairs. Exam Strategy & Shortcut: Co-prime Factor Verification: Instead of finding the H.C.F. of the large numbers, look at their multipliers: \{3, 4, 5, 6\}. A pair of numbers will have an H.C.F. of 15 if and only if their multipliers are co-prime. Form pairs from \{3, 4, 5, 6\}: • (3,4) - Yes • (3,5) - Yes • (3,6) - No (share factor 3) • (4,5) - Yes • - No (share factor 2) • (5,6) - Yes Total valid combinations = 4. This completely eliminates large-number factorization. Common Pitfall: A frequent pitfall is assuming that *any* pair formed from the multiples of 15 will have an H.C.F. of 15. Students often calculate the total combinations of 4 items taken 2 at a time: \binom{4}{2} = 6 and incorrectly choose option (d). You must explicitly check and remove pairs like (45, 90) and (60, 90) where the multiplier terms share a common factor. Final Answer: Therefore, the correct answer is 4.

Question 13

The L.C.M. and ratio of four numbers are 630 and 2 : 3 : 5 : 7 respectively. The difference between the greatest and least numbers is

Accepted Answer

Concept & Logic: The L.C.M. of numbers given in a ratio where the ratio terms are mutually coprime is simply the product of those ratio parts multiplied by their Highest Common Factor (H.C.F.). L.C.M. = H.C.F. × (Product of coprime ratio parts) Step-by-Step Solution: • Let the common factor (H.C.F.) be x. The four numbers can be represented as 2x, 3x, 5x, and 7x. • Because the numbers 2, 3, 5, and 7 are prime to each other (coprime), their L.C.M. is the product of these ratios and x. • L.C.M. = 2 × 3 × 5 × 7 × x = 210x. • We are given that the L.C.M. is 630. • Equating the two: 210x = 630. • Solving for x: x = 630 / 210 = 3. • The greatest number is 7x = 7 × 3 = 21. • The least number is 2x = 2 × 3 = 6. • The difference between the greatest and least numbers is 21 - 6 = 15. Exam Strategy & Shortcut: Look at the ratio: the difference between the greatest part (7) and the least part (2) is 5 parts. This means the final answer MUST be a multiple of 5. Scanning the given options (6, 14, 15, 21), 15 is the only multiple of 5. You can confidently select it without calculating the L.C.M. or solving for x. Common Pitfall: A frequent mistake is solving for x = 3 and then mistakenly choosing option (a) which is 6, or option (d) which is 21, by forgetting that the question specifically asks for the *difference* between the two extreme numbers, not just one of the numbers. Final Answer: Therefore, the correct answer is 15.

Question 14

If the product of two numbers is 324 and their H.C.F. is 3, then their L.C.M. will be

Accepted Answer

Concept & Formula: The relationship between the Highest Common Factor (H.C.F.), Lowest Common Multiple (L.C.M.), and the product of two numbers is constant. Product of Two Numbers = H.C.F. × L.C.M. Step-by-Step Solution: • Given: • Product of the two numbers = 324 • H.C.F. = 3 • Let the required L.C.M. be L. • Substitute the given values directly into the standard formula: 324 = 3 × L. • Solve for L: L = (324/3). • Perform the division: 324 / 3 = 108. Exam Strategy & Shortcut: This is a direct application of the formula and requires minimal calculation. To speed it up mentally, break down 324 into 300 + 24. Dividing by 3 gives you 100 + 8 = 108. You should solve this entirely in your head within 5 seconds without touching pen to paper. Common Pitfall: A common trap for students rushing through the exam is misreading "product of two numbers" as one of the numbers, leading them to multiply 324 by 3 (yielding 972, which is cleverly placed as option a). Always pause for one second to ensure you've correctly identified whether you are given the individual numbers or their total product. Final Answer: Therefore, the correct answer is 108.

Question 15

If H.C.F. of $p$ and $q$ is $x$ and $q = xy$, then the L.C.M. of $p$ and $q$ is

Accepted Answer

Concept & Algebraic Logic: The fundamental rule connecting H.C.F. and L.C.M. applies not just to integers, but perfectly to algebraic variables as well. H.C.F. × L.C.M. = First Number × Second Number Step-by-Step Solution: • Given: * The two numbers are p and q. • Their H.C.F. = x. • We are also given a substitution identity: q = xy. • According to the core property: x × L.C.M. = p × q. • Isolate the L.C.M.: L.C.M. = (p × q/x). • Now, substitute the given value q = xy into the equation: L.C.M. = (p × (xy)/x). • Cancel out the common term x from the numerator and denominator. • L.C.M. = p × y = py. Exam Strategy & Shortcut: Since q = xy and x is the H.C.F., y is the leftover coprime factor of q. The L.C.M. must contain all prime factors at their highest powers. It will simply be the first number p multiplied by the leftover coprime factor of the second number y. Hence, p × y = py. Common Pitfall: Students often get intimidated by algebraic variables and try to plug in random numbers to solve the problem. While plugging in numbers (e.g., p=4, q=6) can work, it takes significantly longer than simply trusting the algebraic formula and substituting the variables directly. Final Answer: Therefore, the correct answer is py.

Question 16

The H.C.F. and L.C.M. of two numbers are 21 and 84 respectively. If the ratio of the two numbers is 1 : 4, then the larger of the two numbers is

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Concept & Formula: When two numbers are given in a ratio a : b, the actual numbers can be represented as ax and bx, where the common multiplier x is the Highest Common Factor (H.C.F.) of the two numbers. Numbers = (Ratio Part 1 × H.C.F.) and (Ratio Part 2 × H.C.F.) Step-by-Step Solution: • Given: * H.C.F. = 21 • L.C.M. = 84 • Ratio of numbers = 1 : 4 • Let the two numbers be 1x and 4x. • By definition, the common factor x is the H.C.F. Therefore, x = 21. • The two numbers are: • First number = 1 × 21 = 21 • Second number = 4 × 21 = 84 • The question asks for the larger of the two numbers. • Comparing 21 and 84, the larger number is 84. Exam Strategy & Shortcut: You don't even need the L.C.M. to solve this specific question. The problem explicitly gives the ratio (1:4) and the H.C.F. (21). The larger number corresponds to the larger ratio part (4). Simply multiply the larger ratio part by the H.C.F.: 4 × 21 = 84. You can solve this in 3 seconds mentally. Common Pitfall: A common trap is assuming you need to use all the provided numbers (like the L.C.M. = 84) in a complex formula. While you *could* use the formula Product = H.C.F. × L.C.M., it unnecessarily complicates a straightforward ratio multiplication. Don't over-engineer the solution. Final Answer: Therefore, the correct answer is 84.

Question 17

The product of the L.C.M. and H.C.F. of two numbers is 24. The difference of two numbers is 2. Find the numbers.

Accepted Answer

Concept & Logic: The foundational rule of H.C.F. and L.C.M. states that the product of the Highest Common Factor and the Lowest Common Multiple is exactly equal to the product of the two numbers themselves. Product of Numbers = H.C.F. × L.C.M. Step-by-Step Solution: • Given: * H.C.F. × L.C.M. = 24 • Difference of the two numbers = 2 • Let the two numbers be x and y (where x > y). • From the foundational rule, x × y = 24. • We are given that x - y = 2, which means x = y + 2. • Substitute x into the product equation: (y + 2) × y = 24. • Expand to form a quadratic equation: y^2 + 2y - 24 = 0. • Factor the quadratic equation: (y + 6)(y - 4) = 0. • Since numbers in this context are positive integers, y = 4. • If y = 4, then x = 4 + 2 = 6. • The numbers are 6 and 4. Exam Strategy & Shortcut: Option Elimination is by far the fastest method here. Since Product of Numbers = H.C.F. × L.C.M., we know the product of the two numbers must be 24. Let's test the options: • (a) 2 × 4 = 8 (Incorrect) • (b) 6 × 4 = 24 (Correct, and 6 - 4 = 2) • (c) 8 × 6 = 48 (Incorrect) • (d) 8 × 10 = 80 (Incorrect) You can arrive at option (b) in under 5 seconds without writing a single algebraic equation. Common Pitfall: Wasting time solving the quadratic equation algebraically is a classic pitfall. While mathematically sound, standard algebraic methods are too slow for competitive exams when a simple cross-check of the options takes mere seconds. Always look at the options first when given simple products and differences. Final Answer: Therefore, the correct answer is 6 and 4.

Question 18

The H.C.F. and L.C.M. of two numbers are 50 and 250 respectively. If the first number is divided by 2, the quotient is 50. The second number is

Accepted Answer

Concept & Logic: The relationship between any two numbers and their Highest Common Factor (H.C.F.) and Lowest Common Multiple (L.C.M.) is a constant product. Additionally, we use the basic division rule to find the unknown first number. First Number × Second Number = H.C.F. × L.C.M. Dividend = Divisor × Quotient Step-by-Step Solution: • Given: • H.C.F. = 50 • L.C.M. = 250 • First, determine the value of the first number. The problem states that when it is divided by 2, the quotient is 50. • First Number = 2 × 50 = 100. • Now, substitute the known values into the fundamental product formula: • 100 × Second Number = 50 × 250 • Simplify the right side of the equation: • 100 × Second Number = 12500 • Isolate the second number: • Second Number = (12500/100) = 125. Exam Strategy & Shortcut: Avoid calculating large numbers like 12500. Set up the equation: 100 × N = 50 × 250. Notice the relationship between the numbers on both sides. 100 is exactly twice 50. For the equation to balance out, N must be exactly half of 250. Half of 250 is 125. You can solve this entirely in your head in under 10 seconds. Common Pitfall: Students frequently misread the phrase "divided by 2, the quotient is 50" and hastily assume the first number itself is 50. This trap leads them to calculate (50 × 250/50) = 250, incorrectly choosing option (d). Always calculate the initial dividend carefully. Final Answer: Therefore, the correct answer is 125.

Question 19

The product of two numbers is 1320 and their H.C.F. is 6. The L.C.M. of the numbers is

Accepted Answer

Concept & Formula: There is a strict mathematical relationship between the product of two numbers and the product of their H.C.F. and L.C.M. Product of Two Numbers = H.C.F. × L.C.M. Step-by-Step Solution: • Given: • Product of the numbers = 1320 • H.C.F. = 6 • Substitute these given values directly into the core formula: • 1320 = 6 × L.C.M. • Isolate the L.C.M. by dividing the total product by the H.C.F.: • L.C.M. = (1320/6) • Perform the division: • 1320 / 6 = 220. Exam Strategy & Shortcut: Break down the mental division to save time and reduce errors. Split 1320 into easily divisible chunks: 1200 + 120. Dividing by 6 gives you 200 + 20 = 220. This technique bypasses the need for long division on paper. Common Pitfall: A recurring error is failing to read the word "product" and assuming 1320 is just one of the two numbers. This leads students to multiply 1320 by 6 to find a larger L.C.M., resulting in 7920 (option d). Always pause to verify if the question provides individual numbers or their combined product. Final Answer: Therefore, the correct answer is 220.

Question 20

Product of two co-prime numbers is 117. Their L.C.M. should be

Accepted Answer

Concept & Rule: Co-prime numbers (or relatively prime numbers) are a pair of numbers that share absolutely no common positive factors other than 1. Consequently, their Highest Common Factor (H.C.F.) is exactly 1. For co-prime numbers a and b, L.C.M.(a,b) = a × b Step-by-Step Solution: • Given: The numbers are co-prime, and their product is 117. • By the mathematical definition of co-prime numbers, their H.C.F. must be 1. • Using the standard relationship formula: • Product of Numbers = H.C.F. × L.C.M. • Substitute the known variables into the equation: • 117 = 1 × L.C.M. • Therefore, the L.C.M. is simply 117. Exam Strategy & Shortcut: This question tests pure conceptual knowledge and requires zero calculation. If you see the term "co-prime" and are asked for the L.C.M., instantly select the product provided in the prompt. Do not waste time trying to factorize the number. Common Pitfall: Students who forget the definition of "co-prime" often attempt to factorize 117 (into 9 × 13) and get bogged down looking for missing information. Others might select "cannot be calculated" out of confusion. Trust the definitions of number properties. Final Answer: Therefore, the correct answer is 117.

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