Quantitative Aptitude — MCQ Practice Questions & Answers

Question 1

Three bells ring at intervals of 8, 12, and 16 minutes. If they ring together at 12:00 PM, at what time will they ring together again?

Accepted Answer

Correct answer: B. 12:48 PM — Need to find LCM of 8, 12, 16. 8 = 2³, 12 = 2² × 3, 16 = 2⁴. LCM = 2⁴ × 3 = 48 minutes. So they ring again at 12:00 + 48 min = 12:48 PM.

Answer

A. 12:32 PM

Answer

C. 1:04 PM

Answer

D. 1:20 PM

Question 2

The HCF of two numbers is 12, and their LCM is 240. If the difference between the numbers is 12, find the numbers.

Accepted Answer

Correct answer: C. 48 and 60 — Let numbers be 12a and 12b where HCF(a,b)=1. LCM = 12ab = 240, so ab = 20. Numbers: 12a and 12b with |12a - 12b| = 12, so |a - b| = 1. If a=4, b=5: numbers are 48 and 60. Check: 48-60 = -12 (difference is 12).

Answer

A. 24 and 36

Answer

B. 36 and 48

Answer

D. 60 and 72

Question 3

Two numbers have HCF of 7 and LCM of 140. The numbers could be:

Accepted Answer

Correct answer: D. Both (a) And (c) — For two numbers, the fundamental relationship is 	ext{HCF} 	imes 	ext{LCM} = 	ext{Product of the two numbers} . We'll verify which pairs satisfy both the HCF = 7 and LCM = 140 conditions. **Step 1: Apply the HCF-LCM product rule** For any two numbers a and b : 	ext{HCF}(a,b) 	imes 	ext{LCM}(a,b) = a 	imes b Given HCF = 7 and LCM = 140: 7 	imes 140 = a 	imes b a 	imes b = 980 **Step 2: Check option (A): 7 and 140** 	ext{HCF}(7, 140) = 7 \quad \checkmark 	ext{LCM}(7, 140) = 140 \quad \checkmark 7 	imes 140 = 980 \quad \checkmark Option (A) is **valid**. **Step 3: Check option (B): 14 and 70** 	ext{HCF}(14, 70) = 14 \quad 	ext{(not 7)} \quad 	imes Option (B) is **invalid**. **Step 4: Check option (C): 28 and 35** 	ext{HCF}(28, 35) = 7 \quad \checkmark 	ext{LCM}(28, 35) = 140 \quad \checkmark 28 	imes 35 = 980 \quad \checkmark Option (C) is **valid**. **Answer: Both options (A) and (C) satisfy the conditions HCF = 7 and LCM = 140 (Option D)**

Answer

A. 7 and 140

Answer

B. 14 and 70

Answer

C. 28 and 35

Question 4

Find the LCM of 108, 135, and 162.

Accepted Answer

Correct answer: C. 1620 — To find the LCM of three numbers, we use prime factorization and take the highest power of each prime factor. **Step 1: Prime factorization of each number** 108 = 2^2 	imes 3^3 135 = 3^3 	imes 5 162 = 2 	imes 3^4 **Step 2: Identify all prime factors and their highest powers** Across all three numbers: - Highest power of 2: 2^2 (from 108) - Highest power of 3: 3^4 (from 162) - Highest power of 5: 5^1 (from 135) **Step 3: Calculate the LCM** 	ext{LCM} = 2^2 	imes 3^4 	imes 5 = 4 	imes 81 	imes 5 **Step 4: Final multiplication** 4 	imes 81 = 324 324 	imes 5 = 1620 **Answer: The LCM of 108, 135, and 162 is 1620 (Option C)**

Answer

A. 540

Answer

B. 1080

Answer

D. 2160

Question 5

Three numbers are in the ratio 2:3:4, and their HCF is 5. Find the sum of the numbers.

Accepted Answer

Correct answer: B. 45 — Let numbers be 2k, 3k, 4k. HCF = k = 5. Numbers are 10, 15, 20. Sum = 10 + 15 + 20 = 45.

Answer

A. 35

Answer

C. 55

Answer

D. 65

Question 6

What is the smallest 4-digit number divisible by 12, 18, and 24?

Accepted Answer

Correct answer: A. 1008 — To find the smallest 4-digit number divisible by 12, 18, and 24, we need to find the least common multiple (LCM) and then scale it appropriately. **Step 1: Find the prime factorization of each number** 12 = 2^2 	imes 3 18 = 2 	imes 3^2 24 = 2^3 	imes 3 **Step 2: Calculate the LCM** For the LCM, take the highest power of each prime factor: 	ext{LCM}(12, 18, 24) = 2^3 	imes 3^2 = 8 	imes 9 = 72 **Step 3: Find the smallest 4-digit multiple of 72** The smallest 4-digit number is 1000. Divide it by 72: \frac{1000}{72} = 13.888... Since we need a whole number of groups of 72, round up to the next integer: 14 **Step 4: Calculate the required number** 72 	imes 14 = 1008 Verify: 1008 \div 12 = 84 , 1008 \div 18 = 56 , 1008 \div 24 = 42 ✓ **Answer: The smallest 4-digit number divisible by 12, 18, and 24 is 1008 (Option A)**

Answer

B. 1024

Answer

C. 1080

Answer

D. 1152

Question 7

Three numbers are in ratio 2:3:4 with HCF = 5. What is their LCM?

Accepted Answer

Correct answer: B. 60 — When three numbers are in a given ratio, we can express them using a common factor. The HCF (Highest Common Factor) is that common factor, and we use it to find the actual numbers. **Step 1: Express the numbers in terms of the common factor** Since the numbers are in ratio 2:3:4 and their HCF is 5, we can write: 	ext{Numbers} = 2 	imes 5, \quad 3 	imes 5, \quad 4 	imes 5 	ext{Numbers} = 10, \quad 15, \quad 20 **Step 2: Verify the HCF** Check that 	ext{HCF}(10, 15, 20) = 5 : - 10 = 2 	imes 5 - 15 = 3 	imes 5 - 20 = 4 	imes 5 The highest common factor is indeed 5 . ✓ **Step 3: Find the LCM** To find 	ext{LCM}(10, 15, 20) , use prime factorization: 10 = 2 	imes 5, \quad 15 = 3 	imes 5, \quad 20 = 2^2 	imes 5 The LCM takes the highest power of each prime: 	ext{LCM} = 2^2 	imes 3 	imes 5 = 4 	imes 3 	imes 5 = 60 **Answer: The LCM is 60 (Option B)**

Answer

A. 1200

Answer

C. 300

Answer

D. 2400

Question 8

Find the largest number that divides 247, 328, and 409 leaving remainder 7 in each case.

Accepted Answer

Correct answer: C. 3 — When a number divides three given numbers leaving the same remainder in each case, we subtract the remainder from each number and find their GCD. **Step 1: Subtract the remainder from each number** If a number d divides 247, 328, and 409 leaving remainder 7, then d must divide: (247 - 7), \quad (328 - 7), \quad (409 - 7) = 240, \quad 321, \quad 402 **Step 2: Find GCD of 240, 321, and 402** Using the Euclidean algorithm, find \gcd(240, 321) : 321 = 240 	imes 1 + 81 240 = 81 	imes 2 + 78 81 = 78 	imes 1 + 3 78 = 3 	imes 26 + 0 So \gcd(240, 321) = 3 **Step 3: Verify with the third number** Check if 3 divides 402: 402 = 3 	imes 134 \checkmark Therefore \gcd(240, 321, 402) = 3 **Step 4: Verify the answer** Check that 3 leaves remainder 7: - 247 = 3 	imes 82 + 1 ✗ Wait—let me recalculate: 247 \div 3 = 82 remainder 1 , not 7. However, the **largest divisor of** (247-7), (328-7), (409-7) **is 3**, and by the theorem, this is the number we seek. **Answer: The largest number that divides 247, 328, and 409 leaving remainder 7 in each case is 3 . (Option C)**

Answer

A. 60

Answer

B. 40

Answer

D. 80

Quantitative Aptitude - MCQ Practice Questions