Quantitative Aptitude - MCQ Practice Questions

When a number is expressed in prime factorization form, we use the divisor formula: if $n=p_1^{a_1}\times p_2^{a_2}\times p_3^{a_3}$, then the total number of divisors is $(a_1+1)(a_2+1)(a_3+1)$.

Step 1: Identify the prime factorization

The given number is:

Here, the exponents are: $a_1=3$, $a_2=2$, $a_3=1$

Step 2: Apply the divisor formula

The number of divisors is found by adding 1 to each exponent and multiplying:

Step 3: Calculate

Step 4: Verify with a divisor example

Each divisor has the form $2^a\times 3^b\times 5^c$ where $0\le a\le 3$, $0\le b\le 2$, $0\le c\le 1$. This gives 4 choices for the power of 2, 3 choices for the power of 3, and 2 choices for the power of 5.

**Answer: The total number of divisors is $24$ (Option B)**

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.

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).

Let A's rate = 1/x, B's rate = 1/y.

From equations: 3/x + 2/y = $\frac{1}{4}$ and 2/x + 3/y = $\frac{1}{3}$.

Solving: x = 30 days

Let tens digit = x, units digit = y.

Then x + y = 12 and 10x + y = 6y + 6.

From second: 10x = 5y + 6.

Substituting y = 12-x: 10x = 5(12-x) + 6 = 60 - 5x + 6.

So 15x = 66, x = 8, y = 4.

Number = 84

Using the formula: Product of two numbers = HCF × LCM.

So 180 = 6 × LCM.

Therefore LCM = $\frac{180}{6}$ = 30

We need to find numbers that satisfy two remainder conditions simultaneously. This is a Chinese Remainder Theorem problem.

Step 1: Set up the congruence equations

Given conditions:

From the first condition, we can write:

Step 2: Substitute into the second condition

Substitute $n=9k+4$ into the second congruence:

Step 3: Solve for k

Find the multiplicative inverse of 9 modulo 11. Since $9\times 5=45=44+1\equiv 1(\mathrm{mod}11)$, the inverse is 5.

Multiply both sides by 5:

So $k=11m+5$ for some integer $m$.

Step 4: Find the general solution and values less than 200

Substitute back:

For $n<200$:

Verification:

Answer: The numbers are 49 and 148 (Option B)

To count how many times digit 7 appears in numbers from 1 to 100, we organize by position: units place and tens place.

Step 1: Count 7 in the units place

The digit 7 appears in the units place in: 7, 17, 27, 37, 47, 57, 67, 77, 87, 97

Step 2: Count 7 in the tens place

The digit 7 appears in the tens place in: 70, 71, 72, 73, 74, 75, 76, 77, 78, 79

Step 3: Account for overlaps

The number 77 contains the digit 7 twice (once in units place, once in tens place). We've counted it correctly in both steps above—no double-counting error since we're counting occurrences of the digit, not distinct numbers.

Step 4: Total count

Answer: The digit 7 appears 20 times in numbers from 1 to 100. (Option D)

Last digits of powers of 3: 3¹=3, 3²=9, 3³=27(7), 3⁴=81(1), 3⁵=243(3)...

Pattern: 3,9,7,1 repeats every 4. 2023 = 4×505 + 3, so 3^2023 has same last digit as 3³, which is 7.

First 20 multiples of 3: 3, 6, 9, ..., 60.

This is AP with a=3, d=3, n=20, l=60.

Sum = n(a+l)/2 = 20(3+60)/2 = 20×$\frac{63}{2}$ = 10×63 = 630.

100 = 2² × 5².

Sum of all divisors = (1+2+4)(1+5+25) = 7 × 31 = 217.

Sum excluding 100 = 217 - 100 = 117.

For two numbers, the fundamental relationship is $\text{HCF}\times \text{LCM}=\text{Product of the two numbers}$. 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 $b$:

Given HCF = 7 and LCM = 140:

Step 2: Check option (A): 7 and 140

Option (A) is valid.

Step 3: Check option (B): 14 and 70

Option (B) is invalid.

Step 4: Check option (C): 28 and 35

Option (C) is valid.

Answer: Both options (A) and (C) satisfy the conditions HCF = 7 and LCM = 140 (Option D)

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

Step 2: Identify all prime factors and their highest powers

Across all three numbers:

Step 3: Calculate the LCM

Step 4: Final multiplication

**Answer: The LCM of 108, 135, and 162 is $1620$ (Option C)**

Let numbers be 2k, 3k, 4k. HCF = k = 5.

Numbers are 10, 15, 20.

Sum = 10 + 15 + 20 = 45.

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

Step 2: Calculate the LCM

For the LCM, take the highest power of each prime factor:

Step 3: Find the smallest 4-digit multiple of 72

The smallest 4-digit number is 1000. Divide it by 72:

Since we need a whole number of groups of 72, round up to the next integer: $14$

Step 4: Calculate the required number

Verify: $1008÷12=84$, $1008÷18=56$, $1008÷24=42$ ✓

**Answer: The smallest 4-digit number divisible by 12, 18, and 24 is $1008$ (Option A)**

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:

Step 2: Verify the HCF

Check that $\text{HCF}(10,15,20)=5$:

The highest common factor is indeed $5$. ✓

Step 3: Find the LCM

To find $\text{LCM}(10,15,20)$, use prime factorization:

The LCM takes the highest power of each prime:

**Answer: The LCM is $60$ (Option B)**

Numbers are 7, 14, 21, 28. 7 = 7, 14 = 2 × 7, 21 = 3 × 7, 28 = 2² × 7. LCM = 2² × 3 × 7 = 4 × 3 × 7 × 2 = 1680.

Using Legendre's formula: ⌊$\frac{27}{3}$⌋ + ⌊$\frac{27}{9}$⌋ + ⌊$\frac{27}{27}$⌋ = 9 + 3 + 1 = 13.

LCM(7, 11, 13) = 1001. Next multiple is 1001. 1001 - 1000 = 1. Actually 1000 ÷ 1001 remainder = 1000. Need 1001 - 1000 = 1. Recheck: 1000 mod 1001 = 1000, so add 1 gives 1001. Add 12 gives 1012 = 1001 + 11.

By Wilson's theorem, (p-1)! ≡ -1 (mod p) for prime p. So 16! ≡ -1 (mod 17). 16! = 13! × 14 × 15 × 16. -1 ≡ 13! × 14 × 15 × 16 (mod 17).