Quantitative Aptitude - MCQ Practice Questions

Even numbers from 2 to 50: 2, 4, 6, ..., 50.

This is an AP with first term = 2, last term = 50, common difference = 2.

Number of terms = 25.

Sum = 25(2+50)/2 = 25 × 26 = 650

Let original number = x.

Then x + 25% of x = 500.

So x + 0.25x = 500, 1.25x = 500, x = 400

Let smaller number = x, larger = 2x.

Then 2x - x = 45, so x = 45.

Numbers are 45 and 90.

Let integers be n-1, n, n+1.

Sum = 3n = 81, so n = 27.

The three integers are 26, 27, 28.

Largest = 28

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)

Sum of squares = 1² + 2² + 3² + 4² + 5² = 1 + 4 + 9 + 16 + 25 = 55.

Formula: n(n+1)(2n+1)/6 = 5×6×$\frac{11}{6}$ = 55

51 = 3 × 17 (not prime), 53 is prime (only divisible by 1 and 53), 55 = 5 × 11 (not prime), 57 = 3 × 19 (not prime).

Therefore, 53 is the smallest prime greater than 50.

36 = 2² × 3².

Number of factors = (2+1)(2+1) = 3 × 3 = 9.

The factors are: 1, 2, 3, 4, 6, 9, 12, 18, 36.

12 = 2² × 3, 18 = 2 × 3², 24 = 2³ × 3. LCM = 2³ × 3² = 8 × 9 = 72.

∛512 = ∛(8³) = 8, since 8 × 8 × 8 = 512.

Odd numbers between 10 and 50: 11, 13, 15, ..., 49.

This is an AP with first term 11, last term 49, and common difference 2.

Number of terms = (49-11)/2 + 1 = 19 + 1 = 20.

LCM(9, 11) = 99 (since 9 and 11 are coprime).

Smallest three-digit multiple of 99 is 99 × 2 = 198.

20 = 2² × 5.

Sum of divisors = (1 + 2 + 4)(1 + 5) = 7 × 6 = 42.

Divisors are: 1, 2, 4, 5, 10, 20.

32 ÷ 7 = 4 remainder 4 (No). 33 ÷ 7 = 4 remainder 5 (No). 34 ÷ 7 = 4 remainder 6 (No). 32 = 7×4 + 5 = 28 + 5 (Yes). 32 ÷ 7 gives remainder 5.

The digital root is obtained by repeatedly summing the digits of a number until a single digit remains. Alternatively, the digital root of a positive integer equals the remainder when divided by 9 (or 9 if the remainder is 0).

Step 1: Sum the digits of 9876

Step 2: Sum the digits of 30

Since we've reached a single digit, the process stops.

**Answer: The digital root of 9876 is $3$ (Option D)**

Using HCF × LCM = Product of two numbers: 15 × 180 = 45 × x.

Therefore, 2700 = 45x, so x = 60.

Perfect squares between 100 and 200: 11² = 121, 12² = 144, 13² = 169, 14² = 196, 15² = 225 (exceeds 200).

So there are 4 perfect squares.

Wait: 10² = 100 (not between), so squares are 121, 144, 169, 196.

Count = 4.

Actually checking: we need squares from 11² to 14², which is 4 numbers.

To find the remainder when $2^{50}$ is divided by 5, we use the Fermat's Little Theorem approach by finding the pattern of powers of 2 modulo 5.

Step 1: Find the pattern of powers of 2 (mod 5)

Calculate successive powers of 2 and their remainders when divided by 5:

The pattern repeats every 4 powers.

**Step 2: Apply the cycle to find $2^{50}(\mathrm{mod}5)$

Since the remainder repeats with period 4, divide the exponent by 4:

This means $2^{50}$ has the same remainder as $2^2$.

Step 3: Calculate the remainder

**Answer: The remainder is $4$ (Option D)**