Quantitative Aptitude - MCQ Practice Questions

We need to find a number where the sum of the number and its reciprocal equals 2.5.

Step 1: Set up the equation

Let the number be $x$. The reciprocal is $\frac{1}{x}$.

Step 2: Clear the fraction

Multiply both sides by $x$:

Rearrange to standard quadratic form:

Step 3: Apply the quadratic formula

Using $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ where $a=1$, $b=-2.5$, $c=1$:

Step 4: Find both solutions

Verification:

**Answer: The number is $2$ or $\frac{1}{2}$ (Option B)**

# Solution: Finding the Value of x Using Exponent Laws

To solve exponential equations, express all terms with the same base and apply exponent division rules.

Step 1: Express 216 as a Power of 6

We need to rewrite 216 in terms of base 6 to simplify the equation.

Step 2: Substitute and Apply Division Rule

Substitute this into the original equation and use the rule that $\frac{a^m}{a^n}=a^{m-n}$.

Step 3: Equate the Exponents

Since the bases are equal, the exponents must be equal.

We are given:

216

6

x

​

=6

Since:

216=6

3

Substitute:

6

3

6

x

​

=6

Using laws of exponents:

6

x−3

=6

1

Therefore,

x−3=1

x=4

So, the value of x is:

4

​

The answer is (C) 4.

To find the sum of all odd divisors of 120, first extract the odd part by removing all factors of 2, then find the sum of divisors of that odd part.

Step 1: Prime factorization of 120

Step 2: Identify the odd part

The odd divisors of 120 come only from the odd prime factors (ignoring all powers of 2):

Step 3: Find all divisors of 15

The divisors of 15 are: $1,3,5,15$

These are exactly the odd divisors of 120 (since multiplying any of these by powers of 2 gives even divisors).

Step 4: Calculate the sum

Alternatively, using the divisor sum formula for $3^1\times 5^1$:

**Answer: The sum of all odd divisors of 120 is $24$ (Option C)**

Sum of first n natural numbers = n(n+1)/2. For n=15: 15×$\frac{16}{2}$ = $\frac{240}{2}$ = 120

12 = 2²×3, 18 = 2×3². LCM = 2²×3² = 4×9 = 36

For divisibility by 6, number must be divisible by both 2 and 3. 104÷6 = 17.33... (not divisible). Others: 72, 84, 90 are all divisible by 6

Let number = x. Then x - x/3 = 30. So 2x/3 = 30, x = 45

Divisors of 16: 1,2,4,8,16. Product = 1×2×4×8×16 = 1024. Or use formula: if n has d divisors, product = n^(d/2). Here d=5, so product = 16^2.5 = 1024

Let numbers be x and x+2. Then x(x+2) = 195. x² + 2x - 195 = 0. Using formula: x = 13. Numbers are 13 and 15. Sum = 28

2^a = 32 = 2^5, so a = 5. And 3^b = 81 = 3^4, so b = 4. Therefore a + b = 5 + 4 = 9

Check 22: 22÷9 = 2 rem 4 ✓, 22÷7 = 3 rem 1 ✗. Check 31: 31÷9 = 3 rem 4 ✓, 31÷7 = 4 rem 3 ✓. Answer is 31 (B, not A). Correction: Option B is correct.

Using Legendre's formula: ⌊$\frac{100}{5}$⌋ + ⌊$\frac{100}{25}$⌋ + ⌊$\frac{100}{125}$⌋ = 20 + 4 + 0 = 24

Let numbers be a and b. a + b = 15 and a² + b² = 117. We know (a+b)² = a² + 2ab + b². So 225 = 117 + 2ab, thus 2ab = 108, ab = 54

Let the number be x. According to the problem: 5x - 3 = 47. Therefore, 5x = 50, so x = 10.

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

We need to find x where x² = 169. Taking square root: x = √169 = 13.

We need a number of the form 8k + 5. Check: 29 = 8(3) + 5 = 24 + 5. ✓

We need to find three consecutive integers that sum to 45, then identify the smallest one.

Step 1: Define the three consecutive integers

Let the smallest integer be $x$. Then the three consecutive integers are:

Step 2: Set up the equation

The sum of these three integers equals 45:

Step 3: Simplify and solve

Combine like terms:

Subtract 3 from both sides:

Divide by 3:

Step 4: Verify

The three consecutive integers are 14, 15, and 16.

**Answer: The smallest integer is $x=14$ (Option B)**

144 = 2⁴ × 3². Odd divisors come only from 3² = (2+1) = 3 odd divisors: 1, 3, 9.

We know that GCD × LCM = Product of the two numbers. So 4 × LCM = 120. Therefore, LCM = $\frac{120}{4}$ = 30.