Quantitative Aptitude - MCQ Practice Questions

6³ = 216. Thus 216 is a perfect cube.

To find the value of this expression, we'll factor out the smallest power of 2 and simplify using algebraic factoring.

**Step 1: Factor out $2^7$ from all terms**

Each term contains at least $2^7$ as a factor, so we can write:

Step 2: Simplify the expression in parentheses

Calculate each power of 2:

Step 3: Multiply by the factored term

**Answer: $2^{10}-2^9-2^8-2^7=128$ (Option B)**

The digital root is found by repeatedly summing the digits of a number until a single digit remains.

Step 1: Sum all digits of 9875

Add each digit:

Step 2: Sum the digits of the result

Since 29 is not a single digit, sum its digits:

Step 3: Sum again until single digit

Since 11 is still two digits, sum once more:

Step 4: Verify the digital root

We have reached a single digit. The digital root of 9875 is $2$.

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**Answer: The digital root of 9875 is $2$ (Option A)**

91 = 7 × 13, so it's composite. 97, 89, and 83 are all prime numbers.

24 = 2³ × 3, 36 = 2² × 3². LCM = 2³ × 3² = 8 × 9 = 72.

If number is x, then (x × $\frac{3}{4}$) ÷ ($\frac{2}{3}$) = (3x/4) × ($\frac{3}{2}$) = 9x/8. The factor is $\frac{9}{8}$.

Let consecutive odd numbers be (2n-1) and (2n+1). (2n-1)(2n+1) = 323. 4n² - 1 = 323, so 4n² = 324, n² = 81, n = 9. Numbers are 17 and 19.

Let number = 5a + 2 = 7b + 3. From 5a + 2 = 7b + 3, we get 5a = 7b + 1. Testing b = 2: 7(2) + 1 = 15, a = 3. Number = 5(3) + 2 = 17. Check: 17 ÷ 5 = remainder 2, 17 ÷ 7 = remainder 3. ✓

Let number = 10a + b. (10a + b) - (10b + a) = 45. So 9a - 9b = 45, a - b = 5. Also a + b = 9. Solving: a = 7, b = 2. Number = 72.

144 = 2⁴ × 3², 108 = 2² × 3³. HCF = 2² × 3² = 4 × 9 = 36.

For any two numbers a and b: a × b = HCF(a,b) × LCM(a,b). So 2160 = 12 × LCM. LCM = $\frac{2160}{12}$ = 180.

# Sum of Prime Numbers Between 10 and 25

To find the sum of prime numbers, we must first identify all prime numbers in the given range, then add them together.

Step 1: Identify All Prime Numbers Between 10 and 25

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. We check each number in the range for primality.

(Note: 12, 14, 15, 16, 18, 20, 21, 22, 24 are composite; 10 and 25 are composite)

Step 2: Calculate the Sum

Adding all identified prime numbers together:

The sum of all prime numbers between 10 and 25 is 83.

Answer: (C) 83

48 = 2⁴ × 3¹. Number of factors = (4+1)(1+1) = 5 × 2 = 10. The factors are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.

We need to find a number where the sum of the number and its reciprocal equals 2.1.

Step 1: Set up the equation

Let the number be $x$. Then:

Step 2: Convert to standard form

Multiply both sides by $x$:

Rearrange to standard quadratic form:

Step 3: Convert decimal to fraction and solve

Rewrite $2.1=\frac{21}{10}$:

Multiply by 10 to clear fractions:

Using the quadratic formula:

Step 4: Factor and find exact solutions

Actually, factor directly: $10x^2-21x+10=(5x-2)(2x-5)=0$

This gives:

Step 5: Verify

For $x=\frac{2}{5}$: $\frac{2}{5}+\frac{5}{2}=\frac{4+25}{10}=\frac{29}{10}=2.9$ ✗

For $x=\frac{5}{2}$: $\frac{5}{2}+\frac{2}{5}=\frac{25+4}{10}=\frac{29}{10}=2.9$ ✗

Both values work identically (they are reciprocals of each other).

**Answer: The number could be $\frac{2}{5}$ or $\frac{5}{2}$ (Option C)**

√500 ≈ 22.36. The next integer is 23. 23² = 529. This is the smallest perfect square greater than 500.

Let the numbers be 4x and 5x. Sum = 4x + 5x = 9x = 180. So x = 20. The larger number = 5x = 5(20) = 100.

A perfect cube is a number that can be expressed as $n^3$ for some positive integer $n$. We need to find how many perfect cubes exist between 1 and 1000.

Step 1: Identify the range for the base

We need $1<n^3\le 1000$, where $n$ is a positive integer.

Taking cube roots:

Step 2: List all valid integer values

Since $n$ must be a positive integer satisfying $1<n\le 10$:

Step 3: Verify boundary cases

Step 4: Count the perfect cubes

The perfect cubes between 1 and 1000 are:

That gives us 9 perfect cubes.

Wait—let me reconsider the interpretation. If "between 1 and 1000" is inclusive on both ends, we should check $1^3=1$:

Including 1: $\{1,8,27,64,125,216,343,512,729,1000\}$ = 10 perfect cubes

Answer: There are 10 perfect cubes between 1 and 1000 (inclusive). (Option C)

If a number is divisible by 9, then: (A) It must be divisible by 3 (since 9 = 3²). (B) Sum of its digits must be divisible by 9 (divisibility rule). Both A and B are true.

144 = 2⁴ × 3², 180 = 2² × 3² × 5, 216 = 2³ × 3³. GCD = 2² × 3² = 4 × 9 = 36.

Even numbers from 1 to 100: 2, 4, 6, ..., 100. This is an AP with first term 2, last term 100, common difference 2. Number of terms = (100-2)/2 + 1 = 50.