Quantitative Aptitude - MCQ Practice Questions

Factoring: x² - 5x + 6 = (x - 2)(x - 3) = 0. Therefore x = 2 or x = 3.

[When a number is divisible by two coprime numbers, it must be divisible by their product.]

Step 1: Identify the Given Conditions

We are told that a number is divisible by both 9 and 11, meaning both divide evenly into this number with no remainder.

Step 2: Check if 9 and 11 are Coprime

Two numbers are coprime if their greatest common divisor (GCD) is 1. Since 9 = 3² and 11 is prime, they share no common factors.

Step 3: Apply the Divisibility Rule for Coprime Numbers

When a number is divisible by two coprime numbers, it must be divisible by their product (by the Fundamental Theorem of Arithmetic).

The answer is (B) 99.

360 = 2³×3²×5¹. Number of divisors = (3+1)(2+1)(1+1) = 4×3×2 = 24.

A number is divisible by 8 if its last 3 digits form a number divisible by 8. 456 ÷ 8 = 57. So 2456 is divisible by 8.

1071 = 462×2 + 147; 462 = 147×3 + 21; 147 = 21×7 + 0. Therefore HCF = 21.

Even numbers: 2, 4, 6, ..., 100. This is AP with a=2, l=100, d=2. n = 50 terms. Sum = (n/2)(a+l) = ($\frac{50}{2}$)(2+100) = 25×102 = 2550.

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)**

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.

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)**

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)

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

We need a number of form LCM(12,15,20) + 7. LCM(12,15,20): 12=2²×3, 15=3×5, 20=2²×5. LCM = 2²×3×5 = 60. So the number = 60k + 7. For k=2: 120+7 = 127.

Let the number be 7k + 3. When divided by 14: if k is even (k=2m), number = 14m + 3 (remainder 3); if k is odd (k=2m+1), number = 14m + 10 (remainder 10). So remainder is either 3 or 10.

Let smaller number be x. Then x(x+12) = 189. x² + 12x - 189 = 0. Factoring: (x+21)(x-9) = 0. Since x must be positive, x = 9.

Sum of digits = 21 means divisible by 3. But divisibility by 9 requires sum = 18, 27, etc. Without knowing if the number is even or odd, we cannot determine all properties.