Competitive Exams : Quantitative Aptitude - Simplification (Set 1) | Concept of Maths

Question 1

1. Evaluate the value of: $$\frac{\sqrt{1296}}{12} + 25 \times 1.4 - 15$$

A) 20

B) 23

C) 25

D) 28

Answer

Correct Answer: B

Solution: Step 1: Calculate $\sqrt{1296}$. Since $30^2 = 900$ and $40^2 = 1600$, and the unit digit is 6, $\sqrt{1296} = 36$. Step 2: Divide: $36 / 12 = 3$. Step 3: Multiply: $25 \times 1.4 = 35$. Step 4: Combine: $3 + 35 - 15 = 38 - 15 = 23$. Correct Answer: 23

Question 2

2. Find the value of: $37\frac{1}{2}\%$ of $800 + 12.5\%$ of $2400$.

A) 450

B) 500

C) 600

D) 650

Answer

Correct Answer: C

Solution: Step 1: Convert percentages to fractions. $37\frac{1}{2}\% = \frac{3}{8}$ and $12.5\% = \frac{1}{8}$. Step 2: Calculate: $(\frac{3}{8} \times 800) + (\frac{1}{8} \times 2400)$. Step 3: $300 + 300 = 600$. Correct Answer: 600

Question 3

3. Solve for $x$: $x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots \infty}}}$

A) 3

B) 2

C) 6

D) Infinite

Answer

Correct Answer: A

Solution: Let $x = \sqrt{6 + x}$. Squaring both sides: $x^2 = 6 + x \Rightarrow x^2 - x - 6 = 0$. Factorizing: $(x-3)(x+2) = 0$. Since the value must be positive, $x = 3$. Correct Answer: 3

Question 4

4. Simplify: $$\frac{0.87 \times 0.87 \times 0.87 + 0.13 \times 0.13 \times 0.13}{0.87 \times 0.87 - 0.87 \times 0.13 + 0.13 \times 0.13}$$

A) 0.74

B) 0.90

C) 1.13

D) 1.00

Answer

Correct Answer: D

Solution: This is in the form $\frac{a^3 + b^3}{a^2 - ab + b^2}$. We know $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. So, the expression simplifies to $a + b$. $0.87 + 0.13 = 1.00$. Correct Answer: 1.00

Question 5

5. If $2^{x+3} \cdot 4^{2x-1} = 8^{x+1}$, then find the value of $x$.

A) 1/2

B) 1

C) 3/2

D) 2

Answer

Correct Answer: B

Solution: Express all terms with base 2: $2^{x+3} \cdot (2^2)^{2x-1} = (2^3)^{x+1}$ $2^{x+3} \cdot 2^{4x-2} = 2^{3x+3}$ Equating exponents: $(x+3) + (4x-2) = 3x+3$ $5x + 1 = 3x + 3 \Rightarrow 2x = 2 \Rightarrow x = 1$. Correct Answer: 1

Question 6

6. Convert the following recurring decimal to a fraction: $0.1\overline{2}$

A) 12/90

B) 12/99

C) 11/90

D) 11/99

Answer

Correct Answer: C

Solution: Rule: (Full number - Non-repeating part) / (9 for each repeating digit, 0 for each non-repeating decimal digit). $0.1\overline{2} = \frac{12 - 1}{90} = \frac{11}{90}$. Correct Answer: 11/90

Question 7

7. Simplify: $\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{5}\right) \dots \left(1 - \frac{1}{n}\right)$

A) 1/n

B) 2/n

C) 2(n-1)/n

D) (n+1)/n

Answer

Correct Answer: B

Solution: The series is: $\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \dots \times \frac{n-1}{n}$. This is a telescoping product where numerators cancel with preceding denominators. The result is $\frac{\text{First Numerator}}{\text{Last Denominator}} = \frac{2}{n}$. Correct Answer: 2/n

Question 8

8. What is the value of $(48)^2 - (32)^2 + \sqrt{x} = 1300$? Solve for $x$.

A) 400

B) 40000

C) 200

D) 20000

Answer

Correct Answer: A

Solution: Step 1: Use $a^2 - b^2 = (a-b)(a+b)$. $(48-32)(48+32) = 16 \times 80 = 1280$. Step 2: $1280 + \sqrt{x} = 1300$. Step 3: $\sqrt{x} = 20$. Step 4: $x = 20^2 = 400$. Correct Answer: 400

Question 9

9. Simplify: $\sqrt{10 + 2\sqrt{21}}$

A) $\sqrt{5} + \sqrt{2}$

B) $\sqrt{8} + \sqrt{2}$

C) $\sqrt{7} + \sqrt{3}$

D) $\sqrt{6} + 2$

Answer

Correct Answer: C

Solution: We need to find two numbers whose sum is 10 and product is 21. The numbers are 7 and 3. $\sqrt{10 + 2\sqrt{21}} = \sqrt{(\sqrt{7})^2 + (\sqrt{3})^2 + 2\sqrt{7}\sqrt{3}} = \sqrt{(\sqrt{7} + \sqrt{3})^2} = \sqrt{7} + \sqrt{3}$. Correct Answer: $\sqrt{7} + \sqrt{3}$

Question 10

10. Find the value of $121 \div 1.1 + 100 \div 0.5 - 20 \times 1.5$.

A) 280

B) 310

C) 250

D) 300

Answer

Correct Answer: A

Solution: Step 1: $121 / 1.1 = 110$. Step 2: $100 / 0.5 = 200$. Step 3: $20 \times 1.5 = 30$. Step 4: $110 + 200 - 30 = 280$. Correct Answer: 280

Competitive Exams : Quantitative Aptitude - Simplification (Set 1) | Concept of Maths