Competitive Exams : Quantitative Aptitude - Simplification (Set 2) | Concept of Maths

Question 1

1. A student was asked to multiply a number by $\frac{3}{2}$ but he divided the number by $\frac{3}{2}$ instead. His result was 10 less than the correct answer. What was the original number?

A) 15

B) 18

C) 12

D) 20

Answer

Correct Answer: C

Let the number be $x$. Correct result: $\frac{3}{2}x$ Incorrect result (division): $x \div \frac{3}{2} = \frac{2}{3}x$ Difference: $\frac{3}{2}x - \frac{2}{3}x = 10$ $$\frac{9x - 4x}{6} = 10$$ $$\frac{5x}{6} = 10 \implies 5x = 60 \implies x = 12$$

Question 2

2. Simplify the expression based on algebraic identities: $$\frac{0.729 + 0.027}{0.81 - 0.27 + 0.09}$$

A) 1.2

B) 0.6

C) 0.9

D) 1.0

Answer

Correct Answer: A

Observe that $0.729 = (0.9)^3$ and $0.027 = (0.3)^3$. The expression is in the form $\frac{a^3 + b^3}{a^2 - ab + b^2}$. We know that $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. Therefore, $\frac{(a+b)(a^2 - ab + b^2)}{a^2 - ab + b^2} = a + b$. Here $a = 0.9$ and $b = 0.3$. Result: $0.9 + 0.3 = 1.2$.

Question 3

3. A man spends $\frac{1}{3}$ of his income on food, $\frac{2}{5}$ of the remainder on house rent and $\frac{1}{5}$ of the remainder on clothes. If he still has Rs. 1,600 left, find his total income.

A) Rs. 4,500

B) Rs. 5,000

C) Rs. 6,000

D) Rs. 4,800

Answer

Correct Answer: B

Let income be $I$. Remaining after food: $I - \frac{1}{3}I = \frac{2}{3}I$ Remaining after rent: $\frac{2}{3}I \times (1 - \frac{2}{5}) = \frac{2}{3}I \times \frac{3}{5} = \frac{2}{5}I$ Remaining after clothes: $\frac{2}{5}I \times (1 - \frac{1}{5}) = \frac{2}{5}I \times \frac{4}{5} = \frac{8}{25}I$ Given: $\frac{8}{25}I = 1600$ $I = \frac{1600 \times 25}{8} = 200 \times 25 = 5000$.

Question 4

4. Find the value of $x$ if: $$x = \sqrt{20 + \sqrt{20 + \sqrt{20 + \dots \infty}}}$$

A) 4

B) 20

C) 6

D) 5

Answer

Correct Answer: D

$x = \sqrt{20 + x}$ Squaring both sides: $x^2 = 20 + x$ $x^2 - x - 20 = 0$ $(x - 5)(x + 4) = 0$ Since $x$ must be positive, $x = 5$. Shortcut: Factorize 20 into consecutive integers $4 \times 5$. For $+$, larger factor is the answer.

Question 5

5. If $a + b + c = 0$, what is the simplified value of $\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}$?

A) 3

B) 0

C) 1

D) -1

Answer

Correct Answer: A

The expression is $\frac{a^3 + b^3 + c^3}{abc}$. We know that if $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$. Therefore, $\frac{3abc}{abc} = 3$.

Question 6

6. Simplify: $(0.333\dots) \times (0.444\dots) \div (0.111\dots)$

A) $0.999\dots$

B) $0.1212\dots$

C) $1.333\dots$

D) $0.444\dots$

Answer

Correct Answer: C

$0.333\dots = \frac{1}{3}$ $0.444\dots = \frac{4}{9}$ $0.111\dots = \frac{1}{9}$ Expression: $(\frac{1}{3} \times \frac{4}{9}) \div \frac{1}{9} = \frac{4}{27} \times 9 = \frac{4}{3} = 1.333\dots$

Question 7

7. Which of the following is the largest value?

A) $\sqrt{2}$

B) $\sqrt[3]{3}$

C) $\sqrt[4]{4}$

D) $\sqrt[6]{6}$

Answer

Correct Answer: B

Compare $2^{1/2}, 3^{1/3}, 4^{1/4}, 6^{1/6}$. LCM of denominators (2, 3, 4, 6) is 12. $2^{6/12} = \sqrt[12]{2^6} = \sqrt[12]{64}$ $3^{4/12} = \sqrt[12]{3^4} = \sqrt[12]{81}$ $4^{3/12} = \sqrt[12]{4^3} = \sqrt[12]{64}$ $6^{2/12} = \sqrt[12]{6^2} = \sqrt[12]{36}$ Clearly, $\sqrt[12]{81}$ is the largest, which is $\sqrt[3]{3}$.

Question 8

8. Simplify: $$[2^2 + (3^2 \times 2) - \{4 + (2 \times (3 - 1))\}]$$

A) 12

B) 10

C) 16

D) 14

Answer

Correct Answer: D

Using VBODMAS: 1. Innermost bracket: $(3 - 1) = 2$. 2. Curly bracket: $\{4 + (2 \times 2)\} = \{4 + 4\} = 8$. 3. Round bracket: $(3^2 \times 2) = (9 \times 2) = 18$. 4. Final expression: $[4 + 18 - 8] = [22 - 8] = 14$.

Question 9

9. If $\frac{x}{y} = \frac{4}{5}$, find the value of $\left(\frac{7x + 6y}{29y}\right)$.

A) $\frac{58}{145}$

B) $\frac{1}{2}$

C) $\frac{4}{5}$

D) 1

Answer

Correct Answer: A

Substitute $x = 4k$ and $y = 5k$. $$\frac{7(4k) + 6(5k)}{29(5k)} = \frac{28k + 30k}{145k} = \frac{58k}{145k}$$ Simplifying by dividing by 29: $\frac{2}{5}$. Note: Option A is $\frac{58}{145}$, which simplifies to $\frac{2}{5}$.

Question 10

10. Solve for $x$: $2^{x-1} + 2^{x+1} = 320$.

A) 6

B) 7

C) 8

D) 5

Answer

Correct Answer: B

$2^{x-1} + 2^{x+1} = 320$ $\frac{2^x}{2} + 2 \cdot 2^x = 320$ Taking $2^x$ common: $2^x (\frac{1}{2} + 2) = 320$ $2^x (\frac{5}{2}) = 320$ $2^x = \frac{320 \times 2}{5} = 64 \times 2 = 128$ Since $128 = 2^7$, then $x = 7$.

Competitive Exams : Quantitative Aptitude - Simplification (Set 2) | Concept of Maths