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

Question 1

1. If $x + \frac{1}{x} = 3$, then what is the value of $x^2 + \frac{1}{x^2}$?

A) 9

B) 7

C) 11

D) 6

Answer

Correct Answer: B

Solution: Given $x + \frac{1}{x} = 3$. Squaring both sides: $(x + \frac{1}{x})^2 = 3^2$ $x^2 + 2(x)(\frac{1}{x}) + \frac{1}{x^2} = 9$ $x^2 + 2 + \frac{1}{x^2} = 9$ $x^2 + \frac{1}{x^2} = 9 - 2 = 7$.

Question 2

2. Simplify: $\sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ$.

A) 1

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

C) $\frac{\sqrt{3}}{2}$

D) 0

Answer

Correct Answer: A

Solution: Method 1 (Values): $\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1$. Method 2 (Identity): Use $\sin(A+B) = \sin A \cos B + \cos A \sin B$. $\sin(60^\circ + 30^\circ) = \sin 90^\circ = 1$.

Question 3

3. If $a+b+c=0$, find the value of $\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}$.

A) 0

B) 1

C) 3

D) -3

Answer

Correct Answer: C

Solution: Expression: $\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = \frac{a^3 + b^3 + c^3}{abc}$. We know that if $a+b+c=0$, then $a^3 + b^3 + c^3 = 3abc$. Substituting this: $\frac{3abc}{abc} = 3$.

Question 4

4. Evaluate $\cos 1^\circ \cos 2^\circ \cos 3^\circ \dots \cos 179^\circ$.

A) 1

B) -1

C) $\frac{1}{\sqrt{2}}$

D) 0

Answer

Correct Answer: D

Solution: The series contains the term $\cos 90^\circ$. Since $\cos 90^\circ = 0$, the entire product becomes $0$ regardless of the other values.

Question 5

5. Find the remainder when $x^3 - 3x^2 + 4x - 5$ is divided by $(x - 2)$.

A) 1

B) -1

C) 3

D) 0

Answer

Correct Answer: B

Solution: By Remainder Theorem, the remainder is $P(2)$. $P(2) = (2)^3 - 3(2)^2 + 4(2) - 5$ $P(2) = 8 - 12 + 8 - 5$ $P(2) = 16 - 17 = -1$.

Question 6

6. If $\tan \theta = \frac{3}{4}$, then find the value of $\frac{1 - \sin \theta}{1 + \sin \theta}$.

A) $\frac{1}{4}$

B) $\frac{1}{7}$

C) $\frac{1}{9}$

D) $\frac{4}{9}$

Answer

Correct Answer: A

Solution: $\tan \theta = \frac{3}{4} \implies \text{Opposite} = 3, \text{Adjacent} = 4$. $\text{Hypotenuse} = \sqrt{3^2 + 4^2} = 5$. So, $\sin \theta = \frac{3}{5}$. Expression: $\frac{1 - 3/5}{1 + 3/5} = \frac{2/5}{8/5} = \frac{2}{8} = \frac{1}{4}$.

Question 7

7. If $x + y = 12$ and $xy = 32$, find $x^2 + y^2$.

A) 80

B) 70

C) 144

D) 64

Answer

Correct Answer: A

Solution: $(x+y)^2 = x^2 + y^2 + 2xy$ $12^2 = x^2 + y^2 + 2(32)$ $144 = x^2 + y^2 + 64$ $x^2 + y^2 = 144 - 64 = 80$.

Question 8

8. If $5\tan \theta = 4$, find the value of $\frac{5\sin \theta - 3\cos \theta}{5\sin \theta + 2\cos \theta}$.

A) $\frac{2}{3}$

B) $\frac{1}{4}$

C) $\frac{1}{6}$

D) $\frac{5}{6}$

Answer

Correct Answer: C

Solution: Divide numerator and denominator by $\cos \theta$: $\frac{5\tan \theta - 3}{5\tan \theta + 2}$ Given $5\tan \theta = 4$. Substituting this: $\frac{4 - 3}{4 + 2} = \frac{1}{6}$.

Question 9

9. Simplify: $\sqrt{7 + 4\sqrt{3}}$.

A) $2 - \sqrt{3}$

B) $2 + \sqrt{3}$

C) $4 + \sqrt{3}$

D) 7

Answer

Correct Answer: B

Solution: We want to express $7 + 4\sqrt{3}$ as a perfect square $(a+b)^2 = a^2 + b^2 + 2ab$. Let $2ab = 4\sqrt{3} \implies ab = 2\sqrt{3}$. Try $a=2, b=\sqrt{3}$. $a^2 + b^2 = 2^2 + (\sqrt{3})^2 = 4 + 3 = 7$ (Matches). So, $\sqrt{7 + 4\sqrt{3}} = \sqrt{(2 + \sqrt{3})^2} = 2 + \sqrt{3}$.

Question 10

10. The angle of elevation of the top of a tower from a point on the ground, which is 30m away from the foot of the tower, is $30^\circ$. Find the height of the tower.

A) $30\sqrt{3}$ m

B) 15 m

C) $20\sqrt{3}$ m

D) $10\sqrt{3}$ m

Answer

Correct Answer: D

Solution: Let height be $h$. $\tan 30^\circ = \frac{h}{30}$ $\frac{1}{\sqrt{3}} = \frac{h}{30}$ $h = \frac{30}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3}$ m.

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