# Algebra Questions for SSC CGL Tier 2 PDF

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## Algebra Questions for SSC CGL Tier 2 PDF

Download SSC CGL Tier 2 Algebra Questions PDF. Top 15 SSC CGL Tier 2 questions based on asked questions in previous exam papers very important for the SSC exam.

Question 1: If x = √5+ 2, then the value $\frac{2x^2-3x-2}{3x^2-4x-3}$ is equal to

a) 0.185

b) 0.525

c) 0.625

d) 0.785

Question 2: The value of $\frac{(243)^\frac{n}{5}\times3^{2n+1}}{9^n\times3^{n-1}}$ is

a) 3

b) 9

c) 6

d) 12

Question 3: The value of 204 × 197 is

a) 40218

b) 40188

c) 40212

d) 39812

Question 4: What number must be added to the expression $16a^{2} -12a$ to make it a perfect square ?

a) $\frac{9}{4}$

b) $\frac{11}{2}$

c) $\frac{13}{2}$

d) $16$

Question 5: The simplest form of the expression $\frac{p^2-p}{2p^3+6p^2}+\frac{p^2-1}{p^2+3p}+\frac{p^2}{p+1}$

a) $2p^{2}$

b) $\frac{1}{2p^2}$

c) $p+3$

d) $\frac{1}{p+3}$

Question 6: If $\frac{5x}{2x^2+5x+1}=\frac{1}{3}$ then the value of $(x+\frac{1}{2x})$ is

a) 15

b) 10

c) 20

d) 5

Question 7: If $x = \frac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}}$, then the value of $\frac{x+\sqrt{2}}{x-\sqrt{2}} + \frac{x+\sqrt{3}}{x-\sqrt{3}}$ is

a) $\sqrt{2}$

b) $\sqrt{3}$

c) $\sqrt{6}$

d) $2$

Question 8: If $x^2 – 3x + 1= 0$ and x > 1, then the value of $(x – \frac{1}{x})$

a) √5 only

b) 1

c) ­√5 only

d) ±√5

Question 9: If $(x-3)^2 + (y – 5)^2 + (z-4)^2 = 0$ then the value of $\frac{x^2}{9} + \frac{y^2}{25} + \frac{z^2}{16}$

a) 12

b) 9

c) 3

d) 1

Question 10: Coefﬁcient of x in (x + 9)(8 – 5x) is

a) 37

b) -53

c) -37

d) 53

Question 11: Coefficient of x in (x + 8)(6 – 3x) is

a) 18

b) 30

c) -18

d) -30

Question 12: If $3x^{2} = 10^{2} – 5^{2}$, find the value of x?

a) 7

b) 5

c) 9

d) 11

Question 13: Simplify 437bxaz / 23ab.

a) 17xzb

b) 9xz

c) 19xz

d) 19ab

Question 14: ­ If 2(3x ­-4) -­ 2 < 4x -2 ≥ 2x ­- 4; then the value of x is

a) 2

b) 5

c) ­4

d) -5

Question 15: If a * b = 2a + 3b – ab, then the value of (3 * 5 + 5 * 3) is

a) 10

b) 6

c) 4

d) 2

Given , x = (√5+ 2)
$\frac{2x^2-3x-2}{3x^2-4x-3}$ = $\frac{2(\sqrt{5}+ 2)^2-3(\sqrt{5}+ 2)-2}{3(\sqrt{5}+ 2)^2-4(\sqrt{5}+ 2)-3}$
= $\frac{21.18}{33.88}$
= 0.625

=$\frac{(3)^\frac{5n}{5}\times3^{2n+1}}{3^2n\times3^{n-1}}$

=$\frac{(3)^{3n+1}}{3^{3n-1}}$

=${(3)^{3n+1-3n+1}}$

=$(3)^{2}$ = 9 (B)

$200\times197=(200+4)(200-3)=(200)^{2}+4(200)-3(200)-12=40000+200-12=40188$

so the answer is option B.

Using the rule, $a^2 – 2ab + b^2 = (a-b)^2$

=> $16a^2 – 12a = (4a)^2 – 2*4a*\frac{3}{2}$

= $(4a)^2 – 2*4a*\frac{3}{2} + (\frac{3}{2})^2 – \frac{9}{4}$

= $(4a – \frac{3}{2})^2$

=> $\frac{9}{4}$ should be added to make the above expression a perfect square.

$\frac{p^2-p}{2p^3+6p^2}$ = $\frac{p(p-1)}{2p^2(p+3)}$ = $\frac{(p-1)}{2p(p+3)}$

$\frac{p^2-1}{p^2+3p}$ = $\frac{(p-1)(p+1)}{p(p+3)}$

$\frac{p^2}{p+1}$ = $\frac{p^2}{p+1}$

$\frac{(p-1)}{2p(p+3)}$ + $\frac{(p-1)(p+1)}{p(p+3)}$ + $\frac{p^2}{p+1}$ = $\frac{1}{2p^2}$

Expression : $\frac{5x}{2x^2+5x+1}=\frac{1}{3}$

=> $2x^2 + 5x + 1 = 15x$

=> $2x^2 + 1 = 10x$

To find : $(x+\frac{1}{2x})$

= $\frac{2x^2 + 1}{2x}$

= $\frac{10x}{2x}$

= 5

Given : $x = \frac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}}$

=> $x = \frac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\times(\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2})$

=> $x=\frac{2\sqrt6(\sqrt3-\sqrt2)}{3-2}$

=> $x=2\sqrt{18}-2\sqrt{12}$

=> $x=6\sqrt2-4\sqrt3$ —————(i)

To find : $\frac{x+\sqrt{2}}{x-\sqrt{2}} + \frac{x+\sqrt{3}}{x-\sqrt{3}}$

= $\frac{6\sqrt2-4\sqrt3+\sqrt{2}}{6\sqrt2-4\sqrt3-\sqrt{2}} + \frac{6\sqrt2-4\sqrt3+\sqrt{3}}{6\sqrt2-4\sqrt3-\sqrt{3}}$     [Using (i)]

= $\frac{7\sqrt{2}-4\sqrt3}{5\sqrt{2}-4\sqrt3} + \frac{6\sqrt2-3\sqrt{3}}{6\sqrt2-5\sqrt{3}}$

= $\frac{(84-35\sqrt6-24\sqrt6+60)+(60-15\sqrt6-24\sqrt6+36)}{60-25\sqrt6-24\sqrt6+60}$

= $\frac{240-98\sqrt6}{120-49\sqrt6}$

= $\frac{2(120-49\sqrt6)}{120-49\sqrt6}=2$

=> Ans – (D)

Expression : $x^2 – 3x + 1= 0$

=> $x^2 + 1 = 3x$

Dividing by $(x)$ on both sides

=> $x + \frac{1}{x} = 3$

$\because (x – \frac{1}{x})^2 = (x + \frac{1}{x})^2 – 4$

=> $x – \frac{1}{x} = \sqrt{9 – 4}$

=> $(x – \frac{1}{x}) = \pm\sqrt{5}$

Expression : $(x-3)^2 + (y – 5)^2 + (z-4)^2 = 0$

Since, all the terms are positive, the only way the sum can be ‘0’ is, if each term is equal to 0.

=> $(x – 3)^2 = 0$

=> $x = 3$

Similarly, $y = 5$ & $z = 4$

To find : $\frac{x^2}{9} + \frac{y^2}{25} + \frac{z^2}{16}$

= $\frac{3^2}{9} + \frac{5^2}{25} + \frac{4^2}{16}$

= 1 + 1 + 1 = 3

A coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression. Eg : In $ax^2$, coefficient is $a$

Expression : $(x + 9)(8 – 5x)$

= $8x – 5x^2 + 72 – 45x$

= $-5x^2 – 37x + 72$

$\therefore$ Coefficient of $x$ = -37

=> Ans – (C)

A coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression. Eg : In $ax^2$, coefficient is $a$

Expression : $(x + 8)(6 – 3x)$

= $6x – 3x^2 + 48 – 24x$

= $-3x^2 – 18x + 48$

$\therefore$ Coefficient of $x$ = -18

=> Ans – (C)

Expression : $3x^{2} = 10^{2} – 5^{2}$

=> $3x^2 = 100 – 25$

=> $3x^2 = 75$

=> $x^2 = \frac{75}{3} = 25$

=> $x = \sqrt{25} = 5$

=> Ans – (B)

Expression : $\frac{437 b x a z}{23 a b}$

= $\frac{437}{23} \times \frac{abxz}{ab}$

= $19 x z$

=> Ans – (C)

Expression 1 : 2(3x ­-4) -­ 2 < 4x -2

=> $6x-8-2$ < $4x-2$

=> $6x-4x$ < $-2+10$

=> $2x$ < $8$

=> $x$ < $4$ ————(i)

Expression 2 : 4x -2 ≥ 2x ­- 4

=> $4x-2x \geq -4+2$

=> $2x \geq -2$

=> $x \geq -1$ ————(ii)

Combining inequalities (i) and (ii), we get : $-1 \leq x$ < $4$

The only value that $x$ can take among the given options = 2

=> Ans – (A)