Algebra Questions for SSC CHSL and MTS | Download PDF

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Algebra Questions for SSC CHSL & MTS 2022
Algebra Questions for SSC CHSL & MTS 2022

Algebra Questions for SSC CHSL and MTS PDF

Here you can download SSC CHSL & MTS 2022 – important SSC CHSL & MTS Algebra Questions PDF by Cracku. Very Important SSC CHSL & MTS 2022 and These questions will help your SSC CHSL & MTS preparation. So kindly download the PDF for reference and do more practice.

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Question 1: If $a^{2}+b^{2}-c^{2}=0$, then the value of $\frac{2(a^{6}+b^{6}-c^{6})}{3a^{2}b^{2}c^{2}}$ is:

a) 3

b) 1

c) 0

d) 2

Question 2: If $a+\frac{1}{a}=5$ then $a^{3} + \frac{1}{a^{3}}$ is:

a) 110

b) 10

c) 80

d) 140

Question 3: The coefficient of x in $(x – 3y)^{3}$ is :

a) $3 y^{2}$

b) $27 y^{2}$

c) $-27 y^{2}$

d) $-3 y^{2}$

Question 4: Expand $\left(\frac{x}{3} + \frac{y}{5}\right)^3$

a) $\frac{x^3}{27} + \frac{x^2y}{25} + \frac{xy^2}{25} + \frac{y^3}{125}$

b) $\frac{x^3}{25} + \frac{x^2y}{15} + \frac{xy^2}{25} + \frac{y^3}{125}$

c) $\frac{x^3}{27} + \frac{xy}{15} + \frac{xy^2}{25} + \frac{y^3}{125}$

d) $\frac{x^3}{27} + \frac{x^2y}{15} + \frac{xy^2}{25} + \frac{y^3}{125}$

Question 5: If $a^2 + b^2 + c^2 = 300$ and $ab + bc + ca = 50$, then what is the value of $a + b + c$ ? (Given that a, b and c are all positive.)

a) 22

b) 20

c) 25

d) 15

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Question 6: If x + y + z = 10 and xy + yz + zx = 15, then find the value of $x^3 + y^3 + z^3 — 3xyz$.

a) 660

b) 525

c) 550

d) 575

Question 7: If $x^{2} – 4x+4=0 $, then the value of 16$(x^{4} – \frac{1}{x^{4}})$ is

a) 127

b) 255

c) $\frac{127}{16}$

d) $\frac{255}{16}$

Question 8: If $a^{3}+\frac{1}{a^{3}} = 52$ then the value of $2\left(a + \frac{1}{a}\right)$ is :

a) 8

b) 2

c) 6

d) 4

Question 9: If $b + c = ax, c + a = by, a + b = cz$, then the value $\frac{1}{9}\left[\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right]$ is:

a) $\frac{1}{9}$

b) 1

c) 0

d) $\frac{1}{3}$

Question 10: Find the product of $(a + b + 2c)(a^{2} + b^{2} + 4c^{2} – ab – 2bc – 2ca)$

a) $a^{3} + b^{3} + 8c^{3} – 2abc$

b) $a^{3} + b^{3} + 8c^{3} – abc$

c) $a^{3} + b^{3} + 6c^{3} – 6abc$

d) $a^{3} + b^{3} + 8c^{3} – 6abc$

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Question 11: $25a^{2}-9$ is factored as

a) $(5a + 3)(5a – 3)$

b) $(5a + 1)(5a – 9)$

c) $(5a – 3)^{2}$

d) $(25a + 1)(a -9)$

Question 12: If $a^{4} + \frac{1}{a^{4}} = 50$, then find the value of $a^{3} + \frac{1}{a^{3}}$

a) $\sqrt{2(1+\sqrt{3})}+(-1+2\sqrt{13})$

b) $\sqrt{2(1+\sqrt{3})}(3-2\sqrt{13})$

c) $\sqrt{2(\sqrt{13}+1)}(3+2\sqrt{13})$

d) $\sqrt{2(1-\sqrt{3})}(-1+2\sqrt{13})$

Question 13: $(a + b – c + d)^2 – (a – b + c – d)^2 = ?$

a) $4a(b + d – c)$

b) $2a(a + b – c)$

c) $2a(b + c – d)$

d) $4a(b – d + c)$

Question 14: The value of $27a^3 – 2\sqrt{2}b^3$ is equal to:

a) $(3a – \sqrt{2}b)(9a^2 – 2b^2 + 6\sqrt{2}ab)$

b) $(3a – \sqrt{2}b)(9a^2 + 2b^2 + 6\sqrt{2}ab)$

c) $(3a – \sqrt{2}b)(9a^2 + 2b^2 + 3\sqrt{2}ab)$

d) $(3a – \sqrt{2}b)(9a^2 – 2b^2 – 3\sqrt{2}ab)$

Question 15: If $x+3y+2=0$ then value of $x^{3}+27y^{3}+8-18xy$ is:

a) -2

b) 2

c) 1

d) 0

Question 16: If $p+q=7$ and $pq=5$, then the value of $p^{3}+q^{3}$ is:

a) 34

b) 238

c) 448

d) 64

Question 17: If $30x^2 – 15x + 1 = 0$, then what is the value of $25x^2 + (36x^2)^{-1}$?

a) $\frac{9}{2}$

b) $6\frac{1}{4}$

c) $\frac{65}{12}$

d) $\frac{55}{12}$

Question 18: If a + b + c = 7 and ab + bc + ca = -6, then the value of $a^3 + b^3 + c^3 – 3abc$ is:

a) 469

b) 472

c) 463

d) 479

Question 19: The given table represents the revenue (in ₹ crores) of a company from the sale of four products A, B, C and D in 6 years. Study the table carefully and answer the question that follows. By what percentage is the total revenue of the company from the sale of products A, B and D in 2012 and 2013 more than the total revenue from the sale of product B in 2013 to 2016?(Correct to one decimal place)

a) 44.5

b) 31.2

c) 43.6

d) 45.4

Question 20: If $P = \frac{x^4 – 8x}{x^3 – x^2 – 2x}, Q = \frac{x^2 + 2x + 1}{x^2 – 4x – 5}$ and $R = \frac{2x^2 + 4x + 8}{x – 5}$, then $(P \times Q) \div R$ is equal to:

a) $\frac{1}{2}$

b) 1

c) 2

d) 4

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Answers & Solutions:

1) Answer (D)

If a + b + c  = 0 then $a^3 + b^3 + c^3 = 3abc$ so,

$a^{6}+b^{6}-c^{6} = 3a^2b^2c^2$

$\frac{2(a^{6}+b^{6}-c^{6})}{3a^{2}b^{2}c^{2}}$

= $\frac{2(3a^{2}b^{2}c^{2})}{3a^{2}b^{2}c^{2}}$ = 2

2) Answer (A)

$a^{3} + \frac{1}{a^{3}}$

= $($a+\frac{1}{a})^3 – 3(a+\frac{1}{a})$

($\because (a + b)^3 = a^3 + b^3 + 3ab(a + b))$

= $5^3 – 3(5)$ = 110

3) Answer (B)

$(x – 3y)^{3} = x^3 – (3y)^3 – 3x.3y(x – 3y)$

($(a – b)^3 = a^3 – b^3 – 3ab(a – b)$)

= $x^3 – 27y^3 – 9xy(x – 3y)$

= $x^3 – 27y^3 – 9x^2y – 27xy^2$

The coefficient of x = 27$y^2$

4) Answer (D)

$\left(\frac{x}{3} + \frac{y}{5}\right)^3$

($\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)$)

=  $(\frac{x}{3})^3 + (\frac{y}{5})^3 + 3(\frac{x}{3})(\frac{y}{5})(\frac{x}{3} + \frac{y}{5}) $

= $\frac{x^3}{27} + \frac{y^3}{125} + \frac{xy}{5} (\frac{x}{3} + \frac{y}{5}) $

= $\frac{x^3}{27} + \frac{y^3}{125} + \frac{x^2y}{15} + \frac{xy^2}{25} $

5) Answer (B)

$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$

$(a + b + c)^2 = 300 + 2(50)$

$(a + b + c)^2 = 400$

a + b + c = 20

6) Answer (C)

$x^3 + y^3 + z^3 — 3xyz = (x + y + z)(x^2 + y^2 + z^2 – xy – yz – xz)$

x + y + z = 10

Taking square on both sides,

$(x + y + z)^2 = 100$

$x^2 + y^2 + z^2 + 2(xy + yz + xz) = 100$

$x^2 + y^2 + z^2 = 100 – 2\times 15 = 00 – 30 = 70$

$x^3 + y^3 + z^3 — 3xyz = (10)(70 – 15) = 10 \times 55 = 550$

7) Answer (B)

$x^{2} —4x+4=0 $

$x^{2} —2x – 2x+4=0 $

$x(x — 2) – 2(x — 2)=0 $

$(x — 2)(x — 2)=0 $

x = 2

now,

16$(x^{4}-\frac{1}{x^{4}})$

= 16$(2^{4}-\frac{1}{2^{4}})$

= 16$(16-\frac{1}{16})$

= $16^2 – 1$ = 255

8) Answer (A)

$a^{3}+\frac{1}{a^{3}} = 52$

$(a + \frac{1}{a})^3 – 3.a.\frac{1}{a}(a + \frac{1}{a}) = 52$

$(\because a^3 + b^3 = (a + b)^3 – 3ab(a + b))$

$(a + \frac{1}{a})^3 – 3(a + \frac{1}{a}) = 52$

From the option A) –

Put the value of $2(a + \frac{1}{a}) = 8$,

$(a + \frac{1}{a}) = 4$

L.H.S.,

$4^3 – 3 \times 4$ = 52

= R.H.S.

$\therefore$ The value of $2\left(a + \frac{1}{a}\right)$ is 8.

9) Answer (A)

$b + c = ax, c + a = by, a + b = cz$

x =$\frac{b + c}{a}$

y =$\frac{c + a}{b}$

z =$\frac{a + b}{c}$

Now,

$\frac{1}{9}\left[\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right]$

x + 1 = $\frac{b + c}{a}$ + 1 = $\frac{a + b + c}{a}$

y + 1 = $\frac{c + a}{b}$ + 1 = $\frac{a + b + c}{b}$

z + 1 = $\frac{a + b}{c}$ + 1 = $\frac{a + b + c}{c}$

= $\frac{1}{9}\left[\frac{1}{\frac{a + b + c}{a}}+\frac{1}{\frac{a + b + c}{b}}+\frac{1}{\frac{a + b + c}{c}}\right]$

$\frac{1}{9}\left[\frac{a}{a + b + c}+\frac{b}{a + b + c}+\frac{c}{a + b + c}\right]$

$\frac{1}{9}[\frac{a + b + c}{a + b + c}]$ = 1/9

10) Answer (D)

$(a + b + 2c)(a^{2} + b^{2} + 4c^{2} – ab – 2bc – 2ca)$

$= a^3 + b^3 + (2c)^3 – 3 \times a \times b \times 2c$

$(\because a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca))$

$= a^3 + b^3 + 8c^3 – 6abc$

11) Answer (A)

$25a^{2}-9$

= $(5a)^2 – (3)^2$
= ($\because a^2 – b^2 = (a + b)(a – b)$)

= (5a + 3)(5a – 3)

12) Answer (C)

$a^{4} + \frac{1}{a^{4}} = 50$

$a^{4} + \frac{1}{a^{4}} + 2= 50 + 2$

$(a^2+\frac{1}{a^2})^2=52$

$(a^2+\frac{1}{a^2})=\sqrt{52}$

$a^2+\frac{1}{a^2} + 2 = \sqrt{52} + 2$

$(a + \frac{1}{a})^2 = \sqrt{52} + 2$

$(a + \frac{1}{a}) = \sqrt{\sqrt{52} + 2}$

$a^{3} + \frac{1}{a^{3}} = (a + b)^3 + 3ab(a + b)$

=$(\sqrt{\sqrt{52} + 2})^3 + \sqrt{\sqrt{52} + 2}$

=$(\sqrt{2\sqrt{13} + 2})^3 + \sqrt{2\sqrt{13} + 2}$

=$\sqrt{2\sqrt{13} + 2}(1 + (\sqrt{2\sqrt{13} + 2})^2)$

=$\sqrt{2\sqrt{13} + 2}(1 + {2\sqrt{13} + 2})$

=$\sqrt{2(\sqrt{13} + 1})(3 + {2\sqrt{13}})$

13) Answer (A)

$(a + b – c + d)^2 – (a – b + c – d)^2$

= [(a + b – c + d) + (a – b + c – d)][(a + b – c + d) – (a – b + c – d)]

($\because a^2 – b^2 = (a + b)(a – b)$)

= (2a)(2b-2c + 2d)

= 4a(b – c + d)

14) Answer (C)

$27a^3 – 2\sqrt{2}b^3 = (3a – \sqrt{2}b)(9a^2 + 2b^2 + 6\sqrt{2}ab)$

($\because a^3 – b^3 = (a – b)(a^2 + ab + b^2)$)

here,

a = 3a

b = $\sqrt{2}b$

15) Answer (D)

$x+3y+2=0$

x + 3y = -2

Taking cube both sides,

$(x + 3y)^3 = -8$

$x^3 + 27y^3 + 3x.3y(x + 3y) = -8$

$x^3 + 27y^3 + 9xy(-2) = -8 $

$x^{3}+27y^{3} -18xy = -8$

$x^{3}+27y^{3}+8-18xy$ = 0

16) Answer (B)

$p^{3}+q^{3} = (p + q)^3 – 3pq(p + q)$

=$7^3 – 3 \times 5(7)$

= 343 – 105 = 238

17) Answer (D)

$30x^2 – 15x + 1 = 0$

Dividing by x,

$30x – 15 + \frac{1}{x} = 0$

$5x – 15/6 + \frac{1}{6x} = 0$

$5x  + \frac{1}{6x} = 5/2$

taking square both side,

$(5x + \frac{1}{6x})^2 = 25/4$

$25x^2 + \frac{1}{36x^2} + 2 \times 5x \times \frac{1}{6x} = 25/4$

$25x^2 + \frac{1}{36x^2} = 25/4 – 5/3$

$25x^2 + \frac{1}{36x^2} = \frac{55}{12} $

18) Answer (A)

We know that,

$a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^2 + b^2 + c^2 – (ab + bc + ac))$

a + b + c = 7

Squaring both sides,

$(a + b + c)^2 = 49$

$a^2 + b^2 + c^2 + 2(ab + bc + ac) = 49$

$a^2 + b^2 + c^2 = 49 + 12 = 61$

$a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^2 + b^2 + c^2 – (ab + bc + ac))$

= 7(61 – (-6)) = 7 $\times$ 67 = 469

19) Answer (D)

Total revenue of the company from the sale of products A, B and D in 2012 and 2013 = 98 + 74 + 74 + 94 + 96 + 102 = 538

Total revenue from the sale of product B in 2013 to 2016 = 96 + 92 + 84 + 98 = 370

Required percentage = $\frac{538 – 370}{370} \times 100$ = 45.4%

20) Answer (A)

$P = \frac{x^4 – 8x}{x^3 – x^2 – 2x}$

$Q = \frac{x^2 + 2x + 1}{x^2 – 4x – 5}$

= $\frac{(x + 1)^2}{x^2 – 4x – 5 + 9 – 9}$

$(P \times Q) \div R$

= $(\frac{x^4 – 8x}{x^3 – x^2 – 2x} \times \frac{x^2 + 2x + 1}{x^2 – 4x – 5}) \div \frac{2x^2 + 4x + 8}{x – 5}$

= $\frac{x^4 – 8x}{x^3 – x^2 – 2x} \times \frac{x^2 + 2x + 1}{x^2 – 4x – 5} \times \frac{x – 5}{2x^2 + 4x + 8}$

= $\frac{x(x^3 – 8)}{x^3 – x^2 – 2x} \times \frac{x^2 + 2x + 1}{x^2 – 4x – 5} \times \frac{x – 5}{2(x^2 + 2x + 4)}$

= $\frac{x(x – 2)(x^2 + 2x – 4)}{x(x^2 – x – 2)} \times \frac{(x + 1)^2}{x^2 – 5x + x – 5} \times \frac{x – 5}{2(x^2 + 2x + 4)}$

= $\frac{(x – 2)}{(x^2 – 2x + x- 2)} \times \frac{(x + 1)^2}{(x +1)(x – 5)} \times \frac{x – 5}{2}$

= $\frac{(x – 2)}{(x – 2)(x + 1)} \times \frac{(x + 1)}{2}$

= $\frac{1}{2}$

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