SSC MTS Algebra Questions [With Solutions PDF]

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SSC MTS ALGEBRA Questions PDF
SSC MTS ALGEBRA Questions PDF

Algebra Questions for SSC MTS

Here you can download the Algebra Questions for SSC MTS PDF with solutions by Cracku. These are the most important Algebra questions PDF prepared by various sources also based on previous year’s papers. Utilize this PDF to Algebra for SSC MTS. You can find a list of Algebra in this PDF which help you to test yourself and practice. So you can click on the below link to download the PDF for reference and do more practice.

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Question 1: What is the coefficient of $x^2$ in the expansion of $\left(5-\frac{x^2}{3}\right)^3$?

a) -25

b) $-\frac{25}{3}$

c) 25

d) $-\frac{5}{3}$

1) Answer (A)

Solution:

$\left(5-\frac{x^2}{3}\right)^3$ = $\left(5-\frac{x^2}{3}\right)\left(5-\frac{x^2}{3}\right)^2$

= $\left(5-\frac{x^2}{3}\right)\left(25+\frac{x^4}{9}-\frac{10x^2}{3}\right)$

= $125+\frac{5x^4}{9}-\frac{50x^2}{3}-\frac{25x^2}{3}-\frac{x^6}{27}+\frac{10x^4}{9}$

= $-\frac{x^6}{27}+\frac{15x^4}{9}-\frac{75x^2}{3}+125$

= $-\frac{x^6}{27}+\frac{5x^4}{3}-25x^2+125$

The coefficient of $x^2$ in the expansion = -25

Hence, the correct answer is Option A

Question 2: Given that $x^8 – 34x^4 + 1 = 0, x > 0$. What is the value of $(x^3 – x^{-3})$?

a) 14

b) 12

c) 18

d) 16

2) Answer (A)

Solution:

$x^8-34x^4+1=0$

$x^8+1=34x^4$

$x^4+\frac{1}{x^4}=34$

$x^4+\frac{1}{x^4}+2=36$

$\left(x^2+\frac{1}{x^2}\right)^2=36$

$x^2+\frac{1}{x^2}=6$

$x^2+\frac{1}{x^2}-2=4$

$\left(x-\frac{1}{x}\right)^2=4$

$x-\frac{1}{x}=2$……..(1)

$\left(x-\frac{1}{x}\right)^3=8$

$x^3-\frac{1}{x^3}-3.x.\frac{1}{x}\left(x-\frac{1}{x}\right)=8$

$x^3-\frac{1}{x^3}-3\left(2\right)=8$

$x^3-\frac{1}{x^3}-6=8$

$x^3-\frac{1}{x^3}=14$

Hence, the correct answer is Option A

Question 3: If $x^4 – 62 x^2 + 1 = 0$, where $x > 0$, then the value of $x^3 + x^{-3}$ is:

a) 500

b) 512

c) 488

d) 364

3) Answer (C)

Solution:

$x^4-62x^2+1=0$

$x^4+1=62x^2$

$x^2+\frac{1}{x^2}=62$

$x^2+\frac{1}{x^2}+2=64$

$\left(x+\frac{1}{x}\right)^2=64$

$x+\frac{1}{x}=8$…….(1)

$\left(x+\frac{1}{x}\right)^3=512$

$x^3+\frac{1}{x^3}+3.x.\frac{1}{x}\left(x+\frac{1}{x}\right)=512$

$x^3+\frac{1}{x^3}+3\left(8\right)=512$

$x^3+\frac{1}{x^3}+24=512$

$x^3+\frac{1}{x^3}=488$

Hence, the correct answer is Option C

Question 4: If $x + \frac{1}{x} = \frac{17}{4}, x > 1$, then what is the value of $x – \frac{1}{x}?$

a) $\frac{9}{4}$

b) $\frac{3}{2}$

c) $\frac{8}{3}$

d) $\frac{15}{4}$

4) Answer (D)

Solution:

$x+\frac{1}{x}=\frac{17}{4}$

$\left(x+\frac{1}{x}\right)^2=\frac{289}{16}$

$x^2+\frac{1}{x^2}+2=\frac{289}{16}$

$x^2+\frac{1}{x^2}=\frac{289}{16}-2$

$x^2+\frac{1}{x^2}=\frac{257}{16}$

$x^2+\frac{1}{x^2}-2=\frac{257}{16}-2$

$\left(x-\frac{1}{x}\right)^2=\frac{257-32}{16}$

$\left(x-\frac{1}{x}\right)^2=\frac{225}{16}$

$x-\frac{1}{x}=\frac{15}{4}$

Hence, the correct answer is Option D

Question 5: If $2x^2 – 7x + 5 = 0$, then what is the value of $x^3 + \frac{125}{8x^3}$?

a) $12\frac{5}{8}$

b) $16\frac{5}{8}$

c) $10\frac{5}{8}$

d) $18\frac{5}{8}$

5) Answer (B)

Solution:

$2x^2-7x+5=0$

$2x^2-2x-5x+5=0$

$2x\left(x-1\right)-5\left(x-1\right)=0$

$\left(x-1\right)\left(2x-5\right)=0$

$x-1=0$ or $2x-5=0$

$x=1$ or $x=\frac{5}{2}$

When $x=1$,

$x^3+\frac{125}{8x^3}=\left(1\right)^3+\frac{125}{8\left(1\right)^3}=1+\frac{125}{8}=\frac{133}{8}=16\frac{5}{8}$

Hence, the correct answer is Option B

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Question 6: If $x – \frac{1}{x} = 1$, then what is the value of $x^8 + \frac{1}{x^8}?$

a) 3

b) 119

c) 47

d) -1

6) Answer (C)

Solution:

$x-\frac{1}{x}=1$

Squaring on both sides,

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

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

Squaring on both sides,

$x^4+\frac{1}{x^4}+2=9$

$x^4+\frac{1}{x^4}=7$

Squaring on both sides,

$x^8+\frac{1}{x^8}+2=49$

$x^8+\frac{1}{x^8}=47$

Hence, the correct answer is Option C

Question 7: If $x^4 + \frac{1}{x^4} = 727, x > 1$, then what is the value of $\left(x – \frac{1}{x}\right)?$

a) 6

b) -6

c) -5

d) 5

7) Answer (D)

Solution:

$x^4+\frac{1}{x^4}=727$

$x^4+\frac{1}{x^4}+2=729$

$\left(x^2+\frac{1}{x^2}\right)^2=729$

$x^2+\frac{1}{x^2}=27$

$x^2+\frac{1}{x^2}-2=25$

$\left(x-\frac{1}{x}\right)^2=25$

Since $x>1$,

$x-\frac{1}{x}=5$

Hence, the correct answer is Option D

Question 8: If $2x^2 – 8x – 1 = 0$, then what is the value of $8x^3 – \frac{1}{x^3}?$

a) 560

b) 540

c) 524

d) 464

8) Answer (A)

Solution:

$2x^2-8x-1=0$

$2x^2-1=8x$

$2x-\frac{1}{x}=8$……..(1)

Cubing on both sides,

$8x^3-\frac{1}{x^3}-3.2x.\frac{1}{x}\left(2x-\frac{1}{x}\right)=512$

$8x^3-\frac{1}{x^3}-6\left(8\right)=512$  [From (1)]

$8x^3-\frac{1}{x^3}-48=512$

$8x^3-\frac{1}{x^3}=560$

Hence, the correct answer is Option A

Question 9: If $y = 2x + 1$, then what is the value of $(8x^3 – y^3 + 6xy)$?

a) 1

b) -1

c) 15

d) -15

9) Answer (B)

Solution:

$y=2x+1$

$2x-y=-1$…….(1)

Cubing on both sides, we get

$8x^3-y^3-3.2x.y\left(2x-y\right)=-1$

$8x^3-y^3-6xy\left(-1\right)=-1$ [From (1)]

$8x^3-y^3+6xy=-1$

Hence, the correct answer is Option B

Question 10: If $x – \frac{2}{x} = 15$, then what is the value of $\left(x^2 + \frac{4}{x^2}\right)$?

a) 229

b) 227

c) 221

d) 223

10) Answer (A)

Solution:

$x-\frac{2}{x}=15$

Squaring on both sides,

$x^2+\frac{4}{x^2}-2.x.\frac{2}{x}=225$

$x^2+\frac{4}{x^2}-4=225$

$x^2+\frac{4}{x^2}=229$

Hence, the correct answer is Option A

Question 11: If $2x + 3y + 1 = 0$, then what is the value of $\left(8x^3 + 8 + 27y^3 – 18xy \right)$?

a) -7

b) 7

c) -9

d) 9

11) Answer (B)

Solution:

$2x+3y+1=0$

$2x+3y=-1$……..(1)

Cubing on both sides,

$8x^3+27y^3+3.2x.3y\left(2x+3y\right)=-1$

$8x^3+27y^3+18xy\left(-1\right)=-1$

$8x^3+27y^3-18xy+8=-1+8$

$8x^3+27y^3-18xy+8=7$

Hence, the correct answer is Option B

Question 12: If $x + \frac{1}{x} = 7$, then $x^2 + \frac{1}{x^2}$ is equal to:

a) 47

b) 49

c) 61

d) 51

12) Answer (A)

Solution:

$x+\frac{1}{x}=7$

Squaring on both sides,

$x^2+\frac{1}{x^2}+2.x.\frac{1}{x}=49$

$x^2+\frac{1}{x^2}+2=49$

$x^2+\frac{1}{x^2}=47$

Hence, the correct answer is Option A

Question 13: If $(2x + y)^3 – (x – 2y)^3 = (x + 3y)[Ax^2 + By^2 + Cxy]$, then what is the value of $(A + 2B + C)?$

a) 13

b) 14

c) 7

d) 10

13) Answer (D)

Solution:

$(2x+y)^3-(x-2y)^3=(x+3y)[Ax^2+By^2+Cxy]$

$\left[2x+y-\left(x-2y\right)\right]\left[\left(2x+y\right)^2+\left(2x+y\right)\left(x-2y\right)+\left(x-2y\right)^2\right]=(x+3y)[Ax^2+By^2+Cxy]$

$\left[x+3y\right]\left[4x^2+y^2+4xy+2x^2-3xy-2y^2+x^2+4y^2-4xy\right]=(x+3y)[Ax^2+By^2+Cxy]$

$\left(x+3y\right)\left[7x^2+3y^2-3xy\right]=(x+3y)[Ax^2+By^2+Cxy]$

Comparing both sides,

A = 7, B = 3 and C = -3

$A+2B+C\ =\ 7+2\left(3\right)-3$ = 10

Hence, the correct answer is Option D

Question 14: If $9(a^2 + b^2) + c^2 + 20 = 12(a + 2b)$, then the value of $\sqrt{6a + 9b + 2c}$ is:

a) 4

b) 3

c) 6

d) 2

14) Answer (A)

Solution:

$9(a^2+b^2)+c^2+20=12(a+2b)$

$9a^2+9b^2+c^2+20=12a+24b$

$9a^2-12a+9b^2-24b+c^2+20=0$

$9a^2-12a+4-4+9b^2-24b+16-16+c^2+20=0$

$\left(3a-2\right)^2-4+\left(3b-4\right)^2-16+c^2+20=0$

$\left(3a-2\right)^2+\left(3b-4\right)^2+c^2=0$

$3a-2=0,\ 3b-4=0,\ c=0$

$a=\frac{2}{3},\ b=\frac{4}{3},\ c=0$

$\sqrt{6a+9b+2c}=\sqrt{6\left(\frac{2}{3}\right)+9\left(\frac{4}{3}\right)+2\left(0\right)}$

= $\sqrt{4+12}$

= $\sqrt{16}$

= 4

Hence, the correct answer is Option A

Question 15: If $x + \frac{1}{x} = 2\sqrt{5}$, then what is the value of $\frac{\left(x^4 + \frac{1}{x^2}\right)}{x^2 + 1}$?

a) 14

b) 17

c) 20

d) 23

15) Answer (B)

Solution:

$x+\frac{1}{x}=2\sqrt{5}$………..(1)

$\left(x+\frac{1}{x}\right)^3=40\sqrt{5}$

$x^3+\frac{1}{x^3}+3.x.\frac{1}{x}\left(x+\frac{1}{x}\right)=40\sqrt{5}$

$x^3+\frac{1}{x^3}+3\left(2\sqrt{5}\right)=40\sqrt{5}$  [From (1)]

$x^3+\frac{1}{x^3}+6\sqrt{5}=40\sqrt{5}$

$x^3+\frac{1}{x^3}=34\sqrt{5}$………(2)

$\frac{\left(x^4+\frac{1}{x^2}\right)}{x^2+1}=\frac{x\left(x^3+\frac{1}{x^3}\right)}{x\left(x+\frac{1}{x}\right)}$

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

$=\frac{34\sqrt{5}}{2\sqrt{5}}$

$=17$

Hence, the correct answer is Option B

Question 16: If $x^4+x^2y^2+y^4=21$ and $x^2+xy+y^2=3$, then what is the value of $\left(-xy\right)$?

a) -1

b) 2

c) 1

d) -2

16) Answer (B)

Solution:

$x^4+x^2y^2+y^4=21$……(1)

$x^2+xy+y^2=3$

$x^2+y^2=3-xy$

$\left(x^2+y^2\right)^2=\left(3-xy\right)^2$

$x^4+y^4+2x^2y^2=9+x^2y^2-6xy$

$x^4+y^4+x^2y^2=9-6xy$

$21=9-6xy$  [From (1)]

$-6xy=12$

$-xy=2$

Hence, the correct answer is Option B

Question 17: If $(x+6)^3+(2x+3)^3+(3x+5)^3=(3x+18)(2x+3)(3x+5)$, then what is the value of x?

a) $-\frac{5}{3}$

b) $\frac{5}{3}$

c) $-\frac{7}{3}$

d) $\frac{7}{3}$

17) Answer (C)

Solution:

$(x+6)^3+(2x+3)^3+(3x+5)^3=(3x+18)(2x+3)(3x+5)$

$(x+6)^3+(2x+3)^3+(3x+5)^3=\left[3\left(x+6\right)\right](2x+3)(3x+5)$

$(x+6)^3+(2x+3)^3+(3x+5)^3-3\left(x+6\right)(2x+3)(3x+5)=0$

This is in the form of $a^3+b^3+c^3-3abc=0$, where $a\ne b\ne c$ then $a+b+c=0$

$\Rightarrow$  $\left(x+6\right)+\left(2x+3\right)+\left(3x+5\right)=0$

$\Rightarrow$  $6x+14=0$

$\Rightarrow$  $x=-\frac{7}{3}$

Hence, the correct answer is Option C

Question 18: If $x + y + z = 3, xy + yz + zx = -12$ and $xyz = -16$, then the value of $\sqrt{x^3 + y^3 + z^3 + 13}$ is:

a) 9

b) 8

c) 10

d) 11

18) Answer (C)

Solution:

$x+y+z=3$

$x+y=3-z$……..(1)

$\left(x+y\right)^3=\left(3-z\right)^3$

$x^3+y^3+3xy\left(x+y\right)=27-z^3-3.3.z\left(3-z\right)$

$x^3+y^3+3xy\left(3-z\right)=27-z^3-9z\left(x+y\right)$  [From (1)]

$x^3+y^3+9xy-3xyz=27-z^3-9xz-9yz$

$x^3+y^3+z^3=27-9xy-9xz-9yz+3xyz$

$x^3+y^3+z^3=27-9\left(xy+yz+zx\right)+3xyz$

$x^3+y^3+z^3=27-9\left(-12\right)+3\left(-16\right)$

$x^3+y^3+z^3=27+108-48$

$x^3+y^3+z^3=87$…….(2)

$\sqrt{x^3+y^3+z^3+13}=\sqrt{87+13}$

$=\sqrt{100}$

$=10$

Hence, the correct answer is Option C

Question 19: What is the coefficient of x in the expansion of $(3x – 4)^3$?

a) 108

b) -108

c) 144

d) -144

19) Answer (C)

Solution:

$(3x – 4)^3$ = $(3x – 4)(3x – 4)^2$

= $(3x – 4)(9x^2+16-24x)$

= $27x^3+48x-72x^2-36x^2-64+96x$

= $27x^3-108x^2+144x-64$

The coefficient of x in the expansion = 144

Hence, the correct answer is Option C

Question 20: If $x – y = 4$ and $x^3 – y^3 = 316, y > 0$ then the value of $x^4 – y^4$ is:

a) 2500

b) 2320

c) 2401

d) 2482

20) Answer (B)

Solution:

$x-y=4$………..(1)

$\left(x-y\right)^3=64$

$x^3-y^3-3xy\left(x-y\right)=64$

$316-3xy\left(4\right)=64$

$12xy=252$

$xy=21$……….(2)

$x-y=4$

$\left(x-y\right)^2=4^2$

$x^2+y^2-2xy=16$

$x^2+y^2-2\left(21\right)=16$

$x^2+y^2=58$……….(3)

$\left(x+y\right)^2=x^2+y^2+2xy$

$\left(x+y\right)^2=58+2\left(21\right)$

$\left(x+y\right)^2=100$

$x+y=10$……….(4)

$x^4-y^4=\left(x^2+y^2\right)\left(x^2-y^2\right)$

$=\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)$

$=\left(58\right)\left(10\right)\left(4\right)$

$=2320$

Hence, the correct answer is Option B

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