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# Most Important Algebra Questions For CMAT [PDF]

Algebra is one of the most important topics in the CMAT, and also it is an important section. One can utilize this article which consists of the most important questions regarding Algebra. Cracku provides you with the Top 15 very Important Algebra Questions for CMAT based on the questions asked in previous exam papers. Click on the link below to download the Algebra Questions for CMAT PDF with detailed answers.

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

a)Â -25

b)Â $-\frac{25}{3}$

c)Â 25

d)Â $-\frac{5}{3}$

1)Â AnswerÂ (A)

Solution:

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

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

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

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

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

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

Hence, the correct answer is Option A

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

a)Â 14

b)Â 12

c)Â 18

d)Â 16

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

a)Â 500

b)Â 512

c)Â 488

d)Â 364

3)Â AnswerÂ (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:Â If $x + \frac{1}{x} = \frac{17}{4}, x > 1$, then what is the value of $x – \frac{1}{x}?$

a)Â $\frac{9}{4}$

b)Â $\frac{3}{2}$

c)Â $\frac{8}{3}$

d)Â $\frac{15}{4}$

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

a)Â $12\frac{5}{8}$

b)Â $16\frac{5}{8}$

c)Â $10\frac{5}{8}$

d)Â $18\frac{5}{8}$

5)Â AnswerÂ (B)

Solution:

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

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

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

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

$x-1=0$ orÂ $2x-5=0$

$x=1$ orÂ $x=\frac{5}{2}$

WhenÂ $x=1$,

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

Hence, the correct answer is Option B

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

a)Â 3

b)Â 119

c)Â 47

d)Â -1

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

a)Â 6

b)Â -6

c)Â -5

d)Â 5

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

a)Â 560

b)Â 540

c)Â 524

d)Â 464

8)Â AnswerÂ (A)

Solution:

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

$2x^2-1=8x$

$2x-\frac{1}{x}=8$……..(1)

Cubing on both sides,

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

$8x^3-\frac{1}{x^3}-6\left(8\right)=512$Â  [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:Â If $y = 2x + 1$, then what is the value of $(8x^3 – y^3 + 6xy)$?

a)Â 1

b)Â -1

c)Â 15

d)Â -15

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

a)Â 229

b)Â 227

c)Â 221

d)Â 223

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

a)Â -7

b)Â 7

c)Â -9

d)Â 9

11)Â AnswerÂ (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:Â If $x + \frac{1}{x} = 7$, then $x^2 + \frac{1}{x^2}$ is equal to:

a)Â 47

b)Â 49

c)Â 61

d)Â 51

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

a)Â 13

b)Â 14

c)Â 7

d)Â 10

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

a)Â 4

b)Â 3

c)Â 6

d)Â 2

14)Â AnswerÂ (A)

Solution:

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

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

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

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

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

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

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

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

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

=Â $\sqrt{4+12}$

=Â $\sqrt{16}$

= 4

Hence, the correct answer is Option A

Question 15:Â 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)Â 14

b)Â 17

c)Â 20

d)Â 23

15)Â AnswerÂ (B)

Solution:

$x+\frac{1}{x}=2\sqrt{5}$………..(1)

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

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

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