Algebra Questions for MAH-CET [PDF]

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Algebra Questions for MAH-CET [PDF]

Here you can download a free Algebra questions PDF with answers for MAH MBA CET 2022 by Cracku. These are some tricky questions in the MAH MBA CET 2022 exam that you need to find the solutions for the given Algebra questions. These questions will help you to practice and solve the Algebra questions in the MAH MBA CET exam. Utilize this best PDF practice set which is included answers in detail. Click on the below link to download the Algebra MCQ PDF for MBA-CET 2022 for free.

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Question 1: If $\sqrt{x}-\frac{1}{\sqrt{x}}=\sqrt{7}$, then the value of $x^2 + \frac{1}{x^2}$ is:

a) 81

b) 60

c) 79

d) 75

1) Answer (C)

Solution:

$\sqrt{x}-\frac{1}{\sqrt{x}}=\sqrt{7}$

$\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2=\left(\sqrt{7}\right)^2$

$x+\frac{1}{x}-2=7$

$x+\frac{1}{x}=9$

$\left(x+\frac{1}{x}\right)^2=9^2$

$x^2+\frac{1}{x^2}+2=81$

$x^2+\frac{1}{x^2}=79$

Hence, the correct answer is Option C

Question 2: If $(56\sqrt{7}x^3-2\sqrt{2}y^3)\div(2\sqrt{7}x-\sqrt{2}y)=Ax^2+By^2-Cxy$, then find the value of $A + B – \sqrt{14}C$.

a) 38

b) 10

c) 19

d) 58

2) Answer (D)

Solution:

$(56\sqrt{7}x^3-2\sqrt{2}y^3)\div(2\sqrt{7}x-\sqrt{2}y)=Ax^2+By^2-Cxy$

$\frac{\left(2\sqrt{7}x-\sqrt{2}y\right)\left(28x^2+2\sqrt{14}xy+2y^2\right)}{\left(2\sqrt{7}x-\sqrt{2}y\right)}=Ax^2+By^2-Cxy$

$28x^2+2\sqrt{14}xy+2y^2=Ax^2+By^2-Cxy$

Comparing both sides,

A = 28, B = 2, C = $-2\sqrt{14}$

$A+B-\sqrt{14}C=28+2-\sqrt{14}\left(-2\sqrt{14}\right)$

$=30+28$

$=58$

Hence, the correct answer is Option D

Question 3: If $\frac{x}{y} + \frac{y}{x} = 2, (x, y \neq 0)$, then the value of $(x – y)$ is:

a) 1

b) 0

c) 2

d) -2

3) Answer (B)

Solution:

$\frac{x}{y}+\frac{y}{x}=2$

$\frac{x^2+y^2}{xy}=2$

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

$x^2+y^2-2xy=0$

$\left(x-y\right)^2=0$

$x-y=0$

Hence, the correct answer is Option B

Question 4: If $\left(2a+\frac{3}{a}-1\right)=11$, what is the value of $\left(4a^2 + \frac{9}{a^2}\right)?$

a) 121

b) 148

c) 132

d) 110

4) Answer (C)

Solution:

$\left(2a+\frac{3}{a}-1\right)=11$

$2a+\frac{3}{a}=12$

$4a^2+\frac{9}{a^2}+2.2a.\frac{3}{a}=144$

$4a^2+\frac{9}{a^2}+12=144$

$4a^2+\frac{9}{a^2}=132$

Hence, the correct answer is Option C

Question 5: If $a^3 – b^3 = 2349$ and $(a – b) = 9$, then $(a + b)^2 – ab$ is equal to:

a) 280

b) 244

c) 261

d) 229

5) Answer (C)

Solution:

$(a-b)=9$………….(1)

$(a-b)^3=729$

$a^3-b^3-3ab\left(a-b\right)=729$

$2349-3ab\left(9\right)=729$

$27ab=1620$

$ab=60$…………..(2)

$(a-b)=9$

$(a-b)^2=81$

$a^2+b^2-2ab=81$

$a^2+b^2-2\left(60\right)=81$

$a^2+b^2-120=81$

$a^2+b^2=201$……….(3)

$(a+b)^2-ab=a^2+b^2+2ab-ab$

$=a^2+b^2+ab$

$=201+60$

$=261$

Hence, the correct answer is Option C

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Question 6: If $x – \frac{1}{x} = \sqrt{77}$, then one of the values of $x^3 + \frac{1}{x^3}$ is:

a) $80\sqrt{77}$

b) -702

c) $77\sqrt{77}$

d) $3\sqrt{77}$

6) Answer (B)

Solution:

$x – \frac{1}{x} = \sqrt{77}$

$\left(x-\frac{1}{x}\right)^2=77$

$x^2+\frac{1}{x^2}-2=77$

$x^2+\frac{1}{x^2}=79$

$x^2+\frac{1}{x^2}+2=81$

$\left(x+\frac{1}{x}\right)^2=81$

$x+\frac{1}{x}=9$ or $x+\frac{1}{x}=-9$

When $x+\frac{1}{x}=-9$

$x^3+\frac{1}{x^3}+3.x.\frac{1}{x}\left(x+\frac{1}{x}\right)=-729$

$x^3+\frac{1}{x^3}+3\left(-9\right)=-729$

$x^3+\frac{1}{x^3}-27=-729$

$x^3+\frac{1}{x^3}=-702$

Hence, the correct answer is Option B

Question 7: If $\sqrt{x}=\sqrt{3}-\sqrt{5}$, then the value of $x^2-16x+6$ is:

a) 4

b) 0

c) 2

d) -2

7) Answer (C)

Solution:

Given,  $\sqrt{x}=\sqrt{3}-\sqrt{5}$

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

$\Rightarrow$  $x=3+5-2\sqrt{15}$

$\Rightarrow$  $x=8-2\sqrt{15}$ ……………(1)

$\Rightarrow$  $x^2=\left(8-2\sqrt{15}\right)^2$

$\Rightarrow$  $x^2=64+60-32\sqrt{15}$

$\Rightarrow$  $x^2=124-32\sqrt{15}$ ………..(2)

$\therefore\ $ $x^2-16x+6=124-32\sqrt{15}-16\left(8-2\sqrt{15}\right)+6$

$=124-32\sqrt{15}-128+32\sqrt{15}+6$

$=130-128$

$=2$

Hence, the correct answer is Option C

Question 8: If $x=\frac{\sqrt{3}}{2}$, then the value of $\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$ is equal to:

a) $\sqrt 2$

b) $\sqrt 3$

c) 3

d) 2

8) Answer (B)

Solution:

Given, $x=\frac{\sqrt{3}}{2}$

$\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\times\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$

$=\frac{1+x+1-x+2\left(\sqrt{1+x}\right)\left(\sqrt{1-x}\right)}{1+x-\left(1-x\right)}$

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

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

$=\frac{1+\sqrt{1-\left(\frac{\sqrt{3}}{2}\right)^2}}{\frac{\sqrt{3}}{2}}$

$=\frac{1+\sqrt{1-\frac{3}{4}}}{\frac{\sqrt{3}}{2}}$

$=\frac{1+\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

$=\frac{3}{2}\times\frac{2}{\sqrt{3}}$

$=\sqrt{3}$

Hence, the correct answer is Option B

Question 9: If $a^3+b^3=62$ and a + b = 2, then the value of ab is:

a) -6

b) 9

c) 6

d) -9

9) Answer (D)

Solution:

Given,  $a+b=2$

$a^3+b^3=62$

$\Rightarrow$  $\left(a+b\right)\left(a^2-ab+b^2\right)=62$

$\Rightarrow$  $\left(2\right)\left(a^2+2ab+b^2-3ab\right)=62$

$\Rightarrow$  $\left(a+b\right)^2-3ab=31$

$\Rightarrow$  $\left(2\right)^2-3ab=31$

$\Rightarrow$  $4-3ab=31$

$\Rightarrow$  $3ab=-27$

$\Rightarrow$  $ab=-9$

Hence, the correct answer is Option D

Question 10: If $a-b=18$ and $a^3-b^3=324$, then find ab.

a) 105

b) -102

c) -104

d) 103

10) Answer (B)

Solution:

Given,  $a-b=18$ and

$a^3-b^3=324$

$\Rightarrow$  $\left(a-b\right)\left(a^2+ab+b^2\right)=324$

$\Rightarrow$  $\left(18\right)\left(a^2-2ab+b^2+3ab\right)=324$

$\Rightarrow$  $\left(a-b\right)^2+3ab=18$

$\Rightarrow$  $\left(18\right)^2+3ab=18$

$\Rightarrow$  $324+3ab=18$

$\Rightarrow$  $3ab=-306$

$\Rightarrow$  $ab=-102$

Hence, the correct answer is Option B

Question 11: If $x^4+\frac{1}{x^4}=14159$, then the value of $x+\frac{1}{x}$ is:

a) 11

b) 12

c) 9

d) 10

11) Answer (A)

Solution:

Given,  $x^4+\frac{1}{x^4}=14159$

$\Rightarrow$  $x^4+\frac{1}{x^4}+2=14159+2$

$\Rightarrow$  $\left(x^2+\frac{1}{x^2}\right)^2=14161$

$\Rightarrow$  $\left(x^2+\frac{1}{x^2}\right)^2=\left(119\right)^2$

$\Rightarrow$  $x^2+\frac{1}{x^2}=119$

$\Rightarrow$  $x^2+\frac{1}{x^2}+2=119+2$

$\Rightarrow$  $\left(x+\frac{1}{x}\right)^2=121$

$\Rightarrow$   $x+\frac{1}{x}=11$

Hence, the correct answer is Option A

Question 12: If $p+\frac{1}{p}=112$, find $(p-112)^{15}+\frac{1}{p^{15}}$

a) 1

b) 15

c) 10

d) 0

12) Answer (D)

Solution:

Given, $p+\frac{1}{p}=112$

$\Rightarrow$  $p-112=-\frac{1}{p}$

$(p-112)^{15}+\frac{1}{p^{15}}=\left(-\frac{1}{p}\right)^{15}+\frac{1}{p^{15}}$

$=-\frac{1}{p^{15}}+\frac{1}{p^{15}}$

$=0$

Hence, the correct answer is Option D

Question 13: If $\frac{4}{3}\left(x^2+\frac{1}{x^2}\right)=110\frac{2}{3}$, find $\frac{1}{9} \left(x^3 – \frac{1}{x^3}\right)$, where $x$ > 0.

a) 74

b) 76

c) 84

d) 85

13) Answer (C)

Solution:

Given, $\frac{4}{3}\left(x^2+\frac{1}{x^2}\right)=110\frac{2}{3}$

$\Rightarrow$  $\frac{4}{3}\left(x^2+\frac{1}{x^2}\right)=\frac{332}{3}$

$\Rightarrow$  $x^2+\frac{1}{x^2}=83$

$\Rightarrow$  $x^2+\frac{1}{x^2}-2=83-2$

$\Rightarrow$  $\left(x-\frac{1}{x}\right)^2=81$

$\Rightarrow$  $x-\frac{1}{x}=9$

$\Rightarrow$  $\left(x-\frac{1}{x}\right)^3=9^3$

$\Rightarrow$  $x^3-\frac{1}{x^3}-3.x.\frac{1}{x}\left(x-\frac{1}{x}\right)=729$

$\Rightarrow$  $x^3-\frac{1}{x^3}-3\left(9\right)=729$

$\Rightarrow$  $x^3-\frac{1}{x^3}-27=729$

$\Rightarrow$  $x^3-\frac{1}{x^3}=756$

$\Rightarrow$  $\frac{1}{9}\left(x^3-\frac{1}{x^3}\right)=\frac{1}{9}\left(756\right)$

$\Rightarrow$  $\frac{1}{9}\left(x^3-\frac{1}{x^3}\right)=84$

Hence, the correct answer is Option C

Question 14: If $x^3+y^3=16$ and $x+y=4$, then the value of $x^4+y^4$ is:

a) 48

b) 32

c) 64

d) 30

14) Answer (B)

Solution:

Given,  $x^3+y^3=16$ and $x+y=4$

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

$\Rightarrow$  $\left(4\right)\left(x^2+2xy+y^2-3xy\right)=16$

$\Rightarrow$  $\left(x+y\right)^2-3xy=4$

$\Rightarrow$  $\left(4\right)^2-3xy=4$

$\Rightarrow$  $3xy=16-4$

$\Rightarrow$  $3xy=12$

$\Rightarrow$  $xy=4$

$\therefore\ $ $x^4+y^4=x^4+y^4+2x^2y^2-2x^2y^2$

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

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

$=\left[\left(x+y\right)^2-2xy\right]^2-32$

$=\left[\left(4\right)^2-2\left(4\right)\right]^2-32$

$=\left[16-8\right]^2-32$

$=64-32$

$=32$

Hence, the correct answer is Option B

Question 15: If $a^2+\frac{2}{a^2}=16$, then find the value of $\frac{72a^2}{a^4+2+8a^2}$

a) 2

b) 4

c) 1

d) 3

15) Answer (D)

Solution:

Given,  $a^2+\frac{2}{a^2}=16$

$\frac{72a^2}{a^4+2+8a^2}=\frac{72a^2}{a^2\left(a^2+\frac{2}{a^2}+8\right)}$

$=\frac{72}{a^2+\frac{2}{a^2}+8}$

$=\frac{72}{16+8}$

$=\frac{72}{24}$

$=3$

Hence, the correct answer is Option D

Question 16: If a + 3b = 12 and ab = 9, then the value of (a – 3b) is:

a) 9

b) 8

c) 6

d) 4

16) Answer (C)

Solution:

Given, $a+3b=13$

$\Rightarrow$  $\left(a+3b\right)^2=12^2$

$\Rightarrow$  $a^2+9b^2+6ab=144$

$\Rightarrow$  $a^2+9b^2+6\left(9\right)=144$

$\Rightarrow$  $a^2+9b^2+54-6ab+6ab=144$

$\Rightarrow$  $a^2+9b^2-6ab+6ab=90$

$\Rightarrow$  $\left(a-3b\right)^2+6ab=90$

$\Rightarrow$  $\left(a-3b\right)^2+6\left(9\right)=90$

$\Rightarrow$  $\left(a-3b\right)^2+54=90$

$\Rightarrow$  $\left(a-3b\right)^2=36$

$\Rightarrow$  $a-3b=6$

Hence, the correct answer is Option C

Question 17: If $1 + 9r^2 + 81r^4 = 256$ and $1 + 3r + 9r^2 = 32$, then find the value of $1 – 3r + 9r^2$.

a) 4

b) 8

c) 16

d) 12

17) Answer (B)

Solution:

Given that $1 + 3r + 9r^2 = 32$

and let $1 – 3r + 9r^2$ be K

Multiply $1 + 3r + 9r^2$ and $1 – 3r + 9r^2$

we get the product as $1 + 9r^2 + 81r^4$ = 32K

but $1 + 9r^2 + 81r^4$= 256
substitute this value in the product

256 = 32k

K=8

Therefore, answer is option B

Question 18: If $x^3 — 6x^2 + ax + b$ is divisible by $(x^2 — 3x + 2)$, then the values of a and b are:

a) a = -6 and b = -11

b) a = -11 and b = 6

c) a = 11 and b = -6

d) a = 6 and b = 11

18) Answer (C)

Solution:

Given, $x^3—6x^2+ax+b$ is divisible by $(x^2 — 3x + 2)$

Let the quotient when $x^3 — 6x^2 + ax + b$ is divisible by $(x^2 — 3x + 2)$ be $x-p$

$\Rightarrow$ $(x^2—3x+2)\left(x-p\right)=x^3—6x^2+ax+b$

$\Rightarrow$  $x^3-3x^2+2x-px^2+3px-2p=x^3—6x^2+ax+b$

$\Rightarrow$  $x^3-\left(3+p\right)x^2+\left(2+3p\right)x-2p=x^3—6x^2+ax+b$

Comparing both sides,

$-\left(3+p\right)=-6$

$\Rightarrow$  $p=3$

$2+3p=a$

$\Rightarrow$  $2+3\left(3\right)=a$

$\Rightarrow$  $a=11$

$-2p=b$

$\Rightarrow$  $-2\left(3\right)=b$

$\Rightarrow$  $b=-6$

$\therefore\ $ a = 11 and b = -6

Hence, the correct answer is Option C

Question 19: If $x – \frac{1}{x} = 13,$ then the value of $x^2 + \frac{1}{x^2}$ is:

a) 165

b) 171

c) 167

d) 169

19) Answer (B)

Solution:

Given, $x-\frac{1}{x}=13$

$\Rightarrow$ $\left(x-\frac{1}{x}\right)^2=13^2$

$\Rightarrow$ $x^2-2.x.\frac{1}{x}+\frac{1}{x^2}=169$

$\Rightarrow$  $x^2+\frac{1}{x^2}-2=169$

$\Rightarrow$  $x^2+\frac{1}{x^2}=171$

Hence, the correct answer is Option B

Question 20: If $A = \frac{x – 1}{x + 1}$, then the value of $A – \frac{1}{A}$ is:

a) $\frac{-4(2x – 1)}{x^2 – 1}$

b) $\frac{x^2 – 1}{-4(2x – 1)}$

c) $\frac{x^2 – 1}{-4(2x + 1)}$

d) $\frac{-4x}{x^2 – 1}$

20) Answer (D)

Solution:

Given, $A=\frac{x-1}{x+1}$

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

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

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

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

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

Hence, the correct answer is Option D

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