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# Algebra Questions for MAH MBA CET Exam

Download MAH MBA CET Algebra Questions and Answers PDF covering the important questions. Most expected Algebra questions with explanations for MAH MBA CET / MMS CET 2021 exam.

Question 1:Â $\sqrt{8 + \sqrt{57 + \sqrt{38 + \sqrt{108 + \sqrt{169}}}}}$

a)Â 4

b)Â 6

c)Â 8

d)Â 10

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

a)Â 10

b)Â 6

c)Â 4

d)Â 2

Question 3:Â If a * bÂ = $a^{b}$, then the value of 5 * 3 is

a)Â 125

b)Â 243

c)Â 53

d)Â 15

Question 4:Â If $x = 1 + \sqrt{2} + \sqrt{3}$ , then the value of $(2x^4 – 8x^3 – 5x^2 + 26x- 28)$ is __?

a)Â $6\sqrt{6}$

b)Â $0$

c)Â $3\sqrt{6}$

d)Â $2\sqrt{6}$

Question 5:Â If $a^2+b^2+c^2=2(a-2b-c-3)$ then the value of a+b+c is

a)Â 3

b)Â 0

c)Â 2

d)Â 4

Question 6:Â Find the simplest value of $2\sqrt{50} + \sqrt{18} – \sqrt{72}$ is __? $(\sqrt{2} = 1.414)$.

a)Â 9.898

b)Â 10.312

c)Â 8.484

d)Â 4.242

Question 7:Â If $a^{3}-b^{3}-c^{3}=0$ then the value of $a^{9}-b^{9}-c^{9}-3a^{3} b^{3} c^{3}$ is

a)Â 1

b)Â 2

c)Â 0

d)Â -1

Question 8:Â If $\frac{p^2}{q^2}+\frac{q^2}{p^2}$=1 then the value of $(p^{6}+q^{6})$ is

a)Â 0

b)Â 1

c)Â 2

d)Â 3

Question 9:Â If $(m+1) = \sqrt{n}+3$ the value of $\frac{1}{2}(\frac{m^{3}-6m^{2}+12m-8}{\sqrt{n}}-n)$

a)Â 0

b)Â 1

c)Â 2

d)Â 3

Question 10:Â If $x=\frac{a-b}{a+b},y=\frac{b-c}{b+c},z=\frac{c-a}{c+a}$ then $\frac{(1-x)(1-y)(1-z)}{(1+x)(1+y)(1+z)}$ is equal to

a)Â 1

b)Â 0

c)Â 2

d)Â $\frac{1}{2}$

Question 11:Â If $\frac{\sqrt{7}-1}{\sqrt{7}+1}-\frac{\sqrt{7}+1}{\sqrt{7}-1}=a+\sqrt{7} b$ the values of a and b are respectively

a)Â $\sqrt{7},-1$

b)Â $\sqrt{7}, 1$

c)Â $0, -\frac{2}{3}$

d)Â $-\frac{2}{3}, 0$

Question 12:Â If m = – 4, n = – 2, then the value of $m^3 – 3m^2 + 3m + 3n + 3n^2 + n^3$ is

a)Â – 126

b)Â 124

c)Â – 124

d)Â 126

Question 13:Â If $x+\frac{1}{x}=1$ then the value of $\frac{x^2+3x+1}{x^2+7x+1}$

a)Â $1$

b)Â $\frac{3}{7}$

c)Â $\frac{1}{2}$

d)Â 2

Question 14:Â If $x=\sqrt{a^3\sqrt{b}\sqrt{a^3}\sqrt{b}}$ then the value of x is

a)Â $\sqrt[5]{ab^3}$

b)Â $\sqrt[3]{a^3b}$

c)Â $\sqrt[3]{a^5b}$

d)Â  $a^2\sqrt b\sqrt[4]{a}$

Question 15:Â If the cube root of 79507 is 43, then the value of $\sqrt[3]{79.507}+\sqrt[3]{0.079507}+\sqrt[3]{0.000079507}$
is

a)Â 0.4773

b)Â 477.3

c)Â 47.73

d)Â 4.773

Question 16:Â If $\frac{x}{y}$=$\frac{3}{4}$ the ratio of $(2x+3y)$ and $(3y-2x)$ is

a)Â 2 : 1

b)Â 3 : 2

c)Â 1 : 1

d)Â 3 : 1

Question 17:Â If m – 5n = 2, then the vlaue of $(m^{3} – 125n^{3}$ – 30 mn) is

a)Â 6

b)Â 7

c)Â 8

d)Â 9

Question 18:Â If $x+\frac{1}{x}=2$ then the value of $x^{12}+\frac{1}{x^{12}}$ is

a)Â 2

b)Â -4

c)Â 0

d)Â 4

Question 19:Â If 5x + 9y = 5 and $125x^{3}$ + $729y^{3}$ = 120 then the value of the product of x and y is

a)Â $\frac{1}{9}$

b)Â $\frac{1}{135}$

c)Â $45$

d)Â $135$

Question 20:Â What is the value of $\frac{(941+149)^{2}+(941-149)^{2}}{(941\times941+149\times149)}?$

a)Â 10

b)Â 2

c)Â 1

d)Â 100

Start from the root of 169 then second root will reduce to 11, thrid root will reduce to 7, fourth root will reduce to 8, and finally it reduce to value 4

For 3*5 put a=3 and b=5 in given equation
and for 5*3 put a=5 and b=3 in equation

Put a=5 and b=3 in given equation
hence it will be $5^{3}$ = 125

x = 1+ $\sqrt {2} + \sqrt {3}$
$(x-1)^{2}$ = $(\sqrt {2} + \sqrt {3}) ^ {2}$
$x^{2} +1 – 2x = 5 + 2 \sqrt {6}$
$x^{2} – 2x = 4 + 2 \sqrt {6}$ ( eq. (1) )
$(x^{2} – 2x)^{2} = x^{4} + 4x^{2} – 4x^{3} = 40 + 16\sqrt{6}$ eq (2)
Now in $2x^{4} – 8x^{3} – 5x^{2} + 26x – 28$
or $2(x^{4} – 4x^{3}) – 5x^{2} + 26x – 28$ ( putting values from eq (1) and eq (2) )
After solving we will get it reduced to $6\sqrt{6}$

GivenÂ $a^2+b^2+ c^2=2(a-2b-c-3)$,

So, $(a-1)^2+(b+2)^2+(c-1)^2=0$

Hence, a=1, b=-2 and c=1

So, the sum of the equation is

Given equation can be reduced in the form of $10\sqrt2 + 3\sqrt2 – 6\sqrt2 = 7\sqrt2$
HenceÂ Â $7\sqrt2$ will be around 9.898

shortcut :

put c = 0 inÂ Â $a^{3}-b^{3}-c^{3}=0$ $\Rightarrow$Â $a^{3}=b^{3}$

$a^{9}-b^{9}-(0)^{9}-3a^{3} b^{3} (0)^{3}$ =Â $a^{9}-b^{9}$ =Â $(a^{3})^{3}-(b^{3})^{3}$ =Â Â $(a)^{3}-(a)^{3}$ = 0Â  ( $\because$ $a^{3}=b^{3}$ )

so the answer is option C.

normal method :

$a^{3}-b^{3}-c^{3}=0$

$a^{3}=b^{3}+c^{3}$

cubing on both sides,

$(a^{3})^{3}=(b^{3}+c^{3})^{3}$

$a^{9}=b^{9}+c^{9}+3b^{3} c^{3}(b^{3}+c^{3})$

$a^{9}=b^{9}+c^{9}+3b^{3} c^{3}(a^{3})$

$a^{9}-b^{9}-c^{9}-3a^{3}b^{3} c^{3}=0$

so the answer is option C.

Expression : $\frac{p^2}{q^2}+\frac{q^2}{p^2}$ = 1

=> $\frac{p^{4}+q^{4}}{p^2q^2}$ = 1

=> $p^4+q^4 = p^2q^2$ ————–Eqn(1)

Now, to find : $(p^{6}+q^{6})$

=> $(p^2)^3 + (q^2)^3$

Using the formula, $a^3 + b^3 = (a+b)(a^2+b^2-ab)$

=> $(p^2+q^2)(p^4+q^4-p^2q^2)$

From eqn (1), we get :

=> $(p^2+q^2)(p^2q^2-p^2q^2)$

=> $(p^2+q^2)*0$

= 0

If $(m+1) = \sqrt{n}+3$

=> $m-2 = \sqrt{n}$ ————–Eqn(1)

to find : $\frac{1}{2}(\frac{m^{3}-6m^{2}+12m-8}{\sqrt{n}}-n)$

$\because (m-2)^3 = m^{3}-6m^{2}+12m-8$

=> $\frac{1}{2}(\frac{(m-2)^3}{\sqrt{n}}-n)$

Using eqn(1), we get :

=> $\frac{1}{2}(\frac{(\sqrt{n})^3}{\sqrt{n}}-n)$

=> $\frac{1}{2}(n-n)$

= 0

If $x=\frac{a-b}{a+b}$

=> $(1-x) = 1- (\frac{a-b}{a+b})$

=> $(1-x) = \frac{2b}{a+b}$

Similarly, $(1+x) = \frac{2a}{a+b}$

Applying the same method, we get :

=> $(1-y) = \frac{2c}{b+c}$ and => $(1+y) = \frac{2b}{b+c}$

=> $(1-z) = \frac{2a}{c+a}$ and => $(1+z) = \frac{2c}{c+a}$

Putting above values in the equation : $\frac{(1-x)(1-y)(1-z)}{(1+x)(1+y)(1+z)}$

=> $\frac{(\frac{2b}{a+b})(\frac{2c}{b+c})(\frac{2a}{c+a})}{(\frac{2a}{a+b})(\frac{2b}{b+c})(\frac{2c}{c+a})}$

=> $\frac{2a*2b*2c}{2a*2b*2c}$

= 1

$\frac{\sqrt{7}-1}{\sqrt{7}+1}-\frac{\sqrt{7}+1}{\sqrt{7}-1}=a+\sqrt{7} b$

L.H.S. = $\frac{\sqrt{7}-1}{\sqrt{7}+1}-\frac{\sqrt{7}+1}{\sqrt{7}-1}$

= $\frac{(\sqrt{7}-1)^2 – (\sqrt{7}+1)^2}{(\sqrt{7}-1)(\sqrt{7}+1)}$

= $\frac{(7+1-2\sqrt{7})-(7+1+2\sqrt{7})}{7-1}$

= $\frac{-4\sqrt{7}}{6}$

= $\frac{-2\sqrt{7}}{3}$

Now, comparing with R.H.S. $a+\sqrt{7} b$

we get,

$a=0$ and $b=\frac{-2}{3}$

We are given that m = -4 and n = -2

Expression : $m^3 – 3m^2 + 3m + 3n + 3n^2 + n^3$

= $(m^3 – 3m^2 + 3m – 1) + (n^3 + 3n^2 + 3n + 1)$

= $(m-1)^3 + (n+1)^3$

= $(-4-1)^3 + (-2+1)^3$

= $(-5)^3 + (-1)^3$

= $-125 – 1 = -126$

Expression : $x+\frac{1}{x}=1$

=> $x^2 + 1 = x$ ——Eqn(1)

To find : $\frac{x^2+3x+1}{x^2+7x+1}$

= $\frac{(x^2+1) + 3x}{(x^2+1) + 7x}$

Using eqn(1),we get :

= $\frac{x + 3x}{x + 7x} = \frac{4}{8}$

= $\frac{1}{2}$

$x=\sqrt{a^3\sqrt{b}\sqrt{a^3}\sqrt{b}}$.

here we know that $\sqrt{b} \times \sqrt{b}$ = b

and $\sqrt{a^3} = a\sqrt{a}$

hence,$x=\sqrt{a^3\sqrt{b}\sqrt{a^3}\sqrt{b}}$ = $a^2\sqrt b\sqrt[4]{a}$

Since $\sqrt[3]{79507}$ = 43

=> $\sqrt[3]{79.507}$ = 4.3

=> $\sqrt[3]{0.079507}$ = 0.43

=> $\sqrt[3]{0.000079507}$ = 0.043

=> 4.3+0.43+0.043 = 4.773

Let $x = 3k$ and $y = 4k$

=> $\frac{2x + 3y}{3y – 2x}$

= $\frac{6k + 12k}{12k – 6k}$

= $\frac{18}{6}$

= $\frac{3}{1}$ = 3 : 1

Using the formula, $(x-y)^3 = x^3 – y^3 -3xy(x-y)$

=> $(m – 5n)^3 = m^3 – 125n^3 – 15mn(m-5n)$

=> $2^3 = m^3 – 125n^3 – 15mn*2$

=> $m^3 – 125n^3 – 30mn = 8$

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

Squaring both sides

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

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

Cubing both sides

=> $x^6 + \frac{1}{x^6} + 3.x.\frac{1}{x}(x+\frac{1}{x}) = 8$

=> $x^6 + \frac{1}{x^6} = 8-6 = 2$

Again, squaring both sides, we get :

=> $x^{12} + \frac{1}{x^{12}} + 2 = 4$

=> $x^{12} + \frac{1}{x^{12}} = 2$

Expression : $5x + 9y = 5$

Cubing both sides, we get :

=> $(5x + 9y)^3 = 125$

=> $125x^3 + 729y^3 + 135xy(5x+9y) = 125$

=> $125x^3 + 729y^3 + 135xy*5 = 125$

Since, $125x^{3}$ + $729y^{3} = 120$

=> $xy = \frac{5}{5*135} = \frac{1}{135}$

Expression : $\frac{(941+149)^{2}+(941-149)^{2}}{(941\times941+149\times149)}$
= $\frac{(941^2 + 149^2 + 2.941.149) + (941^2 + 149^2 – 2.941.149)}{941^2 + 149^2}$
= $\frac{2 * (941^2 + 149^2)}{941^2 + 149^2}$