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# SSC CGL Algebra Questions

Download Top-20 Algebra questions for SSC CGL exam. Most important Algebra questions based on asked questions in previous exam papers for SSC CGL.

Question 1: If a * b = 2a + 3b – ab, then the value of (3 * 5 + 5 * 3) is

a) 10

b) 6

c) 4

d) 2

Question 2: If a * b = $a^{b}$, then the value of 5 * 3 is

a) 125

b) 243

c) 53

d) 15

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

a) $6\sqrt{6}$

b) $0$

c) $3\sqrt{6}$

d) $2\sqrt{6}$

Question 4: If $a^2+b^2+c^2=2(a-2b-c-3)$ then the value of a+b+c is

a) 3

b) 0

c) 2

d) 4

Question 5: Find the simplest value of $2\sqrt{50} + \sqrt{18} – \sqrt{72}$ is __? $(\sqrt{2} = 1.414)$.

a) 9.898

b) 10.312

c) 8.484

d) 4.242

Question 6: 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) 1

b) 2

c) 0

d) -1

Question 7: 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) 1

b) 2

c) 0

d) -1

Question 8: If x + y + z = 6 and $x^{2}+y^{2}+z^{2}$=20 then the value of $x^{3}+y^{3}+z^{3}$-3xyz is

a) 64

b) 70

c) 72

d) 76

Question 9: If $\frac{p^2}{q^2}+\frac{q^2}{p^2}$=1 then the value of $(p^{6}+q^{6})$ is

a) 0

b) 1

c) 2

d) 3

Question 10: 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) 1

b) 0

c) 2

d) $\frac{1}{2}$

Question 11: 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) $\sqrt{7},-1$

b) $\sqrt{7}, 1$

c) $0, -\frac{2}{3}$

d) $-\frac{2}{3}, 0$

Question 12: If $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ then $x^{3} + \frac{1}{x^{3}}$ is equal to

a) 98

b) 1000

c) 5

d) 970

Question 13: $\sqrt{\sqrt{\sqrt{0.00000256}}}$

a) 0.4

b) 0.02

c) 0.04

d) 0.2

Question 14: If m = – 4, n = – 2, then the value of $m^3 – 3m^2 + 3m + 3n + 3n^2 + n^3$ is

a) – 126

b) 124

c) – 124

d) 126

Question 15: If $x+\frac{1}{x}=1$ then the value of $\frac{x^2+3x+1}{x^2+7x+1}$

a) $1$

b) $\frac{3}{7}$

c) $\frac{1}{2}$

d) 2

Question 16: 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) 0.4773

b) 477.3

c) 47.73

d) 4.773

Question 17: If $\frac{x}{y}$=$\frac{3}{4}$ the ratio of $(2x+3y)$ and $(3y-2x)$ is

a) 2 : 1

b) 3 : 2

c) 1 : 1

d) 3 : 1

Question 18: If m – 5n = 2, then the vlaue of $(m^{3} – 125n^{3}$ – 30 mn) is

a) 6

b) 7

c) 8

d) 9

Question 19: If $x+\frac{1}{x}=2$ then the value of $x^{12}+\frac{1}{x^{12}}$ is

a) 2

b) -4

c) 0

d) 4

Question 20: If 5x + 9y = 5 and $125x^{3}$ + $729y^{3}$ = 120 then the value of the product of x and y is

a) $\frac{1}{9}$

b) $\frac{1}{135}$

c) $45$

d) $135$

For 3*5 put a=3 and b=5 in given equation
and for 5*3 put a=5 and b=3 in equation
now add both values

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 $6\sqrt{6}$

Given $a^2+b^2+ c^2=2(a-2b-c-3)$,

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 $10\sqrt2 + 3\sqrt2 – 6\sqrt2 = 7\sqrt2$
Hence  $7\sqrt2$ will be around 9.898

shortcut :

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

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

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.

shortcut :

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

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

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.

We know that $x^{3}+y^{3}+z^{3}-3xyz = (x + y + z)(x^2 +y^2 + z^2 -xy-yz-xz)$
$x^{3}+y^{3}+z^{3}-3xyz = (6)(20 -xy-yz-xz)$
Hence the solution must be a multiple of 6.
Out of the given options only Option C is a multiple of 6.
Hence Option C is the correct answer.

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

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

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

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 $b=\frac{-2}{3}$

$x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

=> $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} * \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

=> $x= 5+2\sqrt{6}$ —————Eqn(1)

Now, $\frac{1}{x}=\frac{1}{5+2\sqrt{6}}$

=> $\frac{1}{x} = \frac{1}{5+2\sqrt{6}} * \frac{5-2\sqrt{6}}{5-2\sqrt{6}}$

=> $\frac{1}{x}= 5-2\sqrt{6}$ —————Eqn(2)

Now, cubing eqns (1)&(2), we get :

=> $x^3 = 125+72\sqrt{6}+150\sqrt{6}+360 = 485+222\sqrt{6}$

and $\frac{1}{x^3} = 125-72\sqrt{6}-150\sqrt{6}+360 = 485-222\sqrt{6}$

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

= $485+222\sqrt{6} + 485-222\sqrt{6}$

= 970

Expression : $\sqrt{\sqrt{\sqrt{0.00000256}}}$

= $\sqrt{\sqrt{0.0016}}$

= $\sqrt{0.04} = 0.2$

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 :

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

= $\frac{1}{2}$

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

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