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# Algebra Questions For SSC CHSL Set-2

SSC CHSL Algebra Set-2 Questions download PDF based on previous year question paper of SSC exams. 20 Very important Algebra Set-2 questions for SSC CHSL Exam.

Question 1: Find the area $(in cm^2)$. If the circumference of a circle is $110 cm$?

a) 962.5

b) 1024.5

c) 866.5

d) 1925

Question 2: Classify the triangle as a type of triangle if the sides of the triangle are 8, 11 and 17 units.

a) Isoceles triangle

b) Acute angled triangle

c) Obtuse angled triangle

d) Right angled triangle

Question 3: Find the area$(in cm^2)$ of a square whose perimeter is 17 cm.

a) 20.0625

b) 14.0625

c) 18.0625

d) 16.0625

Question 4: find the roots of the equation $x^2 + 16x – 57 = 0$

a) -24, 6

b) -14, -2

c) 14, 2

d) -19, 3

Question 5: If $p + \frac{1}{p} = 3$, then find the value of $p^6 + \large\frac{1}{p^6}$.

a) 322

b) 348

c) 329

d) 342

SSC CHSL Study Material (FREE Tests)

Question 6: If $px^3 + qx^2 – 19x – 30 = 0$ is completely divided by $x^2 + 5x + 6$, find the value of $(p, q)$.

a) (1,0)

b) (2,1)

c) (1,0)

d) (4,1)

Question 7: If $l^{3} + m^{3}$ = -218 and $lm = -35$ , then what is the value of $l + m$?

a) -6

b) -2

c) -4

d) -5

Question 8: When a number is increased by 138, it becomes 123% of itself. Find the number?

a) 600

b) 450

c) 750

d) 900

Question 9: If $l^{3} + m^{3}$ = -316 and $l + m = -4$, then what is the value of $l \times m$?

a) -20

b) -21

c) -22

d) -23

Question 10: If $3[\large\frac{5p}{6} – \frac{7}{3} ]$ $= 2[\large\frac{3p}{4} –$ $9]$, find $p$ ?

a) -13

b) -19

c) -17

d) -11

Question 11: If $p + q = -7$ and $p \times q = 5$, then what is the value of $p^{3} + q^{3}$ ?

a) 448

b) 238

c) -448

d) -238

Question 12: Sum of ten times a fraction and its reciprocal is 7. What is the fraction? {It is known that the fraction is less than 0.5}

a) $\large\frac{1}{3}$

b) $\large\frac{1}{5}$

c) $\large\frac{1}{2}$

d) $\large\frac{1}{7}$

Question 13: If $\large\frac{12p}{7}$ $– 4= \large\frac{4}{7}-\frac{10p}{14}$ , then find $p$ ?

a) $\large\frac{32}{17}$

b) $\large\frac{39}{17}$

c) $\large\frac{28}{17}$

d) $\large\frac{27}{17}$

Question 14: If the roots of the equation $x^{2} + 10x ­-119=0$ are p & q, then find the difference between the roots.

a) $-10$

b) $10$

c) $27$

d) $24$

Question 15: Find the value of $(36)^\frac{3}{2}$ $\times$ $(1024)^\frac{2}{5}$ $\times$ $(343)^\frac{1}{3}$

a) 24192

b) 24882

c) 24912

d) 24921

Question 16: The square root of $27-4\sqrt{35}$ is :

a) $\pm(\sqrt{5}+2\sqrt{7})$

b) $\pm(\sqrt{5}-2\sqrt{7})$

c) $\pm(\sqrt{7}-2\sqrt{5})$

d) $\pm(\sqrt{7}+2\sqrt{5})$

Question 17: If $n-\large\frac{1}{n}$ $=7$, find the value of $\large\frac{n^{2}}{n^{4}+49n^2 +1}$ is

a) $\large\frac{1}{60}$

b) $\large\frac{1}{70}$

c) $\large\frac{1}{100}$

d) $\large\frac{1}{50}$

Question 18: If $n-5=2\sqrt{6}$, then the value of $\large\frac{n^{4}+1}{n^{2}}$ is

a) 98

b) 47

c) 23

d) 79

Question 19: If $9^{0.20} \times 3^{1.2} \times 9^{0.4} \times 3^{0.6} = 81^{n}$. Then find $n$.

a) 0.4

b) 0.6

c) 0.75

d) 0.45

Question 20: If $\large\frac{p}{q}+\frac{q}{p}$ $=-1$, then the value of $p^{3}-q^{3}$ is

a) 64

b) 27

c) 100

d) 0

Given, the circumference of a circle is 110 cm

Let radius of circle = $r$ cm

=> Circumference = $2\pi r=110$

=> $2\times\large\frac{22}{7}\times$ $r=110$

=> $r=110\times\large\frac{7}{44}$

=> $r=\large\large\frac{35}{2}$ $=17.5$ cm

$\therefore$ Area of circle = $\pi r^2$

= $\large\frac{22}{7}$ $\times(17.5)^2$ $cm^2$

=  $962.5$

=> Ans – (A)

Let the sides of $\triangle$ ABC be $a,b,c$, where the largest side = $’c’$

If $c^2=a^2+b^2$, then the angle at $C$ is right angle.

If $c^2<a^2+b^2$, then the angle at $C$ is acute angle.

If $c^2>a^2+b^2$, then the angle at $C$ is obtuse angle.

Given, sides of the triangle are 8, 11 and 17 units

Now, according to ques, => $17^2=289$

and $8^2+11^2=64+121=185$

$\therefore c^2>a^2+b^2$, hence it is an obtuse angled triangle.

=> Ans – (C)

Let side of square = $s$ cm

=>Given, Perimeter = $4s=17$

=> $s=\large\frac{17}{4}$ $=4.25$ cm

$\therefore$ Area = $(4.25)^2=18.0625$ $cm^2$

=> Ans – (C)

Given, equation $x^2 + 16x – 57 = 0$

This can be written as,

$x^2 + 19x – 3x – 57 = 0$

(x+19)(x-3)=0

x = -19 or x = 3

Hence, option A is the right answer.

Given,
$p + \large\frac{1}{p}$ $=$ $3$

Squaring both sides,

$p^2$ $+$ $\large\frac{1}{p^2}$ $+$ $2 \times p \times \large\frac{1}{p}$ $=$ $9$

$p^2 + \large\frac{1}{p^2}$ $=$ $7$

Cubing on both sides,

$p^6 + \large\frac{1}{p^6}$ $+ 3 \times p^2 \times \large\frac{1}{p^2}(p^2 + \large\frac{1}{p^2})$ $= 343$

$p^6 + \large\frac{1}{p^6}$ $+ 3 (p^2 + \large\frac{1}{p^2})$ $=343$

$p^6 + \large\frac{1}{p^6}$ $+ 3 (7)$= $343$

$p^6 + \large\frac{1}{p^6}$  $=$ $343-21$

$p^6 + \large\frac{1}{p^6}$  $=$ $322$

Given, $px^3 + qx^2 – 19x – 30 = 0$ is divisible by $x^2 + 5x + 6$

$x^2 + 5x + 6$ can be factorized as $(x + 2)(x + 3)$

x = -2 and x = -3 should satisfy $px^3 + qx^2 – 19x – 30 = 0$.

Substituting the values,

When, $x = -2$

$-8p + 4q + 8 = 0$ or $2p – q – 2 = 0$

When, $x = -3$

$-27p + 9q + 27 = 0$ or $3p – q – 3 = 0$

Subtracting both, $p = 1$

Substituting the value of $p = 1$, we get $q = 0$

Hence, $(p, q)$= $(1, 0)$.

Given, $l^{3} + m^{3}$ = -218 and $lm = -35$

$(l+m)^{3} = l^{3}+ m^{3} + 3lm(l+m)$

$(l+m)^{3} = -218 + 3(-35)(l+m)$

$(l+m)^{3} = -218 -105(l+m)$

we need to solve 3rd degree equation

so,without solving, by verification, we get $l+m = -2$,

so the answer is option B.

When the number is increased by $138$, it becomes $123$% of itself. Let the number be ‘n’

n + $138$ = $123$% of n

$138$ = $\large\frac{123}{100}$n – n

$138$ = n$(\large\frac{123}{100} – 1)$

$132$ = $\large\frac{23}{100}$

x = $600$.

Hence, option A is the correct answer.

Given,

$l^{3} + m^{3}$ = -316 and $l + m = -4$

We know that, $l^{3}+m^{3}=(l+m)^{3}-3lm(l+m)$

$-316=(-4)^{3}-3lm(-4)$

$-316=-64+12lm$

$12lm=-252$

$lm=-21$

So the answer is option B.

Given,

$3[\large\frac{5p}{6} – \frac{7}{3} ]$ $= 2[\large\frac{3p}{4} -$ $9]$

$\large\frac{5p}{2} –$ $7$ $= [\large\frac{3p}{2} -$ $18]$

$\large\frac{5p}{2} – \frac{3p}{2}$ $= 7 – 18$

$\large\frac{5p-3p}{2}$ $= -11$

$p = -11$

Given, $p + q = -7$ and $p \times q = 5$

$p^{3} + q^{3} = (p+q)^{3} – 3pq(p+q)$

$= (-7)^{3} – 3(5)(-7)$

$= -343 + 105 = -238$

So the answer is option D.

Let that fraction be $\large\frac{1}{f}$

$10(\large\frac{1}{f})$ $+ f = 7$

$\Rightarrow (10+f^{2}) = 7 \times f$

$\Rightarrow f^{2}-7f+10 = 0$

$\Rightarrow(f-2)(f-5)=0$

$\Rightarrow f = 2$ or $5$

$\Rightarrow \text{fraction} = \large\frac{1}{x}$ $= 1/2 \text{ or } 1/5$

{Since, It is known that the fraction is less than 0.5} i.e. $f<\large\frac{1}{2}$

$\therefore f = \large\frac{1}{5}$

$\large\frac{12p}{7}$ $– 4= \large\frac{4}{7}-\frac{10p}{14}$

$\large\frac{12p}{7}$ $– 4= \large\frac{4}{7}-\frac{5p}{7}$

$\frac{12p}{7}$ $+\large\frac{5p}{7}= \frac{4}{7} + 4$

$\large\frac{12p+5p}{7}$ $= \large\frac{4+28}{7}$

$\large\frac{17p}{7}$ $=\large \frac{32}{7}$

$17p = 32$

$p=\large\frac{32}{17}$

Given Expression: $x^2 + 10x – 119$

= $x^2 + 17x – 7x – 119$

= $x(x + 17) – 7(x + 17)$

= $(x + 17)(x – 7)$

= $(x = -17 or 7)$

Therefore the roots are 7 & -17

The difference between the roots is 24(p ~ q)

=> Ans – (D)

$(36)^\frac{3}{2} \times (1024)^\frac{2}{5}\times (343)^\frac{1}{3}$

= $(6)^{2\times\large\frac{3}{2}} \times$ $(2)^{10\times\large\frac{2}{5}}$ $\times (7)^{3\times\large\frac{1}{3}}$

= $216\times16\times7$

= $24192$

We have to find $27-4\sqrt{35}$

We can write it as :

= $\sqrt{{27} – 2 \times 2\times \sqrt{5} \times \sqrt{7}}$

Since, $(a^2 + b^2 – 2ab) = (a-b)^2$

= $\sqrt{(2\sqrt{5})^2 + (\sqrt{7})^2 – 2\times2\sqrt{5}\times\sqrt{7}}$

= $\sqrt{(\sqrt{7} – 2\sqrt{5})^2}$

= $\pm(\sqrt{7} – 2\sqrt{5})$

$\large\frac{n^{2}}{n^{4}+49n^2 +1}$

Given, $n-\large\frac{1}{n}$ $=7$

Squaring on both sides,

$(n-\large\frac{1}{n})^2$ $=7^2$

$(n^2 – 2 +\large\frac{1}{n^2})$ $=7^2$

$(n^2 +\large\frac{1}{n^2})$ $= 49+2$

$(n^2 +\large\frac{1}{n^2})$ $= 51$

$(n^4 +1)$ $= 51n^2$

Adding $49n^2$ on both sides

$n^4 +1$ $+ 49n^2$ $= 51n^2 + 49n^2$

$n^4 + 49n^2 + 1$ $= 100n^2$

$\large\frac{n^4 + 49n^2 + 1}{n^2}$ $= 100$

$\large\frac{n^{2}}{n^{4}+49n^2 +1}$ $=\large\frac{1}{100}$

Given,

$n-5=2\sqrt{6}$

$n= 5+2\sqrt{6}$

$\large\frac{1}{n}=\frac{1}{5+2\sqrt{6}}$

$\large\frac{1}{n}=\frac{1}{5+2\sqrt{6}}$ $\times\large\frac{5-2\sqrt{6}}{5-2\sqrt{6}}$

$\large\frac{1}{n}$ $= 5-2\sqrt{6}$

$(n+\large\frac{1}{n})$ $= 10$

$(n+\large\frac{1}{n})^{2}$ $= 10^2$

$n^{2}+\large\frac{1}{n^{2}}$ $+ 2 = 100$

$n^{2}+\large\frac{1}{n^{2}}$ $= 100-2 = 98$

$\large\frac{n^{4}+1}{n^{2}}$ = $98$

$9^{0.20} \times 3^{1.2} \times 9^{0.4} \times 3^{0.6} = 81^{n}$

$3^{0.40} \times 3^{1.2} \times 3^{0.8} \times 3^{0.6} = 3^{4n}$

$3^{0.40+1.2+0.8+0.6} = 3^{4n}$

$3^{0.40+1.2+0.8+0.6} = 3^{4n}$

$3^{3} = 3^{4n}$

$3 = 4n$

$n=0.75$

Given,

$\large\frac{p}{q}+\frac{q}{p}$ $=-1$

$\large\frac{p^{2}+q^{2}}{pq}$ $= -1$

$p^{2}+q^{2} = -pq$

$p^{2}+q^{2} + pq = 0$

We know $p^{3}-q^{3}={(p-q)}{(p^{2}+q^{2}}{+pq)}$

As $p^{2}-q^{2} + pq = 0$, therefore $p^{3}-q^{3}={(p-q)}{(p^{2}+q^{2}}{+pq)}=0$