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Linear Equation Questions For SSC GD PDF Set – 2

SSC GD Constable Linear Equation Question paper with answers download PDF based on SSC GD exam previous papers. 40 Very important Linear Equation questions for GD Constable.

1500+ Must Solve Questions for SSC Exams (Question bank)

Question 1: Find the number of factors of 13x if (7-$\frac{4x}{3}$)($\frac{3}{2}$)=$\frac{x}{6}$.

a) 12

b) 4

c) 8

d) 6

Question 2: Find the value of $\ \sqrt{30+\sqrt{30}+…}$

a) 5

b) 3

c) 6

d) 7

Question 3: $2\frac{1}{5}x^{2}\$= 2750, find the value of x ?

a) 25

b) $25\sqrt{3}$

c) $25\sqrt{2}$

d) 20

Question 4: Sin A + Sin$^{2}\$A = 1, then the value of cos$^{2}\$A + cos$^{4}\$A is

a) 2

b) $\frac{2}{3}$

c) $1\frac{1}{2}$

d) 1

Question 5: If X = $\ \sqrt[3]{5}\$+ 2, then the value of $\ x^{3}-6x^{2}\$+ 12x – 13 is

a) -1

b) 1

c) 2

d) 0

Question 6: If 2x + $\ \frac{2}{x}\$= 3 then the value of $\ x^{3}+\frac{1}{x^{3}}$+ 2 is

a) -$\frac{9}{8}$

b) -$\frac{25}{8}$

c) $\frac{7}{8}$

d) 11

Question 7: If$\ a^{2}+b^{2}+c^{2}$= 2(a-b-c)-3, then the value of 4a – 3b + 5c is

a) 2

b) 3

c) 5

d) 6

Question 8: The value of$\ \frac{3\sqrt{2}}{(\sqrt{3}+\sqrt{6})}-\frac{4\sqrt{3}}{(\sqrt{6}+\sqrt{2})}+\frac{\sqrt{6}}{(\sqrt{2}+\sqrt{3})}\$is

a) $\sqrt{2}$

b) 0

c) $\sqrt{3}$

d) $\sqrt{6}$

Question 9: The value of $\sqrt{2\sqrt[3]{4}\sqrt{2\sqrt[3]{4}}\sqrt[4]{2\sqrt[3]{4}}…..}$ is

a) $2\sqrt[3]4$

b) $\sqrt{2\sqrt[3]4}$

c) $2\sqrt4$

d) $\sqrt[3]4$

Question 10: The value of cosec$^{2}\ 18^\circ\ – \frac{1}{cot^{2}72^\circ}\$is

a) $\frac{1}{\sqrt{3}}$

b) $\frac{\sqrt{2}}{3}$

c) $\frac{1}{2}$

d) 1

Question 11: If$\ x^{2}$- 3x + 1 = 0, then the value of $\ x^{5}+\frac{1}{x^{5}}\$is equal to

a) 87

b) 123

c) 135

d) 201

Question 12: The value of 0.65 x 0.65 + 0.35 x 0.35+0.70 x 0.65 is

a) 1.75

b) 1.00

c) 1.65

d) 1.55

Question 13: The value of $\frac{(75.8)^{2}-(35.8)^{2}}{40}$ is

a) 121.6

b) 40

c) 160

d) 111.6

Question 14: If $tan(\theta)tan(5\theta)=1$, then what is the value of $sin 2\theta$ ?

a) $0$

b) $\frac{1}{2}$

c) $1\sqrt2$

d) $\frac{\sqrt3}{2}$

Question 15: What is the simplified value of $\ \frac{2sin^{3}\ \theta\ – sin\theta}{cos\theta\ -\ 2\ cos^{3}\ \theta}$?

a) tan θ

b) sin θ

c) cos θ

d) cot θ

SSC

Question 16: If$\ \frac{x-xtan^{2}15^\circ}{1+tan^{2}15^\circ}$= sin $60^\circ\$+ cos 30$^\circ\$, then what is then what is the value of x?

a) 2

b) -1

c) -2

d) 1

Question 17: What is the simplified value of$\ \sqrt{\frac{sec^{2} \theta+cosec^{2} \theta}{4}}$?

a) cosec 2θ

b) sec 2θ

c) cosec θ sec θ

d) tan θ

Question 18: The side QR of ΔPQR is produced to S. If ∠PRS = 105° and ∠Q = (1/2)∠P, then what is the value of ∠P?

a) 45

b) 60

c) 70

d) 75

Question 19: In the given figure, O is the center of the circle, $\angle$CAO = 35$^\circ$. What is the value (in degrees) of $\ \angle$AOB?

a) 90

b) 110

c) 160

d) 130

Question 20: In ΔABC, ∠A : ∠B : ∠C = 3 : 3 : 4. A line parallel to BC is drawn which touches AB and AC at P and Q respectively. What is the value of ∠AQP – ∠APQ?

a) 12

b) 18

c) 24

d) 36

Question 21: If $x = 5 – \frac{1}{x}$, then what is the value of $x^{5} + \frac{1}{x^{5}}$?

a) 625

b) 3125

c) 2525

d) 2500

Question 22: If $x^{3} – y^{3} = 112$ and $x – y = 4$, then what is the value of $x^{2} + y^{2}$?

a) 16

b) 20

c) 24

d) 28

Question 23: If $3x – \frac{1}{3x} = 9$, then what is the value of $x^{2} + \frac{x^2}{81}$?

a) 7

b) 83/9

c) 11

d) 121/9

Question 24: If $x^{4}+\frac{1}{x^{4}}= 198$ and $x>0$, then what is the value of $x^{2}-\frac{1}{x^{2}}$?

a) $14$

b) $2\sqrt{7}$

c) $10\sqrt{2}$

d) $10$

Question 25: If $x + y = 4$, then what is the value of $x^{3} +y^{3} +12xy$?

a) 16

b) 32

c) 64

d) 256

Question 26: If $N=(\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}})$, then what is the value of $\frac{1}{N}$ ?

a) $6-\sqrt{35}$

b) $6+\sqrt{35}$

c) $7+\sqrt{35}$

d) $7-\sqrt{35}$

Question 27: If cosec θ + 3 sec θ = 5 cosec θ, then what is the value of cot θ?

a) 4/3

b) 3/4

c) 1/√3

d) √3

Question 28: What is the simplified value of $\ \frac{7}{sec^{2} \theta}+ \frac{3}{1+cot^{2} \theta}+ 4\ sin^{2} \theta$?

a) 3

b) 4

c) 5

d) 7

Question 29: If √5 tan θ = 5 sin θ, then what is the value of (sin$^{2}$ θ – cos$^{2}$θ)?

a) 3/5

b) 1/5

c) 4/5

d) 2/5

Question 30: If tan $\ \theta\$= $\ \frac{2}{3}\$, then what is the value of $\ \frac{15sin^{2} \theta-3cos^{2} \theta}{5sin^{2} \theta+3cos^{2} \theta}$?

a) $\frac{33}{32}$

b) $\frac{11}{29}$

c) $\frac{33}{47}$

d) $\frac{11}{32}$

Question 31: In an isosceles triangle PQR, ∠P = 130$^\circ$. If I is the in-centre of the triangle, then what is the value (in degrees) of ∠QIR?

a) 130

b) 120

c) 155

d) 165

Question 32: In the given figure, EF = CE = CA, What is the value (in degrees) of $\angle$EAC?

a) 58

b) 64

c) 72

d) 32

Question 33: In the given figure, O is the center of the circle, $\ \angle$PQR = 100$^\circ\$and $\angle$STR = 105$^\circ$. What is the value (in degrees) of $\ \angle$OSP?

a) 95

b) 45

c) 75

d) 65

Question 34: What is the value of $\ \frac{(a^{2}+b^{2})(a-b)-(a-b)^{2}}{a^{2}b-ab^{2}}$?

a) 0

b) 1

c) -1

d) 2

Question 35: If $\ x^{2}\ – 3x + 1 = 0$, then what is the value of $\ x^{4}\$+ $\ \frac{1}{x^{4}}$?

a) 11

b) 18

c) 47

d) 51

Question 36: If $\ \frac{3x-1}{x}+\frac{5y-1}{y}+\frac{7z-1}{z}$= 0, then what is the value of $\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\$?

a) -3

b) 0

c) 15

d) 21

Question 37: If $x^{2}- 2\sqrt{10}x+ 1 = 0$, then what is the value of $x – \frac{1}{x}$?

a) 4

b) 6

c) 3

d) 5

Question 38: If $x^{2} -7x + 1 = 0$, then what is the value of $x + \frac{1}{x}$?

a) 7

b) 3

c) 51

d) 47

Question 39: If the angles of a triangle are (2x – 8)$^{o}$, (2x + 18)$^{o}$and 6x$^{o}$. What is the value of 3x (in degrees)?

a) 17

b) 34

c) 51

d) 60

Question 40: What is the value of $999\frac{1}{3}+999\frac{1}{6}+999\frac{1}{12}+999\frac{1}{20}+999\frac{1}{30}$?

a) $999\frac{1}{6}$

b) $999\frac{5}{6}$

c) $4995\frac{1}{6}$

d) $4995\frac{4}{6}$

solving the equation for x,
We get $\frac{21}{2}$=$2x+\frac{x}{6}$

$\frac{21}{2}$=$\frac{13x}{6}$

$x$=$\frac{63}{13}$

$13x$=63
Factorising 63 we get 63=$3^{2}\times 7$
Number of factors=(2+1)(1+1)
=6

Let X=$\sqrt{30+\sqrt{30}+…}$

Above equation can be written as

X=$\Rightarrow\sqrt{30+X}$

Squaring on both sides

$X^{2}$=30+X

$X^{2}$-X-30=0

$X^{2}$-6X+5X-30=0

X(X-6)+5(X-6)=0

(X-6)(X+5)=0

X=-5,6

Taking positive value

X=6

$2\frac{1}{5}x^{2}=2750$

==> $\frac{11}{5}x^{2}=2750$

==> $x^{2}= \frac{2750\times5}{11}$

==> $x^{2}= 1250$

==> $x = \sqrt{1250}=\sqrt{625\times2}$

==> $x = 25\sqrt{2}$

Given sin A+$sin^{2}$ A=1

==> sin A = 1-$sin^{2}$ A

==> sin A = $cos^{2}$ A ($\because cos^{2}A+sin^{2}A$=1)

$cos^{2}$ A=sin A ==> $cos^{4}A$=$sin^{2}A$

$\therefore cos^{2}A+cos^{4}A=1 ( \because sin A+sin^{2}A=1)$

Given x=\sqrt[3]{5}+2

$\ x^{3}-6x^{2}\$+ 12x – 13

= $(\sqrt[3]{5}+2)^{3}-6(\sqrt[3]{5}+2)^{2}+12(\sqrt[3]{5}+2)-13$

= $(5+8+6\times5^{\frac{2}{3}}+12\times5^{\frac{1}{3}})-6[5^{\frac{1}{3}}+4+4\times5^{\frac{1}{3}}]+12(5^{\frac{1}{3}}+2)-13$

= $13+6\times5^{\frac{2}{3}}+12\times5^{\frac{1}{3}}-6\times5^{\frac{2}{3}}-24-24\times5^{\frac{1}{3}}+12\times5^{\frac{1}{3}}+24-13$

= 0

Given 2x+$\frac{1}{x} =$ 3

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

$\Rightarrow$ x+$\frac{1}{x} = \frac{3}{2}$

Cubing on both sides

(x+$\frac{1}{x})^{3} = \frac{27}{8}$

x$^{3}$+$\frac{1}{x^{3}}$+3$\times$x$\times$ $\frac{1}{x}$(x+$\frac{1}{x}$) $= \frac{27}{8}$

$\Rightarrow x^{3}+\frac{1}{x^{3}}+3(\frac{3}{2}) = \frac{27}{8}$

$\Rightarrow x^{3}+\frac{1}{x^{3}} = \frac{27}{8}-\frac{9}{2} = \frac{-9}{8}$

$x^{3}+\frac{1}{x^{3}}+2 = \frac{-9}{8}+2 = \frac{7}{8}$

$\ a^{2}+b^{2}+c^{2}$= 2(a-b-c)-3
We can write the above equation as

$a^{2}-2a+1+b^{2}+2b+1+c^{2}+2c+1$=0
$\Rightarrow (a-1)^{2}+(b+1)^{2}+(c+1)^{2}$=0

$(a-1)^{2}$=0$\Rightarrow$a=1
$(b+1)^{2}$=0$\Rightarrow$ b=-1
$(c+1)^{2}$=0$\Rightarrow$ c=-1

4a-3b+5c= 4(1)-3(-1)+5(-1)=4+3-5=2

$\frac{3\sqrt{2}}{(\sqrt{3}+\sqrt{6})}-\frac{4\sqrt{3}}{(\sqrt{6}+\sqrt{2})}+\frac{\sqrt{6}}{(\sqrt{2}+\sqrt{3})}$

= $\frac{6\sqrt{6}+18+6\sqrt{2}+6\sqrt{3}-(12\sqrt{2}+24+12\sqrt{3}+12\sqrt{6})+6\sqrt{3}+6+6\sqrt{6}+6\sqrt{2}}{(\sqrt{3}+\sqrt{6})(\sqrt{6}+\sqrt{2})(\sqrt{2}+\sqrt{3})}$

= $\frac{6\sqrt{6}+18+6\sqrt{2}+6\sqrt{3}-12\sqrt{2}-24-12\sqrt{3}-12\sqrt{6}+6\sqrt{3}+6+6\sqrt{6}+6\sqrt{2}}{(\sqrt{3}+\sqrt{6})(\sqrt{6}+\sqrt{2})(\sqrt{2}+\sqrt{3})}$

=0

To find : $y=\sqrt{2\sqrt[3]{4}\sqrt{2\sqrt[3]{4}}\sqrt[4]{2\sqrt[3]{4}}…..}$

Let $2\sqrt[3]4=x$

=> $y=\sqrt{(x)\times(\sqrt{x})\times(\sqrt[4]{x})\times…….}$

=> $y^2=(x)^{[1+\frac{1}{2}+\frac{1}{4}+……+\infty]}$

Now, sum of infinite G.P. = $\frac{a}{(1-r)}$, where first term = $a=1$ and common ratio = $r=\frac{1}{2}$

=> $y^2=(x)^{\frac{1}{1-\frac{1}{2}}}$

=> $y^2=(x)^2$

=> $y=x$

$\therefore$ $\sqrt{2\sqrt[3]{4}\sqrt{2\sqrt[3]{4}}\sqrt[4]{2\sqrt[3]{4}}…..}=2\sqrt[3]4$

=> Ans – (A)

cosec$^{2}\ 18^\circ\ – \frac{1}{cot^{2}72^\circ}\$

= cosec$^{2} 18^\circ – tan^{2} 72^\circ$ ($\because \frac{1}{cot^{2} \ominus}$=$tan^{2}\ominus$)

= cosec$^{2} 18^\circ$ – $tan^{2} (90-72)^\circ$
= cosec$^{2} 18^\circ$ – $sec^{2} 18^\circ$ ($\because sec^{2}\ominus= tan^{2}(90^\circ-\ominus)$)
cosec$^{2} 18^\circ$ – $sec^{2} 18^\circ$=1($\because cosec^{2} \ominus$ – $sec^{2} \ominus$=1)

$x^{2}-3x+1$=0

Taking ‘x’ common

x(x-3+$\frac{1}{x})$=0

$\Rightarrow x+\frac{1}{x}$=$3\rightarrow(1)$

Squaring on both sides

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

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

Cubing equation(1) on both sides

$x^{3}+\frac{1}{x^{3}}+3\times x\times\frac{1}{x}(x+\frac{1}{x})$=27

$x^{3}+\frac{1}{x^{3}}$+$3\times 1\times3$=27($\because x+\frac{1}{x}$=3)

$x^{3}+\frac{1}{x^{3}}$=27-9=$18\rightarrow(3)$

Squaring equation(2) on both sides

$x^{4}+\frac{1}{x^{4}}+2\times x^{2}\times\frac{1}{x^{2}}$=49

$x^{4}+\frac{1}{x^{4}}$=$47\rightarrow(4)$

Multiplying equation(1) and equation(4)

$(x^{4}+\frac{1}{x^{4}})(x+\frac{1}{x}$)=$47\times3$

$x^{5}+\frac{1}{x^{5}}+x^{3}+\frac{1}{x^{3}}$=$47\times3$=141

$x^{5}+\frac{1}{x^{5}}+18$=141($\because x^{3}+\frac{1}{x^{3}}$)

$\therefore x^{5}+\frac{1}{x^{5}}$=123

Expression : $(0.65\times0.65)+(0.35\times0.35)+(0.70\times0.65)$

= $(0.65)^2+(0.35)^2+2(0.35)(0.65)$

Comparing with : $(x)^2+(y)^2+2(x)(y)=(x+y)^2$

= $(0.65+0.35)^2=(1)^2=1$

=> Ans – (B)

$\frac{75.8^{2}-35.8^{2}}{40}$=$\frac{(75.8+35.8)(75.8-35.8)}{40}$

=$\frac{111.6\times40}{40}$=111.6

Given : $tan(\theta)tan(5\theta)=1$

Using, $tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}$

$tan(\theta+5\theta)=\frac{tan(\theta)+tan(5\theta)}{1-tan(\theta)tan(5\theta)}$

=> $tan(6\theta)=\frac{tan(\theta)+tan(5\theta)}{1-1}$

=> $tan(6\theta)=\infty$

=> $tan(6\theta)=tan(90^\circ)$

=> $6\theta=90^\circ$

=> $\theta=\frac{90^\circ}{6}=15^\circ$

$\therefore$ $sin(2\theta)=sin(2\times15^\circ)$

= $sin(30^\circ)=\frac{1}{2}$

=> Ans – (B)

Expression : $\ \frac{2sin^{3}\ \theta\ – sin\theta}{cos\theta\ -\ 2\ cos^{3}\ \theta}$

= $\ \frac{sin\ \theta(2sin^{2}\ \theta\ – 1)}{cos\ \theta(2 -\ 2\ cos^{2}\ \theta)}$

= $\frac{sin\ \theta(cos2\ \theta)}{cos\ \theta(cos2\ \theta)}$

= $\frac{sin\ \theta}{cos\ \theta}=tan\ \theta$

=> Ans – (A)

$tan15^\circ=\frac{\sqrt3-1}{\sqrt3+1}$

Expression = $\ \frac{x-xtan^{2}15^\circ}{1+tan^{2}15^\circ}$= sin $60^\circ\$+ cos 30$^\circ\$

=> $\ \frac{x(1-tan^{2}15^\circ)}{1+tan^{2}15^\circ}= \frac{\sqrt3}{2}+\frac{\sqrt3}{2}$

=> $x\times\frac{1-(\frac{\sqrt3-1}{\sqrt3+1})^2}{1+(\frac{\sqrt3-1}{\sqrt3+1})^2}=\sqrt3$

=> $x\times\frac{(\sqrt3+1)^2-(\sqrt3-1)^2}{(\sqrt3+1)^2+(\sqrt3-1)^2}=\sqrt3$

=> $x\times\frac{(3+1+2\sqrt3)-(3+1-2\sqrt3)}{(3+1+2\sqrt3)+(3+1-2\sqrt3)}=\sqrt3$

=> $x\times\frac{4\sqrt3}{8}=\sqrt3$

=> $x=\frac{8}{4}=2$

=> Ans – (A)

Expression = $\ \sqrt{\frac{sec^{2} \theta+cosec^{2} \theta}{4}}$

= $\ \sqrt{\frac{(\frac{1}{cos^2\ \theta})+(\frac{1}{sin^2\ \theta})}{4}}$

= $\ \sqrt{\frac{(\frac{sin^2\ \theta+cos^2\ \theta)}{sin^2\ \theta.cos^2\ \theta}}{4}}$

= $\sqrt{\frac{1}{4sin^2\ \theta cos^2\ \theta}}$

= $\sqrt{(\frac{1}{2sin\ \theta cos\ \theta})^2}$

= $\frac{1}{sin2\theta}=cosec2\theta$

=> Ans – (A)

Let $\angle Q=x$, => $\angle P=2x$

Using exterior angle property in $\triangle$ PQR,

=> $\angle$ P + $\angle$ Q = $\angle$ PRS

=> $2x+x=105^\circ$

=> $x=\frac{105^\circ}{3}=35^\circ$

$\therefore$ $\angle P=2\times35^\circ=70^\circ$

=> Ans – (C)

Given : ∠A : ∠B : ∠C = 3 : 3 : 4 and PQ is parallel to BC

To find : ∠AQP – ∠APQ = ?

Solution : Let $\angle A=3x$, $\angle B=3x$ and $\angle C=4x$

Thus, in $\triangle$ ABC,

=> $\angle A+\angle B+\angle C=180^\circ$

=> $3x+3x+4x=180^\circ$

=> $x=\frac{180^\circ}{10}=18^\circ$

$\because$ PQ $\parallel$ BC, => $\angle$ APQ = $\angle$ B and $\angle$ AQP = $\angle$ C (Corresponding angles)

$\therefore$ $\angle$ AQP – $\angle$ APQ = $4x-3x=x=18^\circ$

=> Ans – (B)

Given : $x=5-\frac{1}{x}$

=> $x+\frac{1}{x}=5=k$

Now, $x^5+\frac{1}{x^5}=[(x^3+\frac{1}{x^3})\times(x^2+\frac{1}{x^2})]-(x+\frac{1}{x})$

= $[(x+\frac{1}{x})^3-3(x+\frac{1}{x})\times(x+\frac{1}{x})^2-2(x)(\frac{1}{x})]-(x+\frac{1}{x})$

= $[(k^3-3k)\times(k^2-2)]-(k)$

= $[(125-15)\times(25-2)]-(5)$

= $(110\times23)-5$

= $2530-5=2525$

=> Ans – (C)

Given : $x^3-y^3=112$ ————–(i)

Also, $x-y=4$ ————-(ii)

Cubing both sides, we get :

=> $(x-y)3=(4)^3$

=> $(x^3-y^3)-3(x)(y)(x-y)=64$

Substituting values from equations (i) and (ii),

=> $112-3xy(4)=64$

=> $12xy=112-64=48$

=> $xy=\frac{48}{12}=4$ ———–(iii)

Now, squaring equation (ii), we get :

=> $x^2+y^2-2xy=16$

=> $x^2+y^2=16+8=24$

=> Ans – (C)

Given : $3x-\frac{1}{3x}=9$

Dividing both sides by 3, => $x-\frac{1}{9x}=3$

Squaring both sides, we get :

=> $(x-\frac{1}{9x})^2=(3)^2$

=> $x^2+\frac{1}{81x^2}-2(x)(\frac{1}{9x})=9$

=> $(x^2+\frac{1}{81x^2})-\frac{2}{9}=9$

=> $(x^2+\frac{1}{81x^2})=9+\frac{2}{9}$

=> $(x^2+\frac{1}{81x^2})=\frac{83}{9}$

=> Ans – (B)

Given : $\ x^{4}+\frac{1}{x^{4}}\ =198$

=> $(x^2-\frac{1}{x^2})^2+2(x^2)(\frac{1}{x^2})=198$

=> $(x^2-\frac{1}{x^2})^2=198-2=196$

=> $x^2-\frac{1}{x^2}=\sqrt{196}=14$

=> Ans – (A)

Given : $x+y=4$ ———–(i)

Cubing both sides, we get :

=> $(x+y)^3=(4)^3$

=> $x^3+y^3+3xy(x+y)=64$

=> $x^3+y^3+3xy(4)=64$

=> $x^3+y^3+12xy=64$

=> Ans – (C)

Given : $N=\frac{\sqrt7-\sqrt5}{\sqrt7+\sqrt5}$

=> $\frac{1}{N}=\frac{\sqrt7+\sqrt5}{\sqrt7-\sqrt5}$

Rationalizing the denominator, we get :

= $\frac{\sqrt7+\sqrt5}{\sqrt7-\sqrt5}\times\frac{\sqrt7+\sqrt5}{\sqrt7+\sqrt5}$

= $\frac{(\sqrt7+\sqrt5)^2}{(\sqrt7-\sqrt5)(\sqrt7+\sqrt5)}$

= $\frac{7+5+2(\sqrt7)(\sqrt5)}{7-5}$

= $\frac{12+2\sqrt{35}}{2}=6+\sqrt{35}$

=> Ans – (B)

Given : $cosec\theta+3sec\theta=5cosec\theta$

=> $3sec\theta=5cosec\theta-cosec\theta$

=> $3sec\theta=4cosec\theta$

=> $\frac{3}{cos\theta}=\frac{4}{sin\theta}$

=> $\frac{cos\theta}{sin\theta}=\frac{3}{4}$

=> $cot\theta=\frac{3}{4}$

=> Ans – (B)

Expression : $\ \frac{7}{sec^{2} \theta}+ \frac{3}{1+cot^{2} \theta}+ 4\ sin^{2} \theta$

= $7cos^2\ \theta+\frac{3}{cosec^2\ \theta}+4sin^2\ \theta$

= $7cos^2\ \theta+3sin^2\ \theta+4sin^2\ \theta$

= $7cos^2\ \theta+7sin^2\ \theta$

= $7(cos^2\ \theta+sin^2\ \theta)$

= $7\times1=7$

=> Ans – (D)

Given : $\sqrt5tan\ \theta=5sin\ \theta$

=> $\frac{sin\ \theta}{cos\ \theta}=\sqrt5sin\ \theta$

=> $cos\ \theta=\frac{1}{\sqrt5}$

=> $cos^2\ \theta=\frac{1}{5}$ —————(i)

Now, $sin^2\ \theta=1-cos^2\ \theta$

=> $sin^2\ \theta=1-\frac{1}{5}=\frac{4}{5}$ ————–(ii)

Subtracting equation (i) from (ii), we get :

$\therefore$ $(sin^2\ \theta-cos^2\ \theta)=\frac{4}{5}-\frac{1}{5}=\frac{3}{5}$

=> Ans – (A)

Given : $tan\ \theta=\frac{2}{3}$

=> $\frac{sin\ \theta}{cos\ \theta}=\frac{2}{3}$

Let $sin\ \theta=2$ and $cos\ \theta=3$

To find : $\ \frac{15sin^{2} \theta-3cos^{2} \theta}{5sin^{2} \theta+3cos^{2} \theta}$

= $\frac{15(2)^2-3(3)^2}{5(2)^2+3(3)^2}$

= $\frac{60-27}{20+27}$

= $\frac{33}{47}$

=> Ans – (C)

Given : I is the incentre of $\triangle$ PQR and $\angle$ BAC = 130°

To find : $\angle$ QIR = $\theta$ = ?

Incentre of a triangle = $90^\circ+\frac{\angle P}{2}$

=> $\theta=90^\circ+\frac{130^\circ}{2}$

=> $\theta=90^\circ+65^\circ$

=> $\theta=155^\circ$

=> Ans – (C)

Given : EF = CE = CA

=> $\angle$ CAE = $\angle$ CEA = $x$ and $\angle$ ECF = $\angle$ EFC = $y$

To find : $\angle$ EAC = $x=?$

Solution : Using exterior angle property, => $\angle$ CAE + $\angle$ CFE = $\angle$ ACD

=> $x+y=96^\circ$ —————–(i)

Also, $\angle$ CEF = $(180^\circ-2y)=180^\circ-x$

=> $x=2y$ ————(ii)

Substituting it in equation (i), => $2y+y=3y=96^\circ$

=> $y=\frac{96}{3}=32^\circ$

$\therefore$ $x=2\times32=64^\circ$

=> Ans – (B)

Given : $\ \angle$PQR = 100$^\circ\$and $\angle$STR = 105$^\circ$

To find : $\ \angle$OSP = ?

Solution : Quadrilateral PQRS is cyclic quadrilateral, hence opposite angles are supplementary.

=> $\angle$ PQR + $\angle$ PSR = $180^\circ$

=> $\angle$ PSR = $180^\circ-100^\circ=80^\circ$ ————–(i)

Also, angle at the centre is double the angle at any point on the circumference of the circle in the same segment.

=> reflex ($\angle$ SOR) = $2$ $\times$ $\angle$ STR

=> reflex ($\angle$ SOR) = $2\times105^\circ=210^\circ$

Thus, $\angle$ SOR = $360^\circ-210^\circ=150^\circ$

Now, in $\triangle$ SOR, OS = OR = radius

=> $\angle$ OSR = $\angle$ ORS = $15^\circ$ ———-(ii)

Subtracting equation (ii) from (i), we get :

$\therefore$ $\angle$ OSP = $80^\circ-15^\circ=65^\circ$

=> Ans – (D)

Expression : $\ \frac{(a^{2}+b^{2})(a-b)-(a-b)^{2}}{a^{2}b-ab^{2}}$

= $\ \frac{(a-b)[(a^{2}+b^{2})-(a-b)]}{(ab)(a-b)}$

= $\frac{(a^2+b^2)-(a-b)}{ab}$

= $\frac{(a-b)^2+2ab-(a-b)}{ab}$

Given : $x^2-3x+1=0$

Dividing both sides by $’x’$

=> $x+\frac{1}{x}=3$

Squaring both sides, we get :

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

=> $x^2+\frac{1}{x^2}=9-2=7$

Again squaring both sides,

=> $x^4+\frac{1}{x^4}+2(x^2)(\frac{1}{x^2})=49$

=> $x^4+\frac{1}{x^4}=49-2=47$

=> Ans – (C)

Given : $\ \frac{3x-1}{x}+\frac{5y-1}{y}+\frac{7z-1}{z}=0$

=> $(3-\frac{1}{x})+(5-\frac{1}{y})+(7-\frac{1}{z})=0$

=> $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3+5+7$

=> $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=15$

=> Ans – (C)

Given : $x^2-2\sqrt{10}x+1=0$

Dividing both sides by $’x’$

=> $x+\frac{1}{x}=2\sqrt{10}$

Squaring both sides, we get :

=> $x^2+\frac{1}{x^2}+2(x)(\frac{1}{x})=40$

=> $x^2+\frac{1}{x^2}=40-2=38$

=> $(x-\frac{1}{x})^2+2(x)(\frac{1}{x})=38$

=> $(x-\frac{1}{x})^2=38-2=36$

=> $x-\frac{1}{x}=\sqrt{36}=6$

=> Ans – (B)

Given : $x^2-7x+1=0$

Dividing both sides by $’x’$

=> $x-7+\frac{1}{x}=0$

=> $x+\frac{1}{x}=7$

=> Ans – (A)

Sum of angles of a triangle = $180^\circ$

=> $(2x-8)^\circ+(2x+18)^\circ+(6x)^\circ=180^\circ$

=> $10x+10=180$

=> $10x=180-10=170$

=> $x=\frac{170}{10}=17$

$\therefore$ $3x=3\times17=51$

=> Ans – (C)

Expression : $999\frac{1}{3}+999\frac{1}{6}+999\frac{1}{12}+999\frac{1}{20}+999\frac{1}{30}$

= $(999+999+999+999+999)+(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30})$

= $(4995)+(\frac{20+10+5+3+2}{60})$

= $4995+\frac{40}{60}$

= $4995\frac{4}{6}$

=> Ans – (D)

We hope this Linear Equation questions for SSC GD will be highly useful for your preparation.