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.
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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
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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
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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
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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
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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) $4995\frac{4}{6}$
Answers & Solutions:
1) Answer (D)
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
2) Answer (C)
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
3) Answer (C)
$2\frac{1}{5}x^{2}=2750$
==> $\frac{11}{5}x^{2}=2750$
==> $x^{2}= \frac{2750\times5}{11}$
==> $x^{2}= 1250$
==> $x = \sqrt{1250}=\sqrt{625\times2}$
==> $x = 25\sqrt{2}$
4) Answer (D)
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 $
$\therefore cos^{2}A+cos^{4}A=1 ( \because sin A+sin^{2}A=1)$
5) Answer (D)
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
6) Answer (C)
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+$\frac{1}{x}$) $= \frac{27}{8}$
$\Rightarrow x^{3}+\frac{1}{x^{3}}+3(\frac{3}{2}) = \frac{27}{8}$
$\Rightarrow x^{3}+\frac{1}{x^{3}} = \frac{27}{8}-\frac{9}{2} = \frac{-9}{8}$
$x^{3}+\frac{1}{x^{3}}+2 = \frac{-9}{8}+2 = \frac{7}{8}$
7) Answer (A)
$\ a^{2}+b^{2}+c^{2}$= 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
8) Answer (B)
$\frac{3\sqrt{2}}{(\sqrt{3}+\sqrt{6})}-\frac{4\sqrt{3}}{(\sqrt{6}+\sqrt{2})}+\frac{\sqrt{6}}{(\sqrt{2}+\sqrt{3})}$
= $\frac{6\sqrt{6}+18+6\sqrt{2}+6\sqrt{3}-(12\sqrt{2}+24+12\sqrt{3}+12\sqrt{6})+6\sqrt{3}+6+6\sqrt{6}+6\sqrt{2}}{(\sqrt{3}+\sqrt{6})(\sqrt{6}+\sqrt{2})(\sqrt{2}+\sqrt{3})}$
= $\frac{6\sqrt{6}+18+6\sqrt{2}+6\sqrt{3}-12\sqrt{2}-24-12\sqrt{3}-12\sqrt{6}+6\sqrt{3}+6+6\sqrt{6}+6\sqrt{2}}{(\sqrt{3}+\sqrt{6})(\sqrt{6}+\sqrt{2})(\sqrt{2}+\sqrt{3})}$
=0
9) Answer (A)
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)
10) Answer (D)
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)
11) Answer (B)
$x^{2}-3x+1$=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
12) Answer (B)
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)
13) Answer (D)
$\frac{75.8^{2}-35.8^{2}}{40}$=$\frac{(75.8+35.8)(75.8-35.8)}{40}$
=$\frac{111.6\times40}{40}$=111.6
14) Answer (B)
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)
15) Answer (A)
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)
16) Answer (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)
17) Answer (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)
18) Answer (C)
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)
19) Answer (C)
20) Answer (B)
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)
21) Answer (C)
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)
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22) Answer (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)
23) Answer (B)
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)
24) Answer (A)
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)
25) Answer (C)
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)
26) Answer (B)
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)
27) Answer (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)
28) Answer (D)
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)
29) Answer (A)
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)
30) Answer (C)
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)
31) Answer (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)
32) Answer (B)
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)
33) Answer (D)
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)
34) Answer (D)
Expression : $\ \frac{(a^{2}+b^{2})(a-b)-(a-b)^{2}}{a^{2}b-ab^{2}}$
= $\ \frac{(a-b)[(a^{2}+b^{2})-(a-b)]}{(ab)(a-b)}$
= $\frac{(a^2+b^2)-(a-b)}{ab}$
= $\frac{(a-b)^2+2ab-(a-b)}{ab}$
35) Answer (C)
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)
36) Answer (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)
37) Answer (B)
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)
38) Answer (A)
Given : $x^2-7x+1=0$
Dividing both sides by $’x’$
=> $x-7+\frac{1}{x}=0$
=> $x+\frac{1}{x}=7$
=> Ans – (A)
39) Answer (C)
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)
40) Answer (D)
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)
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