SSC CGL Expected Quant Questions 2019

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Question 1: Find the value of $\frac{(\cot \theta -cosec\theta +1)(\tan \theta +\sec \theta +1)}{\cos \theta cosec\theta }?$

a) $2\cos \theta$

b) 2

c) $2cot\theta$

d) $2tan\theta$

Question 2: What is the value of $sec{12}^{0}sin{12}^{0}tan{38}^{0}tan{78}^{0}tan{52}^0$?

a) 1

b) 3

c) $\frac{1}{2}$

d) $\frac{3}{2}$

Question 3: What is the value of $\frac{\cos x+cosy}{sinx+siny}$ ?

a) $\tan \frac{x+y}{2}$

b) $\tan \frac{x-y}{2}$

c) $\cot \frac{x-y}{2}$

d) $\cot \frac{x+y}{2}$

Question 4: If sin $x=\frac{1}{2}$ and cos $y=\frac{1}{2}$, what is the value of sin(x+y)?

a) $\frac{2}{3}$

b) $\frac{4}{9}$

c) $\frac{5}{9}$

d) 1

Question 5: What is the value of $\frac{({\tan }^{2}x-{\sin }^{2}x)}{{\sec }^{2}x}$ ?

a) ${\sin }^{4}x$

b) ${\cos }^{2}x$

c) ${\sin }^{2}x$

d) $co{s}^{4}x$

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Question 6: In the given figure, what is the value of $\mathrm{\angle }1+\mathrm{\angle }2+\mathrm{\angle }3+\mathrm{\angle }4+\mathrm{\angle }5+\mathrm{\angle }6+\mathrm{\angle }7+\mathrm{\angle }8+\mathrm{\angle }9+\mathrm{\angle }10$?

 

a) 600

b) 720

c) 900

d) 1080

Question 7: AB is a tangent to a circle with centre O. If the radius at the circle is 7 cm and the length of AB is 24 cm, the what is the length (in cm.) of OA ?

a) 25

b) 26

c) 28

d) 31

Question 8: In $\mathrm{△}$PQR, S and T are the mid points of sides PQ and PR respectively. If $\mathrm{\angle }$QPR = ${45}^0$ and $\mathrm{\angle }$PRQ = ${55}^0$, then what is the value (in degrees) of $\mathrm{\angle }$QST ?

a) 80

b) 85

c) 90

d) 100

Question 9: A,B and C are the three points on a circle such that $\mathrm{\angle }$ABC = ${35}^0$ and $\mathrm{\angle }$BAC = ${85}^0$. What is the angle (in degrees) subtended by arc AB at the center of the circle?

a) 60

b) 90

c) 135

d) 120

Question 10: A circle passing through points Q and R of triangle PQR, cuts the sides PQ and PR at points X and Y respectively. If PQ = PR, then what is the value (in degrees) of $\mathrm{\angle }$PRQ + $\mathrm{\angle }$QXY ?

a) 120

b) 150

c) 240

d) 180

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Question 11: If $x^2-9x-1=0$, then what is the value of $(x^2-\frac{1}{x^2}+5x-\frac{5}{x})$ ?

a) 115

b) 128

c) 124

d) 133

Question 12: If  $(x-\frac{1}{x})=3$, then what is the value of $(x^3-\frac{1}{x^3})$ ?

a) 36

b) 21

c) 9

d) 27

Question 13: If $(3x^2-9x+3)=0$, then what is the value of $(x+\frac{1}{x}{)}^3$ ?

a) 9

b) 729

c) 81

d) 27

Question 14: If $a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=6$, then what is the value of $(a^2+b^2+c^2)$ ?

a) 3

b) 6

c) -3

d) 2

Question 15: If $x=17-4\sqrt{18}$, then find the value of $(\sqrt{x}+\frac{1}{\sqrt{x}})$  ?

a) $7\sqrt{2}$

b) 9

c) 22

d) 6

Question 16: If the ratio of volume of two cubes is 11 : 13, then what is the ratio of the sides of the two cubes ?

a) 11 : 13

b) 121 : 169

c) $(11{)}^{\frac{1}{2}}:(13{)}^{\frac{1}{2}}$

d) $(11{)}^{\frac{1}{3}}:(13{)}^{\frac{1}{3}}$

Question 17: If the diameter of a sphere is 14 cm., then what is the curved surface area (in $cm{.}^2$) of the sphere?

a) 616

b) 1232

c) 2464

d) 576

Question 18: The difference between circumference and the radius of a circle is 111 cm. What is the area (in $c{m}^2$) of the circle?

a) 469

b) 1386

c) 912

d) 1086

Question 19: Three circles of radius 9 cm are kept touching each other. The string is tightly tied around the three circles. What is the length (in cm.) of the string ?

a) $48+18\pi$

b) $48+24\pi$

c) $54+18\pi$

d) $54+24\pi$

Question 20: The perimetre of a rhombus in 20 cm and one of the diagonal is 8 cm. What is the area (in $c{m}^2$) of the rhombus?

a) 12

b) 24

c) 48

d) 96

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Answers & Solutions:

1) Answer (D)

Expression : $\frac{(\cot \theta -cosec\theta +1)(\tan \theta +\sec \theta +1)}{\cos \theta cosec\theta }$

= $\frac{sin\theta }{cos\theta }\times [(\frac{cos\theta }{sin\theta }-\frac{1}{sin\theta }+1)\times (\frac{sin\theta }{cos\theta }+\frac{1}{cos\theta }+1)]$

= $\frac{sin\theta }{cos\theta }\times [(\frac{cos\theta +sin\theta -1}{sin\theta })\times (\frac{cos\theta +sin\theta +1}{cos\theta })]$

Using, $(x-y)(x+y)=x^2-y^2$, where $x=cos\theta +sin\theta$ and $y=1$

= $\frac{1}{co{s}^2\theta }\times [(cos\theta +sin\theta {)}^2-(1{)}^2]$

= $\frac{1}{co{s}^2\theta }\times [co{s}^2\theta +si{n}^2\theta +2cos\theta .sin\theta -1]$

= $\frac{1}{co{s}^2\theta }\times [1+2cos\theta .sin\theta -1]$

= $\frac{1}{co{s}^2\theta }\times (2cos\theta .sin\theta )$

= $\frac{2sin\theta }{cos\theta }=2tan\theta$

=> Ans – (D)

2) Answer (A)

Expression : $sec{12}^{0}sin{12}^{0}tan{38}^{0}tan{78}^{0}tan{52}^0$

= $\frac{1}{cos({12}^{\circ })}.sin({12}^{\circ }).tan({38}^{\circ }).tan({78}^{\circ }).tan({52}^{\circ })$

= $[tan({12}^{\circ }).tan({78}^{\circ })].[tan({38}^{\circ }).tan({52}^{\circ })]$

Using, $tan({90}^{\circ }-\theta )=cot(\theta )$

= $[tan({12}^{\circ }).cot({12}^{\circ })].[tan({38}^{\circ }).cot({38}^{\circ })]$

Also, $tan(\theta )cot(\theta )=1$

= $1\times 1=1$

=> Ans – (A)

3) Answer (D)

Expression : $\frac{\cos x+cosy}{sinx+siny}$

= $\frac{cos(\frac{x+y}{2})cos(\frac{x-y}{2})}{sin(\frac{x+y}{2})cos(\frac{x-y}{2})}$

= $\frac{cos(\frac{x+y}{2})}{sin(\frac{x+y}{2})}$

= $cot(\frac{x+y}{2})$

=> Ans – (D)

4) Answer (D)

Given : $sinx=\frac{1}{2}$ and $cosy=\frac{1}{2}$

=> $sin(x)=sin({30}^{\circ })$

=> $x={30}^{\circ }$

Similarly, => $cos(y)=cos({60}^{\circ })$

=> $y={60}^{\circ }$

$\therefore$ $sin(x+y)$

= $sin({30}^{\circ }+{60}^{\circ })=sin({90}^{\circ })=1$

=> Ans – (D)

5) Answer (A)

Expression : $\frac{({\tan }^{2}x-{\sin }^{2}x)}{{\sec }^{2}x}$

= $\frac{(\frac{si{n}^{2}x}{co{s}^{2}x}-si{n}^{2}x)}{se{c}^{2}x}$

= $\frac{si{n}^{2}x}{se{c}^{2}x}(\frac{1}{co{s}^{2}x}-1)$

= $si{n}^{2}xco{s}^{2}x(\frac{1-co{s}^{2}x}{co{s}^{2}x})$

= $si{n}^{2}x\times (si{n}^{2}x)=si{n}^{4}x$

=> Ans – (A)

6) Answer (B)

7) Answer (A)

Given : OB is radius of circle = 7 cm and tangent AB = 24 cm

To find : OA = ?

Solution : The radius of a circle intersects the tangent at right angle, => $\mathrm{\angle }OBA={90}^{\circ }$

Thus in $\mathrm{△}$ OAB,

=> $(OA{)}^2=(OB{)}^2+(AB{)}^2$

=> $(OA{)}^2=(7{)}^2+(24{)}^2$

=> $(OA{)}^2=49+576=625$

=> $OA=\sqrt{625}=25$ cm

=> Ans – (A)

8) Answer (D)

Given : $\mathrm{\angle }$QPR = ${45}^{\circ }$ and $\mathrm{\angle }$PRQ = ${55}^{\circ }$

To find : $\mathrm{\angle }$QST = ?

Solution : In triangle, PQR

=> $\mathrm{\angle }P+\mathrm{\angle }Q+\mathrm{\angle }R={180}^{\circ }$

=> ${45}^{\circ }+{55}^{\circ }+\mathrm{\angle }Q={180}^{\circ }$

=> $\mathrm{\angle }Q={180}^{\circ }-{100}^{\circ }={80}^{\circ }$

Now, since ST divides PQ and PR equally, thus ST is parallel to QR.

$\therefore$ Angles on the same side of transversal are supplementary, => $\mathrm{\angle }PQR+\mathrm{\angle }QST={180}^{\circ }$

=> $\mathrm{\angle }QST={180}^{\circ }-{80}^{\circ }={100}^{\circ }$

=> Ans – (D)

9) Answer (D)

Given : $\mathrm{\angle }$ABC = ${35}^{\circ }$ and $\mathrm{\angle }$BAC = ${85}^{\circ }$

To find : $\mathrm{\angle }$ AOB = ?

Solution : In triangle, ABC

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

=> ${85}^{\circ }+{35}^{\circ }+\mathrm{\angle }C={180}^{\circ }$

=> $\mathrm{\angle }C={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

Now, angle subtended by an arc at the centre is double the angle subtended by it at any point on the circle.

=> $\mathrm{\angle }AOB=2\times \mathrm{\angle }ACB$

= $2\times {60}^{\circ }={120}^{\circ }$

=> Ans – (D)

10) Answer (D)

Given : PQR is an isosceles triangle, PQ = PR

To find : $\mathrm{\angle }$PRQ + $\mathrm{\angle }$QXY = ?

Solution : Since, $\mathrm{△}$ PQR is isosceles, we have $\mathrm{\angle }Q=\mathrm{\angle }R$

Now, XY is parallel to QR, and sum of angles on the same side of transversal is supplementary, => $\mathrm{\angle }PQR+\mathrm{\angle }QXY={180}^{\circ }$

=> $\mathrm{\angle }$PRQ + $\mathrm{\angle }$QXY = ${180}^{\circ }$

II method : XYRQ is a cyclic quadrilateral and opposite angles in a cyclic quadrilateral are supplementary.

=> Ans – (D)

11) Answer (B)

Given : $x^2-9x-1=0$

=> $x^2-1=9x$

Dividing both sides by ${}^{\prime }{x}^{\prime }$,

=> $x-\frac{1}{x}=9$ ————–(i)

Squaring both sides, we get :

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

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

=> $x^2-\frac{1}{x^2}=81+2=83$ ———–(ii)

$\therefore$ $(x^2-\frac{1}{x^2}+5x-\frac{5}{x})$

= $(x^2-\frac{1}{x^2})+5(x-\frac{1}{x})$

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

= $83+5(9)=128$

=> Ans – (B)

12) Answer (A)

Given : $(x-\frac{1}{x})=3$ ———-(i)

Cubing both sides, we get :

=> $(x-\frac{1}{x}{)}^3=(3{)}^3$

=> $x^3-\frac{1}{x^3}-3(x)(\frac{1}{x})(x-\frac{1}{x})=27$

=> $x^3-\frac{1}{x^3}-3(1)(3)=27$

=> $(x^3-\frac{1}{x^3})=27+9=36$

=> Ans – (A)

13) Answer (D)

Given : $(3x^2-9x+3)=0$

=> $(3x^2+3)=9x$

Dividing both sides by ${}^{\prime }3x^{\prime }$, we get :

=> $x+\frac{1}{x}=3$ ————-(i)

Cubing both sides,

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

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

=> Ans – (D)

14) Answer (A)

Given : $a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=6$

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

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

$\because$ Sum of three positive terms is zero, hence each term is equal to 0.

=> $(a-\frac{1}{a})=$ $(b-\frac{1}{b})=$ $(c-\frac{1}{c})=0$

=> $\frac{a^2-1}{a}=0$

=> $a^2=1$

Similarly, $b^2=c^2=1$

$\therefore$ $(a^2+b^2+c^2)=1+1+1=3$

=> Ans – (A)

15) Answer (D)

Expression : $x=17-4\sqrt{18}$

=> $x=17-2\sqrt{72}$

=> $x=(\sqrt{9}{)}^2+(\sqrt{8}{)}^2+2(\sqrt{9})(\sqrt{8})$

Using, $a^2+b^2+2ab=(a+b{)}^2$

=> $x=(\sqrt{9}+\sqrt{8}{)}^2$

=> $\sqrt{x}=3+2\sqrt{2}$ ————-(i)

Also, $\frac{1}{\sqrt{x}}=\frac{1}{3+2\sqrt{2}}$

Rationalizing the denominator, we get :

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

=> $\frac{1}{\sqrt{x}}=\frac{3-2\sqrt{2}}{9-8}$

=> $\frac{1}{\sqrt{x}}=3-2\sqrt{2}$ ———(ii)

Adding equation (i) and (ii),

$\therefore$ $(\sqrt{x}+\frac{1}{\sqrt{x}})=(3+2\sqrt{2})+(3-2\sqrt{2})=6$

=> Ans – (D)

16) Answer (D)

Let side of the two cubes be $a$ and $b$ units respectively

Ratio of volumes = $\frac{a^3}{b^3}=\frac{11}{13}$

=> $\frac{a}{b}=(\sqrt[3]{\frac{11}{13}})$

=> $\frac{a}{b}=(11{)}^{\frac{1}{3}}:(13{)}^{\frac{1}{3}}$

=> Ans – (D)

17) Answer (A)

Radius of sphere = 7 cm

Curved surface area = $4\pi r^2$

= $4\times \frac{22}{7}\times (7{)}^2=616$ $c{m}^2$

=> Ans – (A)

18) Answer (B)

Let radius of circle = $r$ cm

=> $2\pi r-r=111$

=> $r(2\times \frac{22}{7}-1)=111$

=> $r\times \frac{44-7}{7}=111$

=> $r=111\times \frac{7}{37}=21$ cm

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

= $\frac{22}{7}\times (21{)}^2=1386$ $c{m}^2$

=> Ans – (B)

19) Answer (C)

20) Answer (B)

Given : ABCD is the rhombus whose diagonals bisect at O and the diagonals of a rhombus bisect each other at right angle. BD = 8 cm

=> OB = 4 cm

Perimeter of rhombus = 20 cm

=> BC = $\frac{20}{4}=5$ cm

Thus, in $\mathrm{△}$ BOC,

=> $(OC{)}^2=(BC{)}^2-(OB{)}^2$

=> $(OC{)}^2=(5{)}^2-(4{)}^2$

=> $(OC{)}^2=25-16=9$

=> $OC=\sqrt{9}=3$ cm

Thus, AC = 6 cm and BD = 8 cm

$\therefore$ Area of rhombus = $\frac{1}{2}\times d_1\times d_2$

= $\frac{1}{2}\times 6\times 8=24$ $c{m}^2$

=> Ans – (B)

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