Special Quant Quiz (Advance) SSC CGL Tier-II 2016

Q1. If the sum of the interior angles of a regular polygon be 1080°, the number of sides of the polygon is
(a) 6
(b) 8
(c) 10
(d) 12

Q2. If the sine of an angle is 1/3, then cosine of that angle is:
(a) equal to 1/3
(b) less than 1/3
(c) greater than 1/3
(d) not known

Q3. The least value of 2sin^2⁡θ+3cos^2⁡θ  is:
(a) 1/2
(b) 1
(c) 2
(d) 3

Q4. If 0 < θ £ π/2, then which of the following is true?
(a) (tan^2θ+cot^2⁡θ)≥2
(b) (tan^2θ+cot^2⁡θ)≤2
(c) (tan^2θ+cot^2⁡θ)≤1
(d) None of these

Q5. If 0<θ<π/2, which of the following is true?
(a) sin^2⁡θ+1/sin^2⁡θ <2
(b) sin^2⁡θ+1/sin^2⁡θ =2
(c) sin^2⁡θ+1/sin^2⁡θ >2
(d) None of these

Q6. If 0° < A < 90° and cosA – sinA > 0 then
cosA + sinA cannot be equal to the:
(a) 1/3
(b) 1/2
(c) 1/√2
(d) √2

Q7. If x=√3+√2, then the value of (x+1/x) is
(a) 2√2
(b) 2√3
(c) 2
(d) 3

Q8. If (x+1/x)^2=3 then the value of (x^72+x^66+x^54+x^6+1)
(a) 0
(b) 1
(c) 84
(d) 206

Q9. If a + b + c = 6, a^2+b^2+c^2=14 and a^3+b^3+c^3=36, then the value of abc is
(a) 3
(b) 6
(c) 9
(d) 12
Q10. Which of the following can’t be the ratio of the sides of a triangle.
(a) 2:3:4
(b)1:3:6
(c) 3:4:5
(d) 3:4:6

Q11. If x : y = 7 : 3 then the value of (xy + y^2)/(x^2  – y^2 ) is
(a) 3/4
(b) 4/3
(c) 3/7
(d) 7/3

Q12. In a regular polygon, the exterior and interior angles are in the ratio 1 : 4. The number of sides of the polygon is
(a) 5
(b) 10
(c) 3
(d) 8

Q13. The sum of interior angles of a regular polygon is 1440°. The number of sides of the polygon is
(a) 10
(b) 12
(c) 6
(d) 8

Q14. The ratio of each interior angle to each exterior angle of a regular polygon is 3 : 1. The number of sides of the polygon is:
(a) 6
(b) 7
(c) 8
(d) 9

Q15. The angles of a quadrilateral are in the ratio 1 : 2 : 3 : 4, the largest angle is:
(a) 120°
(b) 134°
(c) 144°
(d) 150°

Solutions

S1. Ans.(b)
Sol. According to question
Sum of interior angle
= (n – 2) × 180°
Given:
Sum of interior angle = 1080°
(n – 2) × 108° = 1080°
(n-2)=(1080°)/180
(n-2)=6
n=6+2=8
Number of sides n = 8

S2. Ans.(c)
Sol.
cos⁡θ=√(1-sin^2⁡θ )=√(1-1/9)=√(8/9)>√(1/9)
So, cos⁡θ>1/3

S3. Ans.(c)
Sol.
2 sin^2⁡θ+3 cos^2⁡θ  = 2(sin^2⁡θ+cos^2⁡θ)+cos^2⁡θ
=2+cos^2⁡θ≥2
Least value of 2 sin^2⁡θ+3 cos^2⁡θ=2

S4. Ans.(c)
Sol. tan^2⁡θ+cot^2⁡θ=tan^2⁡θ+cot^2⁡θ-2 tan⁡θ.cot⁡θ+2 tan⁡θ.cot⁡θ
=(tan⁡θ-cot⁡θ)^2+2≥2
S5. Ans.(a)
Sol. sin^2⁡θ+1/sin^2⁡θ
=sin^2⁡θ+1/sin^2⁡θ -2+2
=(sin⁡θ-1/sin⁡θ)^2+2>2

S6. Ans.(d)
Sol. cos⁡A-sin⁡A>0
sin⁡A<cos⁡A
tan⁡A<1⇒A<45°
∴sin⁡A+cos⁡A<sin⁡45°+cos⁡45°
So, it will not be equal to the √2

S7. Ans.(b)
Sol. x=√3+√2
1/x=1/(√3+√2)×(√3-√2)/(√3-√2)=√3-√2
x+1/x=√3+√2+√3-√2=2√3

S8. Ans.(a)
Sol. (x+1/x)^2=3
x+1/x=√3
x^3+1/x^3 +3√3=3√3
x^3+1/x^3 =0
x^6+1=0
x^6=-1
x^72+x^66+x^54+x^24+x^6+1
(x^6 )^12+(x^6 )^11+(x^6 )^9+(x^6 )^4+x^6+1
(-1)^12+(-1)^11+(-1)^9+(-1)^4+1
1-1-1+1-1+1=0

S9. Ans.(b)
Sol. a + b + c = 6
=a^2+b^2+c^2=14
a^3+b^3+c^3=36
Put values as
a = 1, b = 2, c = 3
1 + 2+ 3 = 6
1 + 4 + 9 = 14
1 + 8 + 27 = 36
∴ abc = 1 × 2 × 3 = 6
S10. Ans.(b)
Sol.. We know that sum of the two side are always greater than third side.
Now, if we consider option (b)
The ratio of the sides of a triangle , 1 : 3 : 6
side of a triangle x, 3x, & 6x but here x + 3x < 6x
So, 1 : 3 : 6 can’t be the ratio of side of a triangle.

S11. Ans.(a)
Sol.
x : y
7 : 3
∴  (xy +y^2)/(x^2 -y^2)=(21 + 9)/(49 – 9)=30/40=3/4

S12. Ans.(b)
Sol. According to question
(Exterior angle )/(Interior angle )=1/4=x/4x
As we know that
Interior angle + Exteriors angle = 180°
∴ x + 4x = 180°
5x = 180°
x = 36°
∴ Number of sides
=(360°)/(Exterior angle)
Number of sides =(360°)/(36°)=10

S13. Ans.(a)
Sol. If the number of sides of regular polygon be = n
sum of the interior angle = (n – 2) × 180° =1440°
n – 2=8
so,  n=10

S14. Ans.(c)
Sol. Interior angle + exterior angle = 180°
3x + x = 180°
4x = 180°
x = 45°
Each exterior angle = 45°
Number of sides =(360°)/(Exterior angle)
=(360°)/(45°)=8

S15. Ans.(c)
Sol. Angles be x, 2x, 3x, 4x
∴ x + 2x + 3x + 4x = 360° ⇒ 10x = 360° ⇒ x = 36°
∴ largest angle = 4x = 144°

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