# Previous Year Questions of Quant for SSC CGL Tier-II 2016 Exam

Q1. If x + y = 90°, then what is √(cos⁡ x.cosec y-cos ⁡x.sin ⁡y) equal to?
(a) cos x
(b) sin x
(c) √(cos ⁡x)
(d) √(sin ⁡x)

Q2. If p = sin^10⁡x,then which one of the following correct for any value x?
(a) p ≥ 1
(b) 0 ≤ p ≤ 1
(c) 1 ≤ p ≤ 2
(d) None of these

Q3. If (sec⁡θ+tan⁡θ)/(sec⁡θ-tan⁡θ )=2(51/79) then the value of sinθ is
(a) 91/144
(b) 39/72
(c) 65/144
(d) 35/72

Q4. The value of sin^2⁡ 22°+sin^2⁡ 68°+cot^2⁡ 30° is:
(a) 5/4
(b) 3/4
(c) 3
(d) 4

Q5. The value of following is cos⁡24°+cos⁡55°+cos⁡25°+cos⁡204°+cos⁡300° ?
(a) – 1/2
(b) 1/2
(c) 2
(d) 1

Q6. If sin A + cosec A = 3, then find the value of sin^4⁡A+1/sin^2⁡A .
(a) 1
(b) 0
(c) 7
(d) 0

Q7. If sin x + cos x = c, then sin^6⁡x+cos^6⁡x is equal to.
(a) (1+6c^2  – 3c^4)/16
(b) (1 + 6c^2  – 3c^4)/4
(c) (1 + 6c^2+ 3c^4)/16
(d) (1 + 6c^2+ 3c^4)/4

Q8. A train A leaves Delhi at 6 AM and reaches Jaipur at 10 AM. Another train B leaves Jaipur at 8 AM and reaches Delhi at 11:30 AM. At what time do the two trains cross each other?
(a) 8:56 am
(b) 8:30 am
(c) 8:00 am
(d) 8:16 am

Q9. In an election between two candidates, 75% of the voters cast their votes, out of which 4% of the votes were declared invalid. The candidate who won the election got 450 votes more than the candidate who lost the election. The ratio of votes of winner and loser is 3:2. Find the total number of votes enrolled in that election.
(a) 3100
(b) 2250
(c) 3200
(d) 3125

Q10. In an examination, 80% of students passed in English, 85% in Mathematics and 75% in both English and Mathematics. If 40 students failed in both the subjects, find the total number of students.
(a) 500
(b) 200
(c) 400
(d) 300

Q11. A dishonest dealer professes to sell his goods at cost price. But he uses a false weight and thus gains 6(18/47)%. For a kg, he uses a weight of:
(a) 940 gms
(b) 947 gms
(c) 953 gms
(d) 960 gms

Q12. Three containers have their volumes in the ratio 3 : 4 : 5. They are full of mixtures of milk and water. The mixtures contain milk and water in the ratio of (4 : 1), (3 : 1) and (5 : 2) respectively. The contents of all these three containers are poured into a fourth container. The ratio of milk and water in the fourth container is:
(a) 4 : 1
(b) 151 : 48
(c) 157 : 53
(d) 5 : 2

Q13. A sum of Rs 731 is distributed among A, B and C, such that A receives 25% more than B and B receives 25% less than C. What is C’s share in the amount?
(a)Rs 172
(b)Rs 200
(c)Rs262
(d)Rs 272

Q14.A shopkeeper mixed two verities of rice at Rs. 20/kg and Rs. 30/kg in the ratio 2 : 3 and sell the mixture at 10% profit. Find the price per kg at which he sold the mixture?
(a)Rs.26
(b)Rs.28.8
(c)Rs.28
(d)Rs.28.6

Q15. A can finish a work in 24 days, B in 9 days and C in 12 days. B and C start the work but are forced to leave after 3 days. The remaining work was done by A in:
(a) 15 days
(b) 10 days
(c) 12 days
(d) 14 days

Solutions

S1. Ans.(b)
Sol. √(cos⁡x cosec y-cos⁡x .sin⁡y ) (∵ x + y = 90°, given)
=√(cos⁡x  cosec (90°-x)-cos⁡x.sin⁡(90°-x)〗 )
=√(cos⁡x. sec⁡x-cos^2⁡x )
=√(1-cos^2⁡x )=√(sin^2⁡x )=sin⁡x

S2. Ans.(b)
Sol. We know that, 0 < sin^2⁡x≤ 1
⇒0≤sin^10⁡x≤1
⇒ 0 ≤ p ≤ 1 (∵ p = sin^10⁡x)

S3. Ans.(c)
Sol. Given, (sec⁡θ+tan⁡θ)/(sec⁡θ-tan⁡θ )=2(51/79)
⇒(sec⁡θ+tan⁡θ)/(sec⁡θ-tan⁡θ )=209/79
[(by componendo-dividendo -> a/b=c/d,(a+b)/(a-b)=(c+d)/(c-d))]
⇒sec⁡θ/tan⁡θ =288/130
⇒(1/cos⁡θ )/(sin⁡θ/cos⁡θ )=288/130
⇒1/sin⁡θ =288/130
⇒ Therefore, sin θ=130/288
⇒ sin θ=65/144

S4. Ans.(d)

Sol. Given
⇒sin^2⁡ 22°+sin^2⁡ 68°+cot^2⁡ 30°
⇒cos^2⁡ 68°+sin^2⁡ 68°+cot^2⁡ 30°
{(Sin⁡(90-θ)=cos⁡θ ->cos⁡(90-θ)=sin⁡θ )}
⇒ 1 + cot^2⁡ 30°〗
⇒ 1 + (√3)^2
[∵ cot30° = √3]
⇒ 4

S5. Ans.(b)
Sol. The value of cos24° + cos55° + cos125° + cos204° + cos300°
We know that, cos(180° ± θ) = − cosθ
⇒ cos24° + cos55° + cos(180° – 55°) + cos(180° + 24°) + cos(360° – 60°)
⇒ cos24° + cos55° – cos55° – cos24° + cos60°
= cos60° = 1/2

S6. Ans.(c)
Sol. sin A + cosec A = 3
Sin A + 1/sin⁡A  = 3
Squaring both sides
sin^2⁡A+1/sin^2⁡A +2=9
sin^4⁡A+1/sin^2⁡A  = 9 – 2 = 7

S7. Ans.(b)
Sol. sin x + cos x = c
Squaring both sides
sin^2⁡x+cos^2⁡x+2 sin⁡x  cos⁡x=c^2
sin⁡x  cos⁡x=(c^2-1)/2
We know that,
sin^6⁡x+cos^6⁡x=1-3 sin^2⁡x  cos^2⁡x
= 1 – 3 ((c^2-1)/2)^2
= 1 – 3 ((c^4  + 1 – 2c^2)/4)
=(1+6c^2-3c^4)/4

S8. Ans.(a)
Sol. Let the distance between Delhi and Jaipur be x km and let the trains meet y hours after 8 am
Clearly, A covers x km in 4 hrs and B covers x km in (7/2) hrs.
∴ Speed of A = x/4 Kmph, Speed of B = 2x/7 Kmph
Distance covered by A in (y+2) hrs + Distance covered by B in y hrs = x
∴x/4 (y+2)+(2x/7)*y=x
(y+2)/4+2y/7=1
y=14/15 hrs = 56 mins
Hence, the train meets at 8:56 am

S9. Ans.(d)
Sol. Let the winner got ‘3x’ votes and loser got ‘2x’ votes
So, 3x – 2x = 450
x = 450
∴ 75% of 96% of Total votes enrolled by = 1350 + 900
⇔(75/100×96/100× x)=2250 ⇔x=((2250× 100 × 100)/(75 × 96))=3125

S10. Ans.(c)
Sol. Let the total number of students be x.
Let A and B represent the sets of students who passed in English and Mathematics respectively.
Then, number of students passed in one or both the subjects
=n(A∪B)=n(A)+n(B)-n (A∩B)=80% of x + 85% of x – 75% of x
=((80/100)x+(85/100)x-(75/100)x)=(90/100)x=(9/10)x.
∴ Students who failed in both the subjects = (x-(9x/10))=x/10.
So, x/10 = 40 or x = 400. Hence total number of students = 400.

S11. Ans.(a)
Sol. Let error = x gms. Then, x/(1000 – x)×100=6 (18/47)  ⇔100x/(1000-x)=300/47
⇔ 47x = 3 (1000 – x) ⇔ 50x = 3000 ⇔ x = 60.
∴ Weight used = (1000 – 60) = 940 gms.

S12. Ans.(c)
Sol. Let the three containers contain 3x, 4x and 5x litres of mixtures respectively.
Milk in 1st mixture =(3x×4/5) litres =12x/5 litres.
Water in 1st mixture = (3x-12x/5) litres =3x/5 litres.
Milk in 2nd mixture =(4x×3/4) litres = 3x litres.
Water in 2nd mixture = (4x-3x) litres = x lites.
Milk in 3rd mixture =(5x×5/7) litres = 25x/7 litres.
Water in 3rd mixture = (5x-25x/7) litres =10x/7 litres.
Total milk in final mixture =(12x/5+3x+25x/7) litres =314x/35 litres.
Total water in final mixture =(3x/5+x+10x/7) litres =106x/35 litres.
Required ratio of milk and water =314x/35:106x/35=157∶53.

S13. Ans.(d)
Sol.  Let C’s share be Rs x.
Then, B gets =0.75 x.
A gets = 1.25×0.75x
x+0.75x+0.9375x=731
2.6875x=731
x=731⁄2.6875 =272

S14. Ans.(d)
Sol. CP of mixture = (20*2+30*3)/5 =26
SP of the mixture = 1.1×26=28.6

S15. Ans.(b)
Sol. Total Unit of work=LCM(Time taken by A,Time taken by B,Time taken by c)=LCM(24,9,12)=72
A’s 1 day work=72/24=3unit
B’s 1 day work=72/9=8 unit
C’s 1 day work=72/12=6unit
work done by(B+C) in 1 day=(8+6)unit=14 unit
work done by(B+C) in 3 day=14×3=42 unit
Remaining work=(Total work – work done by B+C in 3 days)= (72-42)=30unit
Remaing work done by A in=Remaining work/A’s 1 day work= 30/3=10 days.

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