 # Special Quant Quiz (miscellaneous) SSC CGL Tier-II 2016

Q1. The price of an antique is reduced by 20% and then this price is again reduced by 10%. The total reduction of the price ( in terms of percentage ) is
(a) 30%
(b) 28%
(c) 25%
(d) 23%
S1. Ans.(b)
Sol. Le the price of antique = 100 x
Reduced price = 100x×80/100×90/100
= 72x
Total reduction in price =(100x -72x)/100x

=28/100=28%

Q2. In a partnership business, B’s capital was half of A’s. If after 8 months, B withdraws half of his capital and after 2 months more A withdrew 1/4th of his capital, then the profit ratio of A and B will be
(a) 10 : 23
(b) 23 : 10
(c) 5 : 2
(d) 2 : 5

S2. Ans.(b)
Sol. Let capital of A at the starting = 2x
So, capital of B = x
Profit → A : B
For, 8 month
8 × 2x∶x×8
For, next 2 month
2×2x∶x/2×2
For last two month
2×2×3/4 x∶x/2×2
A : B
Total 23x : 10x
Hence, ratio of profit of A : B = 23 : 10

Q6. The distance between 2 places R and S is 42 km. Anita starts from R with a uniform speed of 4 km/h towards S and at the same time Romita starts from S towards R also with a uniform speed. They meet each other after  6 hours. The speed of Romita is
(a) 6 km/hour
(b) 8 km/hour
(c) 3 km/hour
(d) 20 km/hour

Q7. In a right angled triangle ∆PQR, PR is the hypotenuse of length 20 cm, ∠PRQ = 30°, the area of the triangle is
(a) 100√3  cm^2
(b) 100/√3  cm^2
(c) 50√3  cm^2
(d) 25 √3  cm^2

Q8. The outer and inner diameter of a circular path be 728 m and 700 m respectively. They breadth of the path is
(a) 28 m
(b) 20 m
(c) 7 m
(d) 14 m

S8. Ans.(d)
Sol. Radius of the outer circle  =728/2 = 364 m
Radius of inner circle =700/2 = 350 m
Breadth of the path = (364 – 350) m = 14 m

Q12. The sum of the H.C.F and L.C.M. of two numbers is 680 and the L.C.M. is 84 times the H.C.F. if one of the numbers is 56, the other is:
(a) 96
(b) 48
(c) 92
(d)46

S12. Ans.(a)
Sol. h + l = 680 and l = 84 h
∴ h + 84h = 680
⇒ 85h = 680
⇒ h = 8
∴ l = (84 × 8) = 672
Now, h × l = 56×x
⇒ 8 × 672 = 56×x
⇒ x=((8 × 672))/56=96
The other number is 96.

Q15. The average of three numbers is 135. The largest number is 180 and the difference of the other numbers is 25. The smallest number is
(a) 130
(b) 125
(c) 120
(d) 100
S15. Ans.(d)
Sol. Let the numbers be X, Y and Z
∴(X + Y + Z)/3=135⇒X+Y+Z=405
Let X be the largest number
∴ X = 180 ⇒ Y + Z = 225
Y – Z = 25 ⇒ Y = 125, Z = 100 (smallest number)
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