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

Q1. If the area of base of a right circular cone is 3850 cm2 and its height is 84 cm, then the curved surface area of the cone is:
(a) 10001 cm2
(b) 10010 cm2
(c) 10100 cm2
(d) 11000 cm2

Q2. If the height of a right circular cone is increased by 200% and the radius of the base is reduced by 50%, then the volume of the cone:
(a) Remains unaltered
(b) Decreases by 25%
(c) Increases by 25%
(d) Increases by 50%

Q3. A spherical ball of lead, 3 cm in diameter is melted and recast into three spherical balls. The diameter of two of these are 1.5 cm and 2 cm respectively. The diameter of the third ball is:
(a) 2.5 cm
(b) 2.66 cm
(c) 3 cm
(d) 3.5 cm

Q4. A cylindrical tub of radius 12 cm contains water upto a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm. The radius of the ball is:
(a) 4.5 cm
(b) 6 cm
(c) 7.25 cm
(d) 9 cm

Q5. The capacities of two hemispherical vessels are 6.4 litres and 21.6 litres. The areas of inner curved surfaces of the vessels will be in the ratio of:
(a) √2 ∶ √3
(b) 2 : 3
(c) 4 : 9
(d) 16 : 81

Q6. A hemispherical bowl is filled to the brim with a beverage. The contents of the bowl are transferred into a cylindrical vessel that radius is 50% of  its height. If the diameter is same for both the bowl and the cylinder, the volume of the beverage in the cylindrical vessel is:
(a) 66 2/3%
(b) 78 1/2%
(c) 100%
(d) More than 100% (i.e. some liquid will be left in the bowl).

Q7. The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. find the area of the triangle.
(a) 60 cm^2.
(b) 48 cm^2.
(c) 72 cm^2.
(d) 36 cm^2.

Q8. In measuring the sides of a rectangle, one side is taken 5% in excess, and the other 4% in deficit. Find the error percent in the area calculated from these measurements.
(a) .8%
(b) 8%
(c) 1.8%
(d) .4%

Q10. A man standing at a point P is watching the top of a tower, which makes an angle of elevation of 30° with the man’s eye. The man walks some distance towards the tower to watch its top and the angle of elevation becomes 60°. What is the distance between the base of the tower and the point P?
(a) 4√3 units
(b) 8 units
(d) None of these

Q11. A man buys a watch for Rs. 1950 in cash and sells it for Rs. 2200 at a credit of 1 year. If the rate of interest is 10% per annum, the man:
(a) Gains Rs. 55
(b) Gains Rs. 50
(c) Loses Rs. 30
(d) Gains Rs. 30
S11. Ans.(b)
Sol. S.P. = value . of Rs. 2200 due 1 year hence = Rs. [(2200 × 100)/(100 + (10 × 1) )] = Rs. 2000.
∴ Gain = Rs (2000 – 1950) = Rs. 50.
Q12. A and B take part in a 100 m race. A runs at 5 km per hour. A gives B a start of 8 m and still beats him by 8 seconds. The sped of B is:
(a) 5.15 kmph
(b) 4.14 kmph
(c) 4.25 kmph
(d) 4.4 kmph
S12. Ans.(b)
Sol. A’s speed =(5×5/18) m/sec =25/18 m/sec.
Time taken by to cover 100 m =(100×18/25) sec = 72 sec.
∴ Time taken by B to cover 92 m = (72 + 8) sec = 80 sec.
∴ B’s speed =(92/80×18/5) kmph = 4.14 kmph.
Q13. In a 500 m race, the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 140 m. Then A wins by:
(a) 60 m
(b) 40 m
(c) 20 m
(d) 10 m
S13. Ans.(c)
Sol. To reach the winning post A will have to cover a distance of (500 – 140) m i.e. 360 m. While A covers 3 m, B covers 4 m.
While A covers 360 m. B covers (4/3×360) m = 480 m.
Thus, when A reaches the winning post, B covers 480 m and therefore remains 20 m behind.
∴ A wins by 20 m.
Q14. The areas of the three adjacent faces of a rectangular box which meet in a point are known. The product of these areas is equal to:
(a) The volume of the box
(b) Twice the volume of the box
(c) The square of the volume of the box
(d) The cube root of the volume of the box
S14. Ans.(c)
Sol. Let length = l, breadth = b and height = h. Then,
Product of areas of 3 adjacent faces = (lb × bh × lh) =(lbh)^2=(Volume)^2

Q15. The length of an edge of a hollow cube open at one face is √3 metres. What is the length of the largest pole that it can accommodate?
(a) √3 metres
(b) 3 metres
(c) 3√3 metres
(d) 3/√3  metres
S15. Ans.(b)
Sol. Required length = Diagonal = √3 a=(√3×√3)m=3m.

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