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

Q1. From a point in the interior of an equilateral triangle the perpendicular distance of the sides are √3 cm, 2√3 cm respectively and 5√3 cm. The perimeter (in cm) of the triangle is:
(a) 64
(b) 32
(c) 48
(d) 24

Q2. In a triangle ABC, AB + BC = 12 cm, BC + CA = 14 cm and CA + AB = 18 cm. Find the radius of the circle (in cm) which has the same perimeter as the triangle.

(a) 5/2
(b) 7/2
(c) 9/2
(d) 11/2
S2. Ans.(b)
Sol. Perimeter of triangle ABC = AB + BC + CA
=1/2(AB+BC+BC+CA+CA+AB)
= 1/2 (12 + 14 + 18)
= 22 cm
Given that,
Perimeter of circle = perimeter of triangle
2πr = 22
r = 3.5 cm


Q3. A took 15 seconds to cross a rectangular field diagonally walking at the rate of 52 m/min and B took the same time to cross the same field along its sides walking at the rate of 68 m/min. The area of the field is:
(a) 30 m^2
(b) 40 m^2
(c) 50 m^2
(d) 60 m^2

Q4. Three circles of radius 3.5 cm each are placed in such a way that each touches the other. The area of the portion enclosed by the circles is:
(a) 1.975 cm^2
(b) 1.967 cm^2
(c) 19.67 cm^2
(d) 21.21 cm^2

Q5. The ratio of the 3 angles of a quadrilateral is 13:9:5. The value of the 4th angle of the quadrilateral is 36°. What is the difference between the largest and the second smallest angles of the quadrilateral? 
(a) 104°
(b) 108°
(c) 72°
(d) 96°
S5. Ans.(d)
Sol. Let, the three angles of the quadrilateral be 13x°, 19x° and 5x°, respectively.
Now, according to the question,
13x + 9x + 5x = 360 – 36 = 324
⇒ 27x = 324
∴x=324/27 = 12
∴ Required difference = 13x – 5x = 8x = 8 × 12 = 96°

Q6. From a solid right circular cylinder with height 10 cm and radius of the base 6 cm; a right circular cone of the same height and base is removed. The volume (in cm^3) of the remaining solid is:
(a) 377
(b) 754.3
(c) 1131
(d) None of these

Q7. A right circular cone is exactly fitted inside a cube in such a way that the edges of the base of the cone are touching the edges of one of the faces of the cube and the vertex is one the opposite face of the cube. If the volume of the cube is 343 c.c., then what approximately is the volume of the cone?
(a) 90 c.c.
(b) 75 c.c.
(c) 80 c.c.
(d) 85 c.c.
S7. Ans.(a)
Sol. Height of the cone = edge of the cube
diameter of cone = edge of the cube
edge of the cube = √(Volume of cube)
=√(343 c.c.)
= 7 c.c.
Volume of the cone = 1/3×π×r^2×h
=1/3×22/7×7/2×7/2×7
=539/6
≅90 c.c.

Q8. A rectangular sheet of area 264 cm square and width 11 cm is rolled along its breadth to make a hollow cylinder. The volume of the cylinder is:
(a) 231 c.c.
(b) 230 c.c.
(c) 235 c.c.
(d) 234 c.c.
S8. Ans.(a)
Sol. length of a rectangular sheet = 264/11
= 24 cm
After rolled along the width to make a hollow cylinder,
2πr = 11 cm
r = 7/4 cm
Height = 24 cm
Volume of the cylinder = πr^2 h
=22/7×7/4×7/4×24
= 231 c.c.

Q9. A conical vessel of base radius 2 cm and height 3 cm is filled with kerosene. This liquid leaks through a hole in the bottom and collects in a cylindrical jar of radius 2 cm. The kerosene level in the jar is:
(a) 1.5 cm
(b) π cm
(c) 1 cm
(d) 3 cm


Q10. Two chords of length a mt. and b mt. subtend angle 60° and 90° at the centre of circle respectively, which of the following is true. 
(a) b=√2 a
(b) a=√2 b
(c) a = 2b
(d) b = 2a

Q12. The angle of elevation of the top of an unfinished tower at a point distant 120m from its base is 45°. If the elevation of the top at the same point is to be 60°, the tower must be raised to a height:
(a) 120 (√3 + 1) m
(b) 120 (√3 – 1) m 
(c) 10 (√3 + 1) m
(d) None of these
Q13. A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man’s eye when at a distance of 60 m from the tower. After 5 second, the angle of depression becomes 30°. What is the approximate speed of the boat, assuming it is running in still water?
(a) 32 km/h
(b) 42 km/h
(c) 38 km/h
(d) 36 km/h
S14. Ans.(a)
Sol. OR = OS, OR ⊥ DR and OS ⊥ DS
∴ ORDS is a square
Also, BP = BQ, CQ = CR and DR = DS
∴ BQ = BP = 27 cm ⇒ BC – CQ = 27 cm
⇒ 38 – CQ = 27 cm
⇒ CQ = 11 cm
⇒ CR = 11 cm
⇒ CD – DR = 11
⇒ 25 – DR = 11
⇒ DR = 14 cm
⇒ r = 14 cm.
Q15. A rectangular water reservoir is 15 m by 12 m at the base. Water flows into it through a pipe whose cross-section is 5 cm by 3 cm at the rate of 16 m per second. Find the height to which the water will rise in the reservoir in 25 minutes:
(a) 0.2 m
(b) 2 cm 
(c) 0.5 m
(d) None of these 
S15. Ans.(a) 
Sol. Volume of water which flow in 25 minutes
= 25 × 60 × 0.05 × 0.03 × 16 = 36 meter cube
∴ Rise in water level =36/(15 × 12)=1/5  m=0.2 m  

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