# Important Quant Questions For SSC CHSL 2018

## Quant Questions For SSC CHSL 2018

Q1. The perimeters of a square and a rectangle are equal. If their area be ‘A’ m² and ‘B’ m² then correct statement is
(a) A < B
(b) A ≤ B
(c) A > B
(d) A ≥ B

Q2. The diagonal of a cuboid of length 5 cm, width 4 cm and height 3 cm is
(a) 5√2 cm
(b) 2√5 cm
(c) 12 cm
(d) 10 cm

Q3. A rectangle with one side 4 cm is inscribed in a circle of radius 2.5 cm. The area of the rectangle is:
(a) 8 cm²
(b) 12 cm²
(c) 16 cm²
(d) 20 cm²

Q4. The area of the largest sphere (in cm²) that can be drawn inside a square of side 18 cm is
(a) 972 π
(b) 11664 π
(c) 36 π
(d) 288 π

Q5. The area of the circle with radius Y is W. The difference between the areas of the bigger circle (radius with radius Y) and that of the smaller circle (with radius X) is W’. So X/Y is equal to
(a) √(1-w'/w)
(b) √(1+w'/w)
(c) √(1+w/w' )
(d) √(1-w/w')

Q6. An elephant of length 4 m is at one corner of a rectangular cage of size (16 m × 30 m) and faces towards the diagonally opposite corner. If the elephant starts moving towards the diagonally opposite corner it takes 15 seconds to reach this corner. Find the speed of the elephant
(a) 1 m/sec
(b) 2 m/sec
(c) 1.87 m/sec
(d) 1.5 m/sec

Q7. If the area of three adjacent faces of a rectangular box which meet in the corner are 12 cm², 15 cm² and 20 cm² respectively. Then the volume of the box is
(a) 3600 cm³
(b) 300 cm³
(c) 60 cm³
(d) 180 cm³

Q8. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hour completes one round in 8 minutes, then the area of the park is
(a) 153650 m²
(b) 135600 m²
(c) 153600 m²
(d) 156300 m²

Q9. If the radius of a right circular cylinder open at both the ends, is decreased by 25% and the height of the cylinder is increased by 25%. Then the curved surface area of the cylinder thus formed is:
(a) remains unaltered
(b) is increased by 25%
(c) is increased by 6.25%
(d) is decreased by 6.25%

Q10. A cylindrical pencil of diameter 1.2 cm has one of its end sharpened into a conical shape of height 1.4 cm. The volume of the material removed is
(a) 1.056 cm³
(b) 4.224 cm³
(c) 1.089 cm³
(d) 42.24 cm³

Q11. A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park had been used as a lawn. If the area of the lawn is 2109 m² then the width of the road is
(a) 3 m
(b) 5 m
(c) 6 m
(d) 2 m

Q12. Four circles of equal radii are described about the four corners of a square so that each touches two of the other circles. If each side of the square is 140 cm then area of the space enclosed between the circumference of the circle and the square is
(a) 4200 cm²
(b) 2100 cm²
(c) 7000 cm²
(d) 2800 cm²

Q13. The amount of concrete required to build a concrete cylindrical pillar whose base has a perimeter 8.8 metre and curved surface area 17.6 sq. metre. Is (Take π = 22/7)
(a) 8.325 m³
(b) 9.725 m³
(c) 10.5 m³
(d) 12.32 m³

Q14. A hemispherical bowl of internal radius 9 cm, contains a liquid. This liquid is to be filled into small cylindrical bottles of diameter 3 cm and height 4 cm. Then the number of bottles necessary to empty the bowl is
(a) 18
(b) 45
(c) 27
(d) 54

Q15. A rectangular water tank is 80 m × 40 m. Water flows into it through a pipe of 40 sq. cm at the opening at a speed of 10 km/hr. The water level will rise in the tank in half an hour is(in cms)
(a) 3/2
(b) 4/9
(c) 5/9
(d) 5/8

S1. Ans.(c)
Sol. 4a = 2(l + b)
a=1/2 (l+b)
Area of square, A = a²
= ¼ (l² + b² + 2lb)
Area of Rectangle, B = lb
A > B

S2. Ans.(a)
Sol. Diagonal = √(l²+b²+h² )
=√(25+16+9)
=√50
=5√2 cm

S3. Ans.(b)
Sol.
Diagonal of Rectangle = Diameter of circle
= 5 cm
25 = b² + 16
b = 3 cm
area of rectangle = lb
= 12 cm²

S4. Ans.(a)
Sol. Radius of the largest sphere
=18/2=9 cm
Volume of sphere = 4/3 πr³
=4/3×81×9×π
= 972 π

S5. Ans.(a)
Sol. w = π y²
Bigger Area – Smaller Area = w'
πy²-πx²=w'
1-(πx²)/(πy² )=w'/(πy² )

1-x²/y² =w'/w
x/y=√(1-w'/w)

S6. Ans.(b)
Sol. Diagonal = √(16)²+(30)²
=√(256+900)
=√1156
= 34 m
Distance that elephant has to cover = 34 – 4= 30 m
Speed of elephant = 30/15 m/sec= 2 m/sec

S7. Ans.(c)
Sol. Volume of box
=√(12×15×20)
=√3600  cm³
= 60 cm³

S8. Ans.(c)
Sol. L : B = 3 : 2
Let length ⇒ 3x
Distance covered by cyclist = 2(3x + 2x) = 10x
ATQ,
10x = 12 × 8/60
x = 8/50 km
=8/50×1000 m
= 160 m
Length = 480 m
Area = 480 × 320
= 153600 m²

S9. Ans.(d)
Sol. Curved Surface are = 2πrh
New curved surface are
=2π×75/100×r×125/100×h
=2π×3/4×r×5/4×h
=15/8 πrh
Decrease om area
=2πrh-15/8 πrh
=1/8 πrh
% Decrease = (πrh/8)/2πrh×100
=100/16
= 6.25%

S10. Ans.(b)
Sol. Volume of are removed
= πr²h – 1/3 πr²h
=  2/3 πr² h
=2/3×22/7×1.44 ×1.4
=88.704/21
= 4.224 cm³

S11. Ans.(a)
Sol. Let width of Road be x
Area of Rectangle – Area of Road = 2109
60 × 40 – [60 × x + 40x – x²] = 2109
x² – 100x + 291 = 0
x = 3 m

S12. Ans.(a)
Sol. Area enclosed between square and Circumference of Circle
= 140 × 140 – π × 70 × 70
= 19600 – 15400
= 4200 cm²

S13. Ans.(d)
Sol. 2πr = 8.8
r = 1.4 cm
2πrh = 17.6
2 × 22/7 × 1.4 × h = 17.6
h = 2 m
Volume Required = πr²h = 12.32 m³

S14. Ans.(d)
Sol. No. of bottles = (2/3  πr^3)/(πr^2 h)
=(2×9×9×9)/(3×3/2×3/2×4)
= 54

S15. Ans.(d)
Sol. Let the water rises h meter
l × b × h = Area × speed × time
80×40×h=40/(100×100)×10000×1/2
h=1/160 m
=5/8  cm
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