SSC CGL 2018 Daily Quiz 03 – Quantitative Aptitude “Mensuration”
SSC CGL 2018 Quantitative aptitude Practice test. Daily Quiz for SSC CGL 2018 online practice set. Daily tests with questions and answers for SSC CGL exam. Topic tests for SSC online Tier-1
1. What will be the area of trapezium whose
parallel sides are 22 cm and 16 cm long,
and the distance between them is 11 cm?
A) 209 cm2
B) 282 cm2
C) 265 cm2
D) 179 cm2
E) 302 cm2
2. The perimeter of a rectangle is 42 m. If the
area of the square formed on the diagonal
of the rectangle as its side is 1 1/12 % more
than the area of the rectangle, find the
longer side of the rectangle.
A) 19 m
B) 16 m
C) 9 m
D) 5 m
E) 12 m
3. At the rate of Rs. 2 per sq m, cost of
painting a rectangular floor is Rs 5760. If
the length of the floor is 80% more than its
breadth, then what is the length of the
floor?
A) 25 m
B) 72 m
C) 67 m
D) 56 m
E) 46 m
4. A 7 m wide path is to be made around a
circular garden having a diameter of 7 m.
What will be the area of the path in square
metre?
A) 298
B) 256
C) 308
D) 365
E) 387
5. The perimeter of a rectangle of length 62
cm and breadth 50 cm is four times
perimeter of a square. What will be the
circumference of a semicircle whose
diameter is equal to the side of the given
square?
A) 36 cm
B) 25 cm
6.What is the volume of a cylinder whose
curved surface area is 1408 cm2 and height
is 16 cm?
A) 7715 cm3
B) 9340 cm3
C) 8722 cm3
D) 7346 cm3
E) 9856 cm3
7. A cone with diameter of its base as 30 cm
is formed by melting a spherical ball of
diameter 10 cm. What is the approximate
height of the cone?
A) 6 cm
B) 3 cm
C) 2 m
D) 5 cm
E) None of these
8. A cylinder whose base of circumference is
6 m can roll at a rate of 3 rounds per
second. How much distance will the
cylinder cover in 9 seconds?
A) 125 m
B) 162 m
C) 149 m
D) 173 m
E) 157 m
9. A container is formed by surmounting a
hemisphere on a right circular cylinder of
same radius as that of hemisphere. If the
volume of the container is 576π m3 and
radius of cylinder is 6 m, then find the
height of the container.
A) 14 m
B) 12 m
C) 20 m
D) 18 m
E) 22 m
10. The radii of two cylinders are in the ratio 3
: 2 and their curved surface areas are in the
ratio 3 : 5. What is the ratio of their
volumes?
A) 8 : 11
B) 5 : 9
C) 7 : 4
D) 9 : 10
E) 13 : 7
Answer
1. Option A
Solution:
Area of a trapezium = 1/2 (sum of parallel
sides) * (perpendicular distance between
them) = 1/2 (22 + 16) * (11) = 209 cm2
2. Option E
Solution:
Let the sides of the rectangle be l and b
respectively.
From the given data,
√(l2 + b2) = (1 + 1 1/12) lb
=> l2 + b2 = (1 + 13/12) lb = 25/12 * lb
12(l2 + b2 ) = 25 lb
Adding 24 lb on both sides
12 l2 + 12b2 + 24lb = 25 lb
12(l2 + b2 + 2lb) = 49 lb
12(l + b)2 = 49lb
but 2(l + b) = 42 => l + b = 21
So 12(21)2 = 49lb
Solve, we get lb = 108
Since l + b = 21, longer side = 12 m
3. Option B
Solution:
Let the length and the breadth of the floor
be l m and b m respectively.
l = b + 80% of b = l + 0.8 b = 1.8b
Area of the floor = 5760/2 = 2880 sq m
l*b = 2880 i.e., l * l/1.8 = 2880
l = 72
4. Option C
Solution:
Area of the path = Area of the outer circle –
Area of the inner circle = π{7/2 + 7}2 – π
[7/2]2
= 308 sq m
5. Option B
Solution:
Let the side of the square be a cm.
Parameter of the rectangle = 2(62 + 50) =
224 cm Parameter of the square = 56 cm
i.e. 4a = 56
So a = 14
Diameter, d of the semicircle = 14 cm
Circumference of the semicircle = 1/2(π)(r)
+ d
= 1/2(22/7)(7) + 14 = 25 cm
6. Option E
Solution:
2πrh = 1408, h = 16
Solve both, so r = 14
Volume = π r2h = (22/7) * 14 * 14 * 16 =
9856
7. Option C
Solution:
Radius of cone = 30/2 = 15, radius of ball =
10/2 = 5
Volumes will be equal, so
(1/3) π r2h = (4/3) π R3
152h = 4* 53
So h = 2.2
8. Option B
Solution:
Distance covered in one round = 2 x π x r =
6 m
Distance covered in 1 second = 3 x 6 = 18
m
So distance covered in 9 seconds = 18×9=
162m
9. Volume of the container = Volume of the
cylinder + Volume of the hemisphere
Volume of the container = π 62h + (2/3) π
63 = 576π
= π 36 (h + 4) = 576π
Solving we get h = 12
So the height of the container = 12 + 6 = 18
m
10. Option D
Solution:
r1/r2 = 3/2 or r1 = 3/2 * r2
CSA1/CSA2 = 2πr1h1/2πr2h2 = 3/5
So h1/h2 = 2/5
Volume1/ Volume2 = πr12h1/ πr22h2 =
9/10
