### Advance Mathematics for SSC CGL and 10+2

Practice Advance Mathematics Quiz for SSC CGL and other exams.

1.The height of an equilateral triangle whose perimeter is 27 cm. is :
(a) 9/2 √3 cm
(b) 9/2 cm
(c) 9 cm
(d) none of these
2.Find the distance between the two parallel sides of a trapezium if the area of the trapezium is 250 sq.m and the two parallel sides are equal to 15 m and 10 m respectively.
(a) 15 m
(b) 20 m
(c) 25 m
(d) 22 m

3.The length of a rectangle is increased by 60%. By what percent would the width be decreased so as to maintain the same area?
(a) 37 1/2%
(b) 60%
(c) 75%
(d) 120%

4.The volume of a sphere is 8 times that of another sphere. What is the ratio of their surface areas?
(a) 8 : 1
(b) 4 : 1
(c) 2 : 1
(d) 4 : 3

5.How many litres of water flow out of a pipe having an area of cross-section of 5 cm^2 in one minute, if the speed of water in the pipe is 30 cm/s?
(a) 90 L
(b) 15 L
(c) 9 L
(d) 1.5 L

6.A cylindrical rod of length h is method and cast into a cone of base radius twice that of the cylinder. What is the height of the cone?
(a) 3h/4
(b) 4h/3
(c) 2h
(d) h/2

7. If two numbers are each divided by the same divisor, the remainders are 3 and 4, respectively. If the sum of the two numbers be divided by the same divisor, the remainder is 2. The divisor is
(a) 9
(b) 7
(c) 5
(d) 3

8. It is given that [2^(32) + 1] is exactly divisible by a certain number. Which one of the following is also definitely divisible by the same number?
(a) [2^(96) + 1]
(b) [7 × 2^(33)]
(c) [2^(16) – 1]
(d) [2^(16) + 1]

9. In a question on division, the divisor is 7 times the quotient and 3 times the remainder. If the remainder is 28, then the dividend is

(a) 588
(b) 784
(c) 823
(d) 1036

10. 64329 is divided by a certain number. While dividing, the numbers, 175, 114 and 213 appear as three successive remainders. The divisor is
(a) 184
(b) 224
(c) 234
(d) 296

1.(a) 3a = 27
a = 9cm
height = √3/2 a = 9/2 √3

2.(b) Area = 1/2(sum of parallel sides) * (distance between them)
=> 250 = 1/2(15 + 10) * h
=> h = 20m

3.(a) Let length = width = 100m
If length = 160m, then let width = x m
s.t. 160x = 10000
=> x = 10000/160 = 1000/16 = 62 1/2
so width is reduced to 37 1/2 %

4.(b) ; According to question,
4/3 π(r1)^3 = 8 × 4/3 π(r2)^3
(r1)^3/(r2)^3 =8 => r1/r2 = 2/1
Required ratio =(4π〖r1〗^2)/(4π〖r2〗^2 )= 〖(2/1)〗^(2 ) = 4/1

5.(c); Height of water in a second = 30 cm
Height of water in 60 second = 30 * 60
h = 1800 cm
Area of cross section = πr^2=5sq cm
Volume of water flow in one minute = πr^2 h
= 5 * 1800 = 9000 cu cm = 9000/1000 It = 9 lt

6.(a) ; Let the radius of cylinder be r and radius of cone be R
R = 2r
According to question,
Volume of cylinder = Volume of cone
πr^2 h=1/3 πR^2 H
r^2 h=1/3 π(2r)^2 H
H = 3h/4