SSC CGL 2015 : QUANT QUIZ

1.The ratio of investment of two partners is 11:12 and the ratio of their profits 2:3 . if A invested the money for 8 months. Find for how much time B invested his money?

A. 11 months 

B. 10 months

C. 9 months 

D. 8 months

2. A number lies between the cubes of 15 and 16.If the number is divisible by the square of 12 as well as by 7,what is number?

A. 3469

B. 4032

C. 4045

D. 5249

3. A man deposite a certain sum in a bank.He gets 4% per annum intrest for first 3 years, 5% for next 2 years and 6% for the period beyond that. If he gets Rs.2000 as simple intrest for 8 years, how much money did he deposite in the bank.?

A.Rs.8000

B.Rs.6000

C.Rs.4000

D.Rs.5000

4.The age of shiva is five times that of his son.6 years ago his age was nine times that of his son.what is shiva's present age?

A. 60 years 

B. 55 years 

C. 50 years

D. 45 years

5. Trees are to be planted at equal intervals of 10 ft.around a rectangular lawn whose length is twice its breadth. If the total number of trees planted is 36.find the area of the lawn.

A. 7200 sq.ft

B. 6000 sq.ft

C. 5000 sq.ft

D. 4800 sq.ft

6. A coin box contains 50 paise and 1 rupee coins.The total number of coins is 36. The total value of 50 paise coins equal the total value of repess coins.What is the total amount in the box.

A. Rs.18

B. Rs.20

C. Rs.22

D. Rs.24

7. A hallow sphere of internal and external diameter of 4 c.m.and 8.cm.respectively is melted in to a cone  of base diameter 8cm.The height of the cone is -

A. 14 cm.

B. 12 cm.

C. 15 cm.

D. 18 cm.

8.A man starts walking. He walked 2 km in the first hour. then, he walked two-third of the previous hour in each next hour.If he walked continuously then how long maximum could he walk?

A. 60 km.

B. 6 km.

C. 12 km.

D. 8 km. 

9. The ratio of between the school ages of Neeta and sameer is 5:6 respectively.If the ratio between the one-third  age of Neeta and half of the sameer's age is 5:9, then what is the school age of sameer?

A. 30 years

B. 25 years

C. 36 years.

D. Data indequate

10. A boat covers a distance of 30 km downstream in 2 hours while it takes 6 hours to cover the same distance upstream.If the speed of current is half of the speed of the boat then, what is the speed of boat in km per hour?

A. 15 kmph

B. 5 kmph

C. 10 kmph

4. Data inadequate.

Answers and solution:
1(A) Suppose A invested Rs.11 for 8 months and B invested Rs.12 for y months.
Then ratio of investment of A and B = (11x8) : (12xy)
= 88 : 12y
88/12y = 2/3
24y = 88x3
y = 264/24 =11 months.

2(B)
(15)^3= 3375
(16)^3= 4096
(12)^2= 144
The number lies between 3375 and 4096.
The number is divisible by 144 as well as 7.
L.C.M. of 144 and 7.= 1008
so the number must be a multiple of 1008.
1008x3 = 3024
1008x4 = 4032
4032 lies between 3375 and 4096.

3(D)
I = I1 + I2 + I3
I = p/100(r1t1+r2t2+r3t3)
I = p/100(4x3+5x2+6x3)
P = (100xI)/40
P = (100x200)/4
P == Rs. 5000

4(A)
Let the present age of Shva's son be x years.
so shiva's present age = 5x
6 year ago:
son 's age = (x-6) yrs.
Shiva's age = (5x-6)yrs.
So 5x-6 = 9(x-6)
5x-6 = 9x-54
4x = 48
x = 12
so shiva's present age = 5x = 60 yrs.

5(A)
Perimeter of the lawn =2(l+b)= 36x10 ft.
So l+b = 180
But l = 2b so 3b = 180
b = 60 ft
so l= 120 ft
Area of the lawn = lb
= 120 x 60 sq.ft = 7200 sq.ft

6(D)
7(A)
Volume of cone = volume of the spherical shell
1/3πr^2h = 4/3π(R1^3 - R2^3)
h = 4(R1^3-R2^3)/r^2
= 4(4^3-2^3)/4^2 = 14cm.

8(B)
Required distance
=2((1+(2/3)+(2/3)^2+(2/3)^3+..........)
= 2 x 1/(1-2/3) = 2 x 3 = 6 km.

9(D)
10(C)
Downstream speed = 30/2 =15 km/h
Upstreams speed = 30/6 = 5 km/h
so speed of the boat = 1/2(15+5)
=1/2 x 20 = 10 km/h