Quantitative Aptitude - Arithmetic Ability Questions

Clearly, n(S) = (6 x 6) = 36.

Let E = Event that the sum is a prime number.

Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }

n(E) = 15.

P(E) = n(E)/n(S) = 15/36 = 5/12.

The given number series follows a pattern that,

9

9 x 0.5 + 0.5 = 5

5 x 1 + 1 = 6

6 x 2 + 2 = 14

14 x 4 + 4 = 60 (not equals to 56)

60 x 8 + 8 = 488.

Hence, the odd one in the given number series is 56.

A : B = 100 : 90.

A : C = 100 : 87.

B/C = (B/A*A/C) = (90/100*100/87) = 30/29

When B runs 30 m, C runs 29 m.

When B runs 180 m, C runs (29/30*180)m =174m

 B beats C by (180 - 174) m = 6 m.

Step (i): Let x be the number of boys and y be the number of girls.
Given total number of boys and girls = 100
x+y=100 -------------- (i)

Step (ii): A boy gets Rs. 3.60 and a girl gets Rs. 2.40
The amount given to 100 boys and girls = Rs. 312
3.60x + 2.40y = 312 -------------- (ii)

Step (iii):
Solving (i) and (ii)
3.60x + 3.60y = 360 --------- Multiply (i) by 3.60
3.60x + 2.40y = 312 --------- (ii)
1.20y = 48
y = 48 / 1.20
= 40

 Number of girls = 40

Let the number of correct answers be X.

Number of incorrect answers = (60 – X).

4x – (60 – x) = 130

=> 5x = 190 

=> x = 38

Given are the two AP'S:

15,12,9.... in which a=15, d=-3.............(1) 

-15,-13,-11..... in which a'=-15 ,d'=2.....(2)

now using the nth term's formula,we get

a+(n-1)d = a'+(n-1)d'

substituting the value obtained in eq. 1 and 2,

15+(n-1) x (-3) = -15+(n-1) x 2

=> 15 - 3n + 3 = -15 + 2n - 2

=> 12 - 3n = -17 + 2n

=> 12+17 = 2n+3n

=> 29=5n

=> n= 29/5

Let the edge = a cm   

So increase by 50 % = a + a/2 = 3a/2  

Total surface Area of original cube = 6a2  

TSA of new cube = 63a22 =69a24=  13.5a2   

Increase in area = 13.5a2-6a2 =7.5a2  

 7.5a2 Increase % =7.5a26a2×100 = 125%