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.

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

2.5 × 1 + 1 = 3.5

3.5 × 2 + 2 = 9

9 × 4 + 4 = 40

40 × 8 + 8 = 328