Quantitative Aptitude - Arithmetic Ability Questions

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

The given series is 8, 14, 24, 60, 180

8

8 x 1.5 = 12 (NOT 14)

12 x 2 = 24

24 x 2.5 = 60

60 x 3 = 180

Hence, the odd man in the given series is 12.

The given number series 410, 409, 413, 404, ? , follows a pattern that 

410

$$\begin{gathered} 410 - 13 = 409 \\ 409 + 22 = 413 \\ 413 - 33 = 404 \\ 404 + 42 = 420 \end{gathered}$$

Hence, the next number in the given series is 420.

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%