Quantitative Aptitude - Arithmetic Ability Questions

Let P's speed = x km/hr. Then, Q's speed = (7 - x) km/ hr.

So, 24/x + 24/(7 - x) = 14

x2 - 98x + 168 = 0

(x - 3)(x - 4) = 0 => x = 3 or 4.

Since, P is faster than Q,

so P's speed = 4 km/hr and Q's speed = 3 km/hr.

surface area of a cube= 6 x (8 x 8) = 384 sq.ft  

quantity of paint required=(384/16)=24kg  

cost of painting= 36.5 x 24 = Rs.876

πr2h2πr2h = 924264  

=> r = (24/364) x 2 = 7m 

2πrh = 264 ⇒ h = (264 × 722×12 ×17) = 6m   

Required Rato =2r/h  =14/6 = 7:3

There are seven positions to be filled.

The first position can be filled using any of the 7 letters contained in PROBLEM.

The second position can be filled by the remaining 6 letters as the letters should not repeat.

The third position can be filled by the remaining 5 letters only and so on.

Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! ways.

Passing marks are 0.6x 

so 0.3x + 30 = 0.6x   => x = 100

1.12.91 is the first Sunday of December 91.
So, 3.12.91 is the first Tuesday of the month.
Clearly, 10.12.91, 17.12.91, 24.12.91 and 31.12.91 are also Tuesdays.
So, 24.12.91 is the fourth Tuesday.

Let the ratio be 'R'

Total number of students = 1050

Then,

9R + 5R = 1050

14R = 1050

=> R = 75

Hence, the number of boys = 9R = 9 x 75 = 675

Number of shares = 44558.25= 540.

Face value = Rs. (540 x 10) = Rs. 5400.

Annual income = Rs.12100*5400 = Rs. 648.