Quantitative Aptitude - Arithmetic Ability Questions

surface area = [2(10*4 + 4*3 + 10*3)]sq.cm = 164 sq.cm

Let the side of the square plot be 'a' ft.
Given area of the plot (a x a) = 289 => a = 17
Length of the fence = Perimeter of the plot = 4a = 68 ft.
Cost of building the fence = 68 x 58 = Rs. 3944.

Let the original price of the cycle be 100. After the first reduction the price will be 75.

This new price is then reduced by 20% = 0.8 x 75 = 60

60 represents a reduction of 40 percent on the original.

πr2h2πr2h = 924264  

=> r = (24/364) x 2 = 7m 

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

Required Rato =2r/h  =14/6 = 7:3

In 1hr, the bus covers 60 km without stoppages and 40 km with stoppages.

Stoppage time = time take to travel (60 - 40) km i.e 20 km at 60 km/hr.

stoppage time = 20/60 hrs = 20 min.

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.

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

Passing marks are 0.6x 

so 0.3x + 30 = 0.6x   => x = 100