Quantitative Aptitude - Arithmetic Ability Questions

In this type of question, we will first calculate the distance covered in given time.

Distance covered will be : Number of revolutions x Circumference 

So we will be having distance and time, from which we can calculate the speed. 

Radius of wheel = 70 / 2 = 35 cm

Distance covered in 40 revolutions will be 40*circumferene=40*2πr=40*2*227*35=8800cm = 88 m

Distance covered in 1 sec  = 8810=8.8m/s=8.8*185km/hr=31.68km/hr

Let the sum be Rs. p, rate be R% p.a. and time be T years.

Then,

 P×T×(R+2)100-P×T×R100=108 ⇒2PT= 10800 ...... (1)

 And, 

 P×R×(T+2)100-P×R×T100=180 ⇒2PR = 18000 ......(2)

Clearly, from (1) and (2), we cannot find the value of p
So, the data is not sufficient.

πr2h2πr2h = 924264  

=> r = (24/364) x 2 = 7m 

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

Required Rato =2r/h  =14/6 = 7:3

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.

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.

2172+1832

= 200+172+200-172

It is in the form of a+b2+a-b2=2a2+b2

= 2(40000 + 289) = 80578.