Quantitative Aptitude - Arithmetic Ability Questions

(2272-875) = 1397, is exactly divisible by N.

Now , 1397 = 11 x 127

The required 3-digit number is 127,the sum of digits is 10.

Sum of digits = (5 + 1 + 7 + x + 3 + 2 + 4) = ( 22 + x ), which must be divisible by 3.

When x=2. [(22 + 2) = 24 is divisible by 3]

So, the answer is 2

Let the number be x.Then

60% of 35of x=36.

=> 60100*35*x=36

=> 9x25=36 => x=25*369=100

Required number is 100.

Let the Number be Y.

Then  Y = 296 q + 75

= (37 x 8)q +( 37 x 2) + 1

= 37 (8q + 2) + 1

Thus, when the number is divided by 37, the remainder is 1

Required numbers are 10,15,20,25,...,95
This is an A.P. in which a=10,d=5 and l=95.
Let the number of terms in it be n.Then t=95
So a+(n-1)d=95.
10+(n-1)*5=95,then n=18.
Required sum=n/2(a+l)=18/2(10+95)=945.

(Place value of 7)-(face value of 7)
=7000-7=6993.

Let the present ages of son and father be x and (60 -x) years respectively.

Then, (60 - x) - 5= 4(x - 5)

55 - x = 4x - 20

5x = 75  => x = 15

Ramesh : Rajesh =(50,000*36) : (75,000*30) =4 : 5.

Therefore, Rajesh's share=Rs.(27,000*5/9)= Rs.15,000.