Quantitative Aptitude - Arithmetic Ability Questions

52.5% of 800 + 30.5% of 2800 = ? + 87.30

420 + 854 - 87.30 = ?

1274 - 87.30 = ?

? = 1186.70

Time = 2 years 4 months = 2(4/12) years = 2(1/3) years.
Amount = Rs'. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)]
=Rs. [8000 * (23/20) * (23/20) * (21/20)]
= Rs. 11109. .
:. C.I. = Rs. (11109 - 8000) = Rs. 3109.

Saving  = 50% of (100 - 40)% of (100 - 30)% of Rs. 18,400 = Rs. (50/100 * 60/100 * 70/100 * 18400) = Rs. 3864.

Given the number series is 4, 6, 12, 30, 81, 315.

Here it follows a pattern that

4

4 x 1.5 = 6

6 x 2 = 12

12 x 2.5 = 30

30 x 3 = 90 not equals to 81

90 x 3.5 = 315

Hence the odd one in the series is 81.

51 + 52 + 53 + ...........+ 100

 = (1 + 2 + 3 + .... + 100) - (1 + 2 + 3 + ...... + 50)

 = It is in the form of n(n+1)2 series summation

= n1 = 100 , n2 = 50

=100100+12 - 5050+12

 =  (5050 - 1275) = 3775

The series given follow the pattern

+2, +2+4, +2+4+6, +2+4+6+8, +2+4+6+8+10,...

Here ? = 7 + 2 + 4 = 13.

From the given statements A, B & C

Let the present age of Teja = P

Raju present age = P + 4

Sita's present age = 2(P + 4)

Average of ages of Sita & Teja = 19

=> (2P + 8 + P)/2 = 19

= 3P + 8 = 38

=> P = 30/3 = 10

Hence, the present age of Teja = P = 10 yrs.

Therefore, by using all the three statements A, B & C we can find the age of Teja.

Let the numbers be x and (100-x).

Then,x100-x=5*495

 =>  x2-100x+2475=0

 =>  (x-55) (x-45) = 0

 =>  x = 55 or x = 45

  The numbers are 45 and 55

Required difference = (55-45) = 10