Quantitative Aptitude - Arithmetic Ability Questions

Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15 = 15C3  =15*14*133*2*1= 455.  

Let E = event of getting all the 3 red balls.

 n(E) = 5C3 = 5*42*1 = 10.

 => P(E) = n(E)/n(S) = 10/455 = 2/91.

Let the son's present age be x years .Then, (38 - x) = x => x= 19. 

Son's age 5 years back = (19 - 5) = 14 years

Gain % = 100+common gain %2100-100%           

           =100+102100-100

            = 21%

Total number of outcomes possible, n(S) = 10 + 25 = 35

Total number of prizes, n(E) = 10

 P(E)=n(E)n(S)=1035=27

Given number series is 12  9  13.5  30.3  ?  341.7

Here the pattern followed by the series is

12 

12 x 0.75 = 9

9 x 1.5 = 13.5

13.5 x 2.25 = 30.3

30.37 x 3 = 91.12

91.12 x 3.75 = 341.71

Hence, the missing number in the sequence is 91.12.

Ratio of initial investments = 1/2 : 1/3 : 1/4 = 6 : 4 : 3.

Let their initial investments be 6x, 2x and 3x respectively.

A : B : C = (6x * 2 + 3x * 10) : (4x * 12) : (3x * 12) = 42 : 48 : 36 = 7 : 8 : 6.

B's share =  Rs. (378 * 8/21) = Rs. 144.

Given that the student got 125 marks and still he failed by 40 marks

=> The minimum pass mark = 125 + 40 = 165

Given that minimum pass mark = 33% of the total mark

=> total mark =33/100 =165

=> total mark = 16500/33 = 500

Purchase price = Rs.87481-101003 = Rs. [8748 * (10/9) * (10/9 )* (10/9)] = Rs.12000