Quantitative Aptitude - Arithmetic Ability Questions

Part filled in 2 hours = 2/6=1/3

Remaining part =1-13=23

 (A + B)'s 7 hour's work = 2/3

(A + B)'s 1 hour's work = 2/21

C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }

=1/6-2/21 = 1/14

C alone can fill the tank in 14 hours.

sum = x. amount =7x/6.

S.I. =(7x/6 – x) = x/6 ; time = 3 years

Rate =[ (100 * x) / (x * 6 * 3)]% = 5.5%

Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years = 90 years.

Sum of the present ages of wife and child (20 x 2 + 5 x 2) years  = 50 years.

Husband's present age = (90 - 50) years = 40 years.

Time taken in walking both ways = 7 hours 45 minutes                     --------(i)

Time taken in walking one way and riding back= 6 hours 15 minutes-------(ii)

By equation (ii)*2 -(i), we have

Time taken to man ride both ways, = 12 hours 30 minutes - 7 hours 45 minutes

                                                        = 4 hours 45 minutes

Let,   B1 : B2 : B3 = 3x : 4x : 5x 

again  B1 : B2 : B3 = 5y : 4y : 3y

Since there is increase in no.of oranges in first two baskets only, it means the no. of oranges remains constant in the third basket

Therefore,   5x = 3y

Hence    3x : 4x : 5x   =>   9y5:12y5:15y5 = 9y:12y:15y

and       5y : 4y : 3y   =>   25x : 20x : 15x

Therfore, increment in first basket = 16

Increment in second basket = 8

Thus, required ratio = 16/8 = 2:1

P(first letter is not vowel) = 24

P(second letter is not vowel) = 13

So, probability that none of letters would be vowels is = 24×13=16