Quantitative Aptitude - Arithmetic Ability Questions

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

Required number of ways=3C1*6C2+3C2*6C1+3C3 = (45 + 18 + 1) =64

A has to pay = Present worth of Rs220 due 1yr hence  

= Rs.[(220 x 100)/100+(10 x 1)] = Rs.200 

A actually pays = Rs.110+PW  of Rs.110 due 2 yrs hence 

= [110+(110 x 100)] / [100+(10 x 2)] = Rs.192.66  

So, A gains Rs. (200-192.66) = Rs.7.34

Here the ratio of mixtures( i.e milk , water) doesnot matter. But the important point is that whether the total amount ( either pure or mixture ) being transferred is equal or not.Since the total amount ( i.e 5 cups) being transferred from each one to another , hence A =B.

Let the pipe B be closed after 'K' minutes.

30/16 - K/24 = 1 => K/24 = 30/16 - 1 = 14/16

=> K = 14/16 x 24 = 21 min.

In a simultaneous throw of two dice, n(S) = 6 x 6 = 36

Let E = event of getting a doublet = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

 P(E)=n(E)n(S)=636=16

A                   B                 C

3x                 4x                5x 

(3x+x)          2x               (5x+x)         (B Gives 1/4 to A and 1/4 to C) 

=4x              2x                 6x 

(4x+x)          2x                 5x              ( C Gives 1/6 to A) 

=5x              2x                 5x

Therefore, 5 : 2 : 5

If we want to pack the drinks in the least number of cans possible, then each can should contain the maximum numbers of liters possible.As each can contains the same number liters of a drink, the number of liters in each can is a comman factor for 80,144 and 368; and it is also the highest such factor, as we need to store the maximum number of liters in each can.

So, the number of liters in each can  = HCF of 80,144 and 368 = 16 liters.

Now, number of cans of Maaza = 80/16 = 5

Number of cans of Pepsi = 144/16 = 9

Number of cans of Sprite = 368/16 = 23

Thus, the total number of cans required = 5 + 9 + 23 = 37