Quantitative Aptitude - Arithmetic Ability Questions

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 

16 : 56 :: 32 : ?

Let the required number be 'n'

The given numbers follow a logic that,

16/56 = 2/7

=> 32/n = 2/7

=> 32 x 7/2 = n

=> n = 112.

Hence, 16 : 56 :: 32 : 112.

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.

Let x and y be the two parts of 96.

x/7 = y/9 => x:y = 7:9

=> The smallest part is = 7/16 x 96 = 42.

We have to rearrange the equation to make R the subject.

Start by cross multiplying by (r + R); V (r + R) = 12R

Multiply out the bracket Vr + VR = 12R

One day work of 6 boys and 8 girls is given as 6b + 8g = 1/10 -------->(I)

One day work of 26 boys and 48 women is given as 26b + 48w = 1/2 -------->(II)

Divide both sides by 2 in (I) and then multiply both sides by 5

Now we get, 15b + 20g = 1/4.

Therefore, 15 boys and 20 girls can do the same work in 4 days.

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

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.