Quantitative Aptitude - Arithmetic Ability Questions

Clearly, we  have : x = (3y + 3z ) / 6 or 2x = y + z.

There are 7 letters in the word ‘Bengali of these 3 are vowels and 4 consonants.

Considering vowels a, e, i as one letter, we can arrange 4+1 letters in 5! ways in each of which vowels are together. These 3 vowels can be arranged among themselves in 3! ways.

Total number of words = 5! x 3!= 120 x 6 = 720

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.

x128 = 162x

Then x² = 162x128

   = Sqrt of 8² x 6² x 3²

   = 8 x 6 x 3

   x²  = 144.

 x  = 12.

The maximum number of bows will be 4 yards (= 4 x 36 inches) divided by 15 inches.

This gives 9.6. But as a fraction of a bow is no use, we can only make 9 bows.

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.

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 

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