Alligation or Mixture problems
One type of liquid contains 25 % of benzene, the other contains 30% of benzene. A can is filled with 6 parts of the first liquid and 4 parts of the second liquid. Find the percentage of benzene in the new mixture.
Let the capacity of the pot be 'P' litres.Quantity of milk in the mixture before adding milk = 4/9 (P - 8)After adding milk, quantity of milk in the mixture = 6/11 P.6P/11 - 8 = 4/9(P - 8)10P = 792 - 352 => P = 44. The capacity of the pot is 44 liters.
W i n e ( l e f t ) W a t e r ( a d d e d ) = 343 169 It means W i n e ( l e f t ) W i n e ( i n i t i a l a m o u n t ) = 343 512 ? 343 + 169 = 512 Thus , 343 x = 512 x 1 - 15 K 3 ? 343 512 = 7 8 3 = 1 - 15 k 3 => K = 120 Thus the initial amount of wine was 120 liters.
By the rule of alligation: C.P. of 1 kg sugar of 1st kind C.P. of 1 kg sugar of 2nd kind Therefore, Ratio of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3. Let x kg of sugar of 1st kind be mixed with 27 kg of 2nd kind. Then, 7 : 3 = x : 27 or x = (7 x 27 / 3) = 63 kg.
Ratio of milk and water = 2 : 1
Quantity of milk = 60 X 2/3 = 40 litre
Quantity of water = 20 litre
To make ratio, 1: 2, we have to double the water that of milk
So, water should be 80 litre.
That means 80 ? 20 = 60 litre water to be added.
EXAMIANS