Question

The vessels contain mexture of milk and water in the ratio of 8:1 and 1:5 respecively. Find how much must be drawn from each vessel to fill a third vessel of capacity 21 gallons in order that the resulting mixture may be half milk and half water is

Answer 1

Let x=amount that must be drawn from vessel (8:1)

Then 21-x=amount that must be drawn from vessel(1:5)

First vessel contains (8/9) milk and (1/9) water

Second vessel contains (1/6) milk and (5/6) water

Now we know that the amount of pure milk that exists before the mixture takes place ((8/9)*x+(1/6)(21-x)) has to equal the amount of pure milk that exists after the mixture takes place (1/2)*21. Sooooo:

(8/9)x+(21/6)-(1/6)x=21/2 multiply each term by 18

16x+63-3x=189

13x=126

x=9.69 gal-----amount needed from vessel(8:1)

21-x=21-9.69=11.31 gal----amount needed from vessel(1:5)

CK

(8/9)*(9.69)+(1/6)*(11.31)=(21/2)= 10.5

8.61+1.89=10.5