Question 1037120
 A vessel contains a mixture of 16 litres of milk and 8 litres of water and a second vessel contains a mixture of 16 litres of milk and 5 litres of water.
 How much of the mixture of milk and water should be taken from the first and the second vessels respectively and placed in a third vessel, so that the third vessel may contain a mixture of 20 litres of milk and 8 litres of water?
:
the first mixture, contains a total of 16+8 = 24 liters
 it will be 16/24 or 2/3 milk
the second mixture, contains a total of 16+5 = 21 liters
 it will be 16/21 milk
the resulting mixture will be a total of 20+8 = 28 liters
 it will be 20/28 or 5/7 milk
:
let a = amt of the 1st mixture required
the resulting mixture will have 20 liters, therefore
(20 - a) = amt of the 2nd mixture required
:
A mixture equation
{{{2/3}}}a + {{{16/21}}}(20 - a) = {{{5/7}}}(20)
multiply by 21, cancel the denominators, and you have
7(2a) + 16(20-a) = 3(5(20))
14a + 320 - 16a = 300
14a - 16a = 300 - 320
-2a = -20
a = -20/-2
a = +10 liters of mixture 1
then
20 - 10 = 10 liters of mixture 2
:
:
Check this out, see if the amt of water checks out
A is 1/3 water or 3.33 liters
B is 5/21 water or 2.38 liters\
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resulting amt has 2/7 water or 5.71 liters, the sum.