Discussion of Question with ID = 027 under Alligations-and-Mixtures

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Question

A mixture of milk and water contains 47 parts of milk and 6 parts of water. How much fraction of the mixture should be removed and replaced by water so that ratio of water and milk becomes equal?

A

\${41/94}\$.

B

\$1{14/31}\$.

C

\$2{37/96}\$.

D

\$3{35/96}\$.

Soln.
Ans: a

Let the volume of the mixture be 47 + 6 = 53 liters. If x liters of the mixture is removed and replaced by water, the volume of water in the new mixture is \$6 - {6x}/53 + x\$. The volume of the milk in the new mixture would be \$47 - {47x}/53.\$ Equating the two volumes and solving for x we get x = \${53 × 41}/{2 × 47}\$. The fraction that must be removed = \$1/53\$ × \${53 × 41}/{2 × 47}\$, which gives \$41/{2 × 47}\$ = \${41/94}\$.