A beaker is filled with a liquid, 3 parts of which are water and 7 parts some medicine. What part of the mixture should be replaced with water so that that the resultant mixture has water and medicine in a ratio 1:1?
Answer & explanation
Correct answer: option 1
Let the beaker contains 10 litres of liquid.
⇒ Amount of water = 3 L, Amount of Medicine = 7 L,
⇒ Let x litres of this liquid be replace with water.
⇒ Now, Water = $\frac{3x}{10}$ and Medicine = $\frac{7x}{10}$
⇒ Water before replacement + $\frac{7x}{10}$ = Medicine before replacement = $\frac{7x}{10}$ + $\frac{7x}{10}$ = $\frac{7x}{5}$
⇒ $\frac{7}{10}$ part - $\frac{3}{10}$ part = $\frac{7x}{5}$
⇒ $\frac{2}{5}$ part = $\frac{7x}{5}$
⇒ x = $\frac{2}{7}$
Hence, the part of mixture which should be replaced with water is $\frac{2}{7}$.