Question

A bottle contains 3/4 of milk and the rest water. how much of the mixture must be taken away and replaced by equal quantity of water so that the mixture has half milk and half water?

Answer 1

Let the total quantity of mixture = 1 liter.
Now supposing x liters of mixture is withdrawn which contains 3/4x milk and rest water.
So, according to the question 3/4 -3/4x =1/2.
Solving x = 1/3 which means 1/3 of mixture is to be withdrawn to serve the purpose.
So, answer is 33.33%.

Answer 2

$\frac{1}{3}$ of the mixture

$\frac{1}{3}$ of the mixture must be taken away and replaced by equal quantity of water so that the mixture has half milk and half water.

Step-by-step explanation:

Step 1.

Let mixture = 1

Given, milk = $\dfrac{3}{4}$

Then water = 1 - $\dfrac{3}{4}$ = $\dfrac{1}{4}$

Step 2.

Let x amount of mixture be taken away. In x amount of mixture,

milk = $\dfrac{3x}{4}$ and water = $\dfrac{x}{4}$

Then in the remaining mixture,

milk = $\dfrac{3}{4} - \dfrac{3x}{4}$ and water = $\dfrac{1}{4} - \dfrac{x}{4}$

Step 3.

Again, x amount of water is added. Then in the new mixture,

milk = $\dfrac{3}{4} - \dfrac{3x}{4}$ and water = $\dfrac{1}{4} - \dfrac{x}{4} + x$

Step 4.

According to the question,

$\dfrac{3}{4} - \dfrac{3x}{4} : \dfrac{1}{4} - \dfrac{x}{4} + x = \dfrac{1}{2} : \dfrac{1}{2}$

⇒ $\dfrac{3}{4}-\dfrac{3x}{4} : \dfrac{1}{4}+\dfrac{3x}{4}=1:1$

⇒ $\dfrac{3}{4}-\dfrac{3x}{4}=\dfrac{1}{4}+\dfrac{3x}{4}$

• since a : b = c : d gives a × d = b × c

⇒ $\dfrac{3x}{4}+\dfrac{3x}{4}=\dfrac{3}{4}-\dfrac{1}{4}$

⇒ $\dfrac{6x}{4}=\dfrac{2}{4}$

⇒ $6x=2$

⇒ $x=\dfrac{1}{3}$

So, $\dfrac{1}{3}$ of the mixture must be taken away.

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