Question

a vessel is filled with liquid 3 parts of which are water and 5 parts are syrup. how much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?13

Answer 1

Let us suppose that there exists X ml of mixture. Hence the volume of syrup is (5/8)X ml and that of water is (3/8)X ml. Now, remove Y ml of mixture and add Y ml of water to it. Hence, volume of water is (3/8X + 5/8Y) and volume of syrup is (5/8X - 5/8Y). Equating both the values to 1/2X since, both equal to half the volume, we get Y=1/5X. Hence, we need to remove 1/5th part of the mixture and replace it with water

Answer 2

Suppose the vessel initially contains 8 liters of liquid.

Let x liters of this liquid be replaced with water.

Water in new mixture = (3-3x/8+x)

Syrup in new mixture = (5-5x/8)

Then (3-3x/8+x) = (5-5x/8)

5x + 24 = 40 - 5x

10x=16

==>x=8/5

So part of mixture replaced is 8/5*1/8=1/5