Can you solve this algebraically?

There are two beakers A and B ,which are filled with acid and water in the ratio 4:3 and 5:3 respectively.In what ratio ,the mixture A and B to be mixed so that the ratio of acid and water to be 3:2 in the resultant mixture?

Guest Oct 10, 2020

#1**0 **

Add equal parts of water to each beaker as follows:

[4/(3 + w)] + [5/(3 +w)] = 3/2, solve for w

w =3 parts of additional water added to each beaker, so that you would have the ratio of:

Beaker A =4 : 6 and Beaker B =5 : 6 [4/6 + 5/6] =9 : 6 =3 : 2 of acid to water.

Guest Oct 10, 2020

#2**+2 **

Take 7 ml from A (or multiple of 7) and 8 ml (or multiple of 8) from B

A B

4acid + 3 water + 5 acid + 3 water

= 9 acid + 6 water = 3:2 resultant mixture

So 7:8 would be the mix ratio of A :B

ElectricPavlov Oct 10, 2020