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?
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.
+37146 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
