Question 760246
Let the final mix have x ml of 13% solution and y ml of 58% solution.

{{{x + y = 150}}} (it is given that the final mix is 150 ml) {{{ equation (1)}}}

x ml of the mix is equivalent to a 0.13x solution (13% solution of say alcohol means that 100 ml of the solution will have 13 ml of alcohol and the rest as the solvent i.e. water). 

Similarly, y ml of the mix is equivalent of a 0.58x solution and contains 0.58y ml of solute.

But the final mix (150 ml) is 43% strong. i.e it contains 150*0.43 ml of the solute.

Therefore, {{{0.13x + 0.58y = 150 * 0.43 = 64.5}}} {{{ (equation 2) }}}


Solving for simultaneous equations for x and y from equations 1 and 2, we get that x = 50 and y = 100.

(If you don't know how to solve simultaneous equations, the steps are:

eq (1) rewritten as: {{{ 13x + 13y = 150*13 }}} or {{{ 13x + 13y = 1950 }}} eq (3)

eq (2) rewritten as: {{{ 13x + 58y = 6450 }}} eq (4)

Subtracting (3) from (4), {{{ 45y = 4500 }}} or {{{ y = 100 }}}
Since x + y = 150 and y = 100, {{{ x = 50 }}}