Question

A car radiator needs a 80% antifreeze solution. The radiator now holds 15 liters of a 70% solutions. How many liters of this should be drained and replaced with 100% antifreeze to get the desired strength?

Answer 1

The answer is 12 question

Answer 2

So... the radiator has 15 liters of 70% antifreeze.. but needs an 80% antifreeze

well, so, you need to drain some and put some with higher percentage, seems to be, you will end up at the same 15 liters, possible the radiator's capacity, of 80% antifreeze

so, the same amount going out, of 70% is the same amount going in, of 100% antifreeze

now.. let's use the decimal format for the percents, or 70% is 70/100 or 0.7 and so on
$\bf \begin{array}{lccclll} &amount&concentration& \begin{array}{llll} concentration\\ amount \end{array}\\ &-----&-------&-------\\ \textit{70\% sol'n, out}&x&0.7&0.7x\\ \textit{100\% sol'n, in}&x&1.00&x\\ \textit{current 70\% sol'n}&15&0.7&10.5\\ -----&-----&-------&-------\\ mixture&15&0.8&12 \end{array}$

so.. let's subtract, from the current solution, 0.7x and add 1x or x, our antifreeze concentration amount, should be 12 though

10.5 - 0.7x + x = 12

solve for "x"