Respuesta :
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
[tex]\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}[/tex]
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"
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
[tex]\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}[/tex]
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"
