Answer :
now, the percentage figures of say, 70%, 95% and so on, we'll use the decimal format of the, thus 70% is just 70/100 or 0.7 and 95% is just 95/100 or 0.95 and so on
so [tex]\bf \begin{array}{lccclll} &amount&concentration& \begin{array}{llll} concentration\\ amount \end{array}\\ &-----&-------&-------\\ \textit{70\% pure}&x&0.7&0.7x\\ \textit{95\% pure}&y&0.95&0.95y\\ -----&-----&-------&-------\\ mixture&160&0.85&(160)(0.85) \end{array}[/tex]
whatever "x" and "y" amounts are, we know, they must add up to 160 pints
thus x + y = 160
now, whatever the concentrated amount of solute in the mixture, we know they must add up to 160*0.85 or 136
thus 0.7x + 0.95y = 136
thus [tex]\bf \begin{cases} x+y=160\implies \boxed{y}=160-x\\ 0.7x+0.95y=136\\ ----------\\ 0.7x+0.95\left( \boxed{160-x} \right)=136 \end{cases}[/tex]
solve for "x", to see how much of the 70% pure juice will be needed
what about "y"? well, y = 160 - x
so [tex]\bf \begin{array}{lccclll} &amount&concentration& \begin{array}{llll} concentration\\ amount \end{array}\\ &-----&-------&-------\\ \textit{70\% pure}&x&0.7&0.7x\\ \textit{95\% pure}&y&0.95&0.95y\\ -----&-----&-------&-------\\ mixture&160&0.85&(160)(0.85) \end{array}[/tex]
whatever "x" and "y" amounts are, we know, they must add up to 160 pints
thus x + y = 160
now, whatever the concentrated amount of solute in the mixture, we know they must add up to 160*0.85 or 136
thus 0.7x + 0.95y = 136
thus [tex]\bf \begin{cases} x+y=160\implies \boxed{y}=160-x\\ 0.7x+0.95y=136\\ ----------\\ 0.7x+0.95\left( \boxed{160-x} \right)=136 \end{cases}[/tex]
solve for "x", to see how much of the 70% pure juice will be needed
what about "y"? well, y = 160 - x