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aleksley [76]
3 years ago
7

A chemist needs 120 milliliters of a 71% solution but has only 55% and 79% solutions available. Find how many milliliters of eac

h that should be mixed to get the desired solution
Mathematics
1 answer:
irakobra [83]3 years ago
8 0
X = amount of 55% solution in mL
y= amount of 79% solution in mL

\bf \begin{array}{lccclll}
&amount&concentration&\textit{concentration amount}\\
&-----&-----&--------\\
\textit{55\% sol'n}&x&0.55&0.55x\\
\textit{79\% sol'n}&y&0.79&0.79y\\
-----&-----&-----&--------\\
mixture&120&0.71&(120)(0.71)
\end{array}

now.. notice, we use the decimal format for the percents, thus 55% is just 55/100, and 79% is just 79/100 and 71% is well, you guessed it, 71/100 or 0.71  anyway

whatever those amounts of x and y are, they will end up as 120mL total
so, we can say that 

x + y = 120

now, the concentration in each, they'll add up to (120)(0.71)

thus    \bf \begin{cases}
x+y=120\to \boxed{y}=120-x\\\\
0.55x+0.79y=(120)(0.71)\\
----------\\
0.55x+0.79\left( \boxed{120-x} \right)=(120)(0.71)
\end{cases}

solve for "x" to see how much 55% solution will go into the 71% mix

what about the 79% solution? well  y = 120 - x
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