a chemist has one solution is 50%acid and another solution which is 25%acid.how much each should be mixed to make 10litres of 40%acid solution?(use cramers rule)
Answers
Let the amount of 50% acid be x liters.
And the amount of 25% acid be y litres.
To make 10 litres of 40% acid solution,
\begin{lgathered}\begin{array} { c } { 0.5 x + 0.25 y = 4 \ldots \ldots ( i ) } \\\\ { x + y = 10 \ldots \ldots . ( i i ) } \end{array}\end{lgathered}
0.5x+0.25y=4……(i)
x+y=10…….(ii)
[balancing concentrations]
Multiply the equation (i) by 10.
Then the equation (i) becomes,
( i ) \times 10 \rightarrow 5 x + 2.5 y = 40 \rightarrow ( i i i )(i)×10→5x+2.5y=40→(iii)
And multiply the equation (ii) by 5
Then the equation (ii) becomes,
( i i ) \times 5 \rightarrow 5 x + 5 y = 50 \rightarrow ( i v )(ii)×5→5x+5y=50→(iv)
By subtracting, equation (iii) from (iv)
Subtracting,
\begin{lgathered}\begin{array} { c } { 5 x + 2.5 y = 40 } \\\\ { 5 x + 5 y = 50 } \\\\ { - 2.5 y = - 10 } \end{array}\end{lgathered}
5x+2.5y=40
5x+5y=50
−2.5y=−10
Then,
y = \frac { 10 } { 2.5 } = 4y=
2.5
10
=4
Substituting the value of y, (y = 4) in equation (iii) we get,
\begin{lgathered}\begin{array} { c } { 5 x + 2.5 y = 40 } \\\\ { 5 x + 2.5 ( 4 ) = 40 } \\\\ { 5 x + 10 = 40 } \\\\ { 5 x = 30 } \\\\ { x = 6 } \end{array}\end{lgathered}
5x+2.5y=40
5x+2.5(4)=40
5x+10=40
5x=30
x=6
Hence, 6 litres of 50% solution and 4 litres of 25% solution are needed to form 10 litres of 40% of acid solutions.
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