
One type of liquid contains 25 % of benzene, the other contains 30 % of benzene. A can is filled with 6 parts of the first liquid and 4 parts of the second liquid. Find the percentage of benzene in the new mixture?
Answer
575.4k+ views
Hint: start by calculating the individual part of benzene in each type of the liquid for the new mixture, which can be done by multiplying the part value to the percentage of benzene present in each liquid. Total benzene in a new mixture is the sum of both individual parts , Find the percentage of the same by dividing the total part used for the new mixture.
Complete step by step answer:
Benzene in first type of liquid = 25%
Benzene in second type of liquid = 30%
Now , A can is filled with 6 parts of the first type of liquid and 4 parts of the second type of liquid.
Which means benzene from first type of liquid = $6 \times \dfrac{{25}}{{100}}$
benzene from second type of liquid= $4 \times \dfrac{{30}}{{100}}$
Therefore , Total part of benzene present in the can=$6 \times \dfrac{{25}}{{100}} + 4 \times \dfrac{{30}}{{100}}$
$
= \dfrac{{150}}{{100}} + \dfrac{{120}}{{100}} \\
= \dfrac{{270}}{{100}} \\
= 2.7 \\
$
Now , Total part of liquid present in the can = 6 + 4 = 10
Therefore , the percentage of Benzene in new mixture $ = \dfrac{{2.7}}{{10}} \times 100 = \dfrac{{270}}{{10}} = 27\% $
Note: Such problems are generally asked in aptitude tests in various exams. One can use the following alternative solution for faster calculation, this is called the method of allegation. This method is used during adulteration or two different ingredients combined to form a new mixture.
\[{\text{ }}\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{25} \\
{}
\end{array} \searrow } \\
{\begin{array}{*{20}{c}}
{} \\
{30 - x}
\end{array} \swarrow }
\end{array}x\begin{array}{*{20}{c}}
{ \swarrow \begin{array}{*{20}{c}}
{30} \\
{}
\end{array}} \\
{ \searrow \begin{array}{*{20}{c}}
{} \\
{25 - x}
\end{array}}
\end{array}\]
So , we follow the arrow direction and subtract the middle value to get a new value which now will be corresponding to some number or ratio. Solve it and we will have our required values.
$
= \dfrac{{30 - x}}{{25 - x}} = \dfrac{6}{4} = \dfrac{3}{2} \\
\Rightarrow 60 - 2x = 3x - 75 \\
\Rightarrow 5x = 60 + 75 \\
\therefore x = 27\% \\
$
Complete step by step answer:
Benzene in first type of liquid = 25%
Benzene in second type of liquid = 30%
Now , A can is filled with 6 parts of the first type of liquid and 4 parts of the second type of liquid.
Which means benzene from first type of liquid = $6 \times \dfrac{{25}}{{100}}$
benzene from second type of liquid= $4 \times \dfrac{{30}}{{100}}$
Therefore , Total part of benzene present in the can=$6 \times \dfrac{{25}}{{100}} + 4 \times \dfrac{{30}}{{100}}$
$
= \dfrac{{150}}{{100}} + \dfrac{{120}}{{100}} \\
= \dfrac{{270}}{{100}} \\
= 2.7 \\
$
Now , Total part of liquid present in the can = 6 + 4 = 10
Therefore , the percentage of Benzene in new mixture $ = \dfrac{{2.7}}{{10}} \times 100 = \dfrac{{270}}{{10}} = 27\% $
Note: Such problems are generally asked in aptitude tests in various exams. One can use the following alternative solution for faster calculation, this is called the method of allegation. This method is used during adulteration or two different ingredients combined to form a new mixture.
\[{\text{ }}\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{25} \\
{}
\end{array} \searrow } \\
{\begin{array}{*{20}{c}}
{} \\
{30 - x}
\end{array} \swarrow }
\end{array}x\begin{array}{*{20}{c}}
{ \swarrow \begin{array}{*{20}{c}}
{30} \\
{}
\end{array}} \\
{ \searrow \begin{array}{*{20}{c}}
{} \\
{25 - x}
\end{array}}
\end{array}\]
So , we follow the arrow direction and subtract the middle value to get a new value which now will be corresponding to some number or ratio. Solve it and we will have our required values.
$
= \dfrac{{30 - x}}{{25 - x}} = \dfrac{6}{4} = \dfrac{3}{2} \\
\Rightarrow 60 - 2x = 3x - 75 \\
\Rightarrow 5x = 60 + 75 \\
\therefore x = 27\% \\
$
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