Question

Two varieties of oil are mixed in the ratio $4:3$ to produce the first quality and if they are mixed in the ratio $2:3$, the second quality is obtained. How many kg of the first quality be mixed with $10kg$ of the second quality so that a third quality having the two varieties in the ratio $5:4$ may be produced $?$

Answer 1

Let the common ratio of the first quality oil be $x$.

Given, the ratio is $4:3$

Therefore, the total quantity of the first quality oil $=4x+3x=7x$

Total quantity of the second quality oil $=10kg$

Ratio of second quality oil is $2:3$

The quantity of first oil variety $=\frac{2}{5}\times 10=4kg$

The quantity of second oil variety $=\frac{3}{5}\times 10=6kg$

The third ratio is got by mixing the first quality and $10kg$ of second quality.

So,

$\begin{array}{rcl}\frac{4x+4}{3x+6}& =& \frac{5}{4}\\ 4\left(4x+4\right)& =& 5\left(3x+6\right)\\ 16x+16& =& 15x+30\\ x& =& 30-14\\ x& =& 14\end{array}$

Therefore, the total quantity of first quality oil $=7xkg$

$$\begin{gathered} =7\times 14kg \\ =98kg \end{gathered}$$

Hence, $98kg$ of the first quality should be mixed to get the required result.