In a mixture of petrol and kerosene,the quant | vedclass

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(A) Let the total volume of the actual mixture be $V$ litres.Quantity of petrol $= 99 	ext{ litres}$.Concentration of petrol in the actual mixture $=

Vedclass · Accepted Answer

$20$

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(A) Let the total volume of the actual mixture be $V$ litres.Quantity of petrol $= 99 	ext{ litres}$.Concentration of petrol in the actual mixture $= \frac{99}{V} 	imes 100 \%$.In the new mixture,the total volume is $(V - 198) 	ext{ litres}$.Concentration of petrol in the new mixture $= \frac{99}{V - 198} 	imes 100 \%$.According to the problem,the concentration in the actual mixture is $13.33 \%$ (which is $\frac{40}{3} \%$) less than the new mixture:$\frac{99}{V - 198} 	imes 100 - \frac{99}{V} 	imes 100 = \frac{40}{3}$.Dividing by $100$:$99 \left( \frac{1}{V - 198} - \frac{1}{V} ight) = \frac{40}{300} = \frac{2}{15}$.$99 \left( \frac{V - (V - 198)}{V(V - 198)} ight) = \frac{2}{15}$.$99 	imes \frac{198}{V(V - 198)} = \frac{2}{15}$.$V(V - 198) = 99 	imes 99 	imes 15 = 147015$.$V^2 - 198V - 147015 = 0$.Solving the quadratic equation,we get $V = 495$.Actual concentration $= \frac{99}{495} 	imes 100 = \frac{1}{5} 	imes 100 = 20 \%$.

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