In a mixture of petrol and kerosene, petrol is only $99$ litres. If this same quantity of petrol would be presented in another mixture of petrol and kerosene where total volume would be $198$ litres less than the actual mixture then the concentration of petrol in the

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Solution

$L e t s a s s u m e a m o u n t o f p e t r o l = x$
$S o, (x * 100) / 198 = 13.33$
$x = 26.3934$
$N o w s a m e a m o u n t o f p e t r o l i s i n 99 l i t e r s a l s o .$
$s o t h e p e r c e n t a g e i n a c t u a l m i x t u r e i s$
$\frac{26.3934 * 100}{99} = 26.66$

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In a mixture of petrol and kerosene, petrol is only 99 litres. If this same quantity of petrol would be presented in another mixture of petrol and kerosene where total volume would be 198 litres less than the actual mixture then the concentration of petrol in the

Lets assume amount of petrol=xSo,(x∗100)/198=13.33x=26.3934 Now same amount of petrol is in 99 liters also.so the percentage in actual mixture is26.3934∗10099=26.66

In a mixture of petrol and kerosene, petrol is only $99$ litres. If this same quantity of petrol would be presented in another mixture of petrol and kerosene where total volume would be $198$ litres less than the actual mixture then the concentration of petrol in the

$L e t s a s s u m e a m o u n t o f p e t r o l = x$
$S o, (x * 100) / 198 = 13.33$
$x = 26.3934$
$N o w s a m e a m o u n t o f p e t r o l i s i n 99 l i t e r s a l s o .$
$s o t h e p e r c e n t a g e i n a c t u a l m i x t u r e i s$
$\frac{26.3934 * 100}{99} = 26.66$