Solution -
In a mixture of petrol and kerosene,
The petrol = 99
Let,
The kerosene = x
Total mixture of petrol and kerosene = 99 + x
In another mixture of petrol and Kerosene,
Where,
Total volume would be 198 litres less than the actual mixture.
The petrol = 99 (same quantity)
The kerosene = x - 198
Total mixture = 99 + x - 198
= x - 99
Then,
The concentration of petrol in the actual mixture would have been 13.33% point less than that of the new mixture.
A/C
[99 / (x - 99) × 100] - [99 / (x + 99) × 100] = 13.33
[9900 / (x - 99)] - [ 9900 / (x + 99)] = 13.33
9900 [1 / (x - 99) - 1 / (x + 99)] = 13.33
9900 [ (x + 99 - x + 99) / (x² - 99²)] = 13.33
[9900 (198)] / (x² - 99²) = 40/3
x² - 99² = 99² × 15
x² - 9801 = 9801 × 15
x² - 9801 = 147015
x² = 147015 + 9801
x² = 156816
x = (156816)½
x = 396
Thus,
The kerosene = 396 litres
Total mixture = 99 + x
= 99 + 396
= 495 litres
Therefore,
The actual concentration of petrol = [99 / (495)] × 100
= 9900 / 495
= 20%
Hence,
The correct answer is option (a) 20%.
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