Question

A drum contained 200 liters of kerosene. 5 liter of kerosene was lost due to leakage. What percent of kerosene remained in the drum?

Answer 1

Hint: The initial quantity of kerosene is 200 liters and 5 liter of kerosene was last due to leakage. Now, calculate the quantity of the kerosene remaining in the drum. Then, calculate the percentage of the remaining quantity of kerosene using the formula, \[\dfrac{\text{quantity of kerosene remained in the drum}}{\text{initial quantity of kerosene in the drum}}\times 100\] . Now, solve it further and calculate the percentage.

Complete step-by-step answer:
According to the question, it is given that a drum contained 200 liters of kerosene. 5 liter of kerosene was last due to leakage. We have to find the percentage of kerosene remaining in the drum.
The quantity of kerosene that the drum initially contained = 200 liters …………………………(1)
Now, some quantity of kerosene was last due to leakage in the drum.
The quantity of kerosene last due to the leakage in the drum = 5 liters ……………………………..(2)
From equation (1), we have the quantity of kerosene that the drum initially contained, and from equation (2), we have the quantity of kerosene last due to the leakage in the drum.
The remaining quantity of kerosene in the drum = \[\left( 200-5 \right)\] liters = 195 liters ……………………………………(3)
The percentage of the kerosene remained in the drum =
\[\dfrac{\text{quantity of kerosene remained in the drum}}{\text{initial quantity of kerosene in the drum}}\times 100\] ………………………………(4)
Now, putting the value of the remaining quantity of kerosene in the drum from equation (3) and the initial quantity of kerosene in the drum from equation (1), in equation (4), we get
The percentage of the kerosene remained in the drum = \[\dfrac{\text{195}\,\text{liters}}{\text{200}\,\text{liters}}\text{ }\!\!\times\!\!\text{ 100}\] = \[\dfrac{195}{2}\] liters = 97.5 liters.
Therefore, the percentage of the kerosene remained in the drum is 97.5 liters.

Note: In this question, one might make a silly mistake and take the quantity 5 liters as the remaining quantity of kerosene in the drum. Then, calculate the percentage of the remaining quantity of kerosene as \[\dfrac{\text{5}\,\text{liters}}{\text{200}\,\text{liters}}\text{ }\!\!\times\!\!\text{ 100}\] . This is wrong because 5 liters is the quantity of leakage of the kerosene. So, the remaining quantity of kerosene in the drum is equal to \[\left( 200-5 \right)\] liters and the percentage of the kerosene remained in the drum is \[\dfrac{\text{195}\,\text{liters}}{\text{200}\,\text{liters}}\text{ }\!\!\times\!\!\text{ 100}\] .