vinay1983 wrote:A sample of 30 litres of petrol is found to be adulterated to the extent of 30%. How many liters of pure petrol should be added to the above mixture so that the level of impurity stands reduced at 5%
Let's say that the impurity here is water. So, the 30-liter mixture is 30% water. In other words, the mixture has 9 liters of water and 21 liters of petrol.
Let's let x = the number of liters of petrol we're going to add to our mixture.
At this point, I find it useful to quickly sketch a diagram with the two parts of the mixture (petrol and water) separated. This way I know exactly what's going on.
So, as you can see, once we add x liters of petrol, the new mixture still contains 9 liters of water, and the total volume of the new mixture is 30+x liters.
We want the new mixture to be only 5% impure.
In other words, we want 9/(30+x) = 5%
Or, even better, 9/(30+x) = 5/100
Simplify: 9/(30+x) = 1/20
Cross multiply: (9)(20) = (1)(30+x)
Simplify: 180 = 30 + x
Solve: x = 150
So, we must add
150 liters of petrol to reduce the impurity to 5%
Cheers,
Brent
