Problem Solving

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

Hi vinay1983,

In these types of questions, the answers will be numbers, so you can easily TEST THE ANSWERS by plugging them into the given situation and figuring out which answer "matches" the situation.

You'll still need to do some set-up work (in this Q, figuring out that there's 9L of water and 21L of petrol) and you'll need to do some small calculations at the end (if you add 150L of petrol, does the mix become 5%?), but testing the answers can oftentimes be a faster approach to solving the problem.

GMAT assassins aren't born, they're made,
Rich