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Percentage-Mixture

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by vinay1983 » Sun Aug 25, 2013 6:46 am
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%
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by Brent@GMATPrepNow » Sun Aug 25, 2013 7:10 am
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.
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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%

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by [email protected] » Sun Aug 25, 2013 3:47 pm
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.

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by GMATGuruNY » Sun Aug 25, 2013 5:12 pm
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%
Since 30% of the original 30 liters is NON-petrol, the amount of non-petrol = .3(30) = 9 liters.

After pure petrol is added to increase the petrol percentage to 95%, these 9 liters of non-petrol must constitute 5% of the new total:
9 = .05t
t = 9/.05 = 900/5 = 180.

Since the new total = 180 liters, the amount of pure petrol added to the original 30 liters = 180-30 = 150 liters.
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by abhasjha » Mon Aug 26, 2013 7:09 am
here is a shortcut where you can save time vinay ..

petrol percentage in original solution = 100- 30 = 70

petrol percentage in required solution = 100-5 = 95

therefore litres of petrol to be added = solution x (required-present)/ (100-required)
= 30x (95-70) / (100-95)
= 30x (25)/5 = 30x5= 150 ... the answer .

you only need to keep in mind one thing before you apply the formula .. that the percentage component must be same .. that is if in original case impurity is 30 percent and later on it becomes 5 % this means impurity reduces in the final mixture ... which means it is petrol that is getting added and not water. so we need to have percentage of petrol and not water before we apply the formula .. can save your valuable time which u can utilise in later questions or your weaker areas in the test ....
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by mk101 » Mon Aug 26, 2013 4:40 pm
Since the impurity in the solution reduces to factor = 1/6 : the quantity of solution should increase to 6 times its original value
Applying the above logic - we have the new quantity as 30 x 6 = 180
Quantity of pure petrol added = 180 -30 = 150

Remember - we are not changing the quantity of impurity - therefore
Strength of impurity x the quantity = constant = Quantity of impurity
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