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6 litres of pure acid is added to 10 l of a mixture of acid

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by Brent@GMATPrepNow » Mon Jul 24, 2017 10:44 am
Ravi 1596 wrote:6 litres of pure acid is added to 10 litres of a mixture of acid and water with 40% acid.What is the percentage of acid in the new solution
ORIGINAL mixture
Volume = 10 liters
40% acid
40% of 10 liters = 4 liters
So, the ORIGINAL mixture contains 4 liters of pure acid

NEW mixture
6 liters added to 10 liters = 16 liters
So, the TOTAL volume = 16 liters

The amount of pure acid in the NEW mixture = 4 liters + 6 liters = 10 liters

So, 10/16 of the NEW mixture is pure acid

10/16 = 5/8 = [spoiler]62.5%[/spoiler]

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by GMATGuruNY » Mon Jul 24, 2017 12:51 pm
Ravi 1596 wrote:6 litres of pure acid is added to 10 l of a mixture of acid and water with 40% acid.What is the percentage of acid in the new solution
6 liters of 100% acid are combined with 10 liters of 40% acid.
Average acid percentage for all 16 liters = (6*100 + 10*40)/16 = 1000/16 = 125/2 = 62.5.
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by [email protected] » Mon Jul 24, 2017 5:15 pm
Hi Ravi 1596,

Both Brent and Mitch have explained the math behind this question, so I won't rehash any of that here. For future reference though, when posting GMAT questions, you should make sure to post the FULL question (with the 5 answer choices). Sometimes the answer choices are designed to give you a potential 'shortcut' (or help you to avoid a 'long' math approach), so as you continue to study, you should look to take advantage of ALL of the information that you're given to work with - and giving the other site members and Experts access to the full question can help us to show you all of those shortcuts.

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by rsarashi » Tue Jul 25, 2017 9:49 am
Hello Experts,

What am I doing wrong?

10L mixture is given, so x be the acid in this mixture.

Now 6L of pure acid is added. So the equation would be like below.

(x+6/10+6)*100 = 40/100

Please explain if I understood the question in wrong way.

Thanks.
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by Brent@GMATPrepNow » Tue Jul 25, 2017 2:26 pm
rsarashi wrote:Hello Experts,

What am I doing wrong?

10L mixture is given, so x be the acid in this mixture.

Now 6L of pure acid is added. So the equation would be like below.

(x+6/10+6)*100 = 40/100

Please explain if I understood the question in wrong way.

Thanks.
That calculation assumes that the NEW solution is 40% acid (AFTER the 6 liters of pure acid is added), whereas the question tells us that that ORIGINAL solution is 40% acid.
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by Scott@TargetTestPrep » Wed Jul 26, 2017 3:23 pm
Ravi 1596 wrote:6 litres of pure acid is added to 10 l of a mixture of acid and water with 40% acid.What is the percentage of acid in the new solution
The current mixture contains 10 x 0.4 = 4 liters of acid and 6 liters of water. If 6 liters of acid are added, the amount of acid is now 10 liters and the total solution is now 16 liters.

Thus, the percentage of acid is 10/16 x 100 = 62.5%.

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by rsarashi » Thu Jul 27, 2017 8:56 am
That calculation assumes that the NEW solution is 40% acid (AFTER the 6 liters of pure acid is added), whereas the question tells us that that ORIGINAL solution is 40% acid.
Hi Brent ,

Thank you so much for your help sir.

All clear.
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