Solution Problem

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2sun3: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Mon Jul 07, 2008 6:53 am

by 2sun3 » Thu Jul 31, 2008 6:45 am

A chemist has 10 liters of a solution that is 10 percent nitric acid by volume. He wants to dilute the solution to 4 percent strength by adding water. How many liters of water must he add?

If you can give me written explanation behind the step of calculations that would be really helpful.

Thanks

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Source: Beat The GMAT — Problem Solving | Thu Jul 31, 2008 6:45 am

sudhir3127: Legendary Member; Posts: 829; Joined: Mon Jul 07, 2008 10:09 pmFollowed by: 3 members; Location: INDIA; Thanked: 84 times

by sudhir3127 » Thu Jul 31, 2008 6:51 am

my answer is 15L of water...

do let me know if its right,. i will explain u ..

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malolakrupa: Senior | Next Rank: 100 Posts; Posts: 58; Joined: Fri Jul 18, 2008 12:26 pm; Thanked: 2 times

by malolakrupa » Thu Jul 31, 2008 7:00 am

Let x be the total water that needs to be added .

Hence the final volume will be x + 10 as 10 is the initial voulme of the solution

Desired % of nitric acid is = 4 %

Hence the equation would result in

(x+ 10 ) * 0.04 + ( x + 9) = x + 10
0.04x + 0.4 + x+ 9 = x + 10
1.04x + 9.4 = x+ 10
0.04 x = 0.6
x = .6 / 0.04 = 60 / 4 = 15

Thanks,

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VP_RedSoxFan: GMAT Instructor; Posts: 85; Joined: Thu May 01, 2008 12:56 pm; Location: Salt Lake City, UT; Thanked: 24 times; GMAT Score:750+

by VP_RedSoxFan » Thu Jul 31, 2008 7:02 am

If there are 10 L of solution and 10% is nitric acid, then you can say that 1 L is acid and 9 L of water. To have 1 L of acid by 4% of the solution, then you set up this equation:

1L acid/xL solution = 4%

Solve for x and you get 25 L of solution of which 1 L is acid and 24 L is water.

24 L - 9 L = 15L of water added.

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sudhir3127: Legendary Member; Posts: 829; Joined: Mon Jul 07, 2008 10:09 pmFollowed by: 3 members; Location: INDIA; Thanked: 84 times

by sudhir3127 » Thu Jul 31, 2008 7:03 am

i did in a simpler way.. hope its usefull

10L and 10% is nitic acid ...which means 1 lts

now we have 2 make this 1 litre as 4% of the total solution.

hence it is 1/4%= 25lts..

Quantity of water to be added is 25L-10L = 15L.

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2sun3: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Mon Jul 07, 2008 6:53 am

Thanks

by 2sun3 » Thu Jul 31, 2008 9:21 am

Thank you all for your help.

I'll opt for a shorter method as it is my head spins while doing maths. LOL

Let me see if i understood it. If my reasoning is wrong could you point it out to me.

10 liters of solution has 10% of nitric acid, we get 1 litre of nitric acid in 10 liters of solution by 10*10/100= 1

so to make 1 liters as 4% of the total solution 1/4%=25 liters (stating sudhir, which i understood)

since 25 liters is the solution we get for making it 4%, since we already had 10 liters, we need to substract 10 from 25 to get the liters of water added.

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wmensah: Newbie | Next Rank: 10 Posts; Posts: 3; Joined: Sat Feb 28, 2009 12:24 pm

by wmensah » Wed Mar 11, 2009 8:32 am

visit this blog
https://800gmatblog.zoxic.com/page/2

from the mixtures problem, you're taught how to use a table. I applied it to this problem and surprisingly I got the right answer. It was really easy too. Here:

amount % total
sol'n 10 10 100 (ie. 10 * 10)
water x-10 0 0
mixture x 4 4x

% for water is 0 because if he wants to dilute it then he probably wants to add something which doesn't have any of the chemicals in it. We don't know what the total mixture is, so we use x to represent it.

setting up an equation with the last column, we get
100 = 4x
therefore x = 25 (total amount/volume of the mixture)

and therefore water = 25-10 = 15.

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