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# How much water (in grams) should be added to the

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VP
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29 Oct 2008, 01:07
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How much water (in grams) should be added to the 35%-solution of acid to obtain the 10%-solution?

1. There are 50 grams of the 35%-solution
2. In the 35%-solution the ratio of acid to water is 7:13

what is the best way to approach mixture problems ?
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29 Oct 2008, 01:27
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(A)

It is easier to approach the mixture problems by forming grids just like 2D set problems.
(1) gives the value of y and hence x can be calculated.

0.35y = 0.1(x+y)

(2) States that there is 35% of acid in the solution. Doesn't help to find the value of x.

Use this website for the basic mixture problems. Actually, jallenmorris pointed me to the purplemath website.
http://www.purplemath.com/modules/mixture.htm

This will guide you through the methodology and you can google for mixture problems for practice (this is how I learnt how to solve these type of problems- though this kind still haunts me )

HTH.
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Last edited by leonidas on 29 Oct 2008, 01:50, edited 1 time in total.

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Director
Joined: 23 May 2008
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29 Oct 2008, 01:38
leonidas wrote:
(A)

It is easier to approach the mixture problems by forming grids just like 2D set problems.
(1) gives the value of y and hence x can be calculated.

0.35y = 0.1(x+y)

(2) States that there is 35% of acid in the solution. Doesn't help to find the value of x.

Use this website for the basic mixture problems. Actually, jallenmorris pointed me to the purplemath website.
http://www.purplemath.com/modules/mixture.htm

This will guide you through the methodology and you can google for mixture problems for practice (this is how I learnt how to solve these type of problems- though this kind still haunt me )

HTH.

i used this method also but to get D i did this:
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Book1.xls [13.5 KiB]

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Senior Manager
Joined: 29 Mar 2008
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29 Oct 2008, 01:49
bigtreezl wrote:
leonidas wrote:
(A)

It is easier to approach the mixture problems by forming grids just like 2D set problems.
(1) gives the value of y and hence x can be calculated.

0.35y = 0.1(x+y)

(2) States that there is 35% of acid in the solution. Doesn't help to find the value of x.

Use this website for the basic mixture problems. Actually, jallenmorris pointed me to the purplemath website.
http://www.purplemath.com/modules/mixture.htm

This will guide you through the methodology and you can google for mixture problems for practice (this is how I learnt how to solve these type of problems- though this kind still haunt me )

HTH.

i used this method also but to get D i did this:

Here the ratio of acid:water in the 35% solution is 7:13. How did you get 13 gms of liquid?
If there is any information about the weight of acid we can use (7/20)* total = (weight of acid) to find the weight of 35% solution (i.e, y in my case). Since, we are not given any weight (gms), (2) is insuff.
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To find what you seek in the road of life, the best proverb of all is that which says:
"Leave no stone unturned."
-Edward Bulwer Lytton

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VP
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29 Oct 2008, 02:07
OA is A.

Thanks leonidas +1!
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Re: DS : Mixtures   [#permalink] 29 Oct 2008, 02:07
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