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# One fourth of a solution that was 10 percent sugar by weight

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CEO
Joined: 21 Jan 2007
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One fourth of a solution that was 10 percent sugar by weight [#permalink]

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17 Oct 2007, 13:19
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

One fourth of a solution that was 10 percent sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight. The second solution was what percent sugar by weight ?

(A) 34%

(B) 24%

(C) 22%

(D) 18%

(E) 8.5%

How do I set this one up using the algebra?
Intern
Joined: 25 Apr 2007
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17 Oct 2007, 13:42
I think its A.

The way i calculated is as follows;

original solution had 10% sugar by weight ( so 100 gms contains 10 g sugar)
you replace 1/4 of it so thats 25 gms. (we can also find out that 25gms contains 2.5gms of sugar)

Now the resulting solution has 16% - 100gms contains 16gms.

75 gms = 7.5 g
100 gms = 16 g

So new solution of 25 gms has 8.5 gms of sugar which is 34% for 100 g.

Hence A.
VP
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17 Oct 2007, 15:29
see attachment.

since the old ratio is 3:1 (one fourth was replaced) and the new ratio is (x-16):6 then --->

3/1 = (x-16)/6

18 = x-16

34 = x

Attachments

solution.GIF [ 2.98 KiB | Viewed 748 times ]

Director
Joined: 08 Jun 2007
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17 Oct 2007, 20:30
bmwhype2 wrote:
One fourth of a solution that was 10 percent sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight. The second solution was what percent sugar by weight ?

(A) 34%

(B) 24%

(C) 22%

(D) 18%

(E) 8.5%

How do I set this one up using the algebra?

My way :

Assume 100 gm
replaced 25 gm

weight of sugar in 75 gm + weight of sugar in adding 25 gm solution = total weight in 100 gm

10/100 * 75 + x/100 * 25 = 16/100 * 100
x=34
CEO
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17 Oct 2007, 22:16
bmwhype2 wrote:
One fourth of a solution that was 10 percent sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight. The second solution was what percent sugar by weight ?

(A) 34%

(B) 24%

(C) 22%

(D) 18%

(E) 8.5%

How do I set this one up using the algebra?

Prolly more complicated than it needs to be, this one took me bout 15minutes =( I hate mixtures.

1/4x*1/10 --> 1/40x weight of sugar in the 1/4 mixture.

1/40x is replaced with something that yeilds 16/100x. 16/100x is the total weight of the sugar in x. Now to find the weight of the replacement we can take 3/40x (the amount that wasn't replaced, 3/40x is essentially 10% of 3/4x) and subtract it from the total.

16/100x -3/40x ---> 32/200x-15/200x ---> 17/200x equals weight of sugar in the new mixture.

We want weight of sugar over new mixture so 17/200x/1/4x the x's cancel and were left w/ 68/200 ---> .34

A
Re: Mixture   [#permalink] 17 Oct 2007, 22:16
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