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proportions

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VP
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27 Jan 2008, 23:19
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Please, some help on this one

Some of 50%-intensity red paint is replaced with 25% solution of red paint such that the new paint intensity is 30%. What fraction of the original paint was replaced?
 1/30
 1/5
 2/3
 3/4
 4/5
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28 Jan 2008, 02:59
Same as mixture and alligations

Like 50% is mixed with 25%
got 30% intensity paint

ratio in which the original one mixed with the other one willbe

(50-30)%/(30-25)% = 4:1 ratio

so fraction of original paint will be 4/5
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VP
Joined: 22 Nov 2007
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28 Jan 2008, 23:17
hksingh83 wrote:
Same as mixture and alligations

Like 50% is mixed with 25%
got 30% intensity paint

ratio in which the original one mixed with the other one willbe

(50-30)%/(30-25)% = 4:1 ratio

so fraction of original paint will be 4/5

can you explain me why it is (50-30)%/(30-25)%? thanks
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Schools: Chicago Booth 2011
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29 Jan 2008, 09:48
1
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marcodonzelli wrote:
Please, some help on this one

Some of 50%-intensity red paint is replaced with 25% solution of red paint such that the new paint intensity is 30%. What fraction of the original paint was replaced?
 1/30
 1/5
 2/3
 3/4
 4/5

Let H = amt of high intensity paint, L = amt of low intensity paint

We want H/L --- What is the ratio of H to L in the final mixture?

(0.5H + 0.25L)/(H + L) = 0.3 ---> 0.5H + 0.25L = 0.3H + 0.3L

0.2H = 0.05L ----> H/L = 1/4 ... so for every 1 part H we have 4 part L in the mixture

We now have 4 L for every H left in the can, Thus 4/5 of the H was replaced by L

(We could have solved for L/H = 4, thus 4 L for every 1, same thing)
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29 Jan 2008, 23:18
marcodonzelli wrote:
hksingh83 wrote:
Same as mixture and alligations

Like 50% is mixed with 25%
got 30% intensity paint

ratio in which the original one mixed with the other one willbe

(50-30)%/(30-25)% = 4:1 ratio

so fraction of original paint will be 4/5

can you explain me why it is (50-30)%/(30-25)%? thanks

Check the PDF attachment from Killersquirrel on how to attack any mixture type problems. It explains the logic for above.

Thanks Killersquirrel for this excellent document

http://gmatclub.com/forum/7-p415108
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30 Jan 2008, 03:50
I hope to offer a different,but straight forward solution as:
some part of a 50 % conc. solution was replaced by 25% conc. solution to make it an overall 30% solution.
But da end result is that the solution contained something which had a 30% conc. of two different concentrations(say X and Y) in different percentages..and what were da percents..they are: 50% and 25%
=> 50(X)/100 + 25(Y)/100 = 30(X+Y)/100
=>4X = Y =>Y/X = 4/1 which means that required ratio of fnal 25% liq in the total solution is Y/X+Y which is equal to 4/5..
its not a long method..but the explanation makes it look long..!
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31 Jan 2008, 12:31
marcodonzelli wrote:
Please, some help on this one

Some of 50%-intensity red paint is replaced with 25% solution of red paint such that the new paint intensity is 30%. What fraction of the original paint was replaced?
 1/30
 1/5
 2/3
 3/4
 4/5

i hate those alligation ratio approaches. stragihtforward algebra with a clear understanding of concentration, amounts, and total are much better.

x = replaced
let final amount = 1 unit (1lb, 1 gram,. whatever...)

.5(1-x) + .25x = .30(1)
.5 - .5x + .25x = .3
.2 = .25x
20 = 25x
20/25 = x
4/5
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Re: proportions   [#permalink] 31 Jan 2008, 12:31
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