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# Some of 50%-intensity red paint is replaced with 25%

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Director
Joined: 23 Jan 2013
Posts: 546
Schools: Cambridge'16
Followers: 1

Kudos [?]: 38 [0], given: 40

Re: Some of 50%-intensity red paint is replaced with 25% [#permalink]

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08 Nov 2015, 06:05
Equation

1/2x+1/4y=3/10(x+y)

x/y=1/4

Differential approach

50------30---25

20x=5y

x/y=1/4

x=1/5 and y=4/5 of final solution

E
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Joined: 23 Sep 2015
Posts: 238
Location: France
GMAT 1: 690 Q47 V38
GMAT 2: 700 Q48 V38
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Kudos [?]: 38 [0], given: 48

Re: Some of 50%-intensity red paint is replaced with 25% [#permalink]

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09 Nov 2015, 03:15
The ratio for the 50% solution is 1:1, for the 25% is 1;3 and for the 30% mixture is 3;7

I just do the weighted average to find the weight of the 50% solution in the new 30% solution

$$\frac{x}{2}$$+$$\frac{1(1-x)}{4}$$=$$\frac{3}{10}$$

x = $$\frac{1}{5}$$

So $$\frac{4}{5}$$of the old 50% solution was replaced by 4/5 of the 25% solution

we get 0.2*50 + 0.8*25 = 30
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Joined: 07 Dec 2014
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Some of 50%-intensity red paint is replaced with 25% [#permalink]

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11 Nov 2015, 19:49
x=fraction of solution replaced
.5-.5x+.25x=.3
x=4/5
Manager
Joined: 05 Jul 2015
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GMAT 1: 600 Q33 V40
GPA: 3.3
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Re: Some of 50%-intensity red paint is replaced with 25% [#permalink]

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19 Nov 2015, 16:21
My way was so quick that I had to triple check to make sure I didn't fall for a trick.

50-----30-------25
---20------5-----

20/25 = 4/5
Re: Some of 50%-intensity red paint is replaced with 25%   [#permalink] 19 Nov 2015, 16:21

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