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# Some part of the 50% solution of acid was replaced with the

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Some part of the 50% solution of acid was replaced with the [#permalink]

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18 Nov 2006, 20:56
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Question Stats:

83% (00:46) correct 17% (01:06) wrong based on 303 sessions

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Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution f acid was obtained. what part of the original solution was replaced?

A. 1/5
B. 1/4
C. 1/2
D. 3/4
E. 4/5

OPEN DISCUSSION OF THIS QUESTION IS HERE: some-part-of-a-50-solution-of-acid-was-replaced-with-an-68805.html
[Reveal] Spoiler: OA

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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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19 Nov 2006, 07:50
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perhaps the fastest way is :
1/2*x + 3/10* x = 0.4
->x= 1/2

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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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22 Jun 2009, 09:43
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Expert's post
parimal wrote:
Hi all.. I thought the answer was within the question itself. 50% of original solution was replace hence the answer is 1/2 or 50%

Wow - you registered in 2006 and this is your first post! Better late than never!
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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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12 Nov 2009, 17:19
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Hermione wrote:
Does anyone have a quick and easy way of solving this problem? I had to pick from the answer choices but I want to know how to solve it outright.

Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution f acid was obtained. what part of the original solution was replaced?

1/5
1/4
1/2
3/4
4/5

Let X = Part of acid solution replaced, then 1-X will be the parts not replaced.

(1-X)*0.5 + 0.3*x = 0.4
0.5 - 0.5X +0.3X = 0.4
-0.2X = -0.1
X=1/2

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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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07 Nov 2010, 06:59
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Expert's post
greatchap wrote:
I am using a chart to solve this problem. Can anyone point out the error I am making?

See image attached below.

x/2 + 3x/10 is the total stuff which is equal to 4
Solving x/2 + 3x/10 = 4 gives x = 5

Since x is the amount you assumed was replaced out of 10, you get 50%.
Nothing wrong with what you did.

Note: Simply put, this question says that some 50% solution was mixed with some 30% solution to get 40% solution. Using weighted averages, you can find out that they must have been mixed in equal quantities or in other words, 50% of the original solution must have been replaced.
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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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18 Nov 2006, 21:22
X be the amount of 50% solution
n be the amount taken out from 50% solution and the same amount of 30% solution is added later
In the resultant solution, acid amount is X/2 â€“ n/2 + 3n/10 = 2X/5
N = Â½

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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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18 Nov 2006, 21:28
Suppose originally there was x litres of solution. Out of which y litres was removed.

After removing the amount y, x-y litres of solution is left.
Amount of acid in x-y = x-y/2

y litres of solution is replaced by y litres of 30 % solution. Amount of acid in y = 3y/10.

In the new mixture, total acid = (x-y)/2 + 30y/100

(x-y)/2 + 3y/10 = (40 % of x)
Solving the equation gives y= x/2

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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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20 Nov 2006, 08:18
cool, short way to solve it, tobiastt! Thanks everyone
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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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20 Jun 2009, 09:38
Does anyone know how to set this problem up using the box method?
Concent.
Amt

Thanks.

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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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22 Jun 2009, 08:20
Hi all.. I thought the answer was within the question itself. 50% of original solution was replace hence the answer is 1/2 or 50%

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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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23 Jun 2009, 09:12
Thanks.. yes.. I have been hiding in the lurks but hoping to be an active participant now.. cheers!

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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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15 Aug 2009, 19:50
50..................40
.........30.........
10..................20

10/20=1/2
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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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12 Nov 2009, 14:08
Let's assume original amount = 1L

Equation is :

0,5 - 0,5x + 0,3x = 0,4
=> x = 0,5 = 1/2

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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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12 Nov 2009, 15:04
I3igDmsu wrote:
Does anyone know how to set this problem up using the box method?
Concent.
Amt

Thanks.

O R A D
VOLUME X Y Y X
CONCENTRATION 50 50 30 40

50X-50Y+30Y=40X, Y=X/2

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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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02 Feb 2010, 05:19
pierrealexandre77 wrote:
Let's assume original amount = 1L

Equation is :

0,5 - 0,5x + 0,3x = 0,4
=> x = 0,5 = 1/2

Thanks! This explanation really worked for me

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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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09 Apr 2010, 16:04
Wow the box method makes these problems so easy. can you explain a bit more on the box method. ORAD?can we use this for most of the misture problems .any limitations?

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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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16 Apr 2010, 13:37
Initially
50 ml acid and 100 ml of total
suppose x part of it is removed.
thus we are left with 50(1-x) of acid and 100(1-x) of total

if we add x part of 30% acid solution (30ml acid and 100ml total)
acid will be 30x and total = 100x

so total acid = 50(1-x)+ 30x = 50-20x
total solution = 100(1-x)+100x = 100

since if it equal to 40%acid solution(40ml acid and 100ml total)

=> (50-20x)/100 = 40/100 => 50-20x = 40 => 20x=10 => x=1/2

This looks lengthy but we can easily skip these steps while solving it.
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Last edited by gurpreetsingh on 16 Apr 2010, 23:57, edited 1 time in total.

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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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16 Apr 2010, 14:31
mst wrote:
Suppose originally there was x litres of solution. Out of which y litres was removed.

After removing the amount y, x-y litres of solution is left.
Amount of acid in x-y = x-y/2

y litres of solution is replaced by y litres of 30 % solution. Amount of acid in y = 3y/10.

In the new mixture, total acid = (x-y)/2 + 30y/100

(x-y)/2 + 3y/10 = (40 % of x)
Solving the equation gives y= x/2

Great explanation! Thank you.

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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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17 Apr 2010, 07:04
tobiastt wrote:
perhaps the fastest way is :
1/2*x + 3/10* x = 0.4
->x= 1/2

cool, it's nice way to answer quickly .

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Re: Some part of the 50% solution of acid was replaced with the [#permalink]

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07 Nov 2010, 00:47
I am using a chart to solve this problem. Can anyone point out the error I am making?

See image attached below.

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Re: Some part of the 50% solution of acid was replaced with the   [#permalink] 07 Nov 2010, 00:47

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