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If after 200 grams of water were added to the 24% solution

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gracie
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Re: If after 200 grams of water were added to the 24% solution [#permalink] New post   Updated on: 07 Jan 2020, 15:44
If after 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased by one-third, how much of the 24%-solution was used?

A. 180 grams
B. 220 grams
C. 250 grams
D. 350 grams
E. 400 grams

let x=total grams in 24% solution
.76x+200=.84(x+200)→
x=400 grams
E

Originally posted by gracie on 23 Oct 2015, 19:16.
Last edited by gracie on 07 Jan 2020, 15:44, edited 2 times in total.
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Re: If after 200 grams of water were added to the 24% solution [#permalink] New post   10 Jun 2017, 04:16
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bibha wrote:
If after 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased by one-third, how much of the 24%-solution was used?

A. 180 grams
B. 220 grams
C. 250 grams
D. 350 grams
E. 400 grams

1. Let the quantity of the solution be x.
2. (Quantity of 24% solution)* (Strength of the solution )+ ( Quantity of water added)*(strength of water)/ (total quantity)=strength of the resultant solution
3. (x*0.24 ) + (200*0)/ (x+200) =16/100
x=400
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oa7
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Re: If after 200 grams of water were added to the 24% solution [#permalink] New post   12 Jun 2017, 17:23
Is there a standard way to attack mixture problems? There are so many different ways of solving the problem its confusing.
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gracie
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Re: If after 200 grams of water were added to the 24% solution [#permalink] New post   12 Jun 2017, 19:13
bibha wrote:
If after 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased by one-third, how much of the 24%-solution was used?

A. 180 grams
B. 220 grams
C. 250 grams
D. 350 grams
E. 400 grams


let x=grams of 24% solution used
.76x+200=.84(x+200)
x=400 grams
E
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Re: If after 200 grams of water were added to the 24% solution [#permalink] New post   12 Jun 2017, 19:25
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oa7 wrote:
Is there a standard way to attack mixture problems? There are so many different ways of solving the problem its confusing.

You are right. Most of the problems in a topic can be solved by using a common approach. I try to use such an approach. So even if for some problems there may be quicker solutions than a standard solution, trying to think of such quick solutions take time. In a standard approach, you automatically apply the same steps for seemingly different types of problems.
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Re: If after 200 grams of water were added to the 24% solution [#permalink] New post   20 Jun 2017, 07:33
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bibha wrote:
If after 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased by one-third, how much of the 24%-solution was used?

A. 180 grams
B. 220 grams
C. 250 grams
D. 350 grams
E. 400 grams


Since 1/3 of 24 is 8, we know that after 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased to 16%. We can let the original amount of solution = x, and thus the original amount of alcohol = 0.24x. We can create the following equation and solve for x:

0.24x/(x + 200) = 0.16

0.24x = 0.16(x + 200)

0.24x = 0.16x + 32

0.08x = 32

x = 32/0.08 = 3200/8 = 400

Answer: E
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Re: If after 200 grams of water were added to the 24% solution [#permalink] New post   22 Sep 2018, 22:29
Simple Equation:
.24x/(.76x+200)=16/84
=> x=400 grams
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Re: If after 200 grams of water were added to the 24% solution [#permalink] New post   13 Dec 2018, 21:01
24% of x = (24%*2/3) of (x+200)
Solving, x = 400

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Re: If after 200 grams of water were added to the 24% solution [#permalink] New post   24 Jan 2020, 10:12
bibha wrote:
If after 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased by one-third, how much of the 24%-solution was used?

A. 180 grams
B. 220 grams
C. 250 grams
D. 350 grams
E. 400 grams


alcohol_after=alcohol_before(volume_initial/volume_final)
0.24*2/3=0.24(x/x+200)
0.16x+200(0.16)=0.24x
0.08x=200(0.16)
x=200*2=400

Ans (E)
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Re: If after 200 grams of water were added to the 24% solution [#permalink] New post   14 Apr 2022, 09:47
Using weighted average:

Solution has : 24% alcohol and 76% water: ratio A/W=24/76=6/19; we do not know the grams (x) so we have 6x/19x

We add 200 grams of water and the concentration in alcohol decreases by 1/3 : % alcohol= 24*2/3 = 16
New ratio A/W = 16/84 = 4/21.
So A/(W+200) = 4/21

Now to put together both ratios:
6x/(19x+200) = 4/21 ==> x=16

Our first ratio has: 6x+19x = 25x (25 parts).
25*16 = 400 grams
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Re: If after 200 grams of water were added to the 24% solution [#permalink] New post   23 Feb 2023, 04:57
Let alcohol be Bottle A , and pure water be Bottle B
Bottle A contains 24 % alcohol, and
Bottle B contains 0% alcohol.

The percentage of Alcohol in the resulting mixture is 24*(1-1/3) = 16 { As the alcohol concentration is reduced by one-third .. one third of 24 is 8 , 24 - 8 = 16}.

24 0
\ /
16
/ \
16: 8
i.e in the Ratio of 2:1.

if one part is 200 grams, then 2 part is 400 grams.
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Re: If after 200 grams of water were added to the 24% solution [#permalink] New post   04 Mar 2024, 20:25
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Re: If after 200 grams of water were added to the 24% solution [#permalink]
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