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

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Re: If after 200 grams of water were added to the 24% solution  [#permalink]

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New post 10 Jun 2017, 04:16
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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Re: If after 200 grams of water were added to the 24% solution  [#permalink]

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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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Re: If after 200 grams of water were added to the 24% solution  [#permalink]

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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]

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New post 12 Jun 2017, 19:25
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]

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New post 14 Jun 2017, 04:11
1
oa7 wrote:
Is there a standard way to attack mixture problems? There are so many different ways of solving the problem its confusing.


Usually, weighted averages will help.

Check these three posts:

https://www.veritasprep.com/blog/2011/0 ... -averages/
https://www.veritasprep.com/blog/2011/0 ... -mixtures/
https://www.veritasprep.com/blog/2012/0 ... -mixtures/

They will help you sort out most of the mixture problems.
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Re: If after 200 grams of water were added to the 24% solution  [#permalink]

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New post 20 Jun 2017, 07:33
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]

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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]

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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] 13 Dec 2018, 21:01

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