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ration problem

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ration problem [#permalink]

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New post 29 Sep 2010, 00:19
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I am sure this will be easy for you guys.. but i had issues in understanding the explaination in OG, here goes the question:


qs 335 og10) the ratio, by volume, of soap to alcohol to water in a certain solution is 2.5.100. The solution will be altered so that the ratio of soap to alcohol is doubled while the ratio of soap to water is halfed. if the altered solution will contain 100 cubic cetimeters of alcohol, how many centemeters of water will it contain ?

50
200
400
625
800

I am getting 400
.. but the answer says 800 ?? can you explain what am i doing wrong here

S:A:W 2:50:100
new S:A 4:100
new S:W 1:100

if soap is if alcohol is 100 soap is 4 and water is 4*100= 400
[Reveal] Spoiler: OA

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Re: ration problem [#permalink]

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New post 29 Sep 2010, 00:41
vanidhar wrote:
I am sure this will be easy for you guys.. but i had issues in understanding the explaination in OG, here goes the question:


qs 335 og10) the ratio, by volume, of soap to alcohol to water in a certain solution is 2.50.100. The solution will be altered so that the ratio of soap to alcohol is doubled while the ratio of soap to water is halfed. if the altered solution will contain 100 cubic cetimeters of alcohol, how many centemeters of water will it contain ?

50
200
400
625
800

I am getting 400
.. but the answer says 800 ?? can you explain what am i doing wrong here

S:A:W 2:50:100
new S:A 4:100
new S:W 1:100

if soap is if alcohol is 100 soap is 4 and water is 4*100= 400


Given: \(\frac{s}{{\frac{a}{w}}}=\frac{2}{{\frac{50}{100}}}\);

The ratio of soap to alcohol is doubled --> \(\frac{s}{a}=2*\frac{2}{50}=\frac{4}{50}\);

The ratio of soap to water is halved --> \(\frac{s}{w}=\frac{1}{2}*\frac{2}{100}=\frac{1}{100}=\frac{4}{400}\);

New ratio: \(\frac{s}{{\frac{a}{w}}}=\frac{4}{{\frac{50}{400}}}\) --> \(\frac{a}{w}=\frac{50}{400}\) --> if \(a=2*50=100\) then \(w=2*400=800\).

Answer: E.
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600-700 ratios problem [#permalink]

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New post 18 Oct 2010, 11:23
The ratio by volume of soap to alcohol to water is 2:50:100. The solution will be altered so that the ratio of soap to alcohol is doubled while the ratio of soap to water is halved. If the altered solution will contain 100 cubic centimeters of alcohol, how many cubic centimeters of water will it contain?


a 50
b 200
c 400
d 625
e 800

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Re: 600-700 ratios problem [#permalink]

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New post 18 Oct 2010, 12:05
rtaha2412 wrote:
The ratio by volume of soap to alcohol to water is 2:50:100. The solution will be altered so that the ratio of soap to alcohol is doubled while the ratio of soap to water is halved. If the altered solution will contain 100 cubic centimeters of alcohol, how many cubic centimeters of water will it contain?


a 50
b 200
c 400
d 625
e 800


Ratio of S:A:W is 2:50:100

means S/A=2/50=1/25 and A/W= 50/100=1/2 Hence S/W = 1/50

Now in the altered solution S'/A'=(1/25)*2=2/25
While S'/W'=(1/50)*(1/2)=1/100
Hence A'/W'=(1/100)/(2/25)=1/8=25/(8*25) = 25/200

Therefore S':A':W'= 2:25:200

Now let the volume of new solution is x ml.

Alcohol Content ={25/(2+25+200)}*x =100
Hence x =(100*227)/25=908 ml

Hence the water content in new solution = {200/(2+25+200)}*908 = 800

Hence answer is E.

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water [#permalink]

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New post 20 Feb 2011, 08:29
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Re: water [#permalink]

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New post 20 Feb 2011, 08:36
Initial ratio is S:A:W = 2:50:100
After changes, the ratio becomes 4:50:400 (as 2:50 is doubled so S becomes 4 units and 2:100 is halved, so new water units are 400 as new ratios need to be 2:200 or 4:400)
Now, If 50 corresponds to 100 cc, then 400 corresponds to 800 cc - Ans E

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Re: water [#permalink]

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New post 20 Feb 2011, 09:13
2:50:100

\(\frac{2}{50}\) to be doubled \(\frac{2*2}{50} = \frac{4}{50}\)

\(\frac{2}{100}\) is to be halved \(\frac{2}{2*100} = \frac{2}{200} = \frac{2*2}{200*2} = \frac{4}{400}\)

New proportion
Soap:Alcohol:Water
4:50:400

Alcohol is 100cc: Double of 50
Thus entire proportion should be multiplied by 2

8:100:800

Ans: "E"
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Re: water [#permalink]

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New post 20 Feb 2011, 10:14

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Re: ration problem [#permalink]

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Re: ration problem   [#permalink] 14 Dec 2015, 07:16
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