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A cake recipe uses a constant ratio of 2 teaspoons vanilla

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A cake recipe uses a constant ratio of 2 teaspoons vanilla [#permalink] New post 04 Jul 2008, 07:17
A cake recipe uses a constant ratio of 2 teaspoons vanilla extract to 1 ounce chocolate. If the original recipe, which serves four people, is altered proportionally to yield a cake that serves six people, how many ounces of chocolate will be used in the larger cake?

1) The original recipe calls for exactly five teaspoons of vanilla extract.

2) If the original recipe were altered proportionally to yield a cake that serves eight people, ten teaspoons of vanilla extract would be used.

OA to follow. Can you please actually solve the problem to completion. I want to see how the numbers are computed. Thanks!

Last edited by jimmyjamesdonkey on 04 Jul 2008, 07:34, edited 1 time in total.
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Re: Manhattan Ratios [#permalink] New post 04 Jul 2008, 07:30
Jimmy I think you have missed the number of teaspoons of vanilla in second statement. Please edit the question to put the missing information.
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Re: Manhattan Ratios [#permalink] New post 04 Jul 2008, 07:34
Corrected the typo. thanks!
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Re: Manhattan Ratios [#permalink] New post 04 Jul 2008, 07:38
the answer is D. for both cases it's (1/2)*7.5 = 3.75 oz of chocolate
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Re: Manhattan Ratios [#permalink] New post 04 Jul 2008, 07:41
Yes, the OA is D. I was having trouble with Statement 2. The original proportion is V.E/Choc = 2/1 for 4 people. If we wanted to adjust that proportion for 8 people, we would have to double it. Would we multiply the proportion (2/1) * 2 OR would be multiply the proportion (2/1) * (2/2)???
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Re: Manhattan Ratios [#permalink] New post 04 Jul 2008, 07:43
jimmyjamesdonkey wrote:
Yes, the OA is D. I was having trouble with Statement 2. The original proportion is V.E/Choc = 2/1 for 4 people. If we wanted to adjust that proportion for 8 people, we would have to double it. Would we multiply the proportion (2/1) * 2 OR would be multiply the proportion (2/1) * (2/2)???


it's either (1/2)*(1.5*5) or (1/2)*(1.5*10/2). does that help?
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Re: Manhattan Ratios [#permalink] New post 04 Jul 2008, 07:56
Not really...We start with a ratio of 2/1 for 4 people. Statement 2, we start talking about 8 people cake, but we really want to find 6 people cake.

This is how I started to solve (for statement 2, I understand statement 1).

8 people is double 4 people...So I tried to multiply the ratio by 2. (2/1) * 2. Then I would get the amount of chocolate in a 8 person cake. Then I multiple that value by .75 to get the # of chocolate in 6 person cake. Where I botched was multiplying the ratio by 2. I found if I multiplied the numerator and denominator by 2 then this worked fine.

So I guess my question is...How do I double the ratio of 2/1? Do you have to multiply the numerator and denominator of just the numerator by 2?
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Re: Manhattan Ratios [#permalink] New post 04 Jul 2008, 14:40
Quote:
Yes, the OA is D. I was having trouble with Statement 2. The original proportion is V.E/Choc = 2/1 for 4 people. If we wanted to adjust that proportion for 8 people, we would have to double it. Would we multiply the proportion (2/1) * 2 OR would be multiply the proportion (2/1) * (2/2)???


We don’t need to multiply the proportion 2 to 1. It remains the same. It’s the amount of VE and Choco that changes. So it is this amount we need to multiply.

So we have:
St 1.
2/1 = 5/(choco originally)=> choco originally = 5/2. And this amount we need to multiply by 6/4 => 5/2*6/4 = 15/4

St 2.
2/1 = 10/(choco for 8) => choco for 8 = 5. Again, multiply this by 6/8
=> 5*6/8 = 15/4.

Alternatively, you can use this logic:
2/1 = given amount/unknown amount. What the unknown amount should be in order the ration will remain 2/1?

So, for St 2.:
10 VE for eight people means that for 6 people, we need to multiply 10 by 6/8 and thus obtain the amount of VE for 6 people: VE(6) = 15/2. And the amount of Choco should be two times less than the amount of VE (this is where we use the proportion 2/1). Thus, Choco(6) = VE(6)/2 = 15/4.
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Re: Manhattan Ratios [#permalink] New post 04 Jul 2008, 17:52
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Re: Manhattan Ratios [#permalink] New post 05 Jul 2008, 06:05
Very cool. Thanks everyone for your help. While we are on the topic I'd like to ask on more question...about adjusting ratios.

If I said double the proportion (2/1)...how would I do it...would it then be 4/1?

If I said half the proportion (2/1)...would it be 1/1?

Can anyone provide a quick blurb on adjusting proportions?
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Re: Manhattan Ratios [#permalink] New post 06 Jul 2008, 07:06
jimmyjamesdonkey wrote:
Very cool. Thanks everyone for your help. While we are on the topic I'd like to ask on more question...about adjusting ratios.

If I said double the proportion (2/1)...how would I do it...would it then be 4/1?

If I said half the proportion (2/1)...would it be 1/1?

Can anyone provide a quick blurb on adjusting proportions?


I think you are correct.
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Re: Manhattan Ratios [#permalink] New post 06 Jul 2008, 08:43
-> You are correct, while doubling and halving, you multiply and divide the ratio by 2 respectively (not both numerator and denominator of the fractions)
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Re: Manhattan Ratios   [#permalink] 06 Jul 2008, 08:43
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