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A dessert recipe calls for 50% melted chocolate and 50% rasp

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A dessert recipe calls for 50% melted chocolate and 50% rasp [#permalink] New post 19 Aug 2013, 18:09
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A dessert recipe calls for 50% melted chocolate and 50% raspberry puree to make a particular sauce. A chef accidentally makes 15 cups of the sauce with 40% melted chocolate and 60% raspberry puree instead. How many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50% of each?

A. 1.5
B. 2.5
C. 3
D. 4.5
E. 5
[Reveal] Spoiler: OA

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Re: A dessert recipe calls for 50% melted chocolate and 50% rasp [#permalink] New post 19 Aug 2013, 22:34
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vaishnogmat wrote:
Q) A dessert recipe calls for 50% melted chocolate and 50% raspberry puree to make a particular sauce. A chef accidentally makes 15 cups of the sauce with 40% melted chocolate and 60% raspberry puree instead. How many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50% of each?

a) 1.5
b) 2.5
c) 3
d) 4.5
e) 5


we have 15 cups os sauce with \(40 %\) choc and \(60 %\) rasb
cups of choc = \(0.4*15 = 6\)
cups of rasb = \(0.6*15 = 9\)
now let say we removed x cup of original mix and replaced with x cups of choc.
therefore final number of cups of choc =\(6-0.4x+x\)
now this number of cup should be 50% of total = \(15/2 = 7.5\)
therefore \(6-0.4x+x= 7.5\)
on solving \(x= 2.5\)

hence B
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Re: A dessert recipe calls for 50% melted chocolate and 50% rasp [#permalink] New post 19 Aug 2013, 23:43
I misread the question, Again!! Nyways another method
40% 50%
\ /
50%
/ \
100% 10%

so 50/10 = 15-x/x => x =2.5 cups
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Re: A dessert recipe calls for 50% melted chocolate and 50% rasp [#permalink] New post 20 Aug 2013, 01:20
Expert's post
vaishnogmat wrote:
A dessert recipe calls for 50% melted chocolate and 50% raspberry puree to make a particular sauce. A chef accidentally makes 15 cups of the sauce with 40% melted chocolate and 60% raspberry puree instead. How many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50% of each?

A. 1.5
B. 2.5
C. 3
D. 4.5
E. 5


Similar questions to practice:
m07-72458.html
a-certain-bread-recipe-calls-for-wheat-white-flour-and-oat-103934.html
miguel-is-mixing-up-a-salad-dressing-regardless-of-the-109740.html
malik-s-recipe-for-4-servings-of-a-certain-dish-requires-123239.html
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Hope it helps.
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Re: A dessert recipe calls for 50% melted chocolate and 50% rasp [#permalink] New post 20 Aug 2013, 05:28
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vaishnogmat wrote:
A dessert recipe calls for 50% melted chocolate and 50% raspberry puree to make a particular sauce. A chef accidentally makes 15 cups of the sauce with 40% melted chocolate and 60% raspberry puree instead. How many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50% of each?

A. 1.5
B. 2.5
C. 3
D. 4.5
E. 5


1. In 15 cups the proper mix should be 50% melted chocolate and 50% raspberry puree but the actual mix made was 40% melted chocolate and 60% raspberry puree.

2. Raspberry puree should be 10% less and melted chocolate should be 10% more in the mixture. 10% is equal to 1.5 cups i.e., you need to have the net effect of taking 1.5 cups of raspberry puree out of the mixture and adding 1.5 cups of melted chocolate to the mixture.

3.The net effect of taking out 1 cup of mixture and replacing it with 1 cup of melted chocolate is that of taking out 0.6 cup of raspberry puree and adding 0.6 cup of melted chocolate.

4. So to achieve the desired net effect as in (2) we need to take out 1.5/0.6 i.e., 2.5 cups of the mixture and replace it with the same amount of melted chocolate.

The answer is therefore 2.5 cups.
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Re: A dessert recipe calls for 50% melted chocolate and 50% rasp [#permalink] New post 20 Aug 2013, 11:02
Bunuel wrote:
vaishnogmat wrote:
A dessert recipe calls for 50% melted chocolate and 50% raspberry puree to make a particular sauce. A chef accidentally makes 15 cups of the sauce with 40% melted chocolate and 60% raspberry puree instead. How many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50% of each?

A. 1.5
B. 2.5
C. 3
D. 4.5
E. 5


Similar questions to practice:
m07-72458.html
a-certain-bread-recipe-calls-for-wheat-white-flour-and-oat-103934.html
miguel-is-mixing-up-a-salad-dressing-regardless-of-the-109740.html
malik-s-recipe-for-4-servings-of-a-certain-dish-requires-123239.html
a-certain-bread-recipe-calls-for-whole-wheat-flour-white-129148.html
a-recipe-requires-2-1-2-cups-of-flour-2-3-4-cups-of-sugar-152952.html
what-is-the-ratio-of-the-number-of-cups-of-flour-to-the-72081.html

Hope it helps.





Bunuel, how would you solve this question using a methodical approach? Thanks.
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Re: A dessert recipe calls for 50% melted chocolate and 50% rasp [#permalink] New post 02 Sep 2013, 09:35
vaishnogmat wrote:
A dessert recipe calls for 50% melted chocolate and 50% raspberry puree to make a particular sauce. A chef accidentally makes 15 cups of the sauce with 40% melted chocolate and 60% raspberry puree instead. How many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50% of each?

A. 1.5
B. 2.5
C. 3
D. 4.5
E. 5


Conentrating only on chocolate. Assuming that one would need to replace x cups of 40% chocolate by 100% chocolate, then-

(15-x)*(50-40)=x*(100-50),

i.e., product of distances (here the number of cups) from the mean concentration (i.e., 50%) of both the mixtures, i.e., the original mixture of 40% concentration of chocolate and pure chocolate respectively would be equal.
Simplifying, 15*10=60x. Hence, x=2.5 cups.
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Re: A dessert recipe calls for 50% melted chocolate and 50% rasp [#permalink] New post 21 Nov 2013, 11:46
I just used quick math and started with C)

removing 3 cups, 60% of which is rasp, so you're removing 1.8, leaving you with 7.2 cups, and the remaining 1.2 comes from choco, leaving you with 4.8, adding 3 back in, you end up with too much choco, so it must be a or b. with b, you remove 2.5, 60% of which is rasp, or 1.5, leaving you with 7.5, and the remaining 1 comes from choco, leaving you with 5. Add 2.5 pure choco you get 7.5/7.5, so B) is the answer.

I think sometimes in the time time-span it would take to read, comprehend, figure out a formula, write it down and solve, you could have easily just plugged in the numbers. Remember, the GMAT doesn't know/care if you solved via basic plug-in math like I use, or some elegant formula. All that matters is if you got it correct.
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Re: A dessert recipe calls for 50% melted chocolate and 50% rasp [#permalink] New post 11 Jan 2014, 04:51
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vaishnogmat wrote:
A dessert recipe calls for 50% melted chocolate and 50% raspberry puree to make a particular sauce. A chef accidentally makes 15 cups of the sauce with 40% melted chocolate and 60% raspberry puree instead. How many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50% of each?

A. 1.5
B. 2.5
C. 3
D. 4.5
E. 5


Remember he is replacing the mixture by pure chocolate so with every cup X of the mixture he replaces he will pour x cups of pure chocolate. So we have:

6+x-2/5x = 9-3/5x
x=2.5

B

Hope it helps
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Last edited by jlgdr on 08 Feb 2014, 06:02, edited 1 time in total.
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Re: A dessert recipe calls for 50% melted chocolate and 50% rasp [#permalink] New post 25 Jan 2014, 06:50
SravnaTestPrep wrote:
vaishnogmat wrote:
A dessert recipe calls for 50% melted chocolate and 50% raspberry puree to make a particular sauce. A chef accidentally makes 15 cups of the sauce with 40% melted chocolate and 60% raspberry puree instead. How many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50% of each?

A. 1.5
B. 2.5
C. 3
D. 4.5
E. 5


1. In 15 cups the proper mix should be 50% melted chocolate and 50% raspberry puree but the actual mix made was 40% melted chocolate and 60% raspberry puree.

2. Raspberry puree should be 10% less and melted chocolate should be 10% more in the mixture. 10% is equal to 1.5 cups i.e., you need to have the net effect of taking 1.5 cups of raspberry puree out of the mixture and adding 1.5 cups of melted chocolate to the mixture.

3.The net effect of taking out 1 cup of mixture and replacing it with 1 cup of melted chocolate is that of taking out 0.6 cup of raspberry puree and adding 0.6 cup of melted chocolate.

4. So to achieve the desired net effect as in (2) we need to take out 1.5/0.6 i.e., 2.5 cups of the mixture and replace it with the same amount of melted chocolate.

The answer is therefore 2.5 cups.


Using the same method as SwarnaTestprep, I tried solving this problem using a table. You might find it easy to visualize what is going on when pure/impure cups of the ingredients are added or removed.
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Re: A dessert recipe calls for 50% melted chocolate and 50% rasp [#permalink] New post 23 Jun 2014, 19:39
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vaishnogmat wrote:
A dessert recipe calls for 50% melted chocolate and 50% raspberry puree to make a particular sauce. A chef accidentally makes 15 cups of the sauce with 40% melted chocolate and 60% raspberry puree instead. How many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50% of each?

A. 1.5
B. 2.5
C. 3
D. 4.5
E. 5


Responding to a pm:

Quote:
Using the Scale method
40% 50% 100%
15-x x
Hence
(15-x)/x = 50/10
I cannot understand how 15- x cups can be equal to 40% of chocolate . where X is the cups of Mixture removed and replaced with Pure Chocolate.

My understanding:
The 15 cups are prepared by mistaken proportions of 40% chocolate and 60% Rasberry . Hence when we remove x cups of mixture from 15 cups of Chocolate + Rasberry Mixture , we are left with chocolate equal to 40% of 15-x
Hence now 40%* ( 15-x) Choco will be mixed with x cups of Choco at 100% to obtain choco at 50%

Is this understanding correct?

Will the concentration of chocolate always be at 40% ,in the 15 Cups prepared by mistaken combination , even if we consider 1 cup or 2 cups or x cups of the mixture?


Yes, we assume that the mix is homogeneous. Otherwise, we will not be able to solve the question.

Look at the question from a different perspective for ease (don't mix it up with algebra):

You have 15 cups of sauce with 40% chocolate. You also have unlimited amount of pure chocolate sauce. Now you need to mix these two in such a way that you get total 15 cups of sauce with 50% chocolate.

Using scale method:

w1/w2 = (100 - 50)/(50 - 40) = 5/1
w1 - Amount of 40% chocolate sauce
w2 - Amount of pure chocolate sauce

So for every 5 cups of 40% chocolate sauce, we need 1 cup of pure chocolate sauce. This will give us 6 cups of 50% chocolate sauce. But we need 15 cups of 50% chocolate sauce.
So we need to mix 5*15/6 = 12.5 cups of 40% chocolate sauce with 1*15/6 = 2.5 cups of pure chocolate sauce.

Hence, when we are replacing, we remove 2.5 cups of 40% chocolate sauce and put 2.5 cups of pure chocolate in it.

Answer (B)

Look at example 1 here: http://www.veritasprep.com/blog/2012/01 ... -mixtures/
It is very similar to this question.
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A dessert recipe calls for 50% melted chocolate and 50% rasp [#permalink] New post 23 Jun 2014, 22:23
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s1 0.4*15=6 cups, element 1 and 0.6*15=9 cups, element 2
s2 Remove x cups of the mix i.e., -0.4x cups element 1 and -0.6x cups element 2
s3 Add x cups element 1

The desired ratio of the elements is 1:1

Thus (6 - 0.4x +x) / (9 - 0.6x) = 1/1
=> x=2.5 cups
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Re: A dessert recipe calls for 50% melted chocolate and 50% rasp [#permalink] New post 24 Jun 2014, 01:23
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Chocolate ............. Raspberry ............ Total

6 .............................. 9 ........................ 15

Say "x" quantity is removed; New Equation is..

\(6 - \frac{6x}{15}\) .................. \(9-\frac{9x}{15}\) ..................... 15-x

Whatever the quantity removed, same amount of chocolate is added

\(6 - \frac{6x}{15} + x\) ............... \(9 - \frac{9x}{15}\) .................. 15 - x + x

Addition should be 50% of the total

Equation can be setup in 2 ways:

\(6 - \frac{6x}{15} + x = \frac{50}{100} * 15\) ................. (1)

OR

\(9 - \frac{9x}{15} = \frac{50}{100} * 15\) ................... (2)

Will got with (2) as it has variable only on one side (Minimal Calculations)

\(\frac{9x}{15} = 7.5\)

x = 2.5

Answer = B
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Re: A dessert recipe calls for 50% melted chocolate and 50% rasp [#permalink] New post 28 Oct 2014, 09:25
I got this wrong, but I definitely understand it.

I think this way is the easiest to think about solving the problem. We need to end up with equal parts of two ingredients. In 15 cups, we need 7.5 cups of each ingredient. So, that means, taking out 1.5 of 9 total cups of the raspberry puree. Let's just compute how many total cups should be removed: 1.5/9 = x/15 or EVEN EASIER: 1/6 = x/15 --> x = 2.5
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Re: A dessert recipe calls for 50% melted chocolate and 50% rasp [#permalink] New post 01 Nov 2014, 12:42
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The best formula ever to solve most of the REMOVE AND REPLACE mixture questions=

Suppose a container contains x of liquid from which y units are taken out and replaced by water.

After n operations, the quantity of pure liquid = \(x ( 1- \frac{y}{x})^n\)

Lets use it here =

\(\frac{1}{2}* 15= 7.5\) which is a desired value.

Hence,

\(7.5 = 9 ( 1 - \frac{x}{15})^1\)

\(\frac{5}{6}= \frac{15-x}{15}\)

\(6x = 15\)

\(x = 2.5\)

P.S = You can use 6 as well for chocolate sauce in this formula.


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Re: A dessert recipe calls for 50% melted chocolate and 50% rasp [#permalink] New post 07 Nov 2014, 14:37
blueseas wrote:
vaishnogmat wrote:
Q) A dessert recipe calls for 50% melted chocolate and 50% raspberry puree to make a particular sauce. A chef accidentally makes 15 cups of the sauce with 40% melted chocolate and 60% raspberry puree instead. How many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50% of each?

a) 1.5
b) 2.5
c) 3
d) 4.5
e) 5


we have 15 cups os sauce with \(40 %\) choc and \(60 %\) rasb
cups of choc = \(0.4*15 = 6\)
cups of rasb = \(0.6*15 = 9\)
now let say we removed x cup of original mix and replaced with x cups of choc.
therefore final number of cups of choc =\(6-0.4x+x\)
now this number of cup should be 50% of total = \(15/2 = 7.5\)
therefore \(6-0.4x+x= 7.5\)
on solving \(x= 2.5\)

hence B


Hi,
I was with you until " final number of cups of choc =\(6-0.4x+x\)"


After I came up with 6 and 9, i proceeded to divide the options in half. What I mean is, for option B, 2.5 -- if we removed 2.5, that means that we would remove half of the 2.5 = 1.25 of chocolate and 1.25 of puree. I'm not sure why you removed 40%(although I can see that 40% represents the chocolate percent). Logically, if we remove the sauce, wouldn't we remove equal parts of puree and equal parts of chocolate?
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Re: A dessert recipe calls for 50% melted chocolate and 50% rasp [#permalink] New post 25 Jan 2015, 09:05
russ9 wrote:
blueseas wrote:
vaishnogmat wrote:
Q) A dessert recipe calls for 50% melted chocolate and 50% raspberry puree to make a particular sauce. A chef accidentally makes 15 cups of the sauce with 40% melted chocolate and 60% raspberry puree instead. How many cups of the sauce does he need to remove and replace with pure melted chocolate to make the sauce the proper 50% of each?

a) 1.5
b) 2.5
c) 3
d) 4.5
e) 5


we have 15 cups os sauce with \(40 %\) choc and \(60 %\) rasb
cups of choc = \(0.4*15 = 6\)
cups of rasb = \(0.6*15 = 9\)
now let say we removed x cup of original mix and replaced with x cups of choc.
therefore final number of cups of choc =\(6-0.4x+x\)
now this number of cup should be 50% of total = \(15/2 = 7.5\)
therefore \(6-0.4x+x= 7.5\)
on solving \(x= 2.5\)

hence B


Hi,
I was with you until " final number of cups of choc =\(6-0.4x+x\)"


After I came up with 6 and 9, i proceeded to divide the options in half. What I mean is, for option B, 2.5 -- if we removed 2.5, that means that we would remove half of the 2.5 = 1.25 of chocolate and 1.25 of puree. I'm not sure why you removed 40%(although I can see that 40% represents the chocolate percent). Logically, if we remove the sauce, wouldn't we remove equal parts of puree and equal parts of chocolate?


Hi, There! I guess I'm a few months too late on this response, but I'll try to give it a go. :)

When we remove cups of the sauce, we're removing parts of both chocolate and puree --- according to their respective percentages.

In this case: we have 15 cups of sauce. The prompt asks us to remove "X" amount of cups from the sauce, and add the same "X" amount of chocolate - to give us an equal 7.5/7.5 split.
We're not splitting the "X" amount.

For Choice B, if we take 2.5 out of 15 ... we have 12.5 cups of Sauce: giving us 5 cups of Chocolate and 7.5 cups of Puree (since we have to take 40% choc. and 60% Puree from the sauce)
Now, adding the same amount, 2.5 back into the Chocolate, we have the perfect split: 7.5/7.5 ... Hence, B is correct.

Now, how did we get there through the method you tried to follow?

First of all, let's think logically: we have 15 cups of Sauce, broken down to as you pointed out, 6 Chocolate and 9 Puree.
Focus only on the Chocolate. We need to raise its initial value UP to 7.5 by removing cups of the sauce, and replacing the SAME AMOUNT with cups of chocolate.

Now, to get 7.5, we need to remove "X" amount of the 15 cups of sauce and MULTIPLY that value by 40% --- giving us the chocolate value of the "reduced" sauce value.
Just like we earlier with Choice B: (15-2.5)(2/5) = 5

Then, we need to add the SAME "X" amount back into the chocolate. And, that's how we get this equation:
7.5 = 2/5 (15 - X) + X
7.5 = 6 - .4X + X
1.5 = .6X
X = 2.5

A good problem, for good practice. Hope that helped!
Re: A dessert recipe calls for 50% melted chocolate and 50% rasp   [#permalink] 25 Jan 2015, 09:05
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