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Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]
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pavan2185 wrote:
If grapes are 92% water and raisins are 20% water, then how much did a quantity of raisins, which currently weighs 10 pounds, weigh when all the raisins were grapes? (Assume that the only difference between their raisin-weight and their grape-weight is water that evaporated during their transformation.)

A. 25 pounds
B. 46 pounds
C. 92 pounds
D. 100 pounds
E. 146 pounds


Spoiler Alert :VeritasPrep Mock Test Question.

Though I could get this question right, I took more than 3 minutes. Is there any simpler way to solve this one?


Responding to a pm:
Actually, the best method for this question is the one given by Bunuel. I used the same method to give the explanation in the mock too. But you can use weighted average if you wish.

Raisins have 20% water. Water is 100% water. You mix these two to make grapes which has 92% water. (This is another way of saying you remove water from grapes to make raisins)
In what ratio will you mix them?

Wr/Ww = (Cw - Cavg)/(Cavg - Cr) = (100 - 92)/(92 - 20) = 8/72 = 1/9

So every 1 unit of raisin will be mixed with 9 units of water to give 10 units of grapes.
So 10 pounds of raisins will mix with 90 pounds of water to give 100 pounds of grapes.

Answer (D)
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Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]
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Bunuel wrote:
pavan2185 wrote:
If grapes are 92% water and raisins are 20% water, then how much did a quantity of raisins, which currently weighs 10 pounds, weigh when all the raisins were grapes? (Assume that the only difference between their raisin-weight and their grape-weight is water that evaporated during their transformation.)

A. 25 pounds
B. 46 pounds
C. 92 pounds
D. 100 pounds
E. 146 pounds


Spoiler Alert :VeritasPrep Mock Test Question.

Though I could get this question right, I took more than 3 minutes. Is there any simpler way to solve this one?


Since only water evaporates, then the weight of pulp (non-water) in grapes and raisins is the same. Thus 0.08*{weight of grapes}=0.8*{weight of raisins} --> 0.08x = 0.8*10 --> x = 100.

Answer: D.

Similar questions to practice:
each-of-the-cucumbers-in-100-pounds-of-cucumbers-is-composed-102354.html
five-kilograms-of-oranges-contained-98-of-water-141401.html
fresh-dates-contain-90-water-while-dry-dates-contain-143245.html

Hope this helps.



(Assume that the only difference between their raisin-weight and their grape-weight is water that evaporated during their transformation.)

Had the above highlighted condition not mentioned, would 46 pounds be the answer.. ?
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Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]
Hi I understand the solution given but why is the answer not 46 pounds? why can't one get the answer by proportionality? what is the flaw in my logic?Would appreciate an answer if possible.
Thanks so much.
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Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]
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Hi sykhong,

Until you show your 'steps', we won't know exactly where you made your mistake. We CAN prove why 46 pounds is NOT the correct answer though....

IF....the grapes weighted 46 pounds total and those grapes were 92% water, then we'd have....

Total = 46 pounds
Water = .92(46) = 42.32 pounds
Not-Water = 3.68 pounds

The part that is NOT-WATER is key here, since that number does NOT change.

The raisins would ALSO contain those 3.68 pounds of not-water, which would represent 80% of the weight (since 20% of the raisins are water).

3.68 = .8(Total Weight)
3.68/.8 = Total Weight
4.6 pounds = Total weight of the raisins

However, the prompt tells us that the raisins are supposed to weigh 10 POUNDS, so the 4.6 pounds is clearly TOO LOW and the grapes must weigh MORE than 46 pounds.

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Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]
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Presenting a slightly alternate approach:

Since the weight of water is the difference between raisins and grapes, let's calculate the weight of water in \(10\) pounds of raisins and water's weight in raisins when they were grapes

Water in raisins
For \(10\) pounds of raisins there is \(20\)% of water. Hence amount of water = \(20\)% of \(10 = 2\) pounds.

Water in grapes before they evaporated
Let \(x\) pounds be the amount of water that evaporated. So when the raisins were grapes,

Total weight of water = \(2 + x\)

Total weight of grapes = \(10 + x\)

When raisins were grapes, \(\frac{Weight of Water}{Weight of Grapes} = 92\)%

i.e. \(\frac{2 + x}{10 + x} = 0.92\).

Solving for \(x\) would give \(x = 90\) pounds.

Hence total weight of 10 pounds of raisins when they were grapes = \(10 + 90 = 100\) pounds

Hope this helps :)

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Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]
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At first glance this questions seems relatively straightforward but I always find these types of questions some what difficult to work out.

But, I think the best way to do it is clearly write down the info you have and use logic to work it out.

So,

Raisins Contain:

20% Water, and
80% Other

For a total of 10lbs.

Therefore, the raisins contain 2lbs of water and 8lbs of other.

Important to note: The question states the other quantity does not change at all.

Grapes contain:

92% water, and
8% other.

8% other is equals to 8lbs. As this 8% is represented by 8lbs of other the 92% must equal 92lbs of water. 92% + 8% = 100%. Therefore, 92lbs + 8lbs = 100lbs. Answer D.
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Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]
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pavan2185 wrote:
If grapes are 92% water and raisins are 20% water, then how much did a quantity of raisins, which currently weighs 10 pounds, weigh when all the raisins were grapes? (Assume that the only difference between their raisin-weight and their grape-weight is water that evaporated during their transformation.)

A. 25 pounds
B. 46 pounds
C. 92 pounds
D. 100 pounds
E. 146 pounds


Spoiler Alert :VeritasPrep Mock Test Question.hough I could get this question right, I took more than 3 minutes. Is there any simpler way to solve this one?

1 The crucial point to understand is that the weight of the non water elements is the same in grapes and raisins. They are just different percentages of different weights but the actual weight is the same.
2. 80% of 10 =8% of x. So x=100.
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If grapes are 92% water and raisins are 20% water, then how [#permalink]
I keep getting thrown off here because I use the following ratios formula:

G/R = 0.08/0.8

Why doesn't this work? The formula works with other ratios (such as when working with similar triangles). Why do we have to make it 0.08G=0.8R instead? Does it have to do with the fact that we're using percentages?
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Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]
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gmat_george wrote:
I keep getting thrown off here because I use the following ratios formula:

G/R = 0.08/0.8

Why doesn't this work? The formula works with other ratios (such as when working with similar triangles). Why do we have to make it 0.08G=0.8R instead? Does it have to do with the fact that we're using percentages?


We are told that 8% of the weight of grapes equal to 80% of the weight of raisins:
0.08*{weight of grapes}=0.8*{weight of raisins}
{weight of grapes}/{weight of raisins}=0.8/0.08
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Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]
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Grapes --- > Raisins
8% solid------8% solid of Grapes
92% water-----20% water

10 pound raisins
means-> water will be 20% of 10 =2
solid will be 10-2 = 8
now 8 is 8% of grapes
Y x 0.08 = 8
Y = 100 OPTION D
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If grapes are 92% water and raisins are 20% water, then how [#permalink]
OFFICIAL SOLUTION

D. Begin this question by accounting for what you know - as grapes, these items were 92% water and so 8% "other". And the "other" portion doesn't change. So if you call the volume 92 units of water and 8 units of "other", then when water in a raisin makes up 20% of the total, the other 80% is made up by the 8 units of "other". Since 8 units = 80%,then each unit is 10%, meaning that the 20% water equates to 2 units.

That means that the weight of a raisin is 2 units water, 8 units "other", for a total of 10 units. And the weight of a grape is 92 units water, 8 units "other" for a total of 100 units. To get from raisin weight to grape weight, then, you'd multiply by 10, meaning that 10 pounds of raisins would have weighed 100 pounds as grapes.
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Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]
pavan2185 wrote:
If grapes are 92% water and raisins are 20% water, then how much did a quantity of raisins, which currently weighs 10 pounds, weigh when all the raisins were grapes? (Assume that the only difference between their raisin-weight and their grape-weight is water that evaporated during their transformation.)

A. 25 pounds
B. 46 pounds
C. 92 pounds
D. 100 pounds
E. 146 pounds


Spoiler Alert :VeritasPrep Mock Test Question.

Though I could get this question right, I took more than 3 minutes. Is there any simpler way to solve this one?


I understood the explanation given in the thread. Although I got the answer as 46, and I still dont understand what went wrong with my approach.

I solved it this way: If some quantity of raisins with 20% water weigh 10 pounds then those same raisins will weigh how much with 92% water content. I cross multiplied and got 46. Can anyone highlight the logical flaw in this?
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Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]
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Hi Chitra657,

Until you show your 'steps', we won't know exactly where you made your mistake. We CAN prove why 46 pounds is NOT the correct answer though....

IF....the grapes weighted 46 pounds total and those grapes were 92% water, then we'd have....

Total = 46 pounds
Water = .92(46) = 42.32 pounds
Not-Water = 3.68 pounds

The part that is NOT-WATER is key here, since that number does NOT change.

The raisins would ALSO contain those 3.68 pounds of not-water, which would represent 80% of the weight (since 20% of the raisins are water).

3.68 = .8(Total Weight)
3.68/.8 = Total Weight
4.6 pounds = Total weight of the raisins

However, the prompt tells us that the raisins are supposed to weigh 10 POUNDS, so the 4.6 pounds is clearly TOO LOW and the grapes must weigh MORE than 46 pounds.

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Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]
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Chitra657 wrote:
pavan2185 wrote:
If grapes are 92% water and raisins are 20% water, then how much did a quantity of raisins, which currently weighs 10 pounds, weigh when all the raisins were grapes? (Assume that the only difference between their raisin-weight and their grape-weight is water that evaporated during their transformation.)

A. 25 pounds
B. 46 pounds
C. 92 pounds
D. 100 pounds
E. 146 pounds


Spoiler Alert :VeritasPrep Mock Test Question.

Though I could get this question right, I took more than 3 minutes. Is there any simpler way to solve this one?


I understood the explanation given in the thread. Although I got the answer as 46, and I still dont understand what went wrong with my approach.

I solved it this way: If some quantity of raisins with 20% water weigh 10 pounds then those same raisins will weigh how much with 92% water content. I cross multiplied and got 46. Can anyone highlight the logical flaw in this?
Experts? VeritasKarishma Bunuel EMPOWERgmatRichC EgmatQuantExpert ScottTargetTestPrep


Chitra657:

What you need to identify is that the thing that remains equal is the pulp, not the water content. I am not sure what you are equating and how, but what you must equate is the pulp (whatever is there other than water remains the same)

80% of 10 pounds = 8% of Grapes
Grapes = 100 pounds
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Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]
Grapes = 92% water
Raisins = 20% water
Raisins lost 80% of water and weigh 10 pounds
Grapes lost 8% of water and weigh -
80% of water --> 10 pounds
8% of water --> 10*8*100/80
= 100 pounds
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Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]
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