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If grapes are 92% water and raisins are 20% water, then how [#permalink]

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02 Sep 2013, 09:59

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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

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

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.

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

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?

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.

Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]

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03 Apr 2014, 23:10

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

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.

Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]

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10 May 2015, 08:43

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Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]

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06 Jun 2015, 19:46

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.

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.

Hi, to look at the flaw , you will have to give your solution. i think where you could be going wrong in proportionality is by reducing the non water portion too in same ratio. It is clearly given that the loss is only of water portion, so non water portion will remain the same....
_________________

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.

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

If grapes are 92% water and raisins are 20% water [#permalink]

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11 Aug 2015, 10:16

If grapes are 92% water and raisins are 20% water by weight, 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.)

If grapes are 92% water and raisins are 20% water by weight, 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.)

If grapes are 92% water and raisins are 20% water, then how [#permalink]

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06 Dec 2015, 09:45

if we take the grape ratio and apply to it 100 pounds then we will have 92 pounds of water and 8 pounds of pure pulp. the stem says also that out of 10 pounds - 8 pounds is pure pulp.

hence we look up and see that this amount of pure pulp (8 pounds) used to have 92 corresponding pounds of water when this was grape that is 100 in total
_________________

Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]

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21 May 2016, 12:26

1

This post received KUDOS

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.

Re: If grapes are 92% water and raisins are 20% water, then how [#permalink]

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08 Sep 2016, 10:34

Dear All

This is how i tried to solve it.

Letting x equal to the number of pounds of grapes

x * 0.92 * 0.2 should be the pounds of water left after 80% of the water in the grapes evaporates

x * 0.92 * 0.2 = 2 should be equal to pounds of water that is in the raisins. so we should be able to solve for x. but this equation does not lead to any solution.

Where i am making the mistake?

Regards

B

gmatclubot

Re: If grapes are 92% water and raisins are 20% water, then how
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08 Sep 2016, 10:34

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