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Five kilograms of oranges contained 98% of water.

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Five kilograms of oranges contained 98% of water. [#permalink]

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New post 27 Oct 2012, 06:30
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Five kilograms of oranges contained 98% of water. If the next day the concentration of water decreased by 2%, what was the new weight of the oranges, in kilograms?

(A) 4.9
(B) 4.8
(C) 2.5
(D) 2
(E) 0.2

I've got the answer (A) but the correct one is (C). How can it be possible? :o Please, share your ideas!
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Re: Five kilograms of oranges contained 98% of water. [#permalink]

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New post 27 Oct 2012, 07:24
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Five kilograms of oranges contained 98% of water. If the next day the concentration of water decreased by 2%, what was the new weight of the oranges, in kilograms?

(A) 4.9
(B) 4.8
(C) 2.5
(D) 2
(E) 0.2

If C is the answer, then the question means that the concentration of water the newt day became 96%.

Out of 5 kilograms 98%, or 4.9 kilograms was water and 0.1 kilograms was non-water.

The next day, after some water evaporated, oranges became 96% water and 4% of non-water, so the next day 0.1 kilograms of non-water composed 4% of oranges, which means that the new weight of the oranges was x*0.04=0.1 --> x=2.5 kilograms.

Answer: C.

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

Hope it helps.
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Re: Five kilograms of oranges contained 98% of water. [#permalink]

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New post 27 Oct 2012, 07:47
Nice explanation! Thank you! Achievement Unlocked. =)
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Re: Five kilograms of oranges contained 98% of water. [#permalink]

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New post 29 Oct 2012, 06:33
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Bunuel wrote:
Five kilograms of oranges contained 98% of water. If the next day the concentration of water decreased by 2%, what was the new weight of the oranges, in kilograms?

(A) 4.9
(B) 4.8
(C) 2.5
(D) 2
(E) 0.2

If C is the answer, then the question means that the concentration of water the newt day became 96%.

Out of 5 kilograms 98%, or 4.9 kilograms was water and 0.1 kilograms was non-water.

The next day, after some water evaporated, oranges became 96% water and 4% of non-water, so the next day 0.1 kilograms of non-water composed 4% of oranges, which means that the new weight of the oranges was x*0.04=0.1 --> x=2.5 kilograms.

Answer: C.

Similar question to practice: each-of-the-cucumbers-in-100-pounds-of-cucumbers-is-composed-102354.html

Hope it helps.


I also thought that it is A, Bunuel do you think that it is real GMAT like question, because i think that it is a little unclear.
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Re: Five kilograms of oranges contained 98% of water. [#permalink]

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New post 29 Oct 2012, 06:36
Expert's post
ziko wrote:
Bunuel wrote:
Five kilograms of oranges contained 98% of water. If the next day the concentration of water decreased by 2%, what was the new weight of the oranges, in kilograms?

(A) 4.9
(B) 4.8
(C) 2.5
(D) 2
(E) 0.2

If C is the answer, then the question means that the concentration of water the newt day became 96%.

Out of 5 kilograms 98%, or 4.9 kilograms was water and 0.1 kilograms was non-water.

The next day, after some water evaporated, oranges became 96% water and 4% of non-water, so the next day 0.1 kilograms of non-water composed 4% of oranges, which means that the new weight of the oranges was x*0.04=0.1 --> x=2.5 kilograms.

Answer: C.

Similar question to practice: each-of-the-cucumbers-in-100-pounds-of-cucumbers-is-composed-102354.html

Hope it helps.


I also thought that it is A, Bunuel do you think that it is real GMAT like question, because i think that it is a little unclear.


I agree, the wording is ambiguous. So, not a GMAT like question. Check the link in my previous post for similar question with better wording.
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Re: Five kilograms of oranges contained 98% of water. [#permalink]

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New post 19 Jun 2013, 19:02
rainbooow wrote:
Five kilograms of oranges contained 98% of water. If the next day the concentration of water decreased by 2%, what was the new weight of the oranges, in kilograms?

(A) 4.9
(B) 4.8
(C) 2.5
(D) 2
(E) 0.2

I've got the answer (A) but the correct one is (C). How can it be possible? :o Please, share your ideas!



0.02*5 = 0.04*x => x=2.5
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Re: Five kilograms of oranges contained 98% of water. [#permalink]

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New post 30 Jun 2013, 07:07
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98% water =4.9 Kg
2% Non Water = .1kg (I)

After evaporation, let new weight of oranges be X kgs.

Water is 2 % less. 96 % of X.
Therefore, non water is 4% of X. (II)

Weight of Non water is same, before and after evaporation,so equate I and II.

4/100 of X kg =.1Kg, Solve for X.

X= .1*100/2.5

Thanks.
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Re: Five kilograms of oranges contained 98% of water. [#permalink]

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New post 18 Jul 2014, 11:18
Hey Bunuel how did you assume that 98% of water implies 98% of the weight share...is it not ambiguous?..Or am I over thinking?
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Re: Five kilograms of oranges contained 98% of water. [#permalink]

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New post 18 Jul 2014, 12:12
Expert's post
JusTLucK04 wrote:
Hey Bunuel how did you assume that 98% of water implies 98% of the weight share...is it not ambiguous?..Or am I over thinking?


As I mentioned here: five-kilograms-of-oranges-contained-98-of-water-141401.html#p1137031 the wording is indeed ambiguous. You can practice better quality questions testing the same concept which are given in my post above.
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Re: Five kilograms of oranges contained 98% of water. [#permalink]

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New post 01 Oct 2014, 19:40
Weight of water in the beginning = 5*0.98

Weight of water after loss of x kg weight = 5*0.98 - x

Weight of Oranges after loss of weight = 5 - x

New concentration = ( 5*0.98 - x ) / ( 5 - x ) x 100 = 96 (since it's 2% less now)

solve for x
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Re: Five kilograms of oranges contained 98% of water. [#permalink]

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Re: Five kilograms of oranges contained 98% of water.   [#permalink] 05 Oct 2015, 13:12
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