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# Five kilograms of oranges contained 98% of water.

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Intern
Joined: 10 Jun 2012
Posts: 8
Five kilograms of oranges contained 98% of water.  [#permalink]

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27 Oct 2012, 05:30
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55% (hard)

Question Stats:

61% (01:43) correct 39% (02:09) wrong based on 709 sessions

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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? Please, share your ideas!
Math Expert
Joined: 02 Sep 2009
Posts: 64239
Re: Five kilograms of oranges contained 98% of water.  [#permalink]

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27 Oct 2012, 06:24
15
12
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.

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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Intern
Joined: 10 Jun 2012
Posts: 8
Re: Five kilograms of oranges contained 98% of water.  [#permalink]

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27 Oct 2012, 06:47
Nice explanation! Thank you! Achievement Unlocked. =)
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Re: Five kilograms of oranges contained 98% of water.  [#permalink]

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29 Oct 2012, 05:33
1
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.

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.
Math Expert
Joined: 02 Sep 2009
Posts: 64239
Re: Five kilograms of oranges contained 98% of water.  [#permalink]

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29 Oct 2012, 05:36
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.

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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19 Jun 2013, 18:02
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? Please, share your ideas!

0.02*5 = 0.04*x => x=2.5
Intern
Joined: 29 May 2013
Posts: 5
Re: Five kilograms of oranges contained 98% of water.  [#permalink]

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30 Jun 2013, 06:07
1
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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18 Jul 2014, 10:18
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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Posts: 64239
Re: Five kilograms of oranges contained 98% of water.  [#permalink]

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18 Jul 2014, 11:12
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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01 Oct 2014, 18:40
1
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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Five kilograms of oranges contained 98% of water.  [#permalink]

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27 Jun 2016, 22:27
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.

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.

I hope the GMAT differentiates clearly between "percentage" and "percentage points"...If it doesn't..then please tell how would it be stated differently..so there are no surprises in the test.
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Re: Five kilograms of oranges contained 98% of water.  [#permalink]

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01 Feb 2019, 17:43
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

The initial weight of the water is 5 x 0.98 = 4.9.

Since only water evaporates, we can let x = the weight of the water that evaporates and create the equation:

(4.9 - x)/(5 - x) = 0.96

4.9 - x = 0.96(5 - x)

4.9 - x = 4.8 - 0.96x

0.1 = 0.04x

4x = 10

x = 2.5

Since 2.5 lbs of water is lost, the new weight of the oranges is 5 - 2.5 = 2.5 lbs.

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

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16 Mar 2019, 07:44
My Take:

Generally:
(A1*W1 + A2*W2) / (W1 + W2) = Aavg

In this case:(Water evaporates)
(A1*W1 - A2*W2) / (W1 - W2) = Aavg

=> (98*5 - 100*n) / (5 - n) = 96; Here water concentration = 100% and weight of water = n; 96% because water concentration decreased by 2%
=> 98*5 - 100n = 96*5 - 96n
=> 5(98-96) = 100n - 96n
=> 10 = 4n
=> n = 2.5
Therefore, new weight of oranges = 5 - 2.5 = 2.5 kgs
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Re: Five kilograms of oranges contained 98% of water.  [#permalink]

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05 Aug 2019, 10:05
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? Please, share your ideas!

for 5kgs water = 4.9 kgs ans .1 kgs is weight
later 4% or say 0.04*x=.1
x=2.5
IMO C
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Re: Five kilograms of oranges contained 98% of water.  [#permalink]

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05 Aug 2019, 23:41
The answer can't be C because that's nearly 50% reduction in the weight of the oranges which can't be possible with 2% reduction in water. The question has some data missing.

Posted from my mobile device
Re: Five kilograms of oranges contained 98% of water.   [#permalink] 05 Aug 2019, 23:41

# Five kilograms of oranges contained 98% of water.

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