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Five kilograms of oranges contained 98% of water.
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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? Please, share your ideas!
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Re: Five kilograms of oranges contained 98% of water.
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27 Oct 2012, 07:24
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 nonwater. The next day, after some water evaporated, oranges became 96% water and 4% of nonwater, so the next day 0.1 kilograms of nonwater 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: eachofthecucumbersin100poundsofcucumbersiscomposed102354.htmlfreshdatescontain90waterwhiledrydatescontain143245.htmlHope it helps.
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Re: Five kilograms of oranges contained 98% of water.
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27 Oct 2012, 07:47
Nice explanation! Thank you! Achievement Unlocked. =)



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Re: Five kilograms of oranges contained 98% of water.
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29 Oct 2012, 06:33
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 nonwater. The next day, after some water evaporated, oranges became 96% water and 4% of nonwater, so the next day 0.1 kilograms of nonwater 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: eachofthecucumbersin100poundsofcucumbersiscomposed102354.htmlHope 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.
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29 Oct 2012, 06:36
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 nonwater. The next day, after some water evaporated, oranges became 96% water and 4% of nonwater, so the next day 0.1 kilograms of nonwater 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: eachofthecucumbersin100poundsofcucumbersiscomposed102354.htmlHope 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.
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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? 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.
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30 Jun 2013, 07:07
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
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Re: Five kilograms of oranges contained 98% of water.
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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.
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18 Jul 2014, 12:12
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: fivekilogramsoforangescontained98ofwater141401.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.
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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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Five kilograms of oranges contained 98% of water.
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27 Jun 2016, 23:27
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 nonwater. The next day, after some water evaporated, oranges became 96% water and 4% of nonwater, so the next day 0.1 kilograms of nonwater 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: eachofthecucumbersin100poundsofcucumbersiscomposed102354.htmlfreshdatescontain90waterwhiledrydatescontain143245.htmlHope 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.
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01 Feb 2019, 18:43
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. Answer: C
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Re: Five kilograms of oranges contained 98% of water.
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16 Mar 2019, 08:44
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(9896) = 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.
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05 Aug 2019, 11:05
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.
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06 Aug 2019, 00:41
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.
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Re: Five kilograms of oranges contained 98% of water.
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