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Five kilograms of oranges contained 98% of water. [#permalink]
27 Oct 2012, 05:30

10

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

50% (02:05) correct
50% (01:23) wrong based on 284 sessions

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!

Re: Five kilograms of oranges contained 98% of water. [#permalink]
27 Oct 2012, 06:24

5

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

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.

Re: Five kilograms of oranges contained 98% of water. [#permalink]
29 Oct 2012, 05:33

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 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.

Re: Five kilograms of oranges contained 98% of water. [#permalink]
29 Oct 2012, 05:36

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.

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. _________________

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

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