Five kilograms of oranges contained 98% of water. If the nex

[rakeshd347]

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

[mevicks]

Day 1:
5 KGs contain 98% Water and 2% Other stuffs.
Water = 4.9 Kgs
Others = 0.1 Kgs

Day 2:
4.9 Kgs of water decreases by 2% i.e it becomes 98% of what it was before
Water = 4.8
Others = 0.1
Total = 4.9
Answer A

In a time crunch situation I'd quickly narrow down to 4.8 and 4.9 as the probable answers (as others are extreme situations in a real world situation, an orange getting reduced to HALF or less than what it was yesterday is very conspicuous !). Moreover, the problem assumes some information (other stuffs dont evaporate etc) I'd be weary of the source for my GMAT practice. ... Just my two cents worth

Regards,
Vivek

[rakeshd347]

Day 1:
5 KGs contain 98% Water and 2% Other stuffs.
Water = 4.9 Kgs
Others = 0.1 Kgs

Day 2:
4.9 Kgs of water decreases by 2% i.e it becomes 98% of what it was before
Water = 4.8
Others = 0.1
Total = 4.9
Answer A

In a time crunch situation I'd quickly narrow down to 4.8 and 4.9 as the probable answers (as others are extreme situations in a real world situation, an orange getting reduced to HALF or less than what it was yesterday is very conspicuous !). Moreover, the problem assumes some information (other stuffs dont evaporate etc) I'd be weary of the source for my GMAT practice. ... Just my two cents worth

Regards,
Vivek

Hi Vivek,

Good try and I thought the same initially but just paused and thought again. This is 700 level question and answer is not A

[melguy]

Out of 5 kilograms 98% (4.9 kilograms) was water and 2% (0.1 kilograms) was non-water.

The next day, oranges became 96% water and 4% of non-water.

So the next day 0.1 kilograms of non-water composed 4% of oranges, i.e. the new weight of the oranges was 0.04x = 0.1 -> x = 2.5 kilograms

In my opinion answer is C

[vinay1983]

Is it D

[theCodeToGMAT]

Originally --> Solution Weight 5 KG
98% Water of 5

2% Solution of 5

Next Day --> Solution Weight X KG
96% of Water

4% of Solution


0.02 * 5 = 0.04 * X
X = 2.5

Answer [spoiler]{C}[/spoiler]

[mevicks]

Hi Vivek,

Good try and I thought the same initially but just paused and thought again. This is 700 level question and answer is not A

Hi Rakesh,

I know according to the "Souce" the correct answer is 2.5 Kgs, but as mentioned in the above post I find it hard to believe that this is the correct answer! Also considering the "other" approaches posted on "other" forums its weird to know that we can have two ambiguous answers (96 % approach posted on GC).

Regards,
Vivek

[rakeshd347]

Hi Vivek,

Good try and I thought the same initially but just paused and thought again. This is 700 level question and answer is not A

Hi Rakesh,

I know according to the "Souce" the correct answer is 2.5 Kgs, but as mentioned in the above post I find it hard to believe that this is the correct answer! Also considering the "other" approaches posted on "other" forums its weird to know that we can have two ambiguous answers (96 % approach posted on GC).

Regards,
Vivek

Hi Vivek,

You need to see this question from this point of view. The amount of water changes from 98% to 96% but the amount of non water part doesn't change.
So in 5 kg of oranges we had 0.1kg of non water part. Now this 0.1kg in the new solution contribute to 4% right.
So it means 0.4x=0.1 or x=0.1/0.4 or x=2.5 which is the final weight.
SO the correct answer is C

[GMATGuruNY]

I received a PM asking me to comment.
The posted problem is akin to the following:

A tank contains 5 gallons of a solution that is 98% percent water by volume. After a certain amount of water evaporates from the tank, the remaining solution is 96% percent water by volume. What is the volume, in gallons, of the remaining solution in the tank?

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

Let x = the volume of the original solution and y = the volume of the remaining solution.

Amount of water in the original solution = 0.98x.
Amount of water in the remaining solution = 0.96y.
Decrease in total volume = x-y.

Since the decrease in total volume is due to the decrease in the amount of water, we get:
x-y = 0.98x - 0.96y
100x - 100y = 98x - 96y
2x = 4y
(1/2)x = y.

Thus, the volume of the remaining solution is equal to 1/2 the original volume:
(1/2) * 5 = 2.5 gallons.

The correct answer is C.

An alternate approach would be to PLUG IN THE ANSWERS, which represent the volume of the remaining solution.
In the original solution, the amount of water = (0.98)(5) = 4.9 gallons.

Answer choice C: 2.5 gallons
Since 2.5 gallons remain in the tank, the amount of water that evaporates = 2.5 gallons.
Thus, the amount of water that remains in the tank = 4.9 - 2.5 = 2.4 gallons.
In the remaining solution, water/total * 100 = (2.4)/(2.5) * 100 = 24/25 * 100 = 96%.
Success!

[Scott@TargetTestPrep]

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