A glass was filled with 10 ounces of water and spirit mixture with the

Question

A glass was filled with 10 ounces of water and spirit mixture with the components in the ratio 2 : 3 respectively. If 1 percent of the initial quantity of water and 3 percent of the initial quantity of spirit evaporated each day during a twenty-day period, what percent of the original amount of mixture evaporated during this period?

(A) 4.4%
(B) 24.4%
(C) 44.0%
(D) 50.0%
(E) 80.0%

(A) 4.4%
(B) 24.4%
(C) 44.0%
(D) 50.0%
(E) 80.0%

Bunuel · Answer

This is a modified question of the the following OG question: https://gmatclub.com/forum/a-glass-was- ... 40242.html

pushpitkc · Answer

Since the glass is filled with 10 ounces of water and spirit mix, in the ratio 2:3
Water(weight) - 4 ounces
Spirit(weight) - 6 ounces

Since 1% of the water evaporates each day for 20 days, 20% of the water will evaporate(0.8 ounces) 
Since 3% of the spirit evaporates each day for 20 days, 60% of the spirit will evaporate(3.6 ounces)

Total amount lost due to evaporation : 0.8 ounces(Water) + 3.6 ounces(Spirit) = 4.4 ounces(44% of the solution) (Option C)

generis · Answer

1. The ratio of water to spirits is  \frac{2x}{3x}

Sum of the ratio is 2x + 3x = 5x, and 5x = 10 ounces of water and spirit mixture

Multiplier x = 2, so

original amount of water = 2x * 2 = 4 ounces
original amount of spirits = 3x * 2 = 6 ounces

2. 1% of original quantity of water evaporated for 20 days

(.01)(4) x 20 = .04 x 20 = .8 ounces evaporated

3. 3% of original quantity of spirits evaporated for 20 days

(.03)(6) x 20 = .18 x 20 = 3.6 ounces evaporated

4. Total amount evaporated is 3.6 + .8 = 4.4 ounces
 
Evaporated amount as percentage of original =  \frac{4.4}{10}  * 100 = 44%

Answer (C)

To check my work, I also calculated what amount remained after 20 days and subtracted from original amount of 10 ounces

Amount of water remaining is (original) 4 - .8 (evaporated) = 3.2 ounces

Amount of spirits remaining is (original) 6 - 3.6 (evaporated) = 2.4 ounces

Total amount remaining is 3.2 ounces + 2.4 ounces = 5.6 ounces

Amount that evaporated is 10 - 5.6 = 4.4 ounces. That works.

gracie · Answer

20 =.44=44.0%
C

MvArrow · Answer

Can you please explain me this passage?
I can't understand the wording of the problem: it states that there are 10 ounces of water, how can I relate them to the solution?
The rest of the problem is crystal clear, I just don't understand how to calculate the ammount of water and of the substance.

Thanks a lot!

generis · Answer

MvArrow Disregard this post's questions posted earlier. I have removed them. You caught a typographical mistake I made! Sorry to confuse you! See above, where I have crossed out "water" and inserted, in underlined text, "water and spirit mixture." Edited original, too. You are correct: the ratios apply to the total mixture/solution, not to the water alone. Nice catch, thanks! :-D

MvArrow · Answer

Thanks a lot! Now I understand

ScottTargetTestPrep · Answer

We are given that a glass has 10 ounces of water and spirit mixture with the components in the ratio 2 : 3, respectively, or water : spirit = 2x : 3x. Thus:
 
5x = 10
 
x = 2, so there are 4 ounces of water and 6 ounces of spirit.
 
Since 1% of water evaporates each day over a 20-day period, the amount of water that evaporates is:
 
20(0.01)(4) = 0.8 ounces 
 
Since 3% of spirit evaporates each day over a 20-day period, the amount of spirit that evaporates is:
 
20(0.03)(6) = 3.6 ounces 
 
Thus, the amount of mixture that evaporated is 3.6 + 0.8 = 4.4 ounces.
 
So, the percentage of mixture that evaporated is 4.4/10 = 44/100 = 44%.
 
Answer: C

srishti246 · Answer

Why aren't we considering the evaporation to happen successively i.e. 4(100%-1%)^20 + 6(100%-3%)^20 ?? How to identify if we are not supposed to assume deduction in succession?

nick1816 · Answer

Look at the question
A glass was filled with 10 ounces of water and spirit mixture with the components in the ratio 2 : 3 respectively.  If 1 percent of the initial quantity  of water and 3 percent of the initial quantity of spirit evaporated each day during a twenty-day period, what percent of the original amount of mixture evaporated during this period?

The amount of water evaporated on a particular day is dependent on initial quantity of water.
If the amount of water evaporated on a particular day is dependent on the quantity of water present in the mixture on that particular day, then 4(100%-1%)^20 is correct.

srishti246 · Answer

Thank you so much! Makes sense now :D

EgmatQuantExpert · Answer

Given

• A glass was filled with 10 ounces of water and spirit mixture with the components in the ratio 2 : 3 respectively.
• 1 percent of the initial quantity of water and 3 percent of the initial quantity of spirit evaporated each day during a twenty-day period.

To Find

• the percent of the original amount of mixture evaporated during this period.

Approach and Working Out

• 1% for 20 days makes it 20% of water.

• 3% for 20 days makes it 60% of spirit.

• Let the water be 40 and spirit 60.
 o 20% of 40 = 8
o 60% of 60 = 36

• Total = 44%
 
  Correct Answer: Option C

Kinshook · Answer

Given: A glass was filled with 10 ounces of water and spirit mixture with the components in the ratio 2 : 3 respectively. 
Asked: If 1 percent of the initial quantity of water and 3 percent of the initial quantity of spirit evaporated each day during a twenty-day period, what percent of the original amount of mixture evaporated during this period?

In 10 ounces of mixture: - 
Water = 4 ounces
Spirit = 6 ounces

1% of initial quantity of water = .04 ounces
3% of the initial quantity of spirit = .18 ounces

Total quantity of water evaporated in 20 days = 20*.04 = .8 ounces
Total quantity of spirit evaporated in 20 days = 20*.18 = 3.6 ounces
Total quantity of mixture evaporated in 20 days = .8 + 3.6 = 4.4 ounces

The percent of the original amount of mixture evaporated during this period = 4.4/10 = 44%

IMO C

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