A bartender removed from a bottle a certain quantity of a drink, which

Question

A bartender removed from a bottle a certain quantity of a drink, which had 25% alcohol content and replaced the same quantity with another drink that had an alcohol content of 10%. The alcohol content in the drink after replacement is 20%. What was the quantity of the drink that he replaced, if there was 1 litre of the drink in the bottle?

A. 1/5 litres
B. 1/4 litres
C. 1/3 litres
D. 1/2 litres
E. 2/3 litres

A. 1/5 litres 
B. 1/4 litres
C. 1/3 litres
D. 1/2 litres
E. 2/3 litres

AkshdeepS · Accepted Answer

Weighted average method :

generis · Answer

The same unknown amount of liquid is removed and replaced
 x  liters = unknown volume
 .25= % alcohol of removed liquid
 .10= % alcohol of replaced liquid
 1L  = volume of liquid at beginning and end

We have 1 liter of 25% alchol liquid.  x  liters are removed and replaced. Removed  x  is 25% alcohol. Replaced  x  is 10% alcohol. Result is 1 liter of 20% alcohol

Just as with other mixture problems, weighted average works

(1L*A%) - (removed*A%) + (replaced*B%) = (1L*C%)

(.25)(1L) - (.25)(x) +(.10)(x)=(.20)(1L) 
 .25L -.25x+.10x=.20L 
 .05L=.15x 
 x=\frac{.05}{.15}L=\frac{5}{15}L=\frac{1}{3}L

Answer C

EgmatQuantExpert · Answer

Solution

Given:  
 • A bartender removed from a bottle a certain quantity of a drink, which had 25% alcohol content and replaced the same quantity with another drink that had an alcohol content of 10%
• The alcohol content in the drink after replacement is 20%
• There was 1 litre of the drink in the bottle

To find:  
 • The quantity of the drink that was replaced

Approach and Working:   
In the original solution,
 • Water = 750 ml and Alcohol = 250 ml

Assume that the quantity that was replaced was 20n.
In that 20n,
 • Water removed = 0.75 * 20n = 15n
• Alcohol removed = 0.25 * 20n = 5n

In the replaced 20n quantity,
 • Water added = 0.9 * 20n = 18n
• Alcohol added = 0.1 * 20n = 2n

Hence, final value of
 • Water = 750 – 15n + 18n = 750 + 3n
• Alcohol = 250 – 5n + 2n = 250 – 3n

As per the question, 
 • 250 – 3n = 20% of 1000 = 200
Or, 3n = 50
Or, n =  \frac{50}{3} 
Or, 20n =  \frac{1000}{3}  ml =  \frac{1}{3}  ltrs 
 
Hence, the correct answer is option C.

Answer: C

chetan2u · Answer

An apt case for weighted average method 
So whatever is replaced is the quantity of 10% and initial solution still left is the quantity of 25%

So 25%...20%..............10%
Quantity of 10%= \frac{25-20}{25-10}=\frac{5}{15}=1/3 
So 1/3 was replaced.

C

adstudy · Answer

Let 'x' litres of quantity be withdrawn / re.

Now we know - 
Alcohol withdrawn = 0.25x (25% of withdrawn quantity)
Hence Alcohol remaining in bottle = 0.25 (1-x) (25% of remaining quantity. 1 Litre is the total quantity in bottle(given) ) -----  (1)

We also know alcohol added is = 0.1x (10% of added quantity. Same quantity of total mixture is replaced) -----  (2)

New total alcohol content = 0.2 (1L) (20% of Total quantity in bottle i.e. 1 Litre) ------  (3)

Equation for alcohol content will be as follows -

Replaced alcohol + Remaining Alcohol = Total alcohol

i.e.  (2) + (1) = (3)

=> 0.25(1-x) + 0.1x = 0.2(1L)
=> 0.25 - 0.25x + 0.1x = 0.2
=> 0.05 = 0.15x

Multiplying both sides by 100

=> 5 = 15x

therefore x = 1/3 Litres (which is the replaced quantity.)

Answer - C

gracie · Answer

let x=litres of drink replaced
.25-.25x+.10x=.20
x=1/3 litres
C

JeffTargetTestPrep · Answer

Originally, we have 0.25 litres of alcohol in the 1-litre bottle of drink.

We can let x = the quantity being replaced. Thus 0.25x litres of alcohol is replaced by 0.1x litres of alcohol. Therefore, we have:

(0.25 - 0.25x + 0.1x)/(1 - x + x) = 0.2

0.25 - 0.15x = 0.2

0.05 = 0.15x

0.05/0.15 = x

1/3 = x

Answer: C

minustark · Answer

Using the allegation method, we can solve the problem under 30 seconds.
10 25
 20
5 10
1 2
So, the ratio of 10% to 25% is 1: 2 in the new mix. The original amount was 1 liter, so the bartender replaced 1/3 liter.
C is the answer.

IIIIIIIIIIIII · Answer

0,2=0,25*(1-x)+0,1*x
-0,05=-0,15x
x=3/15=1/3

adewale223 · Answer

Let x be volume left

0.25x+0.1(1-x)=0.2

0.25x+0.1-0.1x=0.2

0.15x=0.1
x=2/3

Volume added therefore will be 1-2/3 = 1/3

Posted from my mobile device

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

A bartender removed from a bottle a certain quantity of a drink, which

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

A bartender removed from a bottle a certain quantity of a drink, which

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards