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
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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
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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
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Let 'x' litres of quantity be withdrawn / re.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Alcohol withdrawn = 0.25x (25% of withdrawn quantity)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
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let x=litres of drink replaced
.25-.25x+.10x=.20
x=1/3 litres
C

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
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Weighted average method :

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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.

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