A bucket contains a mixture of two liquids A & B

Here the fraction A:B changes from 5:3 to 3:5 when the 16 litres of A is replaced by 16 litres of B

If we take solution to be 5x + 3x = 8x
Initially
A = 5x
B = 3x

As total solution remains same, 16 litres is removed and 16 litres is added. Therefore final in solution

A' = 3x
B' = 5x

A' = A - 16
=> 5x - 16 = 3x
=> 2x = 16
=> x = 8 litres

B = 3x = 3*8 = 24 litres

So, initially 24 litres of liquid B was there in the bucket.

Answer Choice (E)

this is wrong when you back check your answer .

if initially B is 24 A must be 40.
after removal B should be 18 and A 30 .
after adding 16 litre of B , B should be 32 and A 32 which isn't a 3:5 ratio .

let x=total original mixture
3/8*(x-16)+16=5x/8
x=40
3/8*40=15
B

OA:B

Initial Amount of A: \(5x\)
Initial Amount of B: \(3x\)
Total mixture amount: A+B \(= 5x+3x = 8x\)

\(16\) Litre of this mixture is taken out

Amount of A left :\(\frac{5}{8}*(8x-16)=5x-10\)

Amount of B left :\(\frac{3}{8}*(8x-16)=3x-6\)

16 Litre of B added

Final Amount of A :\(\frac{5}{8}*(8x-16)=5x-10\)

Final Amount of B :\(\frac{3}{8}*(8x-16)+16=3x+10\)

According to the question,

\(\frac{5x-10}{3x+10} =\frac{3}{5}\)

\(25x-50=9x+30\)

\(16x=80\)

\(x=\frac{80}{16}=5\)

Initial Amount of B \(= 3x = 3*5 = 15\) Litres

chetan2u - what do you think of this question?

(testing notifications - pardon for the interruption)

BB,

no notifications in the mails. I just checked the notifications on site and saw this...

Dear GMATGuruNY

Can you share your thoughts in this problem?

We can use ALLIGATION.
Let:
S = the original solution
B = the 16 liters of pure B
M = the final mixture
Alligation can be performed only with percentages or fractions.

Step 1: Convert the ratios to FRACTIONS with the same denominator.
S --> Since A:B = 5:3, \(\frac{B}{total} = \frac{3}{8}\)
B --> \(\frac{B}{total} = \frac{16}{16} = \frac{8}{8}\)
M --> Since A:B = 3:5, \(\frac{B}{total} = \frac{5}{8}\)

Step 2: Plot the 3 numerators on a number line, with the numerators for S and B on the ends and the numerator for the mixture in the middle.
S 3------------5-----------8 B

Step 3: Calculate the distances between the numerators.
S 3-----2-----5-----3-----8 B

Step 4: Determine the ratio in the mixture.
The ratio of S to B is equal to the RECIPROCAL of the distances in red.
S:B = 3:2 = 24:16.

The ratio in blue indicates that the mixture is composed of 24 liters of original solution and 16 liters of pure B, implying that the total volume in the bucket = 40 liters.
Since B constitutes \(\frac{3}{8}\) of the original 40 liters in the bucket, we get:
\(\frac{3}{8} * 40 = 15\) liters

by weighted avg approach:
B initial- 3/8
B after - 5/8

applying in the formula- (5-8)/(8-3)
=> 3/5

hence ratio a:b = 3:2
16% of sol was replaced. hence a:b can be 24:16
total sol = 40
after replace B= 3/8 * 40 = 15 (B)

Just wanted to know if there is any wrong with my approach . I am getting the answer as 15.

After removal of 16 L , the ratio will still be 5: 3 for A and B .

Now when 16l is added ,

5X/3X+16 = 3/5

Solving the equation we get X as 3, and 5X as 15.

Thanks for this very clear approach. The fact that the 16 liters are removed from the whole mixture and not from one of the two liquids was a bit confusing.

3/8*(x)-3/8*(16)+16=5/8*(x)

Where X is the total volume available.

3/8*(X)-6+16=5/8*(X)
=> 10=2/8*(X)
=> X=40

Now as B was 3/8 (Initially)

therefore, 3/8*(40)=15.

Notice that the total liters remains the same.

Let's say we have T liters.

When we draw 16 liters, replace all of it with B, the quantity of B is given by:

(T-16)*3/8 + 16*1 = T*5/8, where 5/8 is the concentration obtained in the end.

We find T = 40 l, so the quantity of B was 40*3/8 = 15 l

Given: A bucket contains a mixture of two liquids A & B in the proportion 5 : 3. If 16 litres of the mixture is replaced by 16 litres of liquid B , then the ratio of the two liquids becomes 3 : 5.
Asked: How much of the liquid B was there in the bucket ?

Let the bucket contain liquids A & B as 5x & 3x laters respectively

If 16 litres of mixture is replaced by 16 litres of liquid B
Liquid A becomes = 5x - 10
Liquid B becomes = 3x - 6 + 16 = 3x + 10
Ratio of A & B becomes = (5x-10)/(3x+10) = 3/5
5(5x-10) = 3(3x+10)
25x - 50 = 9x + 30
16x = 80
x = 5
Liquid B was = 3x = 15 liters

IMO B

A bucket contains a mixture of two liquids A & B in the proportion 5 : 3. If 16 litres of the mixture is replaced by 16 litres of liquid B , then the ratio of the two liquids becomes 3 : 5. How much of the liquid B was there in the bucket ?

Using formula, Wr /Wo = (1- R/M)^n, where Wr= % result what is being replaced and Wo= % original, R= replaced, M= Mixture and n= no. of times of replaced.

So, 3/8 * 8/5 = (1- 16/M)^1, M= 40 lt. So, B = 3*40/8= 15 lt.

Ans. B