Question

A jar contains a mixture of two liquids A and B in the ratio \(4:1\). When \(10\) liters of the mixture is taken out and \(10\) liters of liquid B is poured into the jar, the ratio becomes \(2:3\). How many liters of liquid A was contained in the jar initially?

• A. \(12\) l
• B. \(16\) l
• C. \(20\) l
• D. \(24\) l

Hint:

When some mixture is removed, each ingredient is removed in exactly the same ratio as present in the mixture.

Answer 1

The Correct Option is B

Step 1: Set up the starting amounts.
Liquids A and B start in the ratio $4:1$. Let the amounts be \[ A=4x,\qquad B=x, \] so the total mixture is $5x$ liters.

Step 2: Remove 10 liters in the same ratio.
When you scoop out mixture, A and B leave in their current ratio $4:1$. Out of the $10$ liters taken out: \[ A\text{ removed}=10\times\frac{4}{5}=8\text{ liters},\qquad B\text{ removed}=10\times\frac{1}{5}=2\text{ liters}. \]
Step 3: Write the amounts left after removing.
\[ A=4x-8,\qquad B=x-2. \]
Step 4: Add 10 liters of B.
Only B is poured in, so A stays the same and B increases: \[ A=4x-8,\qquad B=(x-2)+10=x+8. \]
Step 5: Use the new ratio $2:3$.
Now the ratio of A to B is $2:3$, so \[ \frac{4x-8}{x+8}=\frac{2}{3}. \] Cross multiply: \[ 3(4x-8)=2(x+8). \]
Step 6: Solve and find the starting amount of A.
Expand both sides: \[ 12x-24=2x+16. \] Bring like terms together: \[ 10x=40\;\Rightarrow\; x=4. \] The initial amount of A is $4x=4\times4=16$ liters. \[ \boxed{16\text{ liters}} \]