Question

A container holds a mixture of three liquids A, B, and C in ratio 2: 3: 5. If 6 liters of liquid A, 12 liters of liquid B, and a certain amount of liquid C are added to the container, the new ratio of liquids A, B, and C becomes 3: 5: 8. Find the quantity (in liters) of liquid C added.

• A. 15 liters
• B. 16 liters
• C. 18 liters
• D. 20 liters

Answer 1

Answer:

C. 18 liters

Let the initial quantities of liquids (A), (B), and (C) be:

2x,;3x,;5x

After adding the liquids:

• (A = 2x+6)

• (B = 3x+12)

• (C = 5x+y), where (y) is the amount of liquid (C) added.

The new ratio is:

(2x+6):(3x+12):(5x+y)=3:5:8

Step 1: Find (x)

Using the ratio of (A) and (B):

$\frac{2x+6}{3x+12}=\frac{3}{5}$

Cross-multiply:

$5(2x+6)=3(3x+12)$
$10x+30=9x+36$
$x=6$

Find (y)

Now,

$A=2(6)+6=18,\qquad C=5(6)+y=30+y$

Since the new ratio is (3:8),

$\frac{18}{30+y}=\frac{3}{8}$

Cross-multiply:

$18\times8=3(30+y)$
$144=90+3y$
$3y=54$
$y=18$

Final Answer
$\boxed{18\text{ liters}}$