Question

Aman has 2 cans of dilute hydrochloric acid. The first can contains 25% water and the rest is acid. The second can contains 50% water. How much acid should he mix from each of the cans so as to get 12 liters of the acid such that the ratio of water to acid is 3:5?
Options

4 liters, 8 liters

6 liters, 6 liters

5 liters, 7 liters

7 liters, 5 liters

Answer 1

Answer:

6 liter is the right answer thanks for giving me opportunity to give the answer

Answer 2

The correct option is '6 liters, 6 liters'

Step-by-step explanation:

Let x liters from first can and y liters from second can are mixed to obtained 12 liters of solution,

i.e. x + y = 12         ....(1),

∵ first can contains 25% water,

Water in first can = 25% of x = 0.25x

∵ Second can contains 50% water,

Water in second can = 50% of y =0.5y

While,

Resultant solution has water and acid in the ratio of 3 : 5,

So, water in resultant solution = $\frac{3}{3+5}(12)=\frac{3}{8}(12)=\frac{3}{2}(3)=\frac{9}{2}=4.5$

Since, water in first can + water in second can = water in resultant solution

i.e. 0.25x + 0.5y = 4.5

25x + 50y = 450

x + 2y = 18         .....(2),

eq (2) - eq (1)

We get,

y = 6

From eq (1),

x + 6 = 12 ⇒ x = 12 - 6 = 6

Hence, 6 liters from first can and 6 liters from second can should be mixed.

#Learn more:

Acid and water are mixed in the ratio 4:3 to make a liquid on adding 10 more liters of acid ,the ratio changed to 3:1 how many liters of acid and water does the liquid contain now

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