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A vessel contained a certain amount of a solution of acid and water. When 2 litres of water was added to it, the new solution had 50% acid concentration. When 15 litres of acid was further added to this new solution, the final solution had 80% acid concentration. The ratio of water and acid in the original solution was

Solution

✅ Correct Option: 2

We define our unknowns:

ww = initial water (in litres)

aa = initial acid (in litres)


When 2 litres of water is added, the solution becomes 50% acid.

New total volume = a+w+2a + w + 2

Acid percentage = acid amounttotal volume=50%\tfrac{\text{acid amount}}{\text{total volume}} = 50\%

Setting up the equation:

aa+w+2=50%=12\frac{a}{a + w + 2} = 50\% = \frac{1}{2}

2a=a+w+22a = a + w + 2

a=w+2...(1)a = w + 2 \quad \text{...(1)}


Now 15 litres of acid is added to the solution from the previous step, making it 80% acid.

New acid amount = a+15a + 15

New total volume = (a+w+2)+15=a+w+17(a + w + 2) + 15 = a + w + 17

Acid percentage = acid amounttotal volume=80%\tfrac{\text{acid amount}}{\text{total volume}} = 80\%

Setting up the equation:

a+15a+w+17=80%=45\frac{a + 15}{a + w + 17} = 80\% = \frac{4}{5}

5(a+15)=4(a+w+17)5(a + 15) = 4(a + w + 17)

5a+75=4a+4w+685a + 75 = 4a + 4w + 68

a+7=4w...(2)a + 7 = 4w \quad \text{...(2)}


From equation (1): a=w+2a = w + 2

Substituting into equation (2):

(w+2)+7=4w(w + 2) + 7 = 4w

w+9=4ww + 9 = 4w

9=3w9 = 3w

w=3w = 3

Therefore: a=w+2=3+2=5a = w + 2 = 3 + 2 = 5


In the original solution:

Water = 3 litres

Acid = 5 litres

Ratio of water to acid = 3 : 5

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