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Question

If 50L of liquid containing 20% spirit is added with 10L of water then what would be the percentage of spirit in resulting mixture?

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
$16\frac{2}{3}\%$

This problem involves calculating the final concentration (percentage) of spirit in a mixture after adding water.

Calculating Spirit Percentage in Mixture

We need to find the new percentage of spirit after adding water to the initial mixture. The amount of spirit remains constant, but the total volume increases.

Step 1: Calculate the Initial Volume of Spirit

The initial mixture has a volume of 50 L and contains 20% spirit. The volume of spirit is:

Initial Spirit Volume = Total Volume $\times$ Percentage of Spirit

Initial Spirit Volume = $50 \, \text{L} \times \frac{20}{100}$

Initial Spirit Volume = $50 \, \text{L} \times 0.20 = 10 \, \text{L}$

Step 2: Calculate the Total Volume of the Resulting Mixture

10 L of water is added to the initial 50 L mixture.

Resulting Mixture Volume = Initial Volume + Added Water

Resulting Mixture Volume = $50 \, \text{L} + 10 \, \text{L}$

Resulting Mixture Volume = $60 \, \text{L}$

Step 3: Calculate the Final Percentage of Spirit

The volume of spirit (10 L) remains the same, but the total volume is now 60 L. The new percentage is:

Final Spirit Percentage = $\left( \frac{\text{Volume of Spirit}}{\text{Resulting Mixture Volume}} \right) \times 100$

Final Spirit Percentage = $\left( \frac{10 \, \text{L}}{60 \, \text{L}} \right) \times 100$

Final Spirit Percentage = $\frac{1}{6} \times 100$

Final Spirit Percentage = $\frac{100}{6} \% = \frac{50}{3} \% = 16\frac{2}{3}\%$

Therefore, the percentage of spirit in the resulting mixture is $16\frac{2}{3}\%$.

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  2. A container contains 20 L mixture in which there is 10% sulphuric acid. Find the quantity of sulphuric acid to be added in it to make the solution to contain 25% sulphuric acid.

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