A petrol pump owner has a 1-liter mixture containing 10% unleaded petrol (the rest is leaded). How much leaded petrol must be added so that the unleaded percentage becomes 5%?

Introduction / Context:
This is a dilution problem where an inert (leaded with 0% unleaded) is added to reduce the proportion of the active component (unleaded). The amount of unleaded remains fixed while total volume increases.

Given Data / Assumptions:

• Initial mixture = 1 L (1000 ml), unleaded = 10% = 0.1 L.
• Add x liters of leaded (0% unleaded).
• Target unleaded% = 5%.

Concept / Approach:
Unleaded volume stays at 0.1 L. New total volume = 1 + x. Set 0.1 / (1 + x) = 0.05 and solve for x.

Step-by-Step Solution:
0.1 / (1 + x) = 0.051 + x = 0.1 / 0.05 = 2 ⇒ x = 1 L = 1000 ml

Verification / Alternative check:
After adding 1 L leaded, total is 2 L; unleaded = 0.1 L, so percentage = 0.1 / 2 = 5%.

Why Other Options Are Wrong:
900 ml or 1800 ml give unleaded percentages not equal to 5%. Only 1000 ml halves the percentage from 10% to 5%.

Common Pitfalls:
Reducing 10% to 5% by “subtracting” rather than diluting; remember the active amount is unchanged while total grows.

Final Answer:
1000 ml