Correct Answer - Option 3 : 55%
Given:
First Alloy = 15 kg
Second Alloy = 10 kg
Calculation:
Mass of steel in 1st alloy = 0.5 × 15
⇒ 7.5 kg.
Mass of steel in 2nd alloy = 0.75 × 10
⇒ 7.5 kg.
20% steel is reduced = 20% of the (7.5 + 7.5) kg
⇒ 0.20 × 15
⇒ 3 kg.
Remaining steel = 12 kg
Total alloy = 25 – 3
⇒ 22 kg
The required percentage of steel = (12/22) × 100 ≈ 55%
∴ The total steel is present in the third alloy is 55%.