Ravi's salary was increased by 50%. After some time, his new salary was also increased by 50%. Find his gain percentage of salary.

Correct — C. Take Ravi's starting salary as ₹100.

first rise of 50%: 100 × 1.50 = ₹150
second rise of 50%, charged on ₹150 and not on ₹100: 150 × 1.50 = ₹225

The question asks for the gain, measured against where he started:
gain = 225 − 100 = ₹125 on a base of ₹100
= 125% → option (c).

By the successive-change formula:
50 + 50 + (50 × 50)⁄100 = 100 + 25 = 125.

Why the others are wrong

• (a)
  200% — 200% would need the salary to end at ₹300, three times the original. Two 50% rises multiply it by 1.5 × 1.5 = 2.25, taking ₹100 only as far as ₹225.
• (b)
  150% — 150% is where the salary stands after the first rise — ₹150 on a ₹100 base — not the gain after both. The second rise carries it on to ₹225.
• (d)
  100% — 100% adds the two rises, 50 + 50. That ignores the second rise being charged on ₹150, which is ₹75 rather than ₹50; the extra ₹25 is precisely the 25 in 125.

A percentage rise is a multiplier: +50% means × 1.5.

Stack two rises and you multiply the multipliers. 1.5 × 1.5 = 2.25, so the salary ends at 225% of where it began.

The word to watch is gain. A question asking for the final salary as a percentage of the original wants 225%. One asking for the gain wants the increase alone, 225 − 100 = 125%.

The identical product runs backwards for discounts, where 0.6 × 0.7 = 0.42.

Key facts

• A rise of p% multiplies the amount by (1 + p⁄100).
• Two successive 50% rises multiply a salary by 2.25, a gain of 125%.
• For successive changes a% and b%, the net change is a + b + ab⁄100.
• The final salary is 225% of the original while the gain is 125% — the two differ by the original 100%.

Common traps

• Adding 50 and 50 to report 100%.
• Reporting 225%, the final salary as a percentage of the original, when the stem asks for the gain.
• Charging the second rise on the original ₹100 instead of on ₹150.

No directly related past PYQ was found.