Rahul borrowed a sum of money at simple interest of 5% per annum for the first 3 years, 7% per annum for the next 5 years, and 9% per annum for the period beyond 8 years. If he pays a total of ₹6800 as interest only at the end of 10 years, how much money did he borrow?

Correct — C. Simple interest is charged on the same principal P every year, so add the rate × years of each band.

First 3 years at 5%: 3 × 5 = 15%
Next 5 years (4th to 8th) at 7%: 5 × 7 = 35%
Beyond 8, up to year 10, at 9%: 2 × 9 = 18%
Add the bands: 15 + 35 + 18 = 68% of P

Set 68% of P = ₹6800
Divide: P = 6800 × 100 ⁄ 68 = ₹10000 → option (c)

Why the others are wrong

• (a)
  ₹8000 — 68% of ₹8000 is ₹5440, which is ₹1360 short of the ₹6800 paid. ₹8000 would need the bands to add to 85%, not 68%.
• (b)
  ₹9000 — 68% of ₹9000 is ₹6120, still ₹680 short of ₹6800. ₹9000 would need a total rate of about 75.6%.
• (d)
  ₹12000 — 68% of ₹12000 is ₹8160, which is ₹1360 more than the interest paid. ₹12000 would fit a total rate of about 56.7%.

Simple interest = P × R × T ⁄ 100, charged on the original principal only. Interest earned in one year never earns interest itself.

So when the rate changes over time, each band contributes its own R × T and the bands simply add. The loan runs 10 years, so 'beyond 8 years' means years 9 and 10: a 2-year band.

Key facts

• Simple interest = P × R × T ⁄ 100.
• With rate bands, total SI = P × (R₁T₁ + R₂T₂ + R₃T₃) ⁄ 100.
• Here the bands add to 15% + 35% + 18% = 68% of the principal.

Common traps

• Giving the 9% band 3 years instead of 2: after the 8th year only the 9th and 10th remain.
• Averaging the rates to 7% for all 10 years, which gives 70% and about ₹9714.

No directly related past PYQ was found.