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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?

A₹8000

B₹9000

C₹10000

D₹12000

Answer:

C. ₹10000

Read Explanation:

Let the principal be (P).

Simple interest is calculated in parts.

Step 1: Break the time periods

  • First 3 years at 5%

  • Next 5 years at 7%

  • Last 2 years (10 − 8) at 9%

Step 2: Compute interest for each part

First 3 years:

SI1=P×5×3100=15P100SI_1 = \frac{P \times 5 \times 3}{100} = \frac{15P}{100}

Next 5 years:

SI2=P×7×5100=35P100SI_2 = \frac{P \times 7 \times 5}{100} = \frac{35P}{100}

Last 2 years:

SI3=P×9×2100=18P100SI_3 = \frac{P \times 9 \times 2}{100} = \frac{18P}{100}

Step 3: Total interest

SI=15P+35P+18P100=68P100SI = \frac{15P + 35P + 18P}{100} = \frac{68P}{100}

Given:
68P100=6800\frac{68P}{100} = 6800

Step 4: Solve for (P)

P=6800×10068=10000P = \frac{6800 \times 100}{68} = 10000

Final Answer:

Borrowed amount = ₹10,000


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