Let's analyze the problem step-by-step using the simple interest formula:
Problem Summary:
- Rs. 725 lent at the beginning of the year at rate R%R%R% per annum.
- After 8 months, Rs. 362.50 more is lent at twice the original rate, i.e., 2R%2R%2R%.
- Total interest earned at the end of the year from both loans is Rs. 33.50.
- Find the original rate of interest RRR.
Step 1: Define variables
- Principal 1, P1=725P_1=725P1=725
- Principal 2, P2=362.50P_2=362.50P2=362.50
- Rate 1, R%R%R%
- Rate 2, 2R%2R%2R%
- Time for first loan, T1=1T_1=1T1=1 year
- Time for second loan, T2=412=13T_2=\frac{4}{12}=\frac{1}{3}T2=124=31 year (since lent after 8 months, so 4 months remain)
Step 2: Calculate interest from both loans
Using the simple interest formula SI=P×R×T100SI=\frac{P\times R\times T}{100}SI=100P×R×T:
- Interest from first loan:
I1=725×R×1100=725R100I_1=\frac{725\times R\times 1}{100}=\frac{725R}{100}I1=100725×R×1=100725R
- Interest from second loan:
I2=362.50×2R×13100=362.50×2R300=725R300I_2=\frac{362.50\times 2R\times \frac{1}{3}}{100}=\frac{362.50\times 2R}{300}=\frac{725R}{300}I2=100362.50×2R×31=300362.50×2R=300725R
Step 3: Total interest equation
Total interest earned is Rs. 33.50, so:
I1+I2=33.50I_1+I_2=33.50I1+I2=33.50
725R100+725R300=33.50\frac{725R}{100}+\frac{725R}{300}=33.50100725R+300725R=33.50
Multiply through by 300 to clear denominators:
300×725R100+300×725R300=33.50×300300\times \frac{725R}{100}+300\times \frac{725R}{300}=33.50\times 300300×100725R+300×300725R=33.50×300
3×725R+725R=100503\times 725R+725R=100503×725R+725R=10050
2175R+725R=100502175R+725R=100502175R+725R=10050
2900R=100502900R=100502900R=10050
Step 4: Solve for RRR
R=100502900=3.46%R=\frac{10050}{2900}=3.46%R=290010050=3.46%
Final Answer:
The original rate of interest is 3.46% per annum
