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