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Two equal sums of money at simple interest at 11% p.a for 3½ years and 4½ years respectively. If the difference in interest for two periods was rs 412.50, then each sum is?
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Two equal sums of money at simple interest at 11% p.a for 3½ years and...
Given Information:

- Two equal sums of money
- Simple interest rate: 11% per annum
- Time periods: 3½ years and 4½ years
- Difference in interest: Rs 412.50

Step 1: Calculate the Interest for each Period

To find the sum of money, we need to calculate the interest for each period separately.

Interest for the first period (3½ years):
I₁ = P × R × T
Where,
I₁ is the interest for the first period,
P is the principal amount (sum of money),
R is the rate of interest, and
T is the time period in years.

Interest for the second period (4½ years):
I₂ = P × R × T

Step 2: Calculate the Difference in Interest

The difference in interest between the two periods is given as Rs 412.50.

Difference in interest: I₂ - I₁ = Rs 412.50

Substituting the values of interest:

P × R × T₂ - P × R × T₁ = Rs 412.50

Step 3: Simplify the Equation

Since the principal amount (P) is the same for both periods, we can factor it out.

P × R × (T₂ - T₁) = Rs 412.50

Step 4: Calculate the Principal Amount

Now, we can solve the equation to find the principal amount (P).

P = Rs 412.50 / (R × (T₂ - T₁))

Step 5: Substitute the Values to Find each Sum

Substituting the given values into the equation:

Principal amount (P) = Rs 412.50 / (0.11 × (4.5 - 3.5))

Simplifying the expression:

Principal amount (P) = Rs 412.50 / (0.11 × 1)

Principal amount (P) = Rs 3750

Since the sums of money are equal, each sum is Rs 3750.
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Two equal sums of money at simple interest at 11% p.a for 3½ years and 4½ years respectively. If the difference in interest for two periods was rs 412.50, then each sum is?
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