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To solve the problem of finding the principal sum of money lent out at simple interest, we proceed with the following steps:
1. Understand the problem statement: We have a sum lent out at a simple interest rate of [tex]\( 11\% \)[/tex] per annum over two different periods: [tex]\( \frac{7}{2} \)[/tex] years (or [tex]\( 3.5 \)[/tex] years) and [tex]\( \frac{9}{2} \)[/tex] years (or [tex]\( 4.5 \)[/tex] years). The difference in interest earned over these two periods is given as [tex]\( 5500 \)[/tex].
2. Define the necessary parameters:
- Let the principal amount be [tex]\( P \)[/tex].
- The interest rate, [tex]\( r \)[/tex], is [tex]\( 11\% \)[/tex] (or [tex]\( 0.11 \)[/tex] in decimal form).
- Time periods [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex] are [tex]\( \frac{7}{2} \)[/tex] years (3.5 years) and [tex]\( \frac{9}{2} \)[/tex] years (4.5 years), respectively.
3. Calculate the interest earned over each period using the formula for simple interest:
The simple interest [tex]\( I \)[/tex] is given by:
[tex]\[ I = P \times r \times t \][/tex]
- For the first period ([tex]\( t_1 = \frac{7}{2} \)[/tex] years):
[tex]\[ I_1 = P \times 0.11 \times \frac{7}{2} \][/tex]
- For the second period ([tex]\( t_2 = \frac{9}{2} \)[/tex] years):
[tex]\[ I_2 = P \times 0.11 \times \frac{9}{2} \][/tex]
4. Write down the difference in interest for these two periods:
[tex]\[ I_2 - I_1 = 5500 \][/tex]
5. Substitute the expressions for [tex]\( I_1 \)[/tex] and [tex]\( I_2 \)[/tex]:
[tex]\[ (P \times 0.11 \times \frac{9}{2}) - (P \times 0.11 \times \frac{7}{2}) = 5500 \][/tex]
6. Factor out the common terms:
[tex]\[ P \times 0.11 \times \left( \frac{9}{2} - \frac{7}{2} \right) = 5500 \][/tex]
7. Simplify inside the parenthesis:
[tex]\[ \frac{9}{2} - \frac{7}{2} = \frac{2}{2} = 1 \][/tex]
8. Complete the equation:
[tex]\[ P \times 0.11 \times 1 = 5500 \][/tex]
9. Solve for [tex]\( P \)[/tex]:
[tex]\[ P \times 0.11 = 5500 \implies P = \frac{5500}{0.11} \][/tex]
10. Perform the division:
[tex]\[ P = 50000 \][/tex]
Therefore, the principal sum of money lent out is [tex]\( 50,000 \)[/tex].
1. Understand the problem statement: We have a sum lent out at a simple interest rate of [tex]\( 11\% \)[/tex] per annum over two different periods: [tex]\( \frac{7}{2} \)[/tex] years (or [tex]\( 3.5 \)[/tex] years) and [tex]\( \frac{9}{2} \)[/tex] years (or [tex]\( 4.5 \)[/tex] years). The difference in interest earned over these two periods is given as [tex]\( 5500 \)[/tex].
2. Define the necessary parameters:
- Let the principal amount be [tex]\( P \)[/tex].
- The interest rate, [tex]\( r \)[/tex], is [tex]\( 11\% \)[/tex] (or [tex]\( 0.11 \)[/tex] in decimal form).
- Time periods [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex] are [tex]\( \frac{7}{2} \)[/tex] years (3.5 years) and [tex]\( \frac{9}{2} \)[/tex] years (4.5 years), respectively.
3. Calculate the interest earned over each period using the formula for simple interest:
The simple interest [tex]\( I \)[/tex] is given by:
[tex]\[ I = P \times r \times t \][/tex]
- For the first period ([tex]\( t_1 = \frac{7}{2} \)[/tex] years):
[tex]\[ I_1 = P \times 0.11 \times \frac{7}{2} \][/tex]
- For the second period ([tex]\( t_2 = \frac{9}{2} \)[/tex] years):
[tex]\[ I_2 = P \times 0.11 \times \frac{9}{2} \][/tex]
4. Write down the difference in interest for these two periods:
[tex]\[ I_2 - I_1 = 5500 \][/tex]
5. Substitute the expressions for [tex]\( I_1 \)[/tex] and [tex]\( I_2 \)[/tex]:
[tex]\[ (P \times 0.11 \times \frac{9}{2}) - (P \times 0.11 \times \frac{7}{2}) = 5500 \][/tex]
6. Factor out the common terms:
[tex]\[ P \times 0.11 \times \left( \frac{9}{2} - \frac{7}{2} \right) = 5500 \][/tex]
7. Simplify inside the parenthesis:
[tex]\[ \frac{9}{2} - \frac{7}{2} = \frac{2}{2} = 1 \][/tex]
8. Complete the equation:
[tex]\[ P \times 0.11 \times 1 = 5500 \][/tex]
9. Solve for [tex]\( P \)[/tex]:
[tex]\[ P \times 0.11 = 5500 \implies P = \frac{5500}{0.11} \][/tex]
10. Perform the division:
[tex]\[ P = 50000 \][/tex]
Therefore, the principal sum of money lent out is [tex]\( 50,000 \)[/tex].
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