Answer :
Sure, let's work through each of these problems step by step using the formula for simple interest [tex]\( I = P \cdot R \cdot T \)[/tex], where:
- [tex]\( I \)[/tex] is the interest,
- [tex]\( P \)[/tex] is the principal,
- [tex]\( R \)[/tex] is the rate of interest per annum, and
- [tex]\( T \)[/tex] is the time in years.
### (a) Find the principal that will yield an interest of Rs. 750 at 3% per annum in 2 years.
Given:
- Interest ([tex]\( I \)[/tex]) = Rs. 750
- Rate ([tex]\( R \)[/tex]) = 3% per annum
- Time ([tex]\( T \)[/tex]) = 2 years
We need to find the principal [tex]\( P \)[/tex].
[tex]\[ I = P \cdot R \cdot T / 100 \][/tex]
Rearranging to find [tex]\( P \)[/tex],
[tex]\[ P = \frac{I \times 100}{R \times T} \][/tex]
Substitute the values:
[tex]\[ P = \frac{750 \times 100}{3 \times 2} = \frac{75000}{6} = 12500 \][/tex]
Therefore, the principal amount is Rs. 12,500.
### (b) What principal, invested at 5% per annum for 3 months, will yield a simple interest of Rs. 25,100?
Given:
- Interest ([tex]\( I \)[/tex]) = Rs. 25,100
- Rate ([tex]\( R \)[/tex]) = 5% per annum
- Time ([tex]\( T \)[/tex]) = 3 months = [tex]\( \frac{3}{12} \)[/tex] years = 0.25 years
We need to find the principal [tex]\( P \)[/tex].
[tex]\[ P = \frac{I \times 100}{R \times T} \][/tex]
Substitute the values:
[tex]\[ P = \frac{25100 \times 100}{5 \times 0.25} = \frac{2510000}{1.25} = 2008000 \][/tex]
Therefore, the principal amount is Rs. 20,08,000.
### (c) If the simple interest on a sum of money invested at 25% per annum for 3 years is Rs. 60.50, find the sum.
Given:
- Interest ([tex]\( I \)[/tex]) = Rs. 60.50
- Rate ([tex]\( R \)[/tex]) = 25% per annum
- Time ([tex]\( T \)[/tex]) = 3 years
We need to find the principal [tex]\( P \)[/tex].
[tex]\[ P = \frac{I \times 100}{R \times T} \][/tex]
Substitute the values:
[tex]\[ P = \frac{60.5 \times 100}{25 \times 3} = \frac{6050}{75} = 80.67 \][/tex]
Therefore, the principal amount is Rs. 80.67.
### (d) A certain amount of money trebles itself in 8 years. Find the rate of interest.
Given:
- Final amount = 3 times the initial amount
- Time ([tex]\( T \)[/tex]) = 8 years
Let the initial amount be [tex]\( P \)[/tex]. Then the final amount [tex]\( A \)[/tex] is [tex]\( 3P \)[/tex].
Using the formula for simple interest:
[tex]\[ A = P \left( 1 + \frac{R \times T}{100} \right) \][/tex]
Substitute [tex]\( A = 3P \)[/tex] and solve for [tex]\( R \)[/tex]:
[tex]\[ 3P = P \left( 1 + \frac{R \times 8}{100} \right) \][/tex]
Divide both sides by [tex]\( P \)[/tex]:
[tex]\[ 3 = 1 + \frac{8R}{100} \][/tex]
Subtract 1 from both sides:
[tex]\[ 2 = \frac{8R}{100} \][/tex]
Multiply both sides by 100:
[tex]\[ 200 = 8R \][/tex]
Divide both sides by 8:
[tex]\[ R = \frac{200}{8} = 25\% \][/tex]
Therefore, the rate of interest is 25% per annum.
### (e) A certain amount of money is invested at a 20% interest rate. Find how many years it will take to double itself.
Given:
- Final amount = 2 times the initial amount
- Rate ([tex]\( R \)[/tex]) = 20% per annum
Let the initial amount be [tex]\( P \)[/tex]. Then the final amount [tex]\( A \)[/tex] is [tex]\( 2P \)[/tex].
Using the formula for simple interest:
[tex]\[ A = P \left( 1 + \frac{R \times T}{100} \right) \][/tex]
Substitute [tex]\( A = 2P \)[/tex] and solve for [tex]\( T \)[/tex]:
[tex]\[ 2P = P \left( 1 + \frac{20T}{100} \right) \][/tex]
Divide both sides by [tex]\( P \)[/tex]:
[tex]\[ 2 = 1 + \frac{20T}{100} \][/tex]
Subtract 1 from both sides:
[tex]\[ 1 = \frac{20T}{100} \][/tex]
Multiply both sides by 100:
[tex]\[ 100 = 20T \][/tex]
Divide both sides by 20:
[tex]\[ T = \frac{100}{20} = 5 \text{ years} \][/tex]
Therefore, it will take 5 years to double the amount.
### (f) A certain amount of money is invested at a 25% interest rate. What will be the amount after 8 years?
Given:
- Rate ([tex]\( R \)[/tex]) = 25% per annum
- Time ([tex]\( T \)[/tex]) = 8 years
Let the initial amount be [tex]\( P \)[/tex]. We need to find the final amount [tex]\( A \)[/tex].
Using the formula for simple interest:
[tex]\[ A = P \left( 1 + \frac{R \times T}{100} \right) \][/tex]
Substitute the values:
[tex]\[ A = P \left( 1 + \frac{25 \times 8}{100} \right) = P \left( 1 + 2 \right) = P \times 3 \][/tex]
Therefore, the final amount after 8 years will be 3 times the initial amount, [tex]\( 3P \)[/tex].
- [tex]\( I \)[/tex] is the interest,
- [tex]\( P \)[/tex] is the principal,
- [tex]\( R \)[/tex] is the rate of interest per annum, and
- [tex]\( T \)[/tex] is the time in years.
### (a) Find the principal that will yield an interest of Rs. 750 at 3% per annum in 2 years.
Given:
- Interest ([tex]\( I \)[/tex]) = Rs. 750
- Rate ([tex]\( R \)[/tex]) = 3% per annum
- Time ([tex]\( T \)[/tex]) = 2 years
We need to find the principal [tex]\( P \)[/tex].
[tex]\[ I = P \cdot R \cdot T / 100 \][/tex]
Rearranging to find [tex]\( P \)[/tex],
[tex]\[ P = \frac{I \times 100}{R \times T} \][/tex]
Substitute the values:
[tex]\[ P = \frac{750 \times 100}{3 \times 2} = \frac{75000}{6} = 12500 \][/tex]
Therefore, the principal amount is Rs. 12,500.
### (b) What principal, invested at 5% per annum for 3 months, will yield a simple interest of Rs. 25,100?
Given:
- Interest ([tex]\( I \)[/tex]) = Rs. 25,100
- Rate ([tex]\( R \)[/tex]) = 5% per annum
- Time ([tex]\( T \)[/tex]) = 3 months = [tex]\( \frac{3}{12} \)[/tex] years = 0.25 years
We need to find the principal [tex]\( P \)[/tex].
[tex]\[ P = \frac{I \times 100}{R \times T} \][/tex]
Substitute the values:
[tex]\[ P = \frac{25100 \times 100}{5 \times 0.25} = \frac{2510000}{1.25} = 2008000 \][/tex]
Therefore, the principal amount is Rs. 20,08,000.
### (c) If the simple interest on a sum of money invested at 25% per annum for 3 years is Rs. 60.50, find the sum.
Given:
- Interest ([tex]\( I \)[/tex]) = Rs. 60.50
- Rate ([tex]\( R \)[/tex]) = 25% per annum
- Time ([tex]\( T \)[/tex]) = 3 years
We need to find the principal [tex]\( P \)[/tex].
[tex]\[ P = \frac{I \times 100}{R \times T} \][/tex]
Substitute the values:
[tex]\[ P = \frac{60.5 \times 100}{25 \times 3} = \frac{6050}{75} = 80.67 \][/tex]
Therefore, the principal amount is Rs. 80.67.
### (d) A certain amount of money trebles itself in 8 years. Find the rate of interest.
Given:
- Final amount = 3 times the initial amount
- Time ([tex]\( T \)[/tex]) = 8 years
Let the initial amount be [tex]\( P \)[/tex]. Then the final amount [tex]\( A \)[/tex] is [tex]\( 3P \)[/tex].
Using the formula for simple interest:
[tex]\[ A = P \left( 1 + \frac{R \times T}{100} \right) \][/tex]
Substitute [tex]\( A = 3P \)[/tex] and solve for [tex]\( R \)[/tex]:
[tex]\[ 3P = P \left( 1 + \frac{R \times 8}{100} \right) \][/tex]
Divide both sides by [tex]\( P \)[/tex]:
[tex]\[ 3 = 1 + \frac{8R}{100} \][/tex]
Subtract 1 from both sides:
[tex]\[ 2 = \frac{8R}{100} \][/tex]
Multiply both sides by 100:
[tex]\[ 200 = 8R \][/tex]
Divide both sides by 8:
[tex]\[ R = \frac{200}{8} = 25\% \][/tex]
Therefore, the rate of interest is 25% per annum.
### (e) A certain amount of money is invested at a 20% interest rate. Find how many years it will take to double itself.
Given:
- Final amount = 2 times the initial amount
- Rate ([tex]\( R \)[/tex]) = 20% per annum
Let the initial amount be [tex]\( P \)[/tex]. Then the final amount [tex]\( A \)[/tex] is [tex]\( 2P \)[/tex].
Using the formula for simple interest:
[tex]\[ A = P \left( 1 + \frac{R \times T}{100} \right) \][/tex]
Substitute [tex]\( A = 2P \)[/tex] and solve for [tex]\( T \)[/tex]:
[tex]\[ 2P = P \left( 1 + \frac{20T}{100} \right) \][/tex]
Divide both sides by [tex]\( P \)[/tex]:
[tex]\[ 2 = 1 + \frac{20T}{100} \][/tex]
Subtract 1 from both sides:
[tex]\[ 1 = \frac{20T}{100} \][/tex]
Multiply both sides by 100:
[tex]\[ 100 = 20T \][/tex]
Divide both sides by 20:
[tex]\[ T = \frac{100}{20} = 5 \text{ years} \][/tex]
Therefore, it will take 5 years to double the amount.
### (f) A certain amount of money is invested at a 25% interest rate. What will be the amount after 8 years?
Given:
- Rate ([tex]\( R \)[/tex]) = 25% per annum
- Time ([tex]\( T \)[/tex]) = 8 years
Let the initial amount be [tex]\( P \)[/tex]. We need to find the final amount [tex]\( A \)[/tex].
Using the formula for simple interest:
[tex]\[ A = P \left( 1 + \frac{R \times T}{100} \right) \][/tex]
Substitute the values:
[tex]\[ A = P \left( 1 + \frac{25 \times 8}{100} \right) = P \left( 1 + 2 \right) = P \times 3 \][/tex]
Therefore, the final amount after 8 years will be 3 times the initial amount, [tex]\( 3P \)[/tex].
