Simple Interest All Competence Aptitude

Let's denote the two parts of the sum as \(x\) and \(y\).

According to the given information:
1. The interest earned on both parts is equal.
2. The interest earned is calculated using the simple interest formula: \(SI = P \times R \times T\), where \(P\) is the principal, \(R\) is the rate of interest, and \(T\) is the time period.

For the first part, \(x\), at 9% per annum for 5 years, the interest earned is \(0.09 \times x \times 5 = 0.45x\).

For the second part, \(y\), at 10% per annum for 6 years, the interest earned is \(0.10 \times y \times 6 = 0.60y\).

According to the given condition, these interests are equal:

\[0.45x = 0.60y\]

We also know that the sum of the two parts is 2800:

\[x + y = 2800\]

Now, we have a system of two equations:

1. \(0.45x = 0.60y\)
2. \(x + y = 2800\)

We can solve this system of equations to find the values of \(x\) and \(y\).

From the first equation, we can express \(y\) in terms of \(x\):

\[y = \dfrac{0.45}{0.60}x = \dfrac{3}{4}x\]

Now, substitute \(y\) in terms of \(x\) into the second equation:

\[x + \dfrac{3}{4}x = 2800\]

\[x \left(1 + \dfrac{3}{4}\right) = 2800\]

\[x \times \dfrac{7}{4} = 2800\]

\[x = \dfrac{2800 \times 4}{7}\]

\[x = 1600\]

Now, we can find \(y\):

\[y = \dfrac{3}{4} \times 1600 = 1200\]

So, the two sums are \(Rs. 1600\) and \(Rs. 1200\).

Therefore, the correct answer is \(\boxed{\text{Rs. 1600}, \text{Rs. 1200}}\).

Your approach is correct. Let's verify the calculations:

Given:
1. Amount after 2 years (\( A_1 \)) = Rs. 720
2. Amount after 7 years (\( A_2 \)) = Rs. 1020

We know that the amount is equal to the principal plus the interest.

Using the given formula, we can write the equations:

1. \( P + SI_1 = A_1 \)
2. \( P + SI_2 = A_2 \)

Where:
- \( P \) is the principal amount
- \( SI_1 \) is the simple interest for 2 years
- \( SI_2 \) is the simple interest for 7 years

Now, let's calculate the simple interest for 5 years:

\[ SI_{\text{5 years}} = SI_2 - SI_1 = 1020 - 720 = 300 \]

Now, we need to find the simple interest for 2 years:

\[ SI_1 = \dfrac{2}{5} \times 300 = 120 \]

Now, let's find the principal amount:

\[ P = A_1 - SI_1 = 720 - 120 = 600 \]

Therefore, the correct answer is \( \boxed{\text{Rs. 600}} \). 

Your solution and calculations are correct.

Thank you for providing the details. Let's proceed with the calculation using the given information:

Given:
- Simple Interest (S.I) = Rs. 1/day
- Rate (R) = 5%
- Time (T) = 1/365 year

We'll use the formula for simple interest:

\[ \text{S.I} = \dfrac{{P \times R \times T}}{{100}} \]

Substituting the given values:

\[ 1 = \dfrac{{P \times 5 \times \dfrac{1}{365}}}{{100}} \]

Now, let's solve for the Principal (P):

\[ 1 = \dfrac{{5P}}{{36500}} \]

\[ 1 \times 36500 = 5P \]

\[ 36500 = 5P \]

\[ P = \dfrac{{36500}}{5} \]

\[ P = 7300 \]

So, the principal which gives Rs. 1 interest per day at a rate of 5% simple interest per annum is Rs. 7300.

Therefore, the correct answer is: 7300

Given:

P = Rs. 800, A = 956 and t = 3 years

SI = 956 – 800 = Rs. 156

SI = P × r × t/100

⇒ 156 = (800 × 3 × r)/100

⇒ r = 156/24

⇒ r = 6.5%

New rate of interest = 6.5 + 4 = 10.5%

P = Rs. 800 and time = 3 years

SI = Prt/100

⇒ SI = (800 × 10.5 × 3)/100

⇒ SI = 252

∴ Amount = 800 + 252 = 1052

Let's denote the amount lent at 4% per annum as \( x \) and the amount lent at 5% per annum as \( 16000 - x \).

The interest earned from the amount lent at 4% per annum is \( \dfrac{x \times 4}{100} \), and the interest earned from the amount lent at 5% per annum is \( \dfrac{(16000 - x) \times 5}{100} \).

According to the problem, the total interest received is $700:

\[ \dfrac{x \times 4}{100} + \dfrac{(16000 - x) \times 5}{100} = 700 \]

Now, let's solve for \( x \):

\[ \dfrac{4x}{100} + \dfrac{80000 - 5x}{100} = 700 \]\[ 4x + 80000 - 5x = 70000 \]\[ -x = -10000 \]\[ x = 10000 \]

So, the amount lent at 4% per annum was $10,000.

Therefore, the correct answer is 10,000.

To find the rate of interest per annum, we can use the formula for simple interest:

\[ \text{Simple Interest} = \dfrac{P \times R \times T}{100} \]

Where:
- \( P \) is the principal amount (the initial sum of money)
- \( R \) is the rate of interest per annum
- \( T \) is the time in years

We're given two scenarios:

1. After 5 years, the amount is $5,200.
2. After 7 years, the amount is $5,680.

Let's denote the principal amount as \( P \) and the rate of interest as \( R \).

Using the first scenario:

\[ \text{Simple Interest} = \dfrac{P \times R \times T}{100} = \dfrac{P \times R \times 5}{100} \]\[ 5200 - P = \dfrac{P \times R \times 5}{100} \]

Using the second scenario:

\[ \text{Simple Interest} = \dfrac{P \times R \times T}{100} = \dfrac{P \times R \times 7}{100} \]\[ 5680 - P = \dfrac{P \times R \times 7}{100} \]

Now, we can set up a system of equations:

\[ \begin{cases} 5200 - P = \dfrac{5PR}{100} \\ 5680 - P = \dfrac{7PR}{100} \end{cases} \]

Let's solve this system of equations:

From the first equation:\[ 5200 - P = \dfrac{5PR}{100} \]\[ 520000 - 100P = 5PR \]\[ 520000 = 5PR + 100P \]\[ 520000 = P(5R + 100) \]\[ P = \dfrac{520000}{5R + 100} \]

From the second equation:\[ 5680 - P = \dfrac{7PR}{100} \]\[ 568000 - 100P = 7PR \]\[ 568000 = 7PR + 100P \]\[ 568000 = P(7R + 100) \]\[ P = \dfrac{568000}{7R + 100} \]

Since both expressions for \( P \) are equal, we can equate them:

\[ \dfrac{520000}{5R + 100} = \dfrac{568000}{7R + 100} \]

Cross-multiply to solve for \( R \):

\[ 520000(7R + 100) = 568000(5R + 100) \]\[ 3640000R + 52000000 = 2840000R + 56800000 \]\[ 3640000R - 2840000R = 56800000 - 52000000 \]\[ 800000R = 4800000 \]\[ R = \dfrac{4800000}{800000} \]\[ R = 6 \]

So, the rate of interest per annum is \( \boxed{6\%} \).

Let's break down the problem step by step:

1. Interest gained from borrowing $5,000 at 4% per annum for 2 years:\[ \text{Interest} = \dfrac{P \times R \times T}{100} = \dfrac{5000 \times 4 \times 2}{100} = 400 \]

2. Total amount after 2 years when the money is borrowed at 4% per annum:\[ \text{Total amount} = \text{Principal} + \text{Interest} = 5000 + 400 = 5400 \]

3. Interest gained from lending $5,000 at 6.25% per annum for 2 years:\[ \text{Interest} = \dfrac{P \times R \times T}{100} = \dfrac{5000 \times \dfrac{25}{4} \times 2}{100} = 625 \]

4. Total amount after 2 years when the money is lent at 6.25% per annum:\[ \text{Total amount} = \text{Principal} + \text{Interest} = 5000 + 625 = 5625 \]

Now, let's calculate the gain in the transaction:\[ \text{Gain} = \text{Total amount received} - \text{Total amount borrowed} \]\[ \text{Gain} = 5625 - 5400 = 225 \]

So, the gain in the transaction is $225.

Therefore, the correct answer is \(\boxed{225}\).

Certainly, here's a condensed version:

1. Equation Setup:
   \[    (x + x \times 3 \times 5\%) + (x + x \times 2 \times 5\%) + (x + x \times 1 \times 5\%) + x = 6450
   \]

2. Solving:
   \[    \begin{align*}
   & (115x) + (110x) + (105x) + x = 6450 \\
   & 430x + x = 6450 \\
   & 431x = 6450 \\
   & x = \dfrac{6450}{431} \\
   & x = \text{Rs. 1500}
   \end{align*}
   \]

Therefore, the annual installment, represented by Rs. x, is Rs. 1500.

To solve this problem, let's denote the initial principal amount as \( P \).

Given that the principal amount triples itself in 15 years, we can express this as:

\[ P + \text{Simple Interest for 15 years} = 3P \]

Using the formula for simple interest \( \text{Simple Interest} = \dfrac{P \times R \times T}{100} \), where \( P \) is the principal amount, \( R \) is the rate of interest, and \( T \) is the time in years, we can rewrite the equation as:

\[ P + \dfrac{P \times R \times 15}{100} = 3P \]

Solving for \( R \):

\[ \dfrac{P \times R \times 15}{100} = 2P \]

\[ R \times 15 = 200 \]

\[ R = \dfrac{200}{15} = \dfrac{40}{3} \]

So, the rate of interest is \( \dfrac{40}{3} \) percent per annum.

Now, we need to find out in how many years the principal amount becomes 5 times itself.

Let's denote this time as \( T' \).

Using the formula for simple interest, we know:

\[ P + \text{Simple Interest for } T' \text{ years} = 5P \]

\[ P + \dfrac{P \times \dfrac{40}{3} \times T'}{100} = 5P \]

Solving for \( T' \):

\[ \dfrac{P \times \dfrac{40}{3} \times T'}{100} = 4P \]

\[ T' \times \dfrac{40}{3} = 400 \]

\[ T' = \dfrac{400 \times 3}{40} = 30 \]

So, the principal amount becomes 5 times itself in 30 years.

Therefore, the correct answer is \(\boxed{\text{30 years}}\).

Let's denote the rate of interest as \( R \).

For person A:
- Principal (\( P_A \)) = Rs. 3000
- Time (\( T_A \)) = 2.5 years

For person B:
- Principal (\( P_B \)) = Rs. 3200
- Time (\( T_B \)) = 2.5 years

Using the formula for simple interest (\( SI = \dfrac{P \times R \times T}{100} \)), let's calculate the interest for each person:

For person A:\[ SI_A = \dfrac{3000 \times R \times 2.5}{100} = 75R \]

For person B:\[ SI_B = \dfrac{3200 \times R \times 2.5}{100} = 80R \]

Given that person B paid Rs. 40 more interest than person A, we can write the equation:

\[ SI_B - SI_A = 80R - 75R = 40 \]

\[ 5R = 40 \]

\[ R = \dfrac{40}{5} = 8 \]

So, the rate of interest is 8%.

Therefore, the correct answer is \(\boxed{8\%}\).