Simple Interest All Competence Aptitude

To find the average rate of interest per annum, we need to calculate the weighted average of the interest rates based on the amounts invested at each rate.

Let's denote:
- The total capital as \( C \)
- The amount invested at \( 10\% \) interest rate as \( \dfrac{1}{2}C \)
- The amount invested at \( 9\% \) interest rate as \( \dfrac{1}{3}C \)
- The amount invested at \( 12\% \) interest rate as \( \dfrac{1}{6}C \)

Now, let's calculate the weighted average:

\[ \text{Average rate of interest} = \dfrac{\left(\dfrac{1}{2}C \times 10\% + \dfrac{1}{3}C \times 9\% + \dfrac{1}{6}C \times 12\%\right)}{C} \]

\[ = \dfrac{\left(\dfrac{5}{10}C + \dfrac{3}{10}C + \dfrac{2}{10}C\right)}{C} \]

\[ = \dfrac{\left(10C\right)}{10C} \]

\[ = 10\% \]

So, the average rate of interest per annum is \( 10\% \).

Therefore, the correct answer is:

\[ \text{The average rate of interest per annum is } 10\% \]

P = Rs. 36, 000
Time = 5th January to 31st May = 146 days.
Rate = 9.5% per annum = 9.5/365 = 0.026% per day.
Interest = PTR/100 = (36000 * 146 * 0.026)/100 = Rs. 1368.

Given:
- Rate of interest (R) = 12% per annum
- Principal amount (P)
- Number of years (N) to double the principal amount

According to the formula for simple interest:

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

Since we want to find the number of years (N) required to double the principal amount, the simple interest earned in N years will be equal to the principal amount (P).

So, we have:

\[ \dfrac{P \times R \times N}{100} = P \]

Now, let's solve for N:

\[ \dfrac{P \times 12 \times N}{100} = P \]

\[ 12N = 100 \]

\[ N = \dfrac{100}{12} \]

\[ N = 8.33 \]

So, it will take approximately 8.33 years for the principal amount to double.

Now, converting 0.33 years to months:

\[ 0.33 \times 12 \approx 4 \text{ months} \]

Therefore, the principal amount will double in approximately 8 years and 4 months.

Hence, the correct answer is:

\[ \boxed{\text{8 yrs. 4 months}} \]

Let's denote the two parts of the principal amount as \(x\) and \(12,000 - x\), where \(x\) represents the amount invested at 12% per annum for 3 years, and \(12,000 - x\) represents the amount invested at 16% per annum for 4.5 years.

The formula for calculating simple interest is:

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

For the first part (\(x\)), with a principal of \(x\), rate \(R = 12%\), and time \(T = 3\) years, the simple interest is:

\[ \text{Simple Interest}_1 = \dfrac{x \times 12 \times 3}{100} = \dfrac{36x}{100} = 0.36x \]

For the second part (\(12,000 - x\)), with a principal of \(12,000 - x\), rate \(R = 16%\), and time \(T = 4.5\) years, the simple interest is:

\[ \text{Simple Interest}_2 = \dfrac{(12,000 - x) \times 16 \times \dfrac{9}{2}}{100} = \dfrac{72(12,000 - x)}{100} = 864 - 0.72x \]

Given that the simple interest on the first part is equal to the simple interest on the second part, we can set up the equation:

\[ 0.36x = 864 - 0.72x \]

Let's solve this equation to find the value of \(x\):

\[ 0.36x + 0.72x = 864 \]

\[ 1.08x = 864 \]

\[ x = \dfrac{864}{1.08} \]

\[ x = 800 \]

So, the first part (\(x\)) is \(8000\), and the second part (\(12000 - x\)) is \(12000 - 8000 = 4000\).

Now, to find the greater part, we compare \(8000\) with \(4000\), and \(8000\) is greater.

Hence, the correct answer is \( \boxed{8000} \).

Let's denote the sum of money lent to X and Y as \( P \). 

For X, at a rate of 7.5% per annum for 4 years, the simple interest (SI) is given by:

\[ \text{SI}_X = \dfrac{P \times 7.5 \times 4}{100} = 0.3P \]

For Y, at a rate of 7.5% per annum for 5 years, the simple interest (SI) is given by:

\[ \text{SI}_Y = \dfrac{P \times 7.5 \times 5}{100} = 0.375P \]

The difference in interest paid by them is given as $150. So, we can set up the equation:

\[ \text{SI}_Y - \text{SI}_X = 0.375P - 0.3P = 150 \]

\[ 0.075P = 150 \]

\[ P = \dfrac{150}{0.075} = 2000 \]

So, the sum lent to each, \( P \), is $2000.

Therefore, the correct answer is \( \boxed{2000} \).

Let's denote the principal amount as \(P\). 

The formula for simple interest is:

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

Given that the rate of interest is \(16 \dfrac{2}{3}\% = \dfrac{50}{3}\%\) per annum, we can write it as \( R = \dfrac{50}{3} \).

We want to find the time (\(T\)) in years such that the simple interest becomes equal to the principal amount.

So, we set up the equation:

\[ \dfrac{P \times \dfrac{50}{3} \times T}{100} = P \]

Canceling out \(P\) from both sides, we have:

\[ \dfrac{50 \times T}{3 \times 100} = 1 \]

\[ \dfrac{50 \times T}{300} = 1 \]

\[ \dfrac{1}{6} \times T = 1 \]

\[ T = 6 \]

So, the simple interest on the sum of money will be equal to the principal after \(6\) years.

Therefore, the correct answer is \( \boxed{6} \) years.

Let's denote the principal amount as \( P \).

The formula for calculating simple interest is:

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

Given that the interest rate for the initial investment is \( R \), the interest earned after \(2\) years would be:

\[ \text{Interest}_1 = \dfrac{P \times R \times 2}{100} = \dfrac{2PR}{100} \]

If the investment had been made at a rate \( 1\% \) higher, the new rate would be \( R + 1 \) and the interest earned after \(2\) years would be:

\[ \text{Interest}_2 = \dfrac{P \times (R + 1) \times 2}{100} = \dfrac{2P(R + 1)}{100} \]

Given that the difference in interest is \(24\), we can set up the equation:

\[ \text{Interest}_2 - \text{Interest}_1 = 24 \]

\[ \dfrac{2P(R + 1)}{100} - \dfrac{2PR}{100} = 24 \]

\[ \dfrac{2P(R + 1 - R)}{100} = 24 \]

\[ \dfrac{2P}{100} = 24 \]

\[ \dfrac{P}{50} = 24 \]

\[ P = 24 \times 50 \]

\[ P = 1200 \]

So, the principal amount is \( \$1200 \).

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

Let's denote the principal amount as \( P \).

The formula for calculating simple interest is:

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

Given that the annual simple interest rate is \( 10\% \) and the time is \( 4 \) years, the interest earned would be:

\[ \text{Interest} = \dfrac{P \times 10 \times 4}{100} = \dfrac{40P}{100} = \dfrac{2P}{5} \]

Now, we know that at the end of four years, the amount invested plus interest earned was \( \$770 \). So, we can set up the equation:

\[ P + \text{Interest} = 770 \]

\[ P + \dfrac{2P}{5} = 770 \]

Multiplying both sides by \(5\) to clear the fraction:

\[ 5P + 2P = 770 \times 5 \]

\[ 7P = 3850 \]

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

\[ P = \dfrac{3850}{7} \]

\[ P = 550 \]

So, the amount invested was \( \$550 \).

Therefore, the correct answer is \( \boxed{550} \).

Let's denote the principal amount as \( P \).

For the first scenario, where the sum doubles in 5 years, let the rate of interest be \( R_1 \). The amount after 5 years will be \( 2P \). Using the formula for simple interest:

\[ \text{Simple Interest}_1 = \dfrac{P \times R_1 \times 5}{100} \]

For the second scenario, where the sum becomes three times in 12 years, let the rate of interest be \( R_2 \). The amount after 12 years will be \( 3P \). Using the formula for simple interest:

\[ \text{Simple Interest}_2 = \dfrac{P \times R_2 \times 12}{100} \]

Given that the sum doubles in 5 years and becomes three times in 12 years, we can set up the equations:

\[ P + \text{Simple Interest}_1 = 2P \]

\[ P + \dfrac{P \times R_1 \times 5}{100} = 2P \]

\[ \dfrac{P \times R_1 \times 5}{100} = P \]

\[ R_1 \times 5 = 100 \]

\[ R_1 = 20\% \]

\[ P + \text{Simple Interest}_2 = 3P \]

\[ P + \dfrac{P \times R_2 \times 12}{100} = 3P \]

\[ \dfrac{P \times R_2 \times 12}{100} = 2P \]

\[ R_2 \times 12 = 200 \]

\[ R_2 = \dfrac{200}{12} \]

\[ R_2 = 16 \dfrac{2}{3}\% \]

So, the lower rate of interest per annum is \( \boxed{16 \dfrac{2}{3}\%} \).

Let's denote the principal amount as \( P \).

Using the formula for simple interest:

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

Given that the simple interest on a sum of money at the rate of \(5\%\) per annum for \(8\) years is \( Rs. 840 \), we can set up the equation:

\[ 840 = \dfrac{P \times 5 \times 8}{100} \]

\[ 840 = \dfrac{40P}{100} \]

\[ 84000 = 40P \]

\[ P = \dfrac{84000}{40} \]

\[ P = 2100 \]

So, the principal amount is Rs. 2100.

Now, let's find the rate of interest for which the same amount of interest can be received on the same sum after \(5\) years.

Using the same principal amount (\(P = 2100\)) and the formula for simple interest, and given that the time is \(5\) years, let the new rate of interest be \( R' \):

\[ \text{Simple Interest} = \dfrac{P \times R' \times 5}{100} \]

Given that the simple interest remains the same (i.e., \( Rs. 840 \)), we can set up the equation:

\[ 840 = \dfrac{2100 \times R' \times 5}{100} \]

\[ 840 = \dfrac{105 \times R'}{100} \]

\[ 84000 = 105 \times R' \]

\[ R' = \dfrac{84000}{105} \]

\[ R' = 800 \]

So, the rate of interest for which the same amount of interest can be received on the same sum after \(5\) years is \(8\%\) per annum.

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