Simple Interest All Competence Aptitude

Let's denote the amount lent at 5% as \(x\) and the amount lent at 4% as \(60000 - x\).

The formula for calculating simple interest is:

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

Given that the total annual interest is \(2560\), we can set up the equation:

\[ \text{Interest at 5%} + \text{Interest at 4%} = 2560 \]

\[ \dfrac{x \times 5 \times 1}{100} + \dfrac{(60000 - x) \times 4 \times 1}{100} = 2560 \]

Solving this equation will give us the value of \(x\), the amount lent at 5%.

\[ \dfrac{5x}{100} + \dfrac{240000 - 4x}{100} = 2560 \]

\[ \dfrac{5x + 240000 - 4x}{100} = 2560 \]

\[ \dfrac{x + 240000}{100} = 2560 \]

\[ x + 240000 = 2560 \times 100 \]

\[ x + 240000 = 256000 \]

\[ x = 256000 - 240000 \]

\[ x = 16000 \]

So, the amount lent at \(5\%\) is \(16000\), and the amount lent at \(4\%\) is \(60000 - 16000 = 44000\).

Therefore, the money lent at \(4\%\) was \( \boxed{44000} \).

To find out the principal amount that will amount to Rs. 7000 in 5 years at 3 and 1/3 % simple interest, we can use the formula for simple interest:

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

Given that the principal amount is \( P \), the rate of interest is \( 3 \dfrac{1}{3}\% = \dfrac{1}{30} \), and the time is \( T = 5 \) years, we can write the equation for the amount as:

\[ A = P + \text{Simple Interest} \]

\[ A = P + \dfrac{P \times \dfrac{1}{30} \times 5}{100} \]

Given that the amount after 5 years is Rs. 7000, we can set up the equation:

\[ 7000 = P + \dfrac{5P}{30 \times 100} \]

\[ 7000 = P + \dfrac{1}{6}P \]

\[ 7000 = \dfrac{7}{6}P \]

To find \( P \), we multiply both sides by \( \dfrac{6}{7} \):

\[ P = \dfrac{7000 \times 6}{7} \]

\[ P = \dfrac{42000}{7} \]

\[ P = 6000 \]

So, the principal amount that will amount to Rs. 7000 in 5 years at 3 and 1/3 % simple interest is \( \boxed{6000} \).

To find the rate of interest per annum, we can first calculate the total simple interest earned for each deposit separately and then sum them up.

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

For the deposit of Rs. 400 for 2 years:\[ \text{Simple Interest} = \dfrac{400 \times R \times 2}{100} = 8R \]

For the deposit of Rs. 550 for 4 years:\[ \text{Simple Interest} = \dfrac{550 \times R \times 4}{100} = 22R \]

For the deposit of Rs. 1200 for 6 years:\[ \text{Simple Interest} = \dfrac{1200 \times R \times 6}{100} = 72R \]

Now, the total simple interest is the sum of the simple interest earned for each deposit:\[ 8R + 22R + 72R = 1020 \]

\[ 102R = 1020 \]

\[ R = \dfrac{1020}{102} \]

\[ R = 10 \]

So, the rate of interest per annum is \( \boxed{10\%} \).

If prinicipal = x and rate = r% p.a., then
1380 = x + x × 3 × r /100   ...........i
1500 = x + x × 5 × r /100   ...........ii
S.I. for two years = 1500 – 1380 = Rs. 120
∴ x × 2 × r /100 = 120
∴ xr /100 = 60  ..............iii
1200 × r /100 = 60
⇒  r = 6000 /1200 = 5% p.a.

Let's denote the sum lent out at 10% per annum for 5 years as \(x\) and the sum lent out at 12.5% per annum for 4 years as \(1500 - x\).

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 \(R = 10%\), and time \(T = 5\) years, the simple interest is:

\[ \text{Simple Interest}_1 = \dfrac{x \times 10 \times 5}{100} = \dfrac{50x}{100} = 0.5x \]

For the second part (\(1500 - x\)), with a principal of \(1500 - x\), rate \(R = 12.5%\), and time \(T = 4\) years, the simple interest is:

\[ \text{Simple Interest}_2 = \dfrac{(1500 - x) \times 12.5 \times 4}{100} = \dfrac{50(1500 - x)}{100} = 0.5(1500 - x) \]

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

\[ 0.5x = 0.5(1500 - x) \]

\[ 0.5x = 750 - 0.5x \]

\[ 0.5x + 0.5x = 750 \]

\[ x + x = 1500 \]

\[ 2x = 1500 \]

\[ x = \dfrac{1500}{2} \]

\[ x = 750 \]

So, the sum lent out at 12.5% per annum is \(1500 - x = 1500 - 750 = \boxed{750}\).

Let's denote the principal amount deposited in both the company and the bank as \(P\).

For the deposit in the company at \(12\%\) per annum for \(4\) years, the simple interest earned is:

\[ \text{Interest from company} = \dfrac{P \times 12 \times 4}{100} = \dfrac{48P}{100} = 0.48P \]

For the fixed deposit in the bank at \(15\%\) per annum for \(5\) years, the simple interest earned is:

\[ \text{Interest from bank} = \dfrac{P \times 15 \times 5}{100} = \dfrac{75P}{100} = 0.75P \]

Given that the difference in interest from the two sources is \(1350\), we can set up the equation:

\[ 0.75P - 0.48P = 1350 \]

\[ 0.27P = 1350 \]

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

\[ P = \dfrac{1350}{0.27} \]

\[ P = 5000 \]

So, the sum deposited in each case is \( \boxed{5000} \).

Let's denote the amounts invested in the three schemes as \(x\), \(y\), and \(z\) respectively.

Given that the time for the first scheme is \(6\) years, the second scheme is \(10\) years, and the third scheme is \(12\) years.

Using the formula for simple interest \(I = \dfrac{P \times R \times T}{100}\), where \(I\) is the interest earned, \(P\) is the principal amount, \(R\) is the rate of interest, and \(T\) is the time in years.

For the first scheme:\[ I_1 = \dfrac{x \times 10 \times 6}{100} = \dfrac{60x}{100} = 0.6x \]

For the second scheme:\[ I_2 = \dfrac{y \times 12 \times 12}{100} = \dfrac{120y}{100} = 1.2y \]

For the third scheme:\[ I_3 = \dfrac{z \times 15 \times 12}{100} = \dfrac{180z}{100} = 1.8z \]

Given that the interest earned is the same for all schemes, we can set up the equation:

\[ 0.6x = 1.2y = 1.8z \]

We can rewrite this equation as ratios:

\[ \dfrac{x}{y} = \dfrac{1.2}{0.6} = 2 \]\[ \dfrac{y}{z} = \dfrac{1.8}{1.2} = 1.5 \]

To find the ratio of \(x\), \(y\), and \(z\), we can multiply each ratio by a suitable number to make them integers. 

\[ \dfrac{x}{y} = 2 \]\[ \dfrac{y}{z} = 1.5 \]

Multiplying the first ratio by \(3\) and the second ratio by \(2\), we get:

\[ \dfrac{x}{y} = 6 \]\[ \dfrac{y}{z} = 3 \]

So, the ratio of investments is \( \boxed{6 : 3 : 2} \).

Let's denote the original sum of money as \(P\) and the original rate of interest as \(R\).

According to the given conditions:
1. If the sum was lent at the original rate \(R\) for 2 years, the interest earned would be \( \dfrac{P \times R \times 2}{100} = \dfrac{2PR}{100} = 0.02PR \).
2. If the sum was lent at a rate \(R + 3\) for 2 years, the interest earned would be \( \dfrac{P \times (R + 3) \times 2}{100} = \dfrac{2P(R + 3)}{100} = 0.02P(R + 3) \).

Given that the difference in interest is \(300\), we can set up the equation:

\[ 0.02P(R + 3) - 0.02PR = 300 \]

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

\[ 0.02P(R + 3 - R) = 300 \]

\[ 0.02P \times 3 = 300 \]

\[ 0.06P = 300 \]

\[ P = \dfrac{300}{0.06} \]

\[ P = 5000 \]

So, the original sum of money was \( \boxed{5000} \).

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

According to the given conditions:
1. If the scheme triples the investment in 8 years, the amount after 8 years would be \(3P\).
2. If the scheme quadruples the investment in \(T\) years, the amount after \(T\) years would be \(4P\).

Using the formula for simple interest:

\[ \text{Amount} = P + \text{Simple Interest} \]

For the first scenario (tripling the investment in 8 years), we can write the equation as:

\[ 3P = P + \dfrac{P \times R \times 8}{100} \]

\[ 3P = P + \dfrac{8PR}{100} \]

\[ 2P = \dfrac{8PR}{100} \]

\[ 2 = \dfrac{8R}{100} \]

\[ R = 25 \]

Now, we know the rate of interest is \(25\%\) per annum. Let's use this to find the time (\(T\)) for the second scenario (quadrupling the investment):

\[ 4P = P + \dfrac{P \times 25 \times T}{100} \]

\[ 4P = P + \dfrac{25PT}{100} \]

\[ 4P = P + \dfrac{PT}{4} \]

\[ 4 = 1 + \dfrac{T}{4} \]

\[ \dfrac{T}{4} = 3 \]

\[ T = 12 \]

So, to quadruple the investment, you have to invest it for \( \boxed{12} \) years.

If a certain sum doubles in 7 years at simple interest, it means that the interest accrued over 7 years is equal to the original principal amount.

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

According to the formula for simple interest:\[ \text{Interest} = \dfrac{P \times R \times T}{100} \]

Given that the sum doubles in 7 years, we have:\[ \dfrac{P \times R \times 7}{100} = P \]

This simplifies to:\[ 7R = 100 \]

\[ R = \dfrac{100}{7} \]

Now, to find out how many years it takes for the same sum to become 4 times under the same interest rate, we can use the formula for simple interest:

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

Given that the sum becomes 4 times, the interest accrued over \(T\) years is equal to 3 times the original principal amount (\(3P\)).

So, we have:\[ \dfrac{P \times R \times T}{100} = 3P \]

Substituting \(R = \dfrac{100}{7}\):\[ \dfrac{P \times \dfrac{100}{7} \times T}{100} = 3P \]

\[ \dfrac{100T}{7} = 300 \]

\[ T = \dfrac{300 \times 7}{100} \]

\[ T = 21 \]

So, the same sum under the same interest rate will become 4 times in \( \boxed{21} \) years.