Simple Interest All Competence Aptitude

Let's denote the latter sum of money as x.

For the first investment of $500 at 12% per annum for 4 years, the interest I₁ can be calculated using the formula for simple interest:

$$ I_1 = P \cdot R \cdot T $$ $$ I_1 = 500 \cdot \dfrac{12}{100} \cdot 4 $$ $$ I_1 = 240 $$ $$ I_2 = X \cdot \dfrac{10}{100} \cdot 4 $$ $$ I_2 = 0.4 X $$ $$ I_1 + I_2 = 240 + 0.4 X = 480 $$ $$ 0.4 X = 480 - 240 $$ $$ 0.4 X = 240 $$ $$ X = \dfrac{240}{0.4} $$ $$ X = 600 $$

Therefore, the latter sum of money is $600.

So, the correct answer is:

C. 600

The formula for calculating simple interest is given by:

$$ I = P \cdot R \cdot T $$ where: * $I$ is the interest, * $P$ is the principal amount (the amount borrowed), * $R$ is the rate of interest per annum, * $T$ is the time in years. Given: $$ I = 5,400, \quad R = 12\%, \quad T = 3 \text{ years} $$ Substitute into the formula: $$ 5,400 = P \cdot \dfrac{12}{100} \cdot 3 $$ Solve for $P$: $$ P = \dfrac{5,400}{\dfrac{12}{100} \cdot 3} $$ $$ P = \dfrac{5,400}{0.36} $$ $$ P = 15,000 $$

Therefore, the principal amount borrowed by him was $15,000.

So, the correct answer is:

D. 15,000

Let P be the principal amount, R be the rate of interest per annum, and T be the time in years.

The formula for simple interest is given by

$$ I = P \cdot R \cdot T $$

Given that:
- After 2 years, the amount is $5,182, so I₁ = P * R * 2
- After 3 years, the amount is $5,832, so I₂ = P * R * 3

Subtracting the first equation from the second gives the interest for the third year:

Interest difference formula $$ I_2 - I_1 = P \cdot R \cdot 3 - P \cdot R \cdot 2 $$ Substitute the amounts $$ 5,832 - 5,182 = P \cdot R \cdot (3 - 2) $$ Simplify $$ 650 = P \cdot R $$ Use the amount after 2 years $$ 5,182 = P + I_1 $$ Substitute $I_1 = P \cdot R \cdot 2$ $$ 5,182 = P + (P \cdot R \cdot 2) $$ Substitute $P \cdot R = 650$ $$ 5,182 = P + (650 \cdot 2) $$ Simplify $$ 5,182 = P + 1,300 $$ Solve for $P$ $$ P = 5,182 - 1,300 $$ $$ \boxed{P = 3,882} $$

Therefore, the principal amount (in USD) is $3,882.

So, the correct answer is:

C. $3,882

Let P be the principal amount, R be the rate of interest per annum, and T be the time in years.

The formula for simple interest is given by I = P*R*T

Given that the simple interest for 6 years is equal to 30% of the principal:

The simple interest formula is: $$ I = P \cdot R \cdot T $$ Given: $$ P \cdot R \cdot 6 = 0.3 \cdot P $$ Canceling $P$ from both sides: $$ R \cdot 6 = 0.3 $$ Solving for $R$: $$ R = \dfrac{0.3}{6} = 0.05 $$ So, the rate of interest per annum is $0.05$ or $5\%$. Let $T'$ be the time it will take for the simple interest to equal the principal: $$ I = P \cdot R \cdot T' $$ Since $I = P$: $$ P = P \cdot R \cdot T' $$ Cancel $P$ from both sides: $$ R \cdot T' = 1 $$ Substitute $R = 0.05$: $$ 0.05 \cdot T' = 1 $$ Solve for $T'$: $$ T' = \dfrac{1}{0.05} = 20 $$ The time it will take for the simple interest to equal the principal is: $$ \boxed{20 \text{ years}} $$

Therefore, it will be equal to the principal after 20 years.

So, the correct answer is:

A. 20 years

Let $P$ be the principal, $R$ the annual rate of interest, and $T$ the time in years. The simple interest formula is: $$ I = P \cdot R \cdot T $$ The sum doubles in $16 \dfrac{2}{3}$ years, so the total amount is: $$ P + I = 2P $$ Substitute $I = P \cdot R \cdot T$: $$ P + P \cdot R \cdot T = 2P $$ Divide through by $P$: $$ 1 + R \cdot T = 2 $$ $$ R \cdot T = 1 $$ Convert $16 \dfrac{2}{3}$ years to an improper fraction: $$ 16 \dfrac{2}{3} = \dfrac{50}{3} \, \text{years} $$ So: $$ R \cdot \dfrac{50}{3} = 1 $$ $$ R = \dfrac{3}{50} = 0.06 $$ $$ R = 0.06 \text{ or } 6\% $$

Therefore, the rate of simple interest per annum is 0.06 or 6%.

So, the correct answer is:

C. 6%

$$ \begin{aligned} &\text{Given: } P = 6000, \; A = 7200, \; T = 4 \text{ years (simple interest)} \\ &\text{Simple Interest: } SI = A - P = 7200 - 6000 = 1200 \\ &SI = \dfrac{P \cdot R \cdot T}{100} \implies 1200 = \dfrac{6000 \cdot R \cdot 4}{100} \\ &1200 = 240 R \implies R = \dfrac{1200}{240} = 5\% \text{ per year} \\ &\text{New rate: } R_{\text{new}} = 1.5 \cdot 5 = 7.5\% \\ &\text{Simple Interest for 5 years at new rate: } SI_{\text{new}} = \dfrac{P \cdot R_{\text{new}} \cdot T_{\text{new}}}{100} = \dfrac{6000 \cdot 7.5 \cdot 5}{100} = 2250 \\ &\text{New Amount: } A_{\text{new}} = P + SI_{\text{new}} = 6000 + 2250 = 8250 \\ &\boxed{A_{\text{new}} = 8250} \end{aligned} $$

The formula for simple interest is given by:

\[ I = P \cdot R \cdot T \]

where:
- \( I \) is the interest,
- \( P \) is the principal amount,
- \( R \) is the rate of interest per annum, and
- \( T \) is the time in years.

In this case, \( P = 3,000 \), \( I = 1,080 \), and \( R = 12\% = 0.12 \). We need to find \( T \).

\[ 1,080 = 3,000 \cdot 0.12 \cdot T \]

Now, solve for \( T \):

\[ T = \dfrac{1,080}{3,000 \cdot 0.12} \]

\[ T = \dfrac{1,080}{360} \]

\[ T = 3 \]

Therefore, it will take \( \mathbf{\text{3 years}} \) for a sum of Rs. 3,000 to yield an interest of Rs. 1,080 at 12% per annum simple interest.

So, the correct answer is:

B. 3 years

Let \( P \) be the principal amount, \( R \) be the rate of interest per annum, and \( T \) be the time in years.

The formula for simple interest is given by \( I = P \cdot R \cdot T \).

In this case, \( P = 1 \), \( I = 0.01 \) (1 cent is equivalent to 0.01 USD), and \( T = \dfrac{1}{12} \) years (1 month is \(\dfrac{1}{12}\) of a year). We need to find \( R \).

\[ 0.01 = 1 \cdot R \cdot \dfrac{1}{12} \]

Now, solve for \( R \):

\[ R = \dfrac{0.01}{\dfrac{1}{12}} \]

\[ R = 0.01 \cdot 12 \]

\[ R = 0.12 \]

Convert the rate to a percentage:

\[ R = 0.12 \times 100 \]

\[ R = 12\% \]

 

So, the correct answer is:

C. 12%

The formula for calculating simple interest is given by:

\[ I = P \cdot R \cdot T \]

where:
- \( I \) is the interest,
- \( P \) is the principal amount,
- \( R \) is the rate of interest per annum, and
- \( T \) is the time in years.

In this case, the interest per day (\( I_{\text{per day}} \)) is given as $2.00, and the rate per annum (\( R \)) is 5%. We need to find the principal amount (\( P \)).

\[ I_{\text{per day}} = P \cdot \dfrac{R}{100} \cdot \dfrac{1}{365} \]

\[ 2.00 = P \cdot \dfrac{5}{100} \cdot \dfrac{1}{365} \]

Now, solve for \( P \):

\[ P = \dfrac{2.00}{\dfrac{5}{100} \cdot \dfrac{1}{365}} \]

\[ P = \dfrac{2.00}{\dfrac{1}{20} \cdot \dfrac{1}{365}} \]

\[ P = \dfrac{2.00}{\dfrac{1}{7300}} \]

\[ P = 2.00 \times 7300 \]

\[ P = 14,600 \]

Therefore, the sum lent at 5% per annum that produces interest of $2.00 a day is \( \mathbf{\text{C. 14,600}} \).

The formula for calculating simple interest is given by:

\[ I = P \cdot R \cdot T \]

where:
- \( I \) is the interest,
- \( P \) is the principal amount,
- \( R \) is the rate of interest per annum, and
- \( T \) is the time in years.

In this case, \( P = 3,000 \), \( I = 1,080 \), and \( R = 12\% = 0.12 \). We need to find \( T \).

\[ 1,080 = 3,000 \cdot 0.12 \cdot T \]

Now, solve for \( T \):

\[ T = \dfrac{1,080}{3,000 \cdot 0.12} \]

\[ T = \dfrac{1,080}{360} \]

\[ T = 3 \]

Therefore, it will take \( \mathbf{\text{3 years}} \) for a sum of Rs. 3,000 to yield a simple interest of Rs. 1,080 at 12% per annum.

So, the correct answer is:

A. 3 years