Percentage All Competence Aptitude

Let's denote the present value of the scooter as \(P\). According to the given information, the value of the scooter decreases by 10% each year.

The value of the scooter after 1 year (\(P_1\)) can be calculated as:\[ P_1 = P - 0.10P = 0.90P \]

After the second year, the value (\(P_2\)) becomes:\[ P_2 = 0.90P - 0.10(0.90P) = 0.81P \]

After the third year, the value (\(P_3\)) becomes:\[ P_3 = 0.81P - 0.10(0.81P) = 0.729P \]

Now, we know the present value \(P\) is Rs. 7,290. Let's substitute this value into the equation:

\[ 0.729P = 7290 \]

Now, solve for \(P\):

\[ P = \dfrac{7290}{0.729} \]

\[ P = 10000 \]

Therefore, the value of the scooter 3 years back was Rs. 10,000. Hence, the correct answer is Rs. 10,000.

Let Raman's last year's salary be denoted by \(X\).

According to the given information, Raman's salary is increased by 5% this year, so the present salary (\(P\)) is given by:

\[ P = X + 0.05X = 1.05X \]

Now, we are given that Raman's present salary is \(1,806\). Substitute this value into the equation:

\[ 1.05X = 1,806 \]

Now, solve for \(X\):

\[ X = \dfrac{1,806}{1.05} \]

\[ X \approx 1,720 \]

Therefore, Raman's last year's salary was approximately \(1,720\). Hence, the correct answer is \(1,720\).

Let's denote the present value of the property as \(P\).

After the first year, the value (\(P_1\)) becomes:\[ P_1 = P - 0.05P = 0.95P \]

After the second year, the value (\(P_2\)) becomes:\[ P_2 = 0.95P - 0.05(0.95P) = 0.9025P \]

After the third year, the value (\(P_3\)) becomes:\[ P_3 = 0.9025P - 0.05(0.9025P) = 0.857375P \]

Given that the present value is \(4,11,540\):

\[ P_3 = 0.857375P = 4,11,540 \]

Now, solve for \(P\):

\[ P = \dfrac{4,11,540}{0.857375} \]

\[ P \approx 4,80,000 \]

Therefore, the value of the property 3 years ago was approximately \(4,80,000\). Hence, the correct answer is \(4,80,000\).

To calculate the population after 2 years with a 10% annual increase, we can use the formula for compound interest:

\[ \text{Future Population} = \text{Present Population} \times (1 + \text{Rate})^{\text{Time}} \]

Given:
- Present Population = 180,000
- Rate of increase = 10% or 0.10 (as a decimal)
- Time = 2 years

\[ \text{Future Population} = 180,000 \times (1 + 0.10)^2 \]

\[ \text{Future Population} = 180,000 \times 1.1^2 \]

\[ \text{Future Population} = 180,000 \times 1.21 \]

\[ \text{Future Population} = 217,800 \]

Therefore, the population after 2 years will be 217,800. Hence, the correct answer is 217,800.

To find the population of the village two years ago, we need to apply a reverse percentage calculation.

Let \( P \) be the population two years ago.

The present population is given as 67,600, and it has been increasing annually at the rate of 4%. This implies that the population two years ago increased by 4% each year for two years.

The formula for reverse percentage change is given by:

\[ P = \dfrac{\text{Present Value}}{(1 + \text{Rate of Change})^{\text{Number of Years}}} \]

Substitute the given values:

\[ P = \dfrac{67,600}{(1 + 0.04)^2} \]

\[ P = \dfrac{67,600}{1.0816} \]

\[ P \approx 62,500 \]

Therefore, the population of the village two years ago was approximately 62,500. Hence, the correct answer is 62,500.

Let's denote the following:

- \( I \): the percentage of illiterate people in the city.
- \( P \): the percentage of poor people in the city.
- \( R \): the percentage of rich people in the city.

Given information:
1. \( I = 40\% \) (illiterate people in the city).
2. \( P = 60\% \) (poor people in the city).
3. Among the rich (\( R = 100\% - P = 40\% \)), \( 10\% \) are illiterate.

We need to find the percentage of the illiterate poor population.

The percentage of illiterate poor people (\( I \cap P \)) can be found using the formula:\[ I \cap P = I - (R \cap I) \]

We know that \( R \cap I \) is the percentage of rich illiterate people, which is \( 10\% \) of \( R \).

\[ R \cap I = 10\% \times R = 10\% \times 40\% = 4\% \]

Now, substitute the values into the formula:

\[ I \cap P = I - (R \cap I) \]

\[ I \cap P = 40\% - 4\% = 36\% \]

Therefore, the percentage of the illiterate poor population is \( 36\% \). Hence, the correct answer is 36.

To find the rate of growth per annum, you can use the formula for compound annual growth rate (CAGR):

\[ CAGR = \left( \dfrac{\text{Ending Value}}{\text{Beginning Value}} \right)^{\dfrac{1}{\text{Number of Years}}} - 1 \]

In this case:
- Beginning Value (BV) is 40,000,
- Ending Value (EV) is 48,400, and
- Number of Years is 2.

Substitute these values into the formula:

\[ CAGR = \left( \dfrac{48,400}{40,000} \right)^{\dfrac{1}{2}} - 1 \]

\[ CAGR = \left( 1.21 \right)^{\dfrac{1}{2}} - 1 \]

\[ CAGR = 0.1 \]

Convert this to a percentage:

\[ CAGR = 10\% \]

Therefore, the rate of growth per annum is 10%. Hence, the correct answer is 10%.

Let's denote the present population of the village as \( P \).

According to the given information, the population increases by 5% annually. Therefore, the population after one year (\( P_1 \)) is given by:\[ P_1 = P + 0.05P = 1.05P \]

After the second year, the population (\( P_2 \)) becomes:\[ P_2 = 1.05P + 0.05(1.05P) = 1.1025P \]

Given that the present population is \( 4410 \), we can set up the equation:\[ 1.1025P = 4410 \]

Now, solve for \( P \):\[ P = \dfrac{4410}{1.1025} \]

\[ P \approx 4000 \]

Therefore, the population 2 years ago was approximately \( 4000 \). Hence, the correct answer is 4000.