Unit - 3 Growth and Depreciation

 Unit - 3 

Growth and Depreciation

Questions and Solutions

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1. Population Growth and Related Formulas

(a) If the present population of a locality is $P$, the population after $T$ years is $P_T$, and the yearly population growth rate is $R\%$, write the formula to find $P_T$.

Solution:
The formula for population growth with a fixed rate $R\%$ per year is:

$$P_T = P \times \left(1 + \frac{R}{100}\right)^T$$

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(b) If the number of tigers in Chitwan National Park in 2079 B.S. is $x$ and the annual growth rate of tigers is $R\%$, then what will be the number of tigers after $N$ years?

Solution:
The formula for the tiger population after $N$ years is:

$$\text{Tiger population} = x \times \left(1 + \frac{R}{100}\right)^N$$

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(c) The growth rate of foreign employment from Nepal for the first year, second year, and third year are respectively $R_1\%$, $R_2\%$, and $R_3\%$. Write the formula to find the number of employees after 3 years.

Solution:
For varying growth rates across multiple years, the formula is:

$$\text{Number of employees after 3 years} = P \times \left(1 + \frac{R_1}{100}\right) \times \left(1 + \frac{R_2}{100}\right) \times \left(1 + \frac{R_3}{100}\right)$$

Where $P$ is the initial number of employees.

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2. Population and Price Growth

(a) If the death rate is less than the birth rate of a country, does the population of the country increase or decrease?

Answer:
If the death rate is less than the birth rate, the population of the country increases because more people are being born than are dying.

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(b) One year ago, the price of a sack of 25 kg rice was Rs. 1500. As the price has increased by 10% p.a., how much does the sack of 25 kg rice cost now?

Solution:
The formula for the new price is:

$$\text{New price} = \text{Old price} \times \left(1 + \frac{\text{Rate of increase}}{100}\right)$$

Substitute the given values:

$$\text{New price} = 1500 \times \left(1 + \frac{10}{100}\right) = 1500 \times 1.1 = 1650$$

The sack of 25 kg rice now costs Rs. 1650.

Questions and Solutions

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3. Growth Rate Calculations

(a) According to the census of 2021, the population of a city was 5,18,452. If the growth rate was 4.5% per annum, what will be the population of the city after 3 years?

Solution:
The formula for population growth is:

$$\text{Population after \(T\) years} = P \times \left(1 + \frac{R}{100}\right)^T$$

Substitute the given values:

$$P = 518452, \, R = 4.5, \, T = 3$$ $$\text{Population after 3 years} = 518452 \times \left(1 + \frac{4.5}{100}\right)^3 = 518452 \times (1.045)^3$$ $$(1.045)^3 = 1.141166125$$ $$\text{Population after 3 years} \approx 518452 \times 1.141166125 = 591,333$$

The population of the city after 3 years will be approximately 591,333.

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(b) A landlord made an agreement with a businessman to increase the rent by 5% per annum. If the rent of a shutter is Rs. 10,000 now, what will be the rent after 3 years?

Solution:
The formula for rent growth is:

$$\text{Rent after \(T\) years} = \text{Current rent} \times \left(1 + \frac{R}{100}\right)^T$$

Substitute the given values:

$$\text{Current rent} = 10000, \, R = 5, \, T = 3$$ $$\text{Rent after 3 years} = 10000 \times \left(1 + \frac{5}{100}\right)^3 = 10000 \times (1.05)^3$$ $$(1.05)^3 = 1.157625$$ $$\text{Rent after 3 years} = 10000 \times 1.157625 = 11,576.25$$

The rent after 3 years will be Rs. 11,576.25.

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(c) The growth rate of bacteria in curd is 10% per hour. If the number of bacteria at 6 a.m. is $4 \times 10^6$, what will be the number of bacteria after 2 hours?

Solution:
The formula for bacterial growth is:

$$\text{Number of bacteria after \(T\) hours} = \text{Initial number of bacteria} \times \left(1 + \frac{R}{100}\right)^T$$

Substitute the given values:

$$\text{Initial number of bacteria} = 4 \times 10^6, \, R = 10, \, T = 2$$ $$\text{Number of bacteria after 2 hours} = 4 \times 10^6 \times \left(1 + \frac{10}{100}\right)^2 = 4 \times 10^6 \times (1.1)^2$$ $$(1.1)^2 = 1.21$$ $$\text{Number of bacteria after 2 hours} = 4 \times 10^6 \times 1.21 = 4.84 \times 10^6$$

The number of bacteria after 2 hours will be $4.84 \times 10^6$.

More Solution Update soon.