EXERCISE 7.3 — DETAILED EXPLANATIONS

1. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum. (i) Find the population in 2001. (ii) What would be its population in 2005?

Given, Population of the place in 2003 = 54,000

It has increased at the rate of 5% per annum.

Here, 5% is a compound rate.

So we use the formula:

\( A = P \times \left(1 + \dfrac{R}{100}\right)^n \)

Here,

A = Population in year 2003 = 54,000

P = Population in year 2001 (to be found)

R = Rate = 5%

n = Number of years = 2003 − 2001 = 2

Putting values in the formula:

\(54000 = P \times \left(1 + \dfrac{5}{100}\right)^2\)

\(54000 = P \times \left(1 + \dfrac{1}{20}\right)^2\)

\(54000 = P \times \left(20 + \dfrac{1}{20}\right)^2\)

\(54000 = P \times \left(\dfrac{21}{20}\right)^2\)

\(54000 = P \times (1.05)^2\)

\(54000 = P \times 1.1025\)

\(P = \dfrac{54000}{1.1025} = 48980\) (approx)

Therefore, the population in 2001 = 48,980.

(ii) Now, to find population in 2005

From 2003 to 2005 = 2 years

Again using formula:

\( A = P \times \left(1 + \dfrac{R}{100}\right)^n \)

Here,

P = Population in 2003 = 54,000

R = 5%

n = 2

\( A = 54000 \times \left(1 + \dfrac{5}{100}\right)^2\)

Simlarly, \(A = 54000 \times (1.05)^2\)

= \(54000 \times 1.1025\)

= 59,535

Therefore, population in 2005 = 59,535.

2. In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.

Given, Initial count of bacteria = 5,06,000

Rate of increase = 2.5% per hour

n = 2 hours

Formula: \( A = P \times \left(1 + \dfrac{R}{100}\right)^n \)

Here,

P = 506000

R = 2.5

n = 2

So, \(A = 506000 \times \left(1 + \dfrac{2.5}{100}\right)^2\)

= \(506000 \times \left(1 + \dfrac{1}{40}\right)^2\)

= \(506000 \times \left(\dfrac{40 + 1}{40}\right)^2\)

= \(506000 \times \left(\dfrac{41}{40}\right)^2\)

= \(506000 \times (1.025)^2\)

= \(506000 \times 1.050625\)

= 531616 (approx)

Therefore, the bacteria count after 2 hours = 5,31,616.

3. A scooter was bought at ₹42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Given, Price of scooter = ₹42,000

Rate of depreciation = 8% per annum

n = 1 year

Formula for depreciation:

\( A = P \times \left(1 - \dfrac{R}{100}\right)^n \)

Here,

P = 42000

R = 8

n = 1

So, \(A = 42000 \times \left(1 - \dfrac{8}{100}\right)\)

= \(42000 \times \left(1 - \dfrac{2}{25}\right)\)

= \(42000 \times \left(\dfrac{25 - 2}{25}\right)\)

= \(42000 \times \left(\dfrac{23}{25}\right)\)

= \(42000 \times (0.92)\)

= 38640

Therefore, the value of the scooter after one year = ₹38,640.