The present population of a town is 200000. The population will increase by 10% in the first year and 15% in the second year. The population of the town after two years will be:

1. 253000

2. 235000

3. 203500

4. 352000

A machine depreciates at the rate of 12% of its value at the beginning of a year. The machine was purchased for ₹ 10,000 and is sold for ₹ 7,744. The number of years, that the machine was used is:

1. 2

2. 4

3. 6

4. 8

Case Study

In 2015, the population of a town was 20000. During 2016 and 2017, it increases by 4% and 5% every year respectively. In 2018, due to an epidemic and migration to cities, the population decreases at 5%. Based on the above information answer the following questions:

1. The population of the town in 2016 was:
   (a) 20800
   (b) 20500
   (c) 20460
   (d) 20300

2. The difference in the population of the town at the end of the years 2017 and 2015 was:
   (a) 1500
   (b) 1700
   (c) 1800
   (d) 1840

3. The population of the town at the end of the year 2018 was:
   (a) 20100
   (b) 20500
   (c) 20748
   (d) 20850

4. Which of the following expressions gives the population at the end of the year 2018?
   
   (a) $20000 \times \dfrac{104}{100} \times \dfrac{105}{100} \times \dfrac{105}{100}$
   
   (b) $20000 \times \dfrac{104}{100} \times \dfrac{105}{100} \times \dfrac{95}{100}$
   
   (c) $20000 \times \dfrac{104}{100} \times \dfrac{95}{100}\times \dfrac{95}{100}$
   
   (d) $20000 \times \dfrac{96}{100} \times \dfrac{95}{100} \times \dfrac{95}{100}$

5. If the population of the town at the end of 2019 was 21,163, then during 2019, the population increases at the rate of:
   (a) 2%
   (b) 3%
   (c) 3.5% (d) 4%

Assertion (A): ₹ 1,00,000 amounts to ₹ 1,21,000 in 1 year and to ₹1,46,410 in 2 years, compounded annually.

Reason (R): When the interest is compounded annually, then, we have A = $P \times \Big(1 + \dfrac{R}{100} \Big)^n$

1. A is true, R is false

2. A is false, R is true

3. Both A and R are true

4. Both A and R are false