# Comparing Quantities

## Exercise 8.3 Part 2

Question 6: Arif took a loan of Rs 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after 1.5 years if the interest is

(i) compounded annually.

Solution: Amount after 1 year =80000+80000 xx10%=80000+8000=88000
Interest for the next 6 months = 88000 xx 5%=4400
Amount after 1.5 years = 88000+4400 = 92400

(ii) compounded half yearly.

Solution: rate of interest will become half and time will be three half years.

Amount =80000(1+(5)/(100))^3

=80000xx(21)/(20)xx(21)/(20)xx(21)/(20)=92610

Question 7: Maria invested Rs 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find

(i) The amount credited against her name at the end of the second year.

Answer: Amount =8000(1+(5)/(100))^2

=9000xx(21)/(20)xx(21)/(20)=8820

Interest after 2 years = 8820-8000=820

(ii) The interest for the 3rd year.

Solution: Interest for the 3rd year = 8820 xx 5%=441

Question 8: Find the amount and the compound interest on Rs 10,000 for 1.5 years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

Answer: Amount =10000(1+(5)/(100))^3

=10000xx(21)/(20)xx(21)/(20)xx(21)/(20)=11576.25

Interest = 11576.25-10000=1576.25
This interest would be more than the interest compounded annually, as interest compounded half yearly is always more than compounded annually at the same rate of interest.

Question 9: Find the amount which Ram will get on Rs 4096, if he gave it for 18 months at 12.5% per annum, interest being compounded half yearly.

Answer: Amount =4096(1+(25)/(400))^3

=4096xx(17)/(16)xx(17)/(16)xx(17)/(16)=4913

10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum

(i) find the population in 2001.

Solution: Let us assume the population to be P in 2001

Then 54000=P(1+(5)/(100))^2

Or, 54000=P\xx(21)/(20)xx(21)/(20)

Or, P=54000xx(20)/(21)xx(20)/(21)=48979.5

which can be said to be 48980 in round figures.

(ii) what would be its population in 2005?

Answer: Population in 2005 =54000xx(21)/(20)xx(21)/(20)=59535

Question 11: 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.

Solution: The bacteria’s growth will be compounded after every hour

Answer: Population in 2005 =506000xx(1+(25)/(1000))^2

=506000xx(41)/(40)xx(41)/(40)=531616(approx)

which is 531616 in round figures

Question 12: A scooter was bought at Rs 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Solution: Value of scooter after 1 year
= 42000-42000 xx 8% = 42000-3360 = 38640