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 × 10% = 80000 + 8000 = 88000
Interest for the next 6 months = 88000 × 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`