Ex 8.3, 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 1/2 years if the interest is (i) compounded annually.
Given,
Principal (P) = 80000
Rate (R) = 10% p.a
Time (N) = 1 1/2 years
Since n is in fraction,
We use the formula
Compound Interest for 1 1/2 Years
= Compound Interest for 1 year
+ Simple Interest for 1/2 year.
Compound Interest for 1 Year
P = 80000
R = 10%
n = 1
Amount = P (1+π /100)^π
= 80000 Γ (1+10/100)^1
= 80000 Γ (1+1/10)
= 80000 Γ ((10 + 1)/10)
= 80000 Γ 11/10
= 88000
Since,
Amount = Principal + Interest
88000 = 80,000 + Interest
88000 β 80000 = Interest
8000 = Interest
Interest = 8000
β΄ Interest for 1st year = Rs 8000
& Amount after 1st Year = Rs 88000
Simple Interest for next π/π year
Principal = Amount in Previous Year
= 88000
Rate = 10% p.a
Time = 1/2 year
Interest = (π Γ π Γ π)/100
= (88000 Γ10 Γ 1/2)/100
= (88000 Γ 5)/100
= 880 Γ 5
= 4400
Simple Interest for next 1/2 Year = Rs 4400
Now,
Interest after 1 1/4 years = Compound interest for 1 year
+ Simple interest for next 1/2 year
= 8000 + 4400
= 12400
Also,
Amount = Principal + Interest
= 80000 + 12400
= 92400
β΄ Amount = Rs 92,400
Ex 8.3, 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 1/2 years if the interest is (ii) compounded half yearly.
Given,
Principal (P) = 80000
Rate(R) = 10% per annum Compound Half Yearly
= 10/2 % per half yearly
= 5 % Per half yearly
Time (N) = 1 1/2 Years
= 3/2 years
= 3/2 Γ 2 half years
= 3 Half years
Now,
Amount = P (1+π /100)^π
= 80000 Γ (1+5/100)^3
= 80000 Γ (1+1/20)^3
= 80000 Γ ((20 + 1)/20)^3
= 80000 Γ (21/20)^3
= 80,000 Γ ((21 Γ 21 Γ 21)/(20 Γ 20 Γ 20))
= 80,000 Γ ((441 Γ 21)/(400 Γ 20)) = 80000 Γ (9261/8000)
= 10 Γ 9261
= 92610
β΄ Amount = Rs 92,610
Difference in Amounts
= Amount when interest is compounded half yearly
β Amount when interest is compounded annually
= 92610 β 92400
= 210
β΄ Difference in Amounts = Rs 210

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.