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 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+𝑅/100)^𝑛
= 80000 × (1+10/100)^1
= 80000 × (1+1/10)
= 80000 × ((10 + 1)/10)
= 80000 × 11/10
= 88000
Since,
Amount = Principal + Interest
88000 = 80,000 + Interest
88000 − 80000 = Interest
8000 = Interest
Interest = 8000
∴ Interest for 1st year = Rs 8000
& Amount after 1st Year = Rs 88000
Simple Interest for next 𝟏/𝟐 year
Principal = Amount in Previous Year
= 88000
Rate = 10% p.a
Time = 1/2 year
Interest = (𝑃 × 𝑅 × 𝑇)/100
= (88000 ×10 × 1/2)/100
= (88000 × 5)/100
= 880 × 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
∴ Amount = Rs 92,400
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 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 × 2 half years
= 3 Half years
Now,
Amount = P (1+𝑅/100)^𝑛
= 80000 × (1+5/100)^3
= 80000 × (1+1/20)^3
= 80000 × ((20 + 1)/20)^3
= 80000 × (21/20)^3
= 80,000 × ((21 × 21 × 21)/(20 × 20 × 20))
= 80,000 × ((441 × 21)/(400 × 20)) = 80000 × (9261/8000)
= 10 × 9261
= 92610
∴ Amount = Rs 92,610
Difference in Amounts
= Amount when interest is compounded half yearly
– Amount when interest is compounded annually
= 92610 − 92400
= 210
∴ Difference in Amounts = Rs 210
Made by
Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo
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