Ex 8.3, 2 Kamala borrowed Rs 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan? (Hint: Find A for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for 4/12 years).
Given,
Principal (P) = 26400
Rate (R) = 15% p.a
Time (n) = 2 Years 4 Months
= 2 4/12 years
= 21/3 years
Since n is in fraction
We use the formula
Compound interest for 2 1/3 years
= Compound Interest for 2 years
+ Simple interest for the next 1/3 years
Compound interest for 2 years
Principal = 26,400
Rate = 15%
Time (n) = 2
Amount = P (1+𝑅/100)^𝑛
= 26400 (1+15/100)^2
= 26400 ((100 + 15)/100)^2
= 26400 (115/100)^2
= 26400 × ((115 × 115)/(100 ×100))^2
= 264 × 13225/100
= (264 × 13225)/100
= 3491400/100
= 34914
∴ Amount = Rs 34914
Now,
Amount = Principal + Interest
34914 = 26400 + Interest
34914 − 26400 = Interest
8514 = Interest
∴ Interest = 8514
∴ Interest for 2 years = Rs 8514
& Amount after 2 years = Rs 34914
Simple interest for next 𝟏/𝟑 year
Principal will be the amount after 1 year
P = Rs 34914
R = 15% p.a
T = 1/3 years
SI = 𝑃𝑅𝑇/100
= (34914 × 15 × 1/3 )/100
= (34914 ×15)/(100 × 3)
= (34914 × 5)/100
= 34914/20
= 17457/10
= 1745.7
∴ Simple interest for 1/3 years = Rs 1745.7
Now,
Compound Interest for 2 1/3 years
= Compound Interest for 2 years + Simple interest for 1/3 year
= 8514 + 1745.7
= 10259.7
Now,
Amount = Principal + interest
= 26400 + 10259.7
= 36659.70
∴ Kamala will Pay Rs 36659.70 to the Bank.

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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