Ex 8.3, 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.
Given,
Principal (P) = Rs 8000
Rate (R) = 5% per annum
Time = 2 Years
We need to find the amount after 2 years
So, we use the formula
A = P ("1 + " 𝑟/100)^𝑛
Putting values
A = 8000 ("1 + " 5/100)^2
= 8000 ("1 + " 1/20)^2 = 8000 ((20 + 1)/20)^2
= 8000 (21/20)^2
= 8000 × ((21 × 21)/(20 × 20))
= 80 × ((21 × 21)/(2 × 2))
= 20 × 21 × 21
= 20 × 441
= 8820
∴ Amount after 2 years = Rs 8,820
Ex 8.3, 7 (Method 1) Maria invested Rs 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find (ii) The interest for the 3rd year.
Interest in 3rd year = Amount after 2 years × Rate
= 8820 × 5/100 %
= 8820 × 5/100
= 8820 × 1/20
= 882/2
= 441
∴ Interest for the 3rd year = Rs 441
Ex 8.3, 7 (Method 2) Maria invested Rs 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find (ii) The interest for the 3rd year.
Now,
Interest for the 3rd year
= Amount after 3 years − Amount after 2 years
Finding Amount after 3 years
Amount after 3 years = P (1+𝑅/100)^𝑛
Putting values
Amount = 8000 (1+5/100)^3
= 8000 (1+1/20)^3
= 8000 ((20 + 1)/20)^3
= 8000 (21/20)^3
= 8000 × (21 × 21 ×21)/(20 × 20 ×20)
= 8000 × (441 × 21)/(400 × 20)
= 8000 × 9261/8000
= 9261
Therefore,
Interest for the 3rd year
= Amount after 3 years − Amount after 2 years
= 9261 − 8820
= 441
∴ Interest for the 3rd year = Rs 441

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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