Question

Find thebrate of interest at which an amount of 12000 will become 15000 at the end of 10 year

Answer 1

Answer:

2.5% is correct answer

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Answer 2

Given,

P = 12000

A = 15000

T = 10 years

SI = (A-P) = 15000-12000 = 3000

Simple interest formula

$s.i = \frac{(prt)}{100}$

substitute values we get

3000 = (12000)R(10)/100

Solve to get value of R

3 = (12 × R)/10

30 = 12 × R

30/12 = R

15/6 = R

5/2 = R

R = 2.5%