Question

suppose there are two projects :

A = 100 + 100(1+r)
B = 20 + 100(1+r) + 100(1+r)^2

Make it A=B, please show me how to derive r=0.118 (11.8%)

Answer 1

(note : IN THE GIVEN QUESTION IT IS SHOWING 100(1+r)^2 .but it is actually 100 / (1+r)^2 .then only answer will come 11.8%.problem is solved based on this assumption)

given A = 100 + 100 / (1+r)

B = 20 + 100 / (1+r) + 100 / (1+r)^2

also A = B

100 + 100 / (1+r) = 20 + 100 / (1+r) + 100 / (1+r)^2

100 - 20 + 100 / (1+r) - 100 / (1+r) = 100 / (1+r)^2

80 + 0 = 100 / (1+r)^2

(1+r)^2 = 100 / 80

(1+r)^2 = 1.25

(1+r) = (1.25)^(1/2)

1+r = 1.118

r = 1.118 - 1

r = 0.118 or 11.8%

(i will also solve as per information given in the question.but answer can never be 11.80% it will be as follows)

100 + 100(1+r) = 20 + 100(1+r) + 100(1+r)^2

100 - 20 + 100(1+r) - 100(1+r) = 100(1+r)^2

80 + 0 = 100(1+r)^2

(1+r)^2 = 80 / 100

(1+r)^2 = 0.8

1+r = (0.8)^(1/2)

1 + r = 0.8944

r = 0.8944 - 1

r = -0.1056 or -10.56% (it will be negative value)

(in case of any further explanation please comment)