At the beginning of 95,

Principal,

Interest,

Final Amount at the end of 95,

At the beginning of 96,

Principal,

Interest,

Final Amount at the end of 96,

At the beginning of 97,

Principal,

Interest,

Final Amount at the end of 97,

So, from here we can safely deduce that after n years, the amount will be

so after 10 years

This however is the long approach to doing things. If you can identify that it is a geometric progression from the question, you can straight use the formula

For the second set of problem

i)

P = 1000(2.8)^0.02t

when t = 0

p = 1000(2.8)^0.02(0)

p = 1000

when p = 2000,

2000 = 1000(2.8)^0.02t

2 = 2.8^0.02t

ln 2 = 0.02t(ln 2.8)

t = ln 2/0.02(ln2.8)

ii)

let the initial population size = X

Population size = (X)(2^n), where n= number of times the bacteria multiplied

Given information, population size is p = 100X

hence,

100X = X(2^n)

n ln 2= ln 100

n = ln 100/ ln 2

n = 6.644 times

therefore time taken is 6.644 x 12 = 79.728 hours