A prize fund is setup with a single investment of £4000 to provide an annual prize of £150. The fund accrues interest at 5% p. A. Paid yearly. If the first prize is awarded one year after the investment, find the number of years for which the full prize can be awarded.

The invested amount, A=£4000

Rate of interest on this amount, r=5% per annum.

Total interest on the amount £4000 after 1 year

.

The total amount after 1 year, P=£4000+£200=£4200

Amount remains after an annual prize of £150=£4200-£150=£4050.

Note that after providing the annual prize the amount remains is £4050 which is more than the invested amount of £4000 as the prize money is less than the interest gained.

Now, for the 2nd year, the 5% interest is on the principal amount of £4050, which will have more interest as compared to the interest of 1st year.

Again, after prize distribution, the total amount remains for the principal amount of 3rd year will be more than the principal amount of 2nd year. This sequence repeated every year and the total amount will increase every year.

So, it will never end, the full prize money can be awarded for infinitely many years.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!