A trader bought x mangoes at the rate of 4 mangoes for 10 naira, five of the mangoes were bad so he sold the remaining at the rate of 5 mangoes for 20 naira and he made a gain of 10 naira how many mangoes did he buy?

  20

Step-by-step explanation:

The cost of the x mangoes was ...

  c = 10(x/4)

The revenue from the sale of mangoes was ...

  r = 20(x-5)/5

The profit is the difference between revenue and cost:

  10 = r - c = 20(x -5)/5 -10(x/4) = 1.5x -20 . . . . substitute and simplify

  30 = 1.5x . . . . . add 20

  20 = x . . . . . . . .divide by 1.5

The trader bought 20 mangoes.

Check

He paid 20(10/4) = 50 naira. He sold the 15 good mangoes for 20/5(15) = 60 naira, so made a 10 naira profit.

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!