Answer:

Explanation:
One half of a number can be written as

three fifth of a number can be written as

three eighth of a number can be written as

We have to find the sum of all three
which is
Sum =

No we have to make the denominator same for this we have to find the LCM of 2,5,8
Factors of 2 = 2
Factors of 5= 5
Factors of 8 = 2 * 2 * 2
LCM = 2 * 5 * 2 * 2
=40
Multiplying and dividing Every term with 40 for making the denominator same
Sum =

Cancelling out the terms
Sum =

Sum =

Adding the terms of numerator
Sum =

Sum =
this is the answer of the required