Sam would have [tex]\frac{7}{8}[/tex]m of rope left
To determine how much rope Sam has left, the amount of rope he had ( [tex]\frac{3}{2}[/tex] ) will be subtracted from the amount of rope he cut for a project ( [tex]\frac{5}{8}[/tex] ). The final value is the how much rope he has left.
Notably, the numerical values to be added are in fractions, and since fraction has to be solved by its numerator and denominator, we therefore find the lowest common denominators (LCD) of both denominators (2 and 8), and then use it to factor out the numerators, the resulting values will be subtracted to get the final value.
Therefore,
The LCD of 2 and 8 = 8
= [tex]\frac{3}{2}[/tex] – [tex]\frac{5}{8}[/tex]
= (4 x 3) – (1 x 5) / 8
= (12) – (5) / 8
= 12 – 5 / 8
= [tex]\frac{7}{8}[/tex]
Therefore, the total amount of rope Sam has left = [tex]\frac{7}{8}[/tex]m
The above math problem is expressed in fraction. In order to solve the math problem, an order of operation i.e. BODMAS had to be applied. Significantly, Multiplication and Subtraction are part of the arms of BODMAS that were used to derive the solution to the given problem.
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