Question

A carpenter cut four lengths of wood. The lengths were 36 3/4th in., 36 3/8th in., 37 1/2 in., and z in. If the mean of the lengths is 36 5/8th in., what was the length of the fourth piece of wood the carpenter cut?

Answer 1

Answer:

$z=35\dfrac{7}{8}\ in$

Step-by-step explanation:

The mean of four numbers $a,\ b,\ c, \ d$ is

$\dfrac{a+b+c+d}{4}$

Given

$a=36\dfrac{3}{4}\ in\\ \\b=36\dfrac{3}{8}\ in\\ \\c=37\dfrac{1}{2}\ in\\ \\d=z\ in\\ \\\text{Mean}=36\dfrac{5}{8}\ in,$

then

$\dfrac{36\dfrac{3}{4}+36\dfrac{5}{8}+37\dfrac{1}{2}+z}{4}=36\dfrac{5}{8}$

Add numbers in the numerator:

$36\dfrac{3}{4}+36\dfrac{5}{8}+37\dfrac{1}{2}=(36+36+37)+\left(\dfrac{3}{4}+\dfrac{3}{8}+\dfrac{1}{2}\right)=109+\left(\dfrac{6}{8}+\dfrac{3}{8}+\dfrac{4}{8}\right)=109+\dfrac{13}{8}=109+1\dfrac{5}{8}=110\dfrac{5}{8}$

Hence,

$\dfrac{110\dfrac{5}{8}+z}{4}=36\dfrac{5}{8}\\ \\110\dfrac{5}{8}+z=36\dfrac{5}{8}\cdot 4\\ \\110\dfrac{5}{8}+z=144+\dfrac{5}{2}\\ \\110\dfrac{5}{8}+z=144+2\dfrac{1}{2}\\ \\110\dfrac{5}{8}+z=146\dfrac{1}{2}\\ \\x=146\dfrac{1}{2}-110\dfrac{5}{8}\\ \\z=146\dfrac{4}{8}-110\dfrac{5}{8}\\ \\z=36-\dfrac{1}{8}\\ \\z=35\dfrac{7}{8}\ in$