Question

A piece of wire$\frac{3}{4}m$ long was cut into three pieces. The first piece was $\frac{3}{20}m$long while the length of the second piece was$\frac{5}{3}$ of the first. How long was the third piece?

Answer 1

Find the length of the third piece.

A piece of wire$\frac{3}{4}m$ long was cut into three pieces and the first piece was $\frac{3}{20}m$long while the length of the second piece was$\frac{5}{3}$ of the first

STEP 1 : Find the length of the second piece.

The length of the second piece was$\frac{5}{3}$ of the first$=\frac{5}{3}\times \frac{3}{20}$

$=\frac{1}{4}m$

$\therefore$The length of the second piece is $=\frac{1}{4}m$

STEP 2 : Find the length of the third piece.

$$\begin{gathered} Totallength=lengthof(firstpiece+secondpiece+thirdpiece) \\ \frac{3}{4}=\frac{3}{20}+\frac{1}{4}+lengthofthirdpiece \\ \frac{3}{4}=\frac{3}{20}+\frac{1\times 5}{4\times 5}+lengthofthirdpiece \\ \frac{3}{4}=\frac{3}{20}+\frac{5}{20}+lengthofthirdpiece \\ \frac{3}{4}=\frac{3+5}{20}+lengthofthirdpiece \\ \frac{3}{4}=\frac{8}{20}+lengthofthirdpiece \\ \therefore lengthofthirdpiece=\frac{3}{4}-\frac{8}{20} \\ =\frac{3\times 5}{4\times 5}-\frac{8}{20} \\ =\frac{15}{20}-\frac{8}{20} \\ =\frac{15-8}{20} \\ =\frac{7}{20} \end{gathered}$$

$\therefore$The length of third piece of wire is$\frac{7}{20}m$