Question

3
3
3. A piece of wire m long was cut into three pieces. The first piece was m long, while
4
20
length of the second plece was of the first. How long was the third piece?
3​

Answer 1

Answer: The length of third piece is \dfrac{7}{20}\ m207 m

Step-by-step explanation:

Since we have given that

Length of a piece of wire = \dfrac{3}{4}\ m43 m

Length of first piece of wire = \dfrac{3}{20}\ m203 m

Length of second piece of wire = \dfrac{5}{3}\times \dfrac{3}{20}=\dfrac{1}{4}\ m35×203=41 m

Let the length of third piece of wire be x.

According to question, it becomes,

\begin{gathered}\dfrac{3}{20}+\dfrac{1}{4}+x=\dfrac{3}{4}\\\\\\\dfrac{3+5}{20}x=\dfrac{3}{4}\\\\\\\dfrac{8}{20}+x=\dfrac{3}{4}\\\\\\x=\dfrac{3}{4}-\dfrac{8}{20}\\\\\\x=\dfrac{15-8}{20}\ m=\dfrac{7}{20}\end{gathered}203+41+x=43203+5x=43208+x=43x=43−208x=2015−8 m=207

Hence, the length of third piece is \dfrac{7}{20}\ m207 m

Answer: The length of third piece is \dfrac{7}{20}\ m207 m

Step-by-step explanation:

Since we have given that

Length of a piece of wire = \dfrac{3}{4}\ m43 m

Length of first piece of wire = \dfrac{3}{20}\ m203 m

Length of second piece of wire = \dfrac{5}{3}\times \dfrac{3}{20}=\dfrac{1}{4}\ m35×203=41 m

Let the length of third piece of wire be x.

According to question, it becomes,

\begin{gathered}\dfrac{3}{20}+\dfrac{1}{4}+x=\dfrac{3}{4}\\\\\\\dfrac{3+5}{20}x=\dfrac{3}{4}\\\\\\\dfrac{8}{20}+x=\dfrac{3}{4}\\\\\\x=\dfrac{3}{4}-\dfrac{8}{20}\\\\\\x=\dfrac{15-8}{20}\ m=\dfrac{7}{20}\end{gathered}203+41+x=43203+5x=43208+x=43x=43−208x=2015−8 m=207

Hence, the length of third piece is \dfrac{7}{20}\ m207 m