Question

3. A piece of wire 3/4 m long was cut into three pieces. The first piece was 3/20
m long, while the length of the second piece was 5/ 3 of the first.How long was the third piece?​

Answer 1

Step-by-step explanation:

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

20

7

m

Step-by-step explanation:

Since we have given that

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

4

3

m

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

20

3

m

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

3

5

×

20

3

=

4

1

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}

20

3

+

4

1

+x=

4

3

20

3+5

x=

4

3

20

8

+x=

4

3

x=

4

3

−

20

8

x=

20

15−8

m=

20

7

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

20

7

m