Question

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

Answer 1

Answer:

7/20 m

Step-by-step explanation:

As per the provided information in the given question, we have :

• A piece of wire 3/4m long was cut into three pieces.
• The first piece was 3/20m long while the length of the second piece was 5/3 of the first.

We are asked to calculate the length of third piece.

Let us assume the length of third piece as x.

According to the question,

$\longmapsto \rm {Total \; length_{(Wire)} = \dfrac{3}{4} \; m}$

_______________________

$\longmapsto \rm {Length_{(1st \; piece)} = \dfrac{3}{20} \; m}$

_______________________

$\longmapsto \rm {Length_{(2nd \; piece)} = \dfrac{5}{3} \; of \; Length_{(1st \; piece)}}$

Substitute the length of third piece of wire.

$\longmapsto \rm {Length_{(2nd \; piece)} = \dfrac{5}{3} \; of \; \dfrac{3}{20} \; m}$

'Of' means 'multiplication'.

$\longmapsto \rm {Length_{(2nd \; piece)} =\Bigg ( \dfrac{5}{3} \times \dfrac{3}{20} \Bigg ) \; m}$

$\longmapsto \rm {Length_{(2nd \; piece)} =\Bigg ( 5 \times \dfrac{1}{20} \Bigg ) \; m}$

$\longmapsto \rm {Length_{(2nd \; piece)} = \dfrac{5}{20} \; m}$

$\longmapsto \rm {Length_{(2nd \; piece)} = \dfrac{1}{4} \; m}$

_______________________

$\longmapsto \rm {Length_{(3rd \; piece)} =x \; m}$

_______________________

Now,

⇒ Length of 1st piece + Length of 2nd piece + Length of 3rd piece = Total length of wire

$\longmapsto \rm { \dfrac{3}{20} + \dfrac{1}{4} + x = \dfrac{3}{4} }$

Now, taking the L.C.M in L.H.S and performing addition of like terms.

$\longmapsto \rm { \dfrac{3 + 5}{20} + x = \dfrac{3}{4} }$

Performing addition in the numerator of the fraction in L.H.S.

$\longmapsto \rm { \dfrac{8}{20} + x = \dfrac{3}{4} }$

Transposing 8/20 from L.h.S to R.H.S, its sign will get changed.

$\longmapsto \rm { x = \dfrac{3}{4} -\dfrac{8}{20} }$

Now, taking the L.C.M in R.H.S and performing subtraction of like terms.

$\longmapsto \rm { x = \dfrac{15-8}{20} }$

Performing subtraction in the numerator of the fraction.

$\longmapsto \rm { x = \dfrac{7}{20} }$

So,

$\longmapsto \bf {Length_{(3rd \; piece)} = \dfrac{7}{20}\; m}$

∴ Length of the third piece is 7/20 m.

Answer 2

Given :

• Length of a piece of wire =$\sf\:{\dfrac{3}{4}}m$

• Length of first piece of wire =$\sf\:{\dfrac{3}{20}}m$

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

${ }$

Find :

• How long was the third piece?

${ }$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:━━━━━━━━━━━━━━━━━━$

${ }$

Solution :

${ }$

☯ Let the length of third piece of wire be x.

${ }$

$\:\:\:\:\:\:\:{\sf{\underline{\bigstar{\bold{According\:to\:the\:given\:Question:}}}}}$

${ }$

$\:\:\:\:\:\:\:\dashrightarrow\sf\:{\dfrac{3}{20}}\:+\:{\dfrac{1}{4}}\:+\:x\:=\:{\dfrac{3}{4}}$

${ }$

$\:\:\:\:\:\:\:\dashrightarrow\sf\:{\dfrac{3\:+\:5}{20}}x\:=\:{\dfrac{3}{4}}$

${ }$

$\:\:\:\:\:\:\:\dashrightarrow\sf\:{\dfrac{8}{20}}\:+\:x\:=\:{\dfrac{3}{4}}$

${ }$

$\:\:\:\:\:\:\:\dashrightarrow\sf\:{x\:=\:{\dfrac{3}{4}}\:-\:{\dfrac{8}{20}}}$

${ }$

$\:\:\:\:\:\:\:\dashrightarrow\sf\:{x\:=\:{\dfrac{15\:-\:8}{20}}}$

${ }$

$\:\:\:\:\:\:\:\dashrightarrow\sf\:{x\:=\:{\dfrac{7}{20}}}$

${ }$

$\:\:\:\:\:\:\therefore\:{\underline{\sf{Hence,\: the\:length\:of\:third\:piece\:is\:{\sf{\bf {\dfrac{7}{20}}m }}}}}.$

${ }$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:━━━━━━━━━━━━━━━━━━$

${ }$