Question

4 Keith is a plumber. He has a 5 m length of pipe.
He cuts off two pieces of pipe.
The first piece is 2 1/4 m long, the second is 1 2/5 m long.
a How long is the piece of pipe that Keith has left?
b Show how to check your answer is correct.​

Answer 1

Length of the pipe with Keith=5 m

Number of pieces into which Keith cut the pipe= 2

Length of the first piece:-

$= 2 \frac{1}{4} \: \: m$

Length of the second piece:-

$= 1 \frac{2}{5} \: \: m$

Total length of both the pieces:-

$= 2\frac{1}{4} + 1 \frac{2}{5}$

$= \frac{9}{4} + \frac{7}{5}$

$= \frac{9 \times 5}{20} + \frac{7 \times 4}{20}$

$= \frac{45}{20} + \frac{28}{20}$

$= \frac{73}{20}$

$= 3 \frac{13}{20}$

Thus, the total length of both the pieces Keith has got:-

$\bold{= 3 \frac{13}{20} } \bold{\: m}$

Which means :-

Length of the piece which Keith has left:-

$= 5 - 3 \frac{13}{20}$

$= \frac{5}{1} - \frac{73}{20}$

$= \frac{5 \times 20}{20} - \frac{73 \times 1}{20}$

$= \frac{100}{20} - \frac{73}{20}$

$= \frac{27}{20}$

$= 1 \frac{7}{20} \: \:$

Thus, the length of the piece of pipe that Keith has left with him:-

$= \bold{1 \frac{7}{20} \: \:m}$

To check whether or not we have found out the correct length of pipe that Keith had left with him,let us add the measure of all the pieces and check whether or not they form 5m:-

$= 2 \frac{1}{4} + 1 \frac{2}{5} + 1 \frac{7}{20}$

$= \frac{9}{4} + \frac{7}{5} + \frac{27}{20}$

$= \frac{9 \times 5}{20} + \frac{7 \times 4}{20} + \frac{27}{20}$

$= \frac{45}{20} + \frac{28}{20} + \frac{27}{20}$

$= \frac{45 + 28 + 27}{20}$

$= \frac{100}{20}$

$= \frac{100 \div 20}{20 \div 20}$

$= \frac{5}{1}$

$=\bold{ 5 \: m}$

As the measure of all three of these pieces together is forming 5m,we can conclude that we have found out the correct measure of the piece of pipe Keith had left with him.

Therefore, the length of pipe Keith has left with him$= \bold{1 \frac{7}{20} \: \:m}$

Answer 2

$\huge {\mathtt {\green {\underline {\underline {Given}}}}}$

• Length of the pipe with Keith=5 m
• Number of pieces into which Keith cut the pipe= 2
• The first piece is 2 1/4 m long
• the second is 1 2/5 m long.

$\huge {\mathtt {\red {\underline {\underline {To\: Find}}}}}$

• How long is the piece of pipe that Keith has left?

$\huge {\mathtt {\purple{\underline {\underline {Solution}}}}}$

Length of the first piece :

$\sf 2 \dfrac{1}{4} m = 2m$

Length of the second piece:-

$\sf 1 \dfrac{2}{5} \: \: m=1m$

Total length of both the pieces:-

$:\implies 2\dfrac{1}{4} + 1 \dfrac{2}{5} \implies \dfrac{9}{4} + \dfrac{7}{5}$

$:\implies \dfrac{9 \times 5}{20} + \dfrac{7 \times 4}{20}\implies \dfrac{45}{20} + \dfrac{28}{20}$

$:\implies \dfrac{73}{20}\implies 3 \dfrac{13}{20}=3$

Thus, the total length of both the pieces Keith has got:-

$\sf{ 3 \dfrac{13}{20} }\: m$

So,

Length of the piece which Keith has left:-

$:\implies 5 - 3 \dfrac{13}{20}$

$:\implies \dfrac{5}{1} - \dfrac{73}{20}\implies \dfrac{5 \times 20}{20} - \dfrac{73 \times 1}{20}$

$:\implies \dfrac{100}{20} - \dfrac{73}{20} \implies \dfrac{27}{20}$

$:\implies 1 \dfrac{7}{20} \: \:$

Thus, the length of the piece of pipe that Keith has left with him:-

$\sf {1 \dfrac{7}{20} \: \:m}$

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we are also asked to :

Show how to check your answer is correct.

Solution :

As the total length is 5m .. if we add the lengths of 3 parts the result should be 5.

$\sf \dfrac{9}{4} + \dfrac{7}{5} + \dfrac{27}{20}$

$\sf \dfrac{9 \times 5}{20} + \dfrac{7 \times 4}{20} + \dfrac{27}{20}$

$\sf \dfrac{45}{20} + \dfrac{28}{20} + \dfrac{27}{20}$

$\sf \dfrac{45 + 28 + 27}{20}$

$\sf \dfrac{100}{20}= 5m$

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☑ Hope it helps uh ⤴️