Question

From a rope 7 metres long, two pieces of length 2⅗ meters and 3 3/10 metres were cut off. What is the length of the remaining rope?
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Answer 1

Cut off cut off from 11m

2 3/5 +3 3/10

13/5 +33/10

(26+33)/10

59/10 m

Length of remaining= 11-59/10

(110 -59)/10

51/10

5 1/10 m ans.

Answer 2

$\color{red}{\large\underline{\underline\mathtt{Question:-}}}$

$From \: a \: rope \: 7 \: metres \: long, \: two \: pieces \: of \: length \: \: 2 \frac{3}{5} \: metres \: and \: 3 \frac{3}{10} \: metres \: were \: cut \: off. \: What \: is \: the \: length \: of \: the \: remaining \: rope? </p><p>$

$\color{red}{\large\underline{\underline\mathtt{To\: find:-}}}$

$What \: is \: the \: length \: of \: the \: remaining \: rope?$

$\color{red}{\large\underline{\underline\mathtt{Solution:-}}}$

$Total \: length \: of \: the \: rope = 7 \: metres$

$Total \: length \: cut \: off = 2 \frac{3}{5} m \: + 3 \frac{3}{10} m \\ = (\frac{2 \times 5 + 3}{5} + \frac{3 \times 10 + 3}{10} )m \ \ = ( \frac{13}{5} + \frac{33}{10} )m$

$= ( \frac{26}{10} + \frac{33}{10} )m \: \: \: = ( \frac{26 + 33}{10} )m = \frac{59}{10} m$

$Length \: of \: the \: remaining \: rope =(7 \ - \frac{59}{10} )m$

$= (\frac{70}{10} - \frac{59}{10} )m \\ =( \frac{70 - 59}{10} )m \\ = \frac{11}{10} m \\ = 1 \frac{1}{10} m$

$Hence, \: the \: length \: of \: \: the \: remaining \: rope \: is = 1 \frac{1}{10} m$

$\color{red}{\large\underline{\underline\mathtt{Additional\: information:-}}}$

• $Sum \: of \: like \: fractions = \frac{ Sum \: of \: numerators }{ Common \: denominator }$
• $Difference \: of \: like \: fractions = \frac{Difference \: of \: numerators }{Common \: denominator }$