Question

From a rope 15 whole1/2m long three pieces of lengths 1whole2/5m 2whole2/3m and 4whole5/9m are cut off. What is the length of the remaining rope answer in fraction​

Answer 1

➤ Correct Question :-

From a rope of $\sf 15 \dfrac{1}{2}$ long, three pieces of lengths $\sf 1 \dfrac{2}{5}$, $\sf 2 \dfrac{2}{3}$ and $\sf 4 \dfrac{5}{9}$ are cut off. What is the length of the remaining rope.

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➤ Solution :-

${\tt 15 \dfrac{1}{2} - \bigg( 1 \dfrac{2}{5} + 2 \dfrac{2}{3} + 4 \dfrac{5}{9} \bigg)}$

Convert all the mixed fractions as improper fractions.

${\tt \leadsto \dfrac{31}{2} - \bigg( \dfrac{7}{5} + \dfrac{8}{3} + \dfrac{41}{9} \bigg)}$

LCM of 5, 3 and 9 is 45.

${\tt \leadsto \dfrac{31}{2} - \bigg( \dfrac{7 \times 9}{5 \times 9} + \dfrac{8 \times 15}{3 \times 15} + \dfrac{41 \times 5}{9 \times 5} \bigg)}$

Multiply the numerators and denominators of all fractions in bracket.

${\tt \leadsto \dfrac{31}{2} - \bigg( \dfrac{63}{45} + \dfrac{120}{45} + \dfrac{205}{45} \bigg)}$

Write all the numerators with a common denominator.

${\tt \leadsto \dfrac{31}{2} - \bigg( \dfrac{63 + 120 + 205}{45} \bigg)}$

Add the numerators in the given bracket.

${\tt \leadsto \dfrac{31}{2} - \dfrac{388}{45}}$

LCM of 2 and 45 is 90.

${\tt \leadsto \dfrac{31 \times 45}{2 \times 45} - \dfrac{388 \times 2}{45 \times 2}}$

Multiply the numerators and denominators of both fractions.

${\tt \leadsto \dfrac{1395}{90} - \dfrac{776}{90}}$

Subtract the fractions to get the answer.

${\tt \leadsto \dfrac{1395 - 776}{90} = \dfrac{619}{45}}$

Write the improper fraction as a mixed fraction.

${\tt \leadsto \dfrac{619}{45} = \pink{\underline{\boxed{\tt 13 \dfrac{34}{45}}}}}$

$\Huge\therefore$ The left over rope is $\tt 13 \dfrac{34}{45}$