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Rational numbers

Word problems involving rational numbers

Five pieces of satin were cut from one bale of satin. They were 1/2 m, 9/2 m, 3/10 m, 29/10 m, and 101/20 m long. Find their total length.

Their total length is:

$\frac{1}{2}+\frac{9}{2}+\frac{3}{10}+\frac{29}{10}+\frac{101}{20}=$

$\frac{10}{20}+\frac{90}{20}+\frac{6}{20}+\frac{58}{20}+\frac{101}{20}=$

$\frac{265}{20}=\frac{53}{20}m$

From a chain 11 m long, two pieces of lengths $3\frac{3}{5}$ m and $2\frac{3}{10}$ are cut off. What is the length of the remaining chain?

Solution:
Original length of chain = 11 m

 Sum of lengths of pieces cut off: $(3\frac{3}{5}+2\frac{3}{10})m$ $=(3+\;\frac{3}{5}+2\;+\;\frac{3}{10})m$ $=(5+\;\frac{3}{5}\;+\;\frac{3}{10})m$ $=(5+\;\frac{6+3}{10})m$ $=(5+\;\frac{9}{10})m$ $=5\frac{9}{10}m$

 Length of piece left: $(11\;-\;5\frac{9}{10})m$ $=(11\;-\;\frac{59}{10})m$ $=(\frac{110-59}{10})m$ $=\frac{51}{10}m$ $=5\frac{1}{10}m$