Question

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

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Hint:The length of remaining rope can be calculated by subtracting the lengths of two pieces of rope from the total original length of the rope.

We are given that from a rope 11m long, two pieces of lengths $2\dfrac{3}{5}$m and $3\dfrac{3}{{10}}$m are cut off. We need a length of rope left.
First, we need to find the lengths of pieces of rope in proper fraction.
This is done by converting mixed fraction into proper fraction as follows –
$a\dfrac{b}{c} = a + \dfrac{b}{c}$
$a + \dfrac{b}{c} = \dfrac{{ac + b}}{c}$……. (1)
Length of piece 1 = $2\dfrac{3}{5}$m
Using the property in equation (1) we get,
$2\dfrac{3}{5} = \dfrac{{2 \times 5 + 3}}{5} = \dfrac{{13}}{5}$
Thus, Length of piece 1 = $\dfrac{{13}}{5}$m…………….. (2)
Similarly, we have to calculate the length of piece 2 in proper fraction.
Length of piece 2 = $3\dfrac{3}{{10}}$m
$3\dfrac{3}{{10}} = \dfrac{{10 \times 3 + 3}}{{10}} = \dfrac{{33}}{{10}}$
Thus, Length of piece 2 = $\dfrac{{33}}{{10}}$m……….. (3)
Now, let the length of third piece left be X(m)………… (4)
Total length of rope = Length of rope 1 + Length of rope 2 + Length of rope 3
From equation (2), (3) and (4) we get the lengths of pieces –
Total length of rope = $\dfrac{{13}}{5}$+$\dfrac{{33}}{{10}}$+$X$
Total length of rope is 11m.
Therefore, we get the equation –
11=$\dfrac{{13}}{5}$+$\dfrac{{33}}{{10}}$+$X$
Subtracting both sides by $\dfrac{{13}}{5}$and $\dfrac{{33}}{{10}}$we get,
$X = 11 - \dfrac{{13}}{5} - \dfrac{{33}}{{10}}$
Taking L.C.M. of 5 and 10 we get 10. Therefore,
$X = \dfrac{{11(10) - 13(2) - 33}}{{10}}$
$X = \dfrac{{110 - 26 - 33}}{{10}}$
Further simplifying we get,
$X = \dfrac{{51}}{{10}}$
Thus the length of the third piece will be $\dfrac{{51}}{{10}}$m.

Note:Make sure that the units of all the length must be the same while calculating the length and if not must be converted to one single unit.