Question

A piece of wire, $2\dfrac{3}{4}$ metres long, broke into two pieces. One piece is $\dfrac{5}{8}$ metre long. How long is the other piece?

Answer 1

Hint:
We can find the length of the remaining part by subtracting the length of the given piece from the total length of the wire. For subtracting fractions, we can first convert the mixed fraction into improper fraction. Then we can make the denominators of the two fractions equal by multiplying with the same number in the numerator and denominator. Then we can subtract the numerators to get the required solution.

Complete step by step solution:
The total length of the wire is given as $2\dfrac{3}{4}$ metres.
Let $T = 2\dfrac{3}{4}$ .
Now we can convert the mixed fraction to improper fraction.
 $ \Rightarrow T = \dfrac{{2 \times 4 + 3}}{4}$
 $ \Rightarrow T = \dfrac{{8 + 3}}{4}$
 $ \Rightarrow T = \dfrac{{11}}{4}$
Now we can multiply the numerator and denominator with 2.
 $ \Rightarrow T = \dfrac{{22}}{8}$
We are given that the length of one piece is $\dfrac{5}{8}$ metre.
Let $l = \dfrac{5}{8}$ .
We can assume x as the length of the other piece.
Then the sum of the length of the two pieces will give the total length of the wire.
 $ \Rightarrow l + x = T$
On rearranging, we get,
 $ \Rightarrow x = T - l$
On substituting the values, we get,
 $ \Rightarrow x = \dfrac{{22}}{8} - \dfrac{5}{8}$
As the denominators are the same, we can subtract the numerators.
 $ \Rightarrow x = \dfrac{{22 - 5}}{8}$
 $ \Rightarrow x = \dfrac{{17}}{8}$
We can convert it to a mixed fraction.
 $ \Rightarrow x = \dfrac{{2 \times 8 + 1}}{8}$
 $\Rightarrow x=2\dfrac{1}{8}$

Therefore, the length of the other piece of the wire is $2\dfrac{1}{8}$ meters.

Note:
Alternate solution to this problem is given by,
We can simply subtract the length of the given piece from the total length of the wire.
 $ \Rightarrow x = 2\dfrac{3}{4} - \dfrac{5}{8}$
We can convert the mixed fraction to improper fraction.
 $ \Rightarrow x = \dfrac{{11}}{4} - \dfrac{5}{8}$
Now we can take the common term $\dfrac{1}{4}$ from both the term
 $ \Rightarrow x = \dfrac{1}{4}\left( {11 - \dfrac{5}{2}} \right)$
Now we can find the LCM of the terms inside the bracket.
 $ \Rightarrow x = \dfrac{1}{4}\left( {\dfrac{{22 - 5}}{2}} \right)$
Now we can subtract the numerators.
 $ \Rightarrow x = \dfrac{1}{4}\left( {\dfrac{{17}}{2}} \right)$
Now we can open the bracket.
 $ \Rightarrow x = \dfrac{{17}}{8}$
 $\Rightarrow x=2\dfrac{1}{8}$
Therefore, the length of the other piece of the wire is $2\dfrac{1}{8}$ meters.