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A shopkeeper had a coil of electric wire measuring $60\ m$ long. He sold $\frac{2}{5}$ of it to the first customer and $10$ whole $\frac{1}{2}$ to theSecond customer what length of electric wire is left with him.
Given: A shopkeeper had a coil of electric wire measuring $60\ m$ long. He sold $\frac{2}{5}$ of it to the first customer and $10$ whole $\frac{1}{2}$ to the Second customer.
To do: to find the length of electric wire is left with him.
Solution:
Total length$=60\ m$
Wire sold to first costumer$=\frac{2}{5}$ of the wire$=\frac{2}{5}\times 60$
$=\frac{120}{5}$
$=24\ m$
Wire sold to second costumer$=10$ whole $\frac{1}{2}$ of the wire$=\frac{10\times2+1}{2}$
$=\frac{21}{2}$
$=10.5\ m$
Left length of the wire$=60-( 24+10.5)$
$=60-( 34.5)$
$=25.5\ m$
Thus $25.5\ m$ wire is left with him.
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