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A trader has three different types of oils of volume $870$ l, $812$ l and $638$ l. Find the least number of containers of equal size required to store all the oil without getting mixed.
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Sol. Least number of containers means maximum volume in each container. (I) ($\frac{1}{2}$ Mark)
$870 = 2 \times 3 \times 5 \times 29$ (II) ($\frac{1}{2}$ Mark)
$812 = 2^2 \times 7 \times 29$ (III) ($\frac{1}{2}$ Mark)
$638 = 2 \times 11 \times 29$
$\therefore$ H.C.F. $(870, 812, 638) = 2 \times 29 = 58$ (IV) ($\frac{1}{2}$ Mark)
Number of containers of different types required $= \frac{870}{58} + \frac{812}{58} + \frac{638}{58}$ (V) ($\frac{1}{2}$ Mark)
$= 15 + 14 + 11$
$= 40$ (VI) ($\frac{1}{2}$ Mark)
$870 = 2 \times 3 \times 5 \times 29$ (II) ($\frac{1}{2}$ Mark)
$812 = 2^2 \times 7 \times 29$ (III) ($\frac{1}{2}$ Mark)
$638 = 2 \times 11 \times 29$
$\therefore$ H.C.F. $(870, 812, 638) = 2 \times 29 = 58$ (IV) ($\frac{1}{2}$ Mark)
Number of containers of different types required $= \frac{870}{58} + \frac{812}{58} + \frac{638}{58}$ (V) ($\frac{1}{2}$ Mark)
$= 15 + 14 + 11$
$= 40$ (VI) ($\frac{1}{2}$ Mark)