Question

Joe bought a bag of oranges on Monday, and
ate one third of them.
On Tuesday he ate half of the remaining
oranges.
On Wednesday he looked in the bag to find that
he only had two oranges left.
How many oranges were originally in the bag?

Answer 1

$Let \:number \: of \: oranges \: Joe$

$bought = x\: --(1)$

$Number \:of \: oranges \: he \: ate \: on$

$Monday = \frac{x}{3} \: --(2)$

$Number \:of \: oranges \: he \: ate \: on$

$Tuesnesday =\frac{1}{2}\Big( x - \frac{x}{3} \Big)$

$= \frac{x}{2} - \frac{x}{6} \: --(3)$

$Number \:of \:oranges \: remaining \: in \: the$

$bag \: on \: Wednesday = 2 \: --(4)$

/* According to the problem given */

$(1) - (2) - (3) = 2$

$\implies x - \frac{x}{3} - \Big( \frac{x}{2} - \frac{x}{6}\Big) = 2$

$\implies x - \frac{x}{3} - \frac{x}{2} + \frac{x}{6}= 2$

$\implies \frac{6x-2x-3x+x}{6} = 2$

$\implies \frac{7x-5x}{6} = 2$

$\implies \frac{2x}{6} = 2$

$\implies \frac{x}{3} = 2$

$\implies x = 2 \times 3$

$= 6$

Therefore.,

$\red{ Number \:of \: oranges \: originally}$

$\red{in \:the \:bag} \green {= 6}$

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