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# fractions

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Bob sells chocolates. He sold $${1 \over 3}{}{}$$ of it and an additional 2 packets in the morning and
$${}{1 \over 4}{}{}$$ of the remainder and an additional 4 packets in the afternoon. If he had 8 packets of
chocolates left, how many packets of chocolates did Bob have at first ?

Aug 8, 2021

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Assuming that Bob have x packets at first.

Morning sale: $${1 \over 3}{}{}$$x + 2

After morning sale: x - ($${1 \over 3}{}{}$$x + 2) = $${2 \over 3}{}{}$$x - 2

Afternoon sale: $${1 \over 4}{}{}$$($${2 \over 3}{}{}$$x - 2) + 4 = $${1 \over 6}{}{}$$x + $${7 \over 2}{}{}$$
After afternoon sale: $${2 \over 3}{}{}$$x - 2 - ($${1 \over 6}{}{}$$x + $${7 \over 2}{}{}$$) = $${x \over 2}{}{}$$-$${11 \over 2}{}{}$$
$${x \over 2}{}{}$$-$${11 \over 2}{}{}$$= 8                            8 packets left

x = 27

Use the conditions given to establish the equivalence

relationship.

So the answer is 27 packets.

Aug 8, 2021
edited by apsiganocj  Aug 8, 2021