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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
 #1
avatar+272 
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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

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