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25 January, 17:05

Over the course of one day, a store owner determined that 80% of customers bought a drink and 30% of customers bought a snack. If the store sold 120 snacks that day, how many more drinks than snacks did the store sell?

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  1. 25 January, 17:20
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    200

    Step-by-step explanation:

    Let the total number of customers that day be "x"

    30% of total customers bought a snack and that is 120.

    So we can say:

    30% of total (x) is 120

    Note: 30% converted to decimal is 30/100 = 0.3

    This can be translated to an algebraic equation as:

    0.3x = 120

    Now, we can solve for x, the total customers:

    0.3x = 120

    x = 120/0.3

    x = 400

    Since, 80% bought drink, we find 80% of 400:

    80% = 80/100 = 0.8

    0.8 * 400 = 320

    The store sold 320 drinks and 120 snacks. So drinks outweigh snacks be:

    320 - 120 = 200

    So,

    200 more drinks than snacks
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