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8 June, 17:57

A drive in theater charges $3.50 per car, the drive in already admitted 100 cars, solve an inequality to find the number of cars the drive in needs to admit to make atleast 500$

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  1. 8 June, 21:20
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    They need to admit 43 more cars to make at least $500

    Solution:

    Given, A drive in theater charges $3.50 per car,

    The drive in already admitted 100 cars,

    Then, money collected until now is $ 3.50 x 100 = $ 350

    We have to solve an inequality to find the number of cars the drive in needs to admit to make atleast 500$

    Now, let the number of cars need to be admitted be "n"

    Then, inequality is:

    collected amount + needed amount ≥ $ 500

    $ 350 + $ 3.50 x n ≥ $ 500

    350 + 3.5n ≥ 500

    3.5n ≥ 500 - 350

    On solving, we get

    3.5n ≥ 150

    n ≥ 42.857

    As number of cars can't be fractions, n = 43

    Hence, they need to admit 43 more cars to make at least $500
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