math

An artist is going to sell two sizes of prints at an art fair. The artist will charge $20 for the small print and $45 for a large print. The artist would like to sell twice as many small prints as large prints. The booth the artist is renting for the day cost $510. How many of each size print must the artist sell in order to break even at the fair?

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  1. number of small prints --- x
    number of large prints --- 2x

    20x + 45(2x) = 510
    20x+90x=510
    110x = 510
    x= 510/110 = 4.6

    but you can't sell .6 of a print, so I would say he should sell 5 small and 10 large prints

    check:
    5(20) + 10(45) = 550 -- a profit
    if he sold only 4 small and 8 large
    income = 4(20) = 8(45) = 440 , not enough to cover rent.

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  2. You switched the large and small print numbers. Selling twice as many small prints would make it:
    20 (2x) + 45 (x) = 510
    If you solve with that formula in mind, x = 6...and that is the break-even number.

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  3. hi I THINK THE ANSWER IS 569

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  4. 20x(2) + 45x = $510

    40x + 45x =$510

    85x / 85 = $510/85

    x = 6

    She sold twice as many small than large. So, small = 2(6) which equals 12

    20(12) + 45(6) = $510

    240 + 270 = $510

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  5. 12 smalls 6 larges

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  6. 12 small 6 large

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