Posted by **Jane** on Wednesday, January 4, 2012 at 5:51pm.

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?

- math -
**Reiny**, Wednesday, January 4, 2012 at 7:46pm
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.

- math -
**choochoo**, Tuesday, January 15, 2013 at 4:37am
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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**samuel**, Wednesday, October 16, 2013 at 7:44pm
hi I THINK THE ANSWER IS 569

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**RR**, Wednesday, July 9, 2014 at 5:03pm
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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**swagmath**, Thursday, October 9, 2014 at 3:15pm
12 smalls 6 larges

- math -
**person**, Tuesday, May 5, 2015 at 10:50am
12 small 6 large

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