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
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
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.
math - samuel, Wednesday, October 16, 2013 at 7:44pm
hi I THINK THE ANSWER IS 569
math - 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
math - 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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