Question 124667: An art dealer sold two artworks for $1520 thereby making a profit of 25% on the first work and 10% on the other, whereas if he had approached any exhibition he would bave sold them together for $1535 with a profit of 10% on the first and 25% on the other artwork. Find the actual cost of each artwork.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! An art dealer sold two artworks for $1520 thereby making a profit of 25% on the first work and 10% on the other, whereas if he had approached any exhibition he would have sold them together for $1535 with a profit of 10% on the first and 25% on the other artwork. Find the actual cost of each artwork.
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Let x = cost of the 1st artwork, Let y = cost of 2nd artwork
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Art dealer: Retail price 1st piece = 1.25(x); Retail price of 2nd = 1.10(y)
1.25x + 1.10y = 1520
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Exhibition: Retail price 1st piece = 1.10(x); Retail price of 2nd = 1.25(y)
1.10x + 1.25y = 1535
:
Use two equation in the elimination method:
multiply the 1st equation by 100, and the 2nd equation by 88,resulting in:
125x + 110y = 152000
96.8x +110y = 135080
----------------------subtracting eliminates y
28.2x + 0y = 16920
x = 16920/28.2
x = $600 is the cost of the 1st piece
:
2nd piece cost using the 1st equation:
1.25(600) + 1.1y = 1520
750 + 1.1y - 1520
1.1y = 1520 - 750
1.1y = 770
y = 770/1.1
y = $700 is the cost of the 2nd piece:
:
:
Check our solutions in the 2nd equation:
1.1(600) + 1.25(700) =
660 + 875 = 1535 as given
:
Did this make sense to you? In case you wondered how I came up 88 as multiplier
in the 2nd equation. I multiplied the 1st equation by 100 to get rid of the
decimal. Then divided 1.25 into 110 and saw it came out to an even 88.
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