Question 1196691: Kathi and Robert Hawn had pottery stand at the annual Skippack Craft FairThey sold some of their pottery at the original price of $13.50 each, but later decreased the price of each by $2.00. If they sold all 91 pieces and took in $1,092, find how many they sold at the original price and how many they sold at the reduced price.
Found 4 solutions by ikleyn, greenestamps, math_tutor2020, josgarithmetic:
Answer by ikleyn(52800)   (Show Source):
You can put this solution on YOUR website! .
Kathi and Robert Hawn had pottery stand at the annual Skippack Craft FairThey sold some of their pottery
at the original price of $13.50 each, but later decreased the price of each by $2.00. If they sold all 91 pieces
and took in $1,092, find how many they sold at the original price and how many they sold at the reduced price.
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As worded, printed, posted and presented, the problem has no solution in integer numbers of their sold items.
It describes an IMPOSSIBLE situation which NEVER may happen.
The input numbers in the post are DEFECTIVE, making the entire problem DEFECTIVE.
Answer by greenestamps(13200)  (Show Source):
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
x = number of items sold at $13.50 each
91-x = remaining items sold at the cheaper price 13.50-2 = 11.50
13.50x = amount made from selling x items at $13.50 each
11.50(91-x) = amount made from selling 91-x items at $11.50 each
They sold all of their items and made $1092, so,
13.50x + 11.50(91-x) = 1092
13.50x + 11.50*91+ 11.50*(-x) = 1092
13.50x + 1046.5 - 11.50x = 1092
2x + 1046.5 = 1092
2x = 1092-1046.5
2x = 45.5
x = 45.5/2
x = 22.75
Unfortunately we don't get a whole number for x, so there is a typo with one or more of the given numbers.
Please ask your teacher for clarification.
Answer by josgarithmetic(39620) (Show Source):
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