SOLUTION: A dealer bought some radios for a total of $1,080. She gave away 9 radios as gifts, sold the rest for $10 more than she paid for each radio, and broke even. How many radios did she
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Question 520618: A dealer bought some radios for a total of $1,080. She gave away 9 radios as gifts, sold the rest for $10 more than she paid for each radio, and broke even. How many radios did she buy? Answer by Maths68(1474) (Show Source):
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She bought = x radios
Price per radio = 1080/x
Number of Radio Gifted = 9
She sold = x-9
She sold $10 more than the original price = 1080/x + 10
Total Amount radios sold = (1080/x+10)*(x-9) = 1080-9720/x+10x-90 = (10x-9720/x+990)
Finally she broke even
Total amount radio sold = Final Amount
10x-9720/x+990 = 1080
10x-9720/x = 1080 - 990
10x-9720/x = 90
Multiply by x both sides
10x^2-9720= 90x
10x^2-90x-9720=0
10*(x^2-9x-972)=0
Divide by 10 both sides
x^2-9x-972=0
x^2-36x+27x-972=0
x(x-36)+27(x-36)
x-36=0 or x+27=0
x=36 or x=-27 (inadmissible)
She bought 36 radios
Price per radio = 1080/36 = 30
She sold = 36-9 = 27 radios
She sold 27 radios for $10 more than she paid for each radio = 30+10=40
Total Amount she received for selling 27 radios @ $40 = 27*40 = $1080
