Jackie decided to sell a CD of her greatest hits of High school
Musical unplugged. The retail price of the case containing the CD
and the CD itself is $1.10. If the case costs $1.00 more than the
CD, how much does the CD cost without CD Case?
2007-11-01 10:18:13 · 8 answers · asked by Chuck Norris 1 in Science & Mathematics ➔ Mathematics
Everyone is missing something... the case doesn't cost $1. It costs $1 *more* than the CD.
Let n be the price of the CD.
Let 1.00 + n be the price of the case.
1.00 + n + n = 1.10
2n = 1.10 - 1.00
2n = 0.10
n = 0.05
The CD costs 5 cents (and the case costs $1.05)
2007-11-01 10:23:22 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
Let CD be the cost of the CD itself. Let CA be the cost of the case.
Then CD+CA = 1.10 and CA = CD+.10
That is two equations for two unknowns so it is solvable.
Substitute CA = CD+.10 for CA in the first equation to get
CD + (CD+.10) = 1.10
Combine the terms to get 2CD = 1.10 - .10
THat gives 2CD = 1.00
Divide by 2 and get CD = .50
So it is a fifty cent cost for the CD itself.
2007-11-01 10:25:30 · answer #2 · answered by Rich Z 7 · 0⤊ 1⤋
10 cents. Total price (1.10) - case (1.00) = 10 cents
2007-11-01 10:21:14 · answer #3 · answered by lzp 1 · 0⤊ 1⤋
1.10 - cd + case
1.00 - case
1.10-1.00 = .10
the cd alone costs 10 cents
2007-11-01 10:23:11 · answer #4 · answered by Anonymous · 0⤊ 1⤋
10 cents
2007-11-01 10:20:34 · answer #5 · answered by Anonymous · 0⤊ 1⤋
0.10 or ten cents. That sounds more like a wholesale price than a retail one though.
2007-11-01 10:21:23 · answer #6 · answered by Anonymous · 0⤊ 1⤋
10 cents?
2007-11-01 10:21:16 · answer #7 · answered by pokemike01 4 · 0⤊ 1⤋
10 cents...i guess
2007-11-01 10:21:14 · answer #8 · answered by Anonymous · 0⤊ 1⤋