500 ps test 22 test#15

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500 ps test 22 test#15

by dunkin77 » Sun Apr 15, 2007 10:27 am
Hi,

I could not find an answer for this question but I think the answer is B) -- Can anyone please tell me B) is correct? :D


15. A certain used-book dealer sells paperback books at 3 times dealer's cost and hardback books at 4 times dealer's cost. Last week the dealer sold a total of 120 books, each of which had cost the dealer $1. If the gross profit (sales revenue minus dealer's cost) on the sale of all of these books was $300, how many of the books sold were paperbacks?
(A) 40
(B) 60
(C) 75
(D) 90
(E) 100

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by jayhawk2001 » Sun Apr 15, 2007 12:10 pm
3p + 4(120-p) = 120 + 300
p = 60

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by dunkin77 » Sun Apr 15, 2007 1:28 pm
Thank you Jay for your response.

What I thought I did was...

3P + 4H = 300
P + H = 120

3P+4(120-p)=300
p=180???

I thought I got B) but now I can't remember how I got p=60.

Can you explain how come you add 120 to 300 ?

:)

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by ns88 » Sun Apr 15, 2007 2:15 pm
paperback=x
hardcover=y


total books= 120=x+y

total cost $120= 3x+4y

total revenue= (3)($1)x+ 4($1)y= 3x+4y

profit= 300 = (3x+4y)- 120

or 420=3x +4y


since 120= x+ y

x=(120-y)

420= [(3)(120-y) +4y]

y=60

x=60

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by jayhawk2001 » Sun Apr 15, 2007 2:33 pm
dunkin77 wrote:Thank you Jay for your response.

What I thought I did was...

3P + 4H = 300
P + H = 120

3P+4(120-p)=300
p=180???

I thought I got B) but now I can't remember how I got p=60.

Can you explain how come you add 120 to 300 ?

:)
The question states that the difference between the revenus
and the cost is 300. The cost is 120 * $1 = 120. Hence 120 + 300
is the total sale price.

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by dunkin77 » Sun Apr 15, 2007 6:16 pm
Okay - thanks!!

now I remember what I did...

3p+4H - (p+H) = 300

P+H = 120

but yours is more efficient way!!

Thank you!

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by Cybermusings » Mon Apr 16, 2007 4:31 am
Suppose the dealer sold x paperback books and y hardback books.
x+y = 120 ---------- 1)
3x+4y-120 = 300 ------- 2)
Solving the 2 you'll get x = 60 books