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# It costs x dollars each to make the first 1,000 copies of a compact

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Math Expert
Joined: 02 Sep 2009
Posts: 51167
It costs x dollars each to make the first 1,000 copies of a compact  [#permalink]

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16 Oct 2017, 08:54
00:00

Difficulty:

15% (low)

Question Stats:

88% (01:48) correct 12% (01:12) wrong based on 42 sessions

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It costs x dollars each to make the first 1,000 copies of a compact disc and y dollars to make each subsequent copy. If z is greater than 1,000, how many dollars will it cost to make z copies of the compact disc?

(A) 1,000x + yz
(B) zx – zy
(C) 1,000 (z – x) + xy
(D) 1,000 (z – y) + xz
(E) 1,000 (x – y) + yz

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Joined: 02 Oct 2017
Posts: 12
It costs x dollars each to make the first 1,000 copies of a compact  [#permalink]

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16 Oct 2017, 09:03
Total no. of copies to be made =z
Given z > 1000
Cost for making first 1000 copies = 1000x
Remaining copies = z-1000
Cost for remaining copies = (z-1000)y
Total cost = 1000x+ (z-1000)y = 1000(x-y)+yz
Option E.

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It costs x dollars each to make the first 1,000 copies of a compact  [#permalink]

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19 Oct 2017, 07:55
Bunuel wrote:
It costs x dollars each to make the first 1,000 copies of a compact disc and y dollars to make each subsequent copy. If z is greater than 1,000, how many dollars will it cost to make z copies of the compact disc?

(A) 1,000x + yz
(B) zx – zy
(C) 1,000 (z – x) + xy
(D) 1,000 (z – y) + xz
(E) 1,000 (x – y) + yz

Let x = \$2
First 1000 discs cost
\$2 * 1000 = \$2000

Let z, total # of discs = 1500

Let y, cost/disc after first 1000 discs = \$1

# of discs after first 1000
1500 - 1000 = 500 left
Cost of those 500 at y=\$1:
500 * 1 = \$500

Overall cost: 2000 + 500 = 2500

Find the answer that = 2500
Use x = 2, y = 1, z = 1500

(A) 1,000x + yz
(1000)(2) + (1)(1500) =
2000+1500 = 3500. Too large. NO

(B) zx – zy
(1500)(2) - (1500)(1) =
3000-1500 = 1500. Too small. NO

(C) 1,000 (z – x) + xy
1000(1500-2) + (2)(1)
1000(1498) + 2 = huge. Stop. NO

(D) 1,000 (z – y) + xz
1000(1500-1) + (2)(1500)=
1000(1499) + 3000 = huge. Stop. NO

(E) 1,000 (x – y) + yz
1000(2-1) + (1)(1500) =
1000(1) + 1500 = 2500. MATCH

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Joined: 09 Aug 2017
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Re: It costs x dollars each to make the first 1,000 copies of a compact  [#permalink]

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23 Oct 2017, 06:14
I got E, but this is one where I, again, need to read the problem better. I misread the question and didn't realize x is the price per copy for the first thousand. I thought X was total cost, which led me to X + (Z-1000)*Y, which thankfully isn't an answer.

So I regrouped and realized X is price per copy.
-1000x is the cost of the first 1,000 copies.
-y*(z-1,000) is the cost for every copy after the first 1,000.
-So I broke expanded outside the parentheses to get 1,000x+yz-1,000y.
-Regrouped the terms to read 1,000x-1,000y+yz.
-Factored out 1,000(x-y)+yz, which is an answer.
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I'd love to hear any feedback or ways to improve my problem solving. I make a lot of silly mistakes. If you've had luck improving on stupid mistakes, I'd love to hear how you did it.

Also, I appreciate any kudos.

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Re: It costs x dollars each to make the first 1,000 copies of a compact  [#permalink]

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21 Jun 2018, 16:09
Bunuel wrote:
It costs x dollars each to make the first 1,000 copies of a compact disc and y dollars to make each subsequent copy. If z is greater than 1,000, how many dollars will it cost to make z copies of the compact disc?

(A) 1,000x + yz
(B) zx – zy
(C) 1,000 (z – x) + xy
(D) 1,000 (z – y) + xz
(E) 1,000 (x – y) + yz

The cost is:

1000x + y(z - 1000)

1000x + yz - 1000y

1000(x - y) + yz

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Re: It costs x dollars each to make the first 1,000 copies of a compact &nbs [#permalink] 21 Jun 2018, 16:09
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