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If an item that originally sold for z dollars was marked up
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Updated on: 09 Oct 2013, 09:12
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72% (01:49) correct 28% (02:04) wrong based on 445 sessions
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If an item that originally sold for z dollars was marked up by x percent and then discounted by y percent, which of the following expressions represents the final price of the item? A. (10,000z + 100z(x – y) – xyz)/10,000 B. (10,000z + 100z(y – x) – xyz)/10,000 C. (100z(x – y) – xyz)/10000 D. (100z(y – x) – xyz)/10000 E. 10000 /(x – y) Actually figured it out as I was typing it in by picking the following numbers:
z=$10, x=20%, y=50% 1.2z*.5= 6
Final price should be $6.
Any other ways?
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Originally posted by GMATT73 on 15 Nov 2006, 08:55.
Last edited by Bunuel on 09 Oct 2013, 09:12, edited 1 time in total.
Edited the question and added the OA.



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z dollars was marked up by x percent:
z+(z*x/100)=(100z+zx)/100
This discounted by y%:
{(100z+zx)/100}{(100z+zx)/100 * (y/100)}
{(100z+zx)/100}{100yz+xyz/100^2}
Solving I get A



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Re: PS XYZ
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15 Nov 2006, 09:34
GMATT73 wrote: If an item that originally sold for z dollars was marked up by x percent and then discounted by y percent, which of the following expressions represents the final price of the item?
(A) [10,000z + 100z(x â€“ y) â€“ xyz]/10,000 (B) [10,000z + 100z(y â€“ x) â€“ xyz]/10,000 (C) [100z(x â€“ y) â€“ xyz]/10,000 (D) [100z(y â€“ x) â€“ xyz]/10,000 (E) 10,000/(100yz + xy) Actually figured it out as I was typing it in by picking the following numbers:
z=$10, x=20%, y=50% 1.2z*.5= 6
Final price should be $6.
Any other ways?
Matt, that is the easier way. Other way is to directly multiply and take the common factors out..but that seems to be time taking in this problem. As you know, whenever its possible, solving by substituting the numbers is the best way to solve problems when we are under time pressure
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Re: PS XYZ
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15 Nov 2006, 12:12
GMATT73 wrote: If an item that originally sold for z dollars was marked up by x percent and then discounted by y percent, which of the following expressions represents the final price of the item?
(A) [10,000z + 100z(x â€“ y) â€“ xyz]/10,000 (B) [10,000z + 100z(y â€“ x) â€“ xyz]/10,000 (C) [100z(x â€“ y) â€“ xyz]/10,000 (D) [100z(y â€“ x) â€“ xyz]/10,000 (E) 10,000/(100yz + xy) Actually figured it out as I was typing it in by picking the following numbers:
z=$10, x=20%, y=50% 1.2z*.5= 6
Final price should be $6.
Any other ways?
If markup is x%, new price will be (100+x)/100 times old price.
Discounted by y% means new price will be (100y)/100 times old price.
So we have final price =z(100+x)(100y)/10000
and the terms rearranged give A.



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To me, for some questions, picking nos. and applying those in each ans choice and finding the right choice seems tedious. I guess it depends on the method of practice and the strength of the individual, as I see in many threads, and of course, the time pressure.



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Re: If an item that originally sold for z dollars was marked up
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09 Oct 2013, 08:57
Solving algebraically: Z * (1+(x/100)) * (1(y/100)) Z* (100+x/100) * (100y/100) (100Z + ZX)/100 * (100y/100) [10,000Z + 100 ZX  100 ZY  XYZ] /10,000 [10,000Z +100Z (X  Y)  XYZ] / 10,000 Choice A
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Re: If an item that originally sold for z dollars was marked up
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24 Nov 2013, 05:59
Assume Z=100, x=10, y=10. So final price is 99. Now put these value each option. Only option A will give right answer.
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Re: If an item that originally sold for z dollars was marked up
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06 Jun 2017, 16:42
GMATT73 wrote: If an item that originally sold for z dollars was marked up by x percent and then discounted by y percent, which of the following expressions represents the final price of the item?
A. (10,000z + 100z(x – y) – xyz)/10,000
B. (10,000z + 100z(y – x) – xyz)/10,000
C. (100z(x – y) – xyz)/10000
D. (100z(y – x) – xyz)/10000
E. 10000 /(x – y) We are given that an item that originally sold for z dollars was marked up by x percent and then discounted by y percent; thus, the final price of the item is: z(1 + x/100)(1  y/100) z[(100 + x)/100][(100  y)/100] z[(10,000 + 100x  100y  xy)/10,000 [(10,000z + 100xz  100yz  xyz)/10,000 [(10,000z + 100z(x  y)  xyz)/10,000 Answer: A
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Re: If an item that originally sold for z dollars was marked up
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07 Nov 2017, 01:30
Ans is A x% rise and y percent fall net change formula is (a + b + ab/100 )% if rise +ve and if fall ve so x is + and y is  net change will be xy  xy/100 and this is % of z so net change ie rise => z( xyxy/100)1/100 and last price was z so new one is z+ z( xy  xy/100)1/100 so becomes A
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Re: If an item that originally sold for z dollars was marked up
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23 Jun 2018, 22:50
Use the rule for consecutive percentage changes. If there are consecutive %changes a and b. Then a+b+(ab/100) % is the net change. Substitute for x and y and solve. The answer is option A.
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Re: If an item that originally sold for z dollars was marked up
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24 Jun 2018, 03:56
GMATT73 wrote: If an item that originally sold for z dollars was marked up by x percent and then discounted by y percent, which of the following expressions represents the final price of the item?
A. (10,000z + 100z(x – y) – xyz)/10,000
B. (10,000z + 100z(y – x) – xyz)/10,000
C. (100z(x – y) – xyz)/10000
D. (100z(y – x) – xyz)/10000
E. 10000 /(x – y) Let z=1, x=0 and y=100. Since the final discount = 100%, the final price = 0. Now plug z=1, x=0 and y=100 into the five numerators to see which yields a final price of 0. Since xyz = 0, this term can be ignored. Only the numerator in A yields a value of 0: 10,000z + 100z(xy) = 10,000*1 + 100*1*(0100) = 10,000  10,000 = 0.
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Re: If an item that originally sold for z dollars was marked up &nbs
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