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In 1985 a company sold a brand of shoes to retailers for a fixed price https://gmatclub.com/forum/in1985acompanysoldabrandofshoestoretailersforafixedprice209046.html 
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Author:  Bunuel [ 24 Nov 2015, 11:04 ] 
Post subject:  In 1985 a company sold a brand of shoes to retailers for a fixed price 
In 1985 a company sold a brand of shoes to retailers for a fixed price per pair. In 1986 the number of pairs of the shoes that the company sold to retailers decreased by 20 percent, while the price per pair increased by 20 percent. If the company’s revenue from the sale of the shoes in 1986 was $3.0 million, what was the approximate revenue from the sale of the shoes in 1985 ? (A) $2.4 million (B) $2.9 million (C) $3.0 million (D) $3.1 million (E) $3.6 million 
Author:  Skywalker18 [ 24 Nov 2015, 11:12 ] 
Post subject:  Re: In 1985 a company sold a brand of shoes to retailers for a fixed price 
Let the number of pairs of shoes sold in 1985 = N let price per pair of shoe sold in 1985 = P In 1986,number of pairs of shoes sold decreased by 20 % = .8N Price per pair increased by 20% = 1.2P Company's revenue from sale of shoes in 1986 = 3 million$ => .8N * 1.2 P = 3 =>.96 NP = 3 => NP = 3/.96 = 3.125 million $ Answer D 
Author:  Abhishek009 [ 24 Nov 2015, 11:39 ]  
Post subject:  Re: In 1985 a company sold a brand of shoes to retailers for a fixed price  
Bunuel wrote: In 1985 a company sold a brand of shoes to retailers for a fixed price per pair. In 1986 the number of pairs of the shoes that the company sold to retailers decreased by 20 percent, while the price per pair increased by 20 percent. If the company’s revenue from the sale of the shoes in 1986 was $3.0 million, what was the approximate revenue from the sale of the shoes in 1985 ? (A) $2.4 million (B) $2.9 million (C) $3.0 million (D) $3.1 million (E) $3.6 million A am using a typical grid method of price, quantity and Revenue matrix From the matrix we get 96 = 3mn 1 = 3/96 mn 100 = (3/96) * 100 => 3.125 Hence answer is D

Author:  girishiyer [ 08 Sep 2017, 03:09 ] 
Post subject:  Re: In 1985 a company sold a brand of shoes to retailers for a fixed price 
Let in :1985 no of pairs=n price/pair=m revenue=mn In 1986: no of pairs=0.8n price/pair=1.2m revenue=0.96mn so in 1985=3/0.96=3.125 (C) 
Author:  ScottTargetTestPrep [ 11 Sep 2017, 15:59 ] 
Post subject:  Re: In 1985 a company sold a brand of shoes to retailers for a fixed price 
Bunuel wrote: In 1985 a company sold a brand of shoes to retailers for a fixed price per pair. In 1986 the number of pairs of the shoes that the company sold to retailers decreased by 20 percent, while the price per pair increased by 20 percent. If the company’s revenue from the sale of the shoes in 1986 was $3.0 million, what was the approximate revenue from the sale of the shoes in 1985 ? (A) $2.4 million (B) $2.9 million (C) $3.0 million (D) $3.1 million (E) $3.6 million We can let the number of pairs of shoes sold in 1985 = n and the price per pair = p. Thus, the revenue in 1985 is np and we can create the following equation for the revenue in 1986: (0.8n)(1.2p) = 3,000,000 0.96np = 3,000,000 np = 3,125,000, which is roughly 3.1 million. Answer: D 
Author:  ScottTargetTestPrep [ 08 Jan 2020, 20:38 ] 
Post subject:  Re: In 1985 a company sold a brand of shoes to retailers for a fixed price 
Bunuel wrote: In 1985 a company sold a brand of shoes to retailers for a fixed price per pair. In 1986 the number of pairs of the shoes that the company sold to retailers decreased by 20 percent, while the price per pair increased by 20 percent. If the company’s revenue from the sale of the shoes in 1986 was $3.0 million, what was the approximate revenue from the sale of the shoes in 1985 ? (A) $2.4 million (B) $2.9 million (C) $3.0 million (D) $3.1 million (E) $3.6 million If we let Q = the number of pairs of shoes sold in 1985, then the number of pairs sold in 1986 is 0.8Q. If we let P = the price per pair in 1985, then 1.2 P is the price per pair in 1986. We use the formula Quantity x Price = Revenue and create the equation: 0.8Q * 1.2P = 3 0.96QP = 3 QP = 3.125 Answer: D 
Author:  brains [ 14 Jan 2020, 09:38 ] 
Post subject:  Re: In 1985 a company sold a brand of shoes to retailers for a fixed price 
20 % increase can be written as equialvent to 5 becomes 6 and 20% decrease can be written as equivalent to 5 becomes 4 Now x be the pairs and y be the price pairs decreased by 20% means if 5x be the original pairs , now it has become 4x and price increased by 20% means if 5y be the original price.now it has become 6y original revenue = 5x*5y = 25xy new revenue = 4x * 6y = 24xy Given , new revenue = 3 million so 24xy = 3 xy = 1/8 so original = 25xy = 25/8 = 3.1 million Hope it helps. 
Author:  ajaymahadev [ 25 Jan 2020, 01:08 ] 
Post subject:  Re: In 1985 a company sold a brand of shoes to retailers for a fixed price 
In 1986, no of pairs of shoes decreased by 20 percent. and price per pair increased by 20%. Revenue in 1986 = 3 M Let the number of pairs of shoes sold in 1985 = x price = p in 1986, no of pairs of shoes = 0.8x price = 1.2p so, (0.8x)(1.2p)=3 or 0.96px=3 we need revenue in 1985, which is px = 3/0.96 which is approximately 3.1 M, Option (D) 
Author:  henilshaht [ 03 Jul 2020, 14:51 ] 
Post subject:  In 1985 a company sold a brand of shoes to retailers for a fixed price 
I have a silly doubt. I assumed the n = 1000, so the cost c came as $3000 in 1985. New n = 800, c = 3600 Revenue in 1986 = 800*3600 = $2880000 = $2.88M This answer is wrong. But can someone explain the flaw in my approach? Bunuel wrote: In 1985 a company sold a brand of shoes to retailers for a fixed price per pair. In 1986 the number of pairs of the shoes that the company sold to retailers decreased by 20 percent, while the price per pair increased by 20 percent. If the company’s revenue from the sale of the shoes in 1986 was $3.0 million, what was the approximate revenue from the sale of the shoes in 1985 ?
(A) $2.4 million (B) $2.9 million (C) $3.0 million (D) $3.1 million (E) $3.6 million 
Author:  CEdward [ 29 Mar 2021, 10:20 ] 
Post subject:  Re: In 1985 a company sold a brand of shoes to retailers for a fixed price 
Can someone explain what I did wrong here? I don't quite follow xy = 3 < 3million revenue x  > 4x/5 y > 6y/5 (4x/5) x (6y/5) = (4x/5) x (6(3/x)/5) = 72/25 < slightly less than 3. 
Author:  deepakbarchha1 [ 03 Oct 2021, 05:00 ] 
Post subject:  In 1985 a company sold a brand of shoes to retailers for a fixed price 
Please observe the answer options here. If one know that by increase in price and decrease in units sold, revenue goes down. Hence 1985 Sales should be higher than sales of 1986. One can eliminate answer options A, B, C. according to problem : 0.8 * 1.2 = 3 (given in the problem) 0.96 = 3 3/0.96 = 3.12 Ans : D ($3.1 m) Hope it helps. 
Author:  Hoozan [ 12 Jan 2022, 23:59 ] 
Post subject:  In 1985 a company sold a brand of shoes to retailers for a fixed price 
IanStewart KarishmaB BrentGMATPrepNow After reading the question, is there any logic that we can use in order to infer that the Revenue in 85 will definitely be > 3Million? (i.e. greater than the revenue in 86) given that the number of pairs sold has gone down by 20% and the price has gone up by the same factor (20%) 
Author:  IanStewart [ 13 Jan 2022, 06:30 ] 
Post subject:  Re: In 1985 a company sold a brand of shoes to retailers for a fixed price 
Hoozan wrote: After reading the question, is there any logic that we can use in order to infer that the Revenue in 85 will definitely be > 3Million? (i.e. greater than the revenue in 86) given that the number of pairs sold has gone down by 20% and the price has gone up by the same factor (20%) Yes, there is  if you ever increase something by x%, then decrease it by x%, then thinking of x as a decimal (so thinking of, say, 20% as 0.2), you're just multiplying by (1 + x)(1  x) = 1  x^2. That must be less than 1 (we're subtracting a positive square from 1), so we always get a decrease. That's what is happening to revenue here, and while this still leaves two answers to choose from, since x = 0.2, we're multiplying by 1  (0.2)^2 = 0.96, so the decrease is only 4% and the answer must be D. 
Author:  KarishmaB [ 13 Jan 2022, 23:28 ] 
Post subject:  Re: In 1985 a company sold a brand of shoes to retailers for a fixed price 
Bunuel wrote: In 1985 a company sold a brand of shoes to retailers for a fixed price per pair. In 1986 the number of pairs of the shoes that the company sold to retailers decreased by 20 percent, while the price per pair increased by 20 percent. If the company’s revenue from the sale of the shoes in 1986 was $3.0 million, what was the approximate revenue from the sale of the shoes in 1985 ? (A) $2.4 million (B) $2.9 million (C) $3.0 million (D) $3.1 million (E) $3.6 million 1985 Revenue = Number * Price 1986 Revenue = (4/5)*Number * (6/5)*Price = (24/25) *Number * Price = (24/25) * (1985 Revenue) = 3 million 1985 Revenue = (25/24) * 3 million = Approx 3.1 million Answer (D) 
Author:  Elite097 [ 24 Dec 2022, 02:18 ] 
Post subject:  Re: In 1985 a company sold a brand of shoes to retailers for a fixed price 
avigutman what would be a way to calculate 3mn/.96 faster? I could narrow to D and E but spent a lot of total  totally 8 mins on this questiin 
Author:  avigutman [ 24 Dec 2022, 09:16 ] 
Post subject:  In 1985 a company sold a brand of shoes to retailers for a fixed price 
Elite097 wrote: avigutman what would be a way to calculate 3mn/.96 faster? I could narrow to D and E but spent a lot of total  totally 8 mins on this questiin Elite097 If you find yourself having to compute something like 3/.96 on the GMAT, you need to redo the problem a different way, or just guess and move on. It’s not appropriate to attempt such computations on this test. The exception is people who are able to perform such computations incredibly quickly (and accurately), but those people are unlikely to be reading this post. For appropriate ways to solve this problem, I recommend the posts by IanStewart and KarishmaB above. 
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