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# In 1985 a company sold a brand of shoes to retailers for a fixed price

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Re: In 1985 a company sold a brand of shoes to retailers for a fixed price [#permalink]
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)
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Re: In 1985 a company sold a brand of shoes to retailers for a fixed price [#permalink]
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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.

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Re: In 1985 a company sold a brand of shoes to retailers for a fixed price [#permalink]
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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

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Re: In 1985 a company sold a brand of shoes to retailers for a fixed price [#permalink]
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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.
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Re: In 1985 a company sold a brand of shoes to retailers for a fixed price [#permalink]
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)
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In 1985 a company sold a brand of shoes to retailers for a fixed price [#permalink]
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
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Re: In 1985 a company sold a brand of shoes to retailers for a fixed price [#permalink]
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.
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In 1985 a company sold a brand of shoes to retailers for a fixed price [#permalink]
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.
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In 1985 a company sold a brand of shoes to retailers for a fixed price [#permalink]
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%)
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Re: In 1985 a company sold a brand of shoes to retailers for a fixed price [#permalink]
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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.
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Re: In 1985 a company sold a brand of shoes to retailers for a fixed price [#permalink]
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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

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Re: In 1985 a company sold a brand of shoes to retailers for a fixed price [#permalink]
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
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In 1985 a company sold a brand of shoes to retailers for a fixed price [#permalink]
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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Re: In 1985 a company sold a brand of shoes to retailers for a fixed price [#permalink]
In 1986, the number of pairs sold decreased by 20%, which means the company sold 80% of x pairs. The price per pair increased by 20%, which means the new price per pair is 120% of y.

The revenue in 1986 can be calculated as the product of the number of pairs sold and the price per pair:

Revenue in 1986 = 0.8x * 1.2y = 0.96xy

Given that the revenue in 1986 is \$3.0 million, we have:

0.96xy = 3.0 million

To find the revenue in 1985, we can divide both sides of the equation by 0.96:

xy = 3.0 million / 0.96 ≈ 3.125 million

Therefore, the approximate revenue from the sale of the shoes in 1985 was \$3.125 million, which is closest to option (D) \$3.1 million.
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Re: In 1985 a company sold a brand of shoes to retailers for a fixed price [#permalink]
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Re: In 1985 a company sold a brand of shoes to retailers for a fixed price [#permalink]
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