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Intern  Joined: 09 Jun 2015
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In an opera theater. there are 300 seats available.  [#permalink]

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Question Stats: 60% (02:54) correct 40% (03:19) wrong based on 342 sessions

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In an opera theater. there are 300 seats available. When all the seats in the theater are sold out, the price of each ticket is $60. For every$1 increase in the price of the ticket, the number of seats sold decreases by 2. How much should the theater owner charge for each seat to make maximum profit?

A)$87 B)$95
C) $105 D)$120
E) $127 Originally posted by itwarriorkarve on 31 Oct 2015, 10:53. Last edited by ENGRTOMBA2018 on 01 Nov 2015, 05:31, edited 1 time in total. Formatted the question ##### Most Helpful Expert Reply Veritas Prep GMAT Instructor V Joined: 16 Oct 2010 Posts: 9784 Location: Pune, India In an opera theater. there are 300 seats available. [#permalink] ### Show Tags 3 2 itwarriorkarve wrote: In an opera theater. there are 300 seats available. When all the seats in the theater are sold out, the price of each ticket is$60. For every $1 increase in the price of the ticket, the number of seats sold decreases by 2. How much should the theater owner charge for each seat to make maximum profit? A)$87
B)$95 C)$105
D) $120 E)$127

As shown by Brent above, we need to maximise the revenue which is given by
(60 + x)*(300 - 2x)
Expand to get: 18000 + 180x - 2x^2

Recognise that this is a quadratic with negative co-efficient of x^2 so it represents a downward facing parabola. It will have the maximum value at x = -b/2a

x = -180/2*(-2) = 45

So the owner should charge 60 + x = 60 + 45 = $105 for each seat. Answer (C) _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > ##### General Discussion GMAT Club Legend  V Joined: 12 Sep 2015 Posts: 4063 Location: Canada Re: In an opera theater. there are 300 seats available. [#permalink] ### Show Tags 1 2 itwarriorkarve wrote: In an opera theater. there are 300 seats available. When all the seats in the theater are sold out, the price of each ticket is$60. For every $1 increase in the price of the ticket, the number of seats sold decreases by 2. How much should the theater owner charge for each seat to make maximum profit? A)$87 B)$95 C)$105 D) $120 E)$127

I'm not a big fan of this question. The only approach (without using techniques that are out of scope for the GMAT) is TEST THE ANSWERS, and the great thing about most GMAT math problems is that they can be solved using more than 1 approach.

To see what I mean, consider this algebraic approach:

Let x = the number of dollars added to the price of the ticket.
So, the new ticket price = 60+x dollars.

For every $1 increase in the price of the ticket, the number of seats sold decreases by 2. So, 300-2x = the number of seats sold. Revenue = (ticket price)(# seats sold) So, revenue = (60+x)(300-2x) Expand to get: revenue = 18,000 + 300x - 120n - 2x² Simplify and rearrange to get: revenue = -2x² + 180x + 18,000 So our goal is to find the value of x that MAXIMIZES the value of -2x² + 180x + 18,000 1) One approach is to use calculus and find the derivative of -2x² + 180x + 18,000 and blah blah blah...(OUT OF SCOPE!) 2) Another approach is to recognize that y = -2x² + 180x + 18,000 represents a downward parabola AND the coordinates of the vertex (at the TOP of the parabola) represent the information that MAXIMIZES profits. So, let's use approach #2 and try to determine the vertex of the parabola described by y = -2x² + 180x + 18,000 We'll use a process called COMPLETING THE SQUARE (which is also beyond the scope of the GMAT) [more on completing the square here http://www.mathsisfun.com/algebra/compl ... quare.html] Start with: y = -2x² + 180x + 18,00 Factor -2 from first two terms to get: y = -2(x² - 90x) + 18,000 Complete the square to get: y = -2(x² - 90x + 2025 - 2025) + 18,000 Expand and remove -2025 from brackets to get: y = -2(x² - 90x + 2025) + 4050 +18,000 NOTE: Through our handwork, x² - 90x + 2025 is a square that can be factored! Simplify and rewrite to get: y = -2(x - 45)² + 22,050 Now that we've rewritten our equation in this nice form, we can see that the vertex of the parabola has coordinates (45, 22050) 22050 is the HIGHEST POINT of the parabola, which means the MAXIMUM revenue is$22,050
45 is the value of x that achieves this maximum revenue
So, to maximize revenue, we must add $45 to the ticket price to get$60 + $45 =$105

Cheers,
Brent
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Re: In an opera theater. there are 300 seats available.  [#permalink]

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GMATPrepNow wrote:
Expand to get: revenue = 18,000 + 300x - 120n - 2x²
Simplify and rearrange to get: revenue = -2x² + 180x + 18,000
So our goal is to find the value of x that MAXIMIZES the value of -2x² + 180x + 18,000

Hi, Could you please explain how you got the equation?

I understand that the revenue will be (60 + x)(300 - 2x). But how/why are you adding 18000 to this?
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Re: In an opera theater. there are 300 seats available.  [#permalink]

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3
2
itwarriorkarve wrote:
In an opera theater. there are 300 seats available. When all the seats in the theater are sold out, the price of each ticket is $60. For every$1 increase in the price of the ticket, the number of seats sold decreases by 2. How much should the theater owner charge for each seat to make maximum profit?

A)$87 B)$95
C) $105 D)$120
E) $127 hi, The above posts have given an algebric way of finding the answer.. BUT even if you do not know these, the Q can be easily solved through SUBSTITUTION.. We should always start with the middle value, as it will save on looking for two values- could be above or below the middle number present revenue = 300*60=18000 C)$105
change in price = 105-60 =45..
so seats = 300 - 2*45 = 210
so total revenue = 210*105 = 22050 > 18000 so OK..

lets see lower value
B)$95 - change in price = 95-60 =35.. so seats = 300 - 2*35 = 230 so total revenue = 230*95 = 21850 < 22050 so no need to check for lower numbers.. now the next higher value to 105.. D)$120

change in price = 120-60 =60..
so seats = 300 - 2*60 = 180
so total revenue = 180*120 = 21600 < 22050 so no need to check for any higher value...

ans C
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Re: In an opera theater. there are 300 seats available.  [#permalink]

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Hi All,

Can we solve this by using below equation :

xy = 18000

putting the value of y = 18000/x in below
(x-2)(y+1)

and then using rule of maxima???

Regards,
Abhijit
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In an opera theater. there are 300 seats available.  [#permalink]

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3
(60+x)(300-2x)>[60+(x+1)][300-2(x+1)]
(60+x)(300-2x)>(61+x)(298-2x)
4x>178
x>44.5
x=45
60+45=$105 Veritas Prep GMAT Instructor V Joined: 16 Oct 2010 Posts: 9784 Location: Pune, India Re: In an opera theater. there are 300 seats available. [#permalink] ### Show Tags 1 AbhijitGoswami wrote: Hi All, Can we solve this by using below equation : xy = 18000 putting the value of y = 18000/x in below (x-2)(y+1) and then using rule of maxima??? Regards, Abhijit How are you defining x and y here? What do these variables stand for e.g. "Number of ..." ? Once you think about this, think what do (x - 2) and (y+1) represent? Will their product be the maximum revenue? _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > Intern  B Joined: 15 Mar 2017 Posts: 3 Re: In an opera theater. there are 300 seats available. [#permalink] ### Show Tags we have GM =<AM (60+x)(300 - 2x ) max? 1/2(120+2x)(300 - 2x) -> leave out the 0.5 and work with the factor including x only SQRT((120+ 2x)(300-2x))ay =< 1/2 (12+300 +2x - 2x ) - a constant. where have a maximum value for SQRT, therefor the profit under. Manager  S Joined: 14 Sep 2016 Posts: 59 Concentration: Finance, Economics Re: In an opera theater. there are 300 seats available. [#permalink] ### Show Tags chetan2u wrote: itwarriorkarve wrote: In an opera theater. there are 300 seats available. When all the seats in the theater are sold out, the price of each ticket is$60. For every $1 increase in the price of the ticket, the number of seats sold decreases by 2. How much should the theater owner charge for each seat to make maximum profit? A)$87
B)$95 C)$105
D) $120 E)$127

hi,
The above posts have given an algebric way of finding the answer..
BUT even if you do not know these, the Q can be easily solved through SUBSTITUTION..
We should always start with the middle value, as it will save on looking for two values- could be above or below the middle number
present revenue = 300*60=18000

C) $105 change in price = 105-60 =45.. so seats = 300 - 2*45 = 210 so total revenue = 210*105 = 22050 > 18000 so OK.. lets see lower value B)$95 -

change in price = 95-60 =35..
so seats = 300 - 2*35 = 230
so total revenue = 230*95 = 21850 < 22050 so no need to check for lower numbers..

now the next higher value to 105..
D) $120 change in price = 120-60 =60.. so seats = 300 - 2*60 = 180 so total revenue = 180*120 = 21600 < 22050 so no need to check for any higher value... ans C Kaplan method states to not start with with middle value. I believe this question is an exception to that rule since it is a parabola that peaks somewhere. Usually you want to start with B or D and then solve for which ever of the two next that you did not choose. In normal questions if you start with B and the answer you need is lower then you know it is A. If the answer is higher test D. By testing D next you know if you are right, too high or too slow and will only ever need to solve for these two. This method makes it so you often will only have to solve for one value but at most two. Starting with C you will always have to test two values unless you hit it on the head with C. Intern  B Joined: 15 Jun 2013 Posts: 45 Schools: Ivey '19 (I) GMAT 1: 690 Q49 V35 GPA: 3.82 In an opera theater. there are 300 seats available. [#permalink] ### Show Tags The easiest way to solve this question is to know the "function derivative" concept: in order to find the maximum of the defined function - just take the first-level derivative. Then this question can be solved in less than a minute: Revenue function will be (60+x)*(300-2x)=60*300+180x-2x^2 Take the first order derivative: -4x+180=0 or x=45 New price= 45+60=105. Answer C Director  D Affiliations: IIT Dhanbad Joined: 13 Mar 2017 Posts: 732 Location: India Concentration: General Management, Entrepreneurship GPA: 3.8 WE: Engineering (Energy and Utilities) Re: In an opera theater. there are 300 seats available. [#permalink] ### Show Tags itwarriorkarve wrote: In an opera theater. there are 300 seats available. When all the seats in the theater are sold out, the price of each ticket is$60. For every $1 increase in the price of the ticket, the number of seats sold decreases by 2. How much should the theater owner charge for each seat to make maximum profit? A)$87
B)$95 C)$105
D) $120 E)$127

The language of the question is not very clear. It says "When all the seats in the theater are sold out, the price of each ticket is $60.", what does this mean. That at the price of$60 the tickets has been sold out. So the maximum price charged by the theater is $60 and then we have to reverse calculate something. Please ask the original source to re frame the question to clearly provide the information. It should have been like " The price of each seat is$60. But to increase the revenue he increases price of the seats by $1 for after every 2 seats sold. itwarriorkarve Please send a message to 800 score to re-frame the question. _________________ CAT 2017 (98.95) & 2018 (98.91) : 99th percentiler UPSC Aspirants : Get my app UPSC Important News Reader from Play store. MBA Social Network : WebMaggu Appreciate by Clicking +1 Kudos ( Lets be more generous friends.) What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish". Intern  B Joined: 05 Mar 2017 Posts: 1 Re: In an opera theater. there are 300 seats available. [#permalink] ### Show Tags VeritasPrepKarishma wrote: itwarriorkarve wrote: In an opera theater. there are 300 seats available. When all the seats in the theater are sold out, the price of each ticket is$60. For every $1 increase in the price of the ticket, the number of seats sold decreases by 2. How much should the theater owner charge for each seat to make maximum profit? A)$87
B)$95 C)$105
D) $120 E)$127

As shown by Brent above, we need to maximise the revenue which is given by
(60 + x)*(300 - 2x)
Expand to get: 18000 + 180x - 2x^2

Recognise that this is a quadratic with negative co-efficient of x^2 so it represents a downward facing parabola. It will have the maximum value at x = -b/2a

x = -180/2*(-2) = 45

So the owner should charge 60 + x = 60 + 45 = $105 for each seat. Answer (C) Karishma, I understand that -b/a is the sum of the two roots, but where do you get 2a and how does this give you the maximum value? Manager  B Joined: 06 Aug 2017 Posts: 78 GMAT 1: 570 Q50 V18 GMAT 2: 610 Q49 V24 GMAT 3: 640 Q48 V29 Re: In an opera theater. there are 300 seats available. [#permalink] ### Show Tags itwarriorkarve wrote: In an opera theater. there are 300 seats available. When all the seats in the theater are sold out, the price of each ticket is$60. For every $1 increase in the price of the ticket, the number of seats sold decreases by 2. How much should the theater owner charge for each seat to make maximum profit? A)$87
B)$95 C)$105
D) $120 E)$127

The answer is C as follows.

The answer to this question can be derived by using the knowledge of maxima for a quadratic equation.

Based on the conditions given in the question, the equation will be as follows

Let x be the increase in price of the ticket.
Total earning --> (60+x)(300-2x) ==> $$-2x^2 + 180x + 18000$$
Since this is an equation of inverted parabola, it will have its maximum value at -b/2a ==> -180/2*(-2)==>45
So price of the ticket will be 60+45 = 105

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Re: In an opera theater. there are 300 seats available.  [#permalink]

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I would disregard this question. The makers of the GMAT specifically write questions that do not lend themselves towards mathematical knowledge outside of the scope of the test. For example, this question can be solved in about 15 seconds with calculus. You just take the derivative and set it equal to zero. The GMAT is (quite frustratingly) not a math test, it's a reasoning test. This is just testing if you know basic calculus.
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Re: In an opera theater. there are 300 seats available.  [#permalink]

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itwarriorkarve wrote:
In an opera theater. there are 300 seats available. When all the seats in the theater are sold out, the price of each ticket is $60. For every$1 increase in the price of the ticket, the number of seats sold decreases by 2. How much should the theater owner charge for each seat to make maximum profit?

A)$87 B)$95
C) $105 D)$120
E) $127 We can let x = the number of$1 increments in price (or the number of 2-seat decrements) and y = total revenue. Thus, the expression (60 + x) is the increased ticket price and (300 - 2x) represents the decreased number of tickets sold at that increased price. Recall that revenue is the product of the ticket price and the number of tickets sold. Thus, we have:

y = (60 + x)(300 - 2x)

Expanding the right-hand side, we have:

y = 18000 + 180x - 2x^2

Notice that the above is a quadratic equation. Let’s rearrange the terms so that it’s in standard ax^2 + bx + c form:

y = -2x^2 + 180x + 18000

We see that the graph of the quadratic equation is a downward parabola, so the maximum revenue is the y-value of the vertex. To get the y-value of the vertex, we must get the x-value of the vertex first, using the formula x = -b/(2a):

x = -180/(2(-2)) = -180/-4 = 45

Thus, to maximize the revenue, the ticket price should be 60 + 45 = \$105.

(Note: We are using the term “revenue” instead of “profit” because the amount of money received from sales is technically called “revenue,” not “profit.” As we know, the term “profit” is the difference between revenue and cost. The problem uses the term “profit” instead of “revenue” perhaps because cost is either zero or negligible.)

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