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At a certain symphonic concert, tickets for the orchestra [#permalink]
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22 Oct 2013, 11:40
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At a certain symphonic concert, tickets for the orchestra level were $50 and tickets for the balcony level were $30. These two ticket types were the only source of revenue for this concert. If R% of the revenue for the concert was from the sale of balcony tickets, and B% of the tickets sold were balcony tickets, then which of the following expresses B in terms of R? (A) \(\frac{200R}{(500 + 3R)}\) (B) \(\frac{300R}{(500  2R)}\) (C) \(\frac{300R}{(200 + 5R)}\) (D) \(\frac{500R}{(300 + 2R)}\) (E) \(\frac{500R}{(200 + 3R)}\)For more practice problems involving percents, some tips about handling percent problems on the GMAT, and the complete explanation of this particular problem, see: http://magoosh.com/gmat/2013/gmatquant ... hpercents Mike
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Re: At a certain symphonic concert, tickets for the orchestra [#permalink]
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22 Oct 2013, 12:17



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Re: At a certain symphonic concert, tickets for the orchestra [#permalink]
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23 Oct 2013, 16:28
mikemcgarry wrote: At a certain symphonic concert, tickets for the orchestra level were $50 and tickets for the balcony level were $30. These two ticket types were the only source of revenue for this concert. If R% of the revenue for the concert was from the sale of balcony tickets, and B% of the tickets sold were balcony tickets, then which of the following expresses B in terms of R? (A) \(\frac{200R}{(500 + 3R)}\) (B) \(\frac{300R}{(500  2R)}\) (C) \(\frac{300R}{(200 + 5R)}\) (D) \(\frac{500R}{(300 + 2R)}\) (E) \(\frac{500R}{(200 + 3R)}\)For more practice problems involving percents, some tips about handling percent problems on the GMAT, and the complete explanation of this particular problem, see: http://magoosh.com/gmat/2013/gmatquant ... hpercents Mike ok here is how i did it i picked numbers of two kinds of tickets as 3 and 5 respectively (for purpose of easy calculation), so we got 50*3=150, and 30*5=150 for balcony tickets. total revenue=150+150=300, so R=(150/300)*100=50 and B=(5/8)*100=500/8 then all we have to do is test following options by inserting R=50 into them and see if we yield B=500/8



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Re: At a certain symphonic concert, tickets for the orchestra [#permalink]
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24 Oct 2013, 00:08



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Re: At a certain symphonic concert, tickets for the orchestra [#permalink]
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03 Jan 2014, 16:07
mikemcgarry wrote: At a certain symphonic concert, tickets for the orchestra level were $50 and tickets for the balcony level were $30. These two ticket types were the only source of revenue for this concert. If R% of the revenue for the concert was from the sale of balcony tickets, and B% of the tickets sold were balcony tickets, then which of the following expresses B in terms of R? (A) \(\frac{200R}{(500 + 3R)}\) (B) \(\frac{300R}{(500  2R)}\) (C) \(\frac{300R}{(200 + 5R)}\) (D) \(\frac{500R}{(300 + 2R)}\) (E) \(\frac{500R}{(200 + 3R)}\)For more practice problems involving percents, some tips about handling percent problems on the GMAT, and the complete explanation of this particular problem, see: http://magoosh.com/gmat/2013/gmatquant ... hpercents Mike Yeah very tough to pick numbers indeed! What approach do you suggest Mike? Cheers! J



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At a certain symphonic concert, tickets for the orchestra [#permalink]
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03 Jan 2014, 16:43
jlgdr wrote: mikemcgarry wrote: At a certain symphonic concert, tickets for the orchestra level were $50 and tickets for the balcony level were $30. These two ticket types were the only source of revenue for this concert. If R% of the revenue for the concert was from the sale of balcony tickets, and B% of the tickets sold were balcony tickets, then which of the following expresses B in terms of R? (A) \(\frac{200R}{(500 + 3R)}\) (B) \(\frac{300R}{(500  2R)}\) (C) \(\frac{300R}{(200 + 5R)}\) (D) \(\frac{500R}{(300 + 2R)}\) (E) \(\frac{500R}{(200 + 3R)}\)For more practice problems involving percents, some tips about handling percent problems on the GMAT, and the complete explanation of this particular problem, see: http://magoosh.com/gmat/2013/gmatquant ... hpercents Mike Yeah very tough to pick numbers indeed! What approach do you suggest Mike? Cheers! J Dear jlgdr, I don't know if you followed that link to the blog article, but in that article, I show a full solution using picking numbers. For starters, we know that if B = 0, then R = 0. Here, all the answers satisfy that. We also know that if B = 100, then R = 100. Right there, we can plug in and eliminate some answers. Those are two particular easy choices to make for numbers to pick. A little more challenging, but not so hard  suppose they sell just as many balcony tickets as orchestra tickets, a 50/50 split. Then, of course, B = 50. For simplicity, suppose they sold 10 of each. That's $500 for the ten orchestra tickets, and $300 for the ten balcony tickets, for a grand total of $800 in revenue. Then, the balcony accounts for 300/800 = 3/8 of the revenue or 37.5 percent. R = 37.5  that's a little harder to plug in, but after we eliminate the first batch of answers, we don't have so many to check. Does all this make sense? Mike
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Re: At a certain symphonic concert, tickets for the orchestra [#permalink]
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Updated on: 10 Jan 2014, 03:27
From what we have so far, we get this:
50x+30y=T (Total Revenue) 30y=RT/100 y= B(x+y)/100
Fro these equations above, we get this: 30y= (50x+30y)R/100 and y= (x+y)B/100 OR
3000y= 30Bx+30By and 3000y= 50Rx+ 30Ry
by subtracting from each other we get this: (*) x/y= (30B30R)/(50R30B) Stop here! We need x/y. Right. We already have this: x/y= {B(x+y)/100}/{(100B)(x+y)/100} by reducing, we get this (**) x/y= B/(100B)!
By replacing (*) with (**) we can easily find the solution:
B= 500R/(2R+300)!!!
Originally posted by aja1991 on 09 Jan 2014, 22:33.
Last edited by aja1991 on 10 Jan 2014, 03:27, edited 1 time in total.



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Re: At a certain symphonic concert, tickets for the orchestra [#permalink]
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10 Jan 2014, 01:50
Let x=no. of orchestra tickets y=no. of Balcony tickets We know,50x+30y=T(Total Revenue) (1)
Also,RT/100=30y OR T/y=3000/R
And B/100*(x+y)=y OR x/y=100B/B
Divide (1) by y
We get, 50*(100B/B) + 30=3000/R
Simplifying,100/B=(300030R+50R)/50R
B=500R/(300+2R)



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Re: At a certain symphonic concert, tickets for the orchestra [#permalink]
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10 Jan 2014, 17:18
I am not sure if this question is worded incorrectly. Refer to " If R% of the revenue for the concert was from the sale of balcony tickets, and B% of the tickets sold were balcony tickets,
Both R and B refer to Balcony tickets hence its adding to the confusion.



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Re: At a certain symphonic concert, tickets for the orchestra [#permalink]
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10 Jan 2014, 17:55
Epex wrote: I am not sure if this question is worded incorrectly. Refer to " If R% of the revenue for the concert was from the sale of balcony tickets, and B% of the tickets sold were balcony tickets,
Both R and B refer to Balcony tickets hence its adding to the confusion. Dear EpexI'm happy to help. The percent R is a percent of money, a percent of revenue, so that's a dollar amount. The percent B is a percent of the number of tickets, so that's a number of tickets. Suppose the symphony sold ten balcony ticket and ten orchestra tickets. That's 20 tickets, and ten are balcony tickets, which is B = 50%. Meanwhile, from those sales, they take in $300 from the balcony tickets and $500 from the orchestra tickets, for a total revenue of $800. Of this $800, $300 came from balcony tickets: that's a R = (3/8)*100 = 37.5%. Does this distinction make sense? Mike
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Re: At a certain symphonic concert, tickets for the orchestra [#permalink]
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11 Jan 2014, 20:42
mikemcgarry wrote: Epex wrote: I am not sure if this question is worded incorrectly. Refer to " If R% of the revenue for the concert was from the sale of balcony tickets, and B% of the tickets sold were balcony tickets,
Both R and B refer to Balcony tickets hence its adding to the confusion. Dear EpexI'm happy to help. The percent R is a percent of money, a percent of revenue, so that's a dollar amount. The percent B is a percent of the number of tickets, so that's a number of tickets. Suppose the symphony sold ten balcony ticket and ten orchestra tickets. That's 20 tickets, and ten are balcony tickets, which is B = 50%. Meanwhile, from those sales, they take in $300 from the balcony tickets and $500 from the orchestra tickets, for a total revenue of $800. Of this $800, $300 came from balcony tickets: that's a R = (3/8)*100 = 37.5%. Does this distinction make sense? Mike Thanks. It does help. "R" used in question stem makes the reader feel that it's related to total revenue not the revenue of sale from B tickets.



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Re: At a certain symphonic concert, tickets for the orchestra [#permalink]
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12 Jan 2014, 20:24
Epex wrote: Thanks. It does help. "R" used in question stem makes the reader feel that it's related to total revenue not the revenue of sale from B tickets. Dear Epex, I don't know about this. It says, " R% of the revenue for the concert was from the sale of balcony tickets." That quite clearly states that R% is a percent of the total revenue. If you mean that the choice of the letter " R" is confusing, then you really have to move past your attachment to the letters used for variables. For example, usually a problem will use D for distance and T for time, but it would mathematically perfectly legal to use T for distance and D for time, as long as that was clearly specified in the problem. The meaning of variables is determined exclusively from how they are mathematically defined, and the letter chosen for the variable is 100% arbitrary. In general, the GMAT usually is not trying to fool testtakers with cheap tricks such as reversing familiar variables (the DT switch above), but in a problem in which there are not standard variables, any letters could be used. Don't be attached to the choice of letter. Always pay attention to the mathematical definition of the variable. Does this make sense? Mike
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Re: At a certain symphonic concert, tickets for the orchestra [#permalink]
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10 Jul 2016, 00:31
mikemcgarry wrote: At a certain symphonic concert, tickets for the orchestra level were $50 and tickets for the balcony level were $30. These two ticket types were the only source of revenue for this concert. If R% of the revenue for the concert was from the sale of balcony tickets, and B% of the tickets sold were balcony tickets, then which of the following expresses B in terms of R? (A) \(\frac{200R}{(500 + 3R)}\) (B) \(\frac{300R}{(500  2R)}\) (C) \(\frac{300R}{(200 + 5R)}\) (D) \(\frac{500R}{(300 + 2R)}\) (E) \(\frac{500R}{(200 + 3R)}\)For more practice problems involving percents, some tips about handling percent problems on the GMAT, and the complete explanation of this particular problem, see: http://magoosh.com/gmat/2013/gmatquant ... hpercents Mike fell for the number picking first...but then algebra worked for me fast.. +1 for the question.
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Re: At a certain symphonic concert, tickets for the orchestra [#permalink]
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13 Nov 2016, 11:34
Hi Mike, are you saying that in order to solve this question you suggest to try an easy technique (testing extremes) and if that does not give us the answer, then try a relatively harder technique of a 50/50 split for # of tickets.
I read your post on how to eliminate some answers using easy technique smart numbers. Using extremes? is that the correct name for the technique you described? I was able to eliminate A and C when I tried r= 100% and b=100%. When I tried r=0% and b=0% gave me nothing to eliminate. Just as you said it would.
Then I tried the little harder technique. Using smart numbers at 50/50 split for number of tickets. In my example I used 1 ticket for O and 1 ticket for B. This little harder work gave me the answer I was looking for.
$PPU # = T O => 50 (1) = 50 B => 30 (1) = 30 Total 2 = 80
R=30/80 = 0.375 =37.5% B = 1/2 = 0.50 = 50%
Only D gives you B in terms of R
(500(37.5))/(300 + (2(37.5))) = 50



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Re: At a certain symphonic concert, tickets for the orchestra [#permalink]
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13 Nov 2016, 18:47
lalania1 wrote: Hi Mike, are you saying that in order to solve this question you suggest to try an easy technique (testing extremes) and if that does not give us the answer, then try a relatively harder technique of a 50/50 split for # of tickets.
I read your post on how to eliminate some answers using easy technique smart numbers. Using extremes? is that the correct name for the technique you described? I was able to eliminate A and C when I tried r= 100% and b=100%. When I tried r=0% and b=0% gave me nothing to eliminate. Just as you said it would.
Then I tried the little harder technique. Using smart numbers at 50/50 split for number of tickets. In my example I used 1 ticket for O and 1 ticket for B. This little harder work gave me the answer I was looking for.
$PPU # = T O => 50 (1) = 50 B => 30 (1) = 30 Total 2 = 80
R=30/80 = 0.375 =37.5% B = 1/2 = 0.50 = 50%
Only D gives you B in terms of R
(500(37.5))/(300 + (2(37.5))) = 50 Dear lalania1I'm happy to respond. Precisely! We use extremes, such as R = 0 and R = 100%, to eliminate "lowhanging fruit"those answer choices that can be easily eliminated. I think you would make you job even easier by using fractions rather than decimals. If we have a percent of (3/8)*100%, there's no law that says we have to convert that to a decimal. We can lever it exactly like that, and all our multiplication is much easier. Does all this make sense? Mike
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At a certain symphonic concert, tickets for the orchestra [#permalink]
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17 Jan 2017, 05:02
mikemcgarry wrote: At a certain symphonic concert, tickets for the orchestra level were $50 and tickets for the balcony level were $30. These two ticket types were the only source of revenue for this concert. If R% of the revenue for the concert was from the sale of balcony tickets, and B% of the tickets sold were balcony tickets, then which of the following expresses B in terms of R? (A) \(\frac{200R}{(500 + 3R)}\) (B) \(\frac{300R}{(500  2R)}\) (C) \(\frac{300R}{(200 + 5R)}\) (D) \(\frac{500R}{(300 + 2R)}\) (E) \(\frac{500R}{(200 + 3R)}\)For more practice problems involving percents, some tips about handling percent problems on the GMAT, and the complete explanation of this particular problem, see: http://magoosh.com/gmat/2013/gmatquant ... hpercents Mike This is my solution. \(R = \frac{(30y)}{(50x+30y)} * 100  (1)\) \(B = \frac{(y)}{(x+y)} * 100 (2)\) From (1) \(R(50x+30y) = 3000y\) \(50Rx + 30Ry = 3000y\) \(50Rx = 3000y  30Ry\) \(x = \frac{(3000y  30Ry)}{50R}\) \(x =\frac{(300y  3Ry)}{5R}\) From (2) \(B = \frac{y}{(x+y)} * 100\) \(B = \frac{100y}{[(300y  3Ry)/(5R) + y]}\) \(B = \frac{100}{[( 3003R)/5R] + 1}\) \(B = \frac{5R*100}{(300  3R + 5R)}\) \(B = \frac{500R}{(300 + 2R)}\)
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