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# An amusement park currently charges the same price for each ticket of

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An amusement park currently charges the same price for each ticket of [#permalink]

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13 Nov 2014, 09:42
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Tough and Tricky questions: Word problems.

An amusement park currently charges the same price for each ticket of admission. If the current price of admission were to be increased by $$3$$, 12 fewer tickets could be bought for $$160$$, excluding sales tax. What is the current price of each ticket?

A. $$3$$
B. $$5$$
C. $$8$$
D. $$20$$
E. $$32$$

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: An amusement park currently charges the same price for each ticket of [#permalink]

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13 Nov 2014, 10:14
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Answer A: 160/3 = 50.33 tickets. 160/6 = 25.66 tickets, difference is not 12
Answer B: 160/5= 32, 160/8 = 20, difference of 12

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Re: An amusement park currently charges the same price for each ticket of [#permalink]

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13 Nov 2014, 11:07
A) 160/3=53.33 not integer (eliminate)
B) 160/5=32 160/2=80 difference is not 12 (eliminate)
C) 160/8=20 160/5=32 32-20=12

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Re: An amusement park currently charges the same price for each ticket of [#permalink]

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13 Nov 2014, 11:36
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Hey! this is how i got the answer.

160/x = (160/x) +12

Solve the equation and we will get x= -8 or x=5, the price of the ticket cannot be negative so x=5 is the correct answer.
B

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Re: An amusement park currently charges the same price for each ticket of [#permalink]

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13 Nov 2014, 20:34
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Price ............. Quantity .................. Total

a ...................... $$\frac{160}{a}$$ ..................... 160 (Assume price = "a")

a+3 ................... $$\frac{160}{a} - 12$$ ............... 160 (Price increased by 3, sales decline by 12)

Solving the equation

$$(a+3)* (\frac{160}{a} - 12) = 160$$

$$a^2 + 3a - 40 = 0$$

a = 5
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Re: An amusement park currently charges the same price for each ticket of [#permalink]

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14 Nov 2014, 02:40
Plug in is the best approach for this question to save time. Answer is B.

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Re: An amusement park currently charges the same price for each ticket of [#permalink]

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14 Nov 2014, 05:03
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Bunuel wrote:

Tough and Tricky questions: Word problems.

An amusement park currently charges the same price for each ticket of admission. If the current price of admission were to be increased by $$3$$, 12 fewer tickets could be bought for $$160$$, excluding sales tax. What is the current price of each ticket?

A. $$3$$
B. $$5$$
C. $$8$$
D. $$20$$
E. $$32$$

Kudos for a correct solution.

Let price be p & no. of tickets be n.

p*n = 160
(p+3)*(n-12) = 160

If price is 3, n = 160/3, n is not a whole no. , so 3 is not the answer
If price is 5, n=32 from 1st eqn...checking with 2nd equation 8*(n-12) = 160 -> 8n = 256 -> n=32. Price $5 satisfies both equations. Hence answer B. Kudos [?]: 39 [1], given: 40 Math Expert Joined: 02 Sep 2009 Posts: 42544 Kudos [?]: 135294 [0], given: 12686 Re: An amusement park currently charges the same price for each ticket of [#permalink] ### Show Tags 14 Nov 2014, 09:32 Expert's post 1 This post was BOOKMARKED Bunuel wrote: Tough and Tricky questions: Word problems. An amusement park currently charges the same price for each ticket of admission. If the current price of admission were to be increased by $$3$$, 12 fewer tickets could be bought for $$160$$, excluding sales tax. What is the current price of each ticket? A. $$3$$ B. $$5$$ C. $$8$$ D. $$20$$ E. $$32$$ Kudos for a correct solution. Official Solution: An amusement park currently charges the same price for each ticket of admission. If the current price of admission were to be increased by $$3$$, 12 fewer tickets could be bought for $$160$$, excluding sales tax. What is the current price of each ticket? A. $$3$$ B. $$5$$ C. $$8$$ D. $$20$$ E. $$32$$ The question asks us to determine the current price of each ticket of admission. If we let $$p$$ equal the current price per ticket, and $$n$$ equal the number of tickets that can be bought for $$160$$, we can set up some equations. First, we know that $$pn = 160$$. Second, we are told that if $$p$$ is increased by 3, then 12 fewer tickets can be bought with $$160$$. This equation also expresses how many tickets can be bought for $$160$$, giving us: $$(p + 3)(n - 12) = 160$$. Solve the first equation for $$n$$, giving: $$n = \frac{160}{p}$$. Substitute this value for $$n$$ into the second equation, solving for $$p$$: $$(p + 3)(\frac{160}{p} - 12) = 160$$ Multiply both sides by $$p$$ to get rid of the fraction: $$(p + 3)(\frac{160}{p} - 12)p = (p + 3)(160 - 12p) = 160p$$. Multiply through to get rid of the parentheses: $$160p - 12p^2 + 480 - 36p = 160p$$. Combine like terms: $$124p - 12p^2 + 480 = 160p$$. Set the equation equal to 0: $$0 = 12p^2 + 36p - 480$$.Divide both sides by 12: $$0 = p^2 + 3p - 40$$. Factor: $$0 = (p - 5)(p + 8)$$. Therefore, $$p = 5$$ or $$p = -8$$. Since the the price cannot be negative, $$p = 5$$. Answer: B. _________________ Kudos [?]: 135294 [0], given: 12686 Non-Human User Joined: 09 Sep 2013 Posts: 14909 Kudos [?]: 287 [0], given: 0 Re: An amusement park currently charges the same price for each ticket of [#permalink] ### Show Tags 23 Jun 2016, 15:35 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Kudos [?]: 287 [0], given: 0 Non-Human User Joined: 09 Sep 2013 Posts: 14909 Kudos [?]: 287 [0], given: 0 Re: An amusement park currently charges the same price for each ticket of [#permalink] ### Show Tags 19 Aug 2017, 19:26 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Kudos [?]: 287 [0], given: 0 VP Joined: 22 May 2016 Posts: 1108 Kudos [?]: 397 [0], given: 640 An amusement park currently charges the same price for each ticket of [#permalink] ### Show Tags 20 Aug 2017, 08:36 Bunuel wrote: Tough and Tricky questions: Word problems. An amusement park currently charges the same price for each ticket of admission. If the current price of admission were to be increased by $$3$$, 12 fewer tickets could be bought for $$160$$, excluding sales tax. What is the current price of each ticket? A. $$3$$ B. $$5$$ C. $$8$$ D. $$20$$ E. $$32$$ Kudos for a correct solution. A 90-second approach using answer choices (maybe quicker -- I only caught this 20 seconds in, after trying C the long way). The$160 total stays constant, but the ticket price goes up by $3. Add$3 to each answer, and you eliminate four answers very quickly: A, C, D, and E.

The new prices for those four choices do not divide evenly into $160.* Total price / price per ticket = # of tickets, which will not be a whole integer at new price. A. ($3-->$6) C. ($8-->$11) D. ($20-->$23) E. ($32-->$35) That leaves B. If brave, mark and move on. If not (raises hand), check.$160/$5 = 32 tickets sold$160/$8 = 20 tickets sold Higher price = 12 fewer tickets. Correct. Answer B * -- C and D (11 and 23) are pretty obvious. --If uncertain about A, add digits of 160; not divisible by 3, so not divisible by 6. -- For E, 35 * 2 = 70, doubled is$140. The extra is \$20 (160-140). That won't work; you can't divide 20 by 35 evenly.

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An amusement park currently charges the same price for each ticket of   [#permalink] 20 Aug 2017, 08:36
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