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S95-24

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S95-24 [#permalink]

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New post 16 Sep 2014, 01:49
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Question Stats:

77% (02:58) correct 23% (02:23) wrong based on 61 sessions

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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. \($32\)
[Reveal] Spoiler: OA

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Re S95-24 [#permalink]

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New post 16 Sep 2014, 01:49
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
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: S95-24 [#permalink]

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New post 10 Jul 2015, 11:35
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I did it by using the answer choices. I will only show it for B, the correct answer.

If the initial price was 5, then 160/5 = 32 tickets.
If the initial price is increased by 3, then 160/8 = 20 tickets.

32 - 20 = 12 fewer tickets.

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Re: S95-24 [#permalink]

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New post 21 Sep 2016, 19:51
agreed... doing the full algebra as described here takes up a too much valuable time. you can set up the equation and plug numbers, using intuition to guide where to start

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Re: S95-24 [#permalink]

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New post 19 Mar 2017, 08:01
mhmmm, true it is faster. Did it the long way and i took me a bit over two minutes

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Re: S95-24   [#permalink] 19 Mar 2017, 08:01
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