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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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New post 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.

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

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New post 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

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

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New post 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

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

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

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New post 13 Nov 2014, 20:34
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Answer = B = 5

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

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New post 14 Nov 2014, 09:32
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.
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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

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

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New post 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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