An Amusement Park Currently Charges the Same Price for each Ticket of Admission GMAT Problem Solving

Question: 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?

1. $3
2. $5
3. $8
4. $20
5. $32

“An amusement park currently charges the same price for each ticket of admission" - is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been borrowed from the book “GMAT Official Guide Quantitative Review”.

To understand GMAT Problem Solving questions, applicants must possess fundamental qualitative skills. Quant tests a candidate's aptitude in reasoning and mathematics. The GMAT Quantitative test's problem-solving phase consists of a question and a list of possible responses. By using mathematics to answer the question, the candidate must select the appropriate response. The problem-solving section of the GMAT Quant topic is made up of very complicated math problems that must be solved by using the right math facts.

Answer: C
Solution and Explanation:
Approach Solution 1:

The question asks us to determine the current price of each ticket for admission.

If we let pp equal the current price per ticket, and nn 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 nn, giving: n = 160 p.

Substitute this value for nn into the second equation, solving for pp:

(p+3)(160p−12)=160

Multiply both sides by pp to get rid of the fraction: (p+3)(160/p−12)p = (p+3)(16012p) = 160p.

Multiply through to get rid of the parentheses: 160p−12p2+480−36p=160p

Combine like terms: 124p−12p2+480=160p. Set the equation equal to 0: 0 = 12 p^2+36p − 480. Divide both sides by 12: 0=p2+3p−400=p2+3p−40.

Factor: 0=(p−5)(p+8)

Therefore, p = 5 or p = −8. Since the price cannot be negative, p=5.

B is the correct answer.

Correct Answer: B

Approach Solution 2:

Price ............. Quantity .................. Total

a ...................... 160a……………….160(Assume price = "a")

a+3 ................... 160/a − 12…………… 160 (Price increased by 3, sales decline by 12)

Solving the equation

(a+3)∗(160/a−12)=160

a^2+3a−40=0

a = 5

B is the correct answer.

Correct Answer: B

Approach Solution 3:

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.

Correct Answer: B

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