TheUltimateWinner wrote:
A certain museum show was attended by members and visitors. The cost of entrance was $8 for the members and $12 for the visitors. What was the average (arithmetic mean) cost of all the tickets sold for the museum show?
(1) The total revenue from sale of all tickets was $620.
(2) The ratio of members to visitors that attended the show was 4 to 7.
Given: Cost of entrance for member = $8, cost of entrance for visitor = $12
Average cost of all the tickets sold =?Let x = number of members, y = number of visitors, and Total revenue = T
Also, T = 8 * x + 12 * y
Average cost = Total revenue/Total of visitors and members entered = \(\frac{T}{x + y}\) =?
Statement 1:
T = $620
There is no data on x and y.
Statement 1 is Not Sufficient.
Statement 2:
x : y = 4 : 7
=> x = \(\frac{4y}{7}\)
There is no information on total revenue.
Statement 2 is also Not Sufficient.
Statement 1 and Statement 2 combined:
T =$620 and x = \(\frac{4y}{7}\)
Also, T = 8x + 12y
Average = \(\frac{T}{x + y}\) = \(\frac{8x + 12y}{x + y}\) = \(\frac{8(x + y)}{x + y}\) + \(\frac{4y}{x + y}\) = 8 + 4 * \(\frac{7}{11}\) \(\approx\) 8.5 [\(\frac{y}{x + y}\) = \(\frac{7}{11}\)
Statement 1 and Statement 2 together are Sufficient.
So, correct answer is option C.P.S. - If we try to solve the equations, we get x = 21.38 and y = 37.41, which is not possible, as number of visitors and members have to be whole numbers. _________________