Suppose that there are two types of tickets to a show: advance and same-day. the combined cost of one advance ticket and one same-day ticket is $75. for one performance, 20 advance tickets and 35 same-day tickets were sold. the total amount paid for the tickets was $2100. what was the price of each kind of ticket?

the price if one advance ticket is $35

the price of one same day ticket is $40

Step-by-step explanation:

Let x represent the price if one advance ticket.

Let y represent the price of one same day ticket.

The combined cost of one advance ticket and one same-day ticket is $75. It means that

x + y = 75

For one performance, 20 advance tickets and 35 same-day tickets were sold. The total amount paid for the tickets was $2100. It means that

20x + 35y = 2100 - - - - - - - - - 1

Substituting x = 75 - y into equation 1, it becomes

20(75 - y) + 35y = 2100

1500 - 20y + 35y = 2100

- 20y + 35y = 2100 - 1500

15y = 600

y = 600/15 = 40

x = 75 - y = 75 - 40

x = 35

solution:

as, mean is defined as sum of all the observations divided by total number of observations.

and , median is mid value of the data whether arranged in ascending order or descending order.

consider, roadsters:

2,3,4,5,6,7,8,9

mean =

also, median= , because number of observations is even, so median is mean of two mid values.

now, coming to bandits

2,3,4,5,6

mean =  

median = 4, as number of observations is odd.

so, roadsters has greater median as well mean than bandits.

→option a is true about the repairs performed on 2 types of cars:

bandits have lower median and mean than roadsters.

47 - 90 = 43

43 - 180 = x

x = 137