Tickets for an American Baseball League game for 3 adults and 3 children cost less than $75, while tickets for 2 adults and 4 children cost less than $62. Could the tickets cost $20 for adults and $8 for children?

No, the cost of 1 adult ticket cannot be $ 20 and the cost of 1 children ticket is not $ 8.

Step-by-step explanation:

Let us assume the cots of ticket for 1 adult = $ x

Let us assume the cots of ticket for 1 children = $ y

So, the cost of ticket for 3 adults = 3 x (Cost of 1 adult ticket) = 3 x

The cost of ticket for 3 children = 3 x (Cost of 1 children ticket) = 3 y

Also, given the combined cost of 3 adult and children ticket is less than $75.

⇒ 3 x + 3 y < 75 ... (1)

Similarly, the cost of ticket for 2 adults = 2 x

The cost of ticket for 4 children = 4 y

⇒ 2 x + 4 y < 62 ... (2)

Now, solving for the values of x and y, we get:

3 x + 3 y < 75 or, x + y < 25 ⇒ x = 25 - y (substitute in 2)

2 x + 4 y < 62 or, x + 2 y < 31

⇒ 25 - y + 2 y < 31

or, y + 25 < 31

or, y < 6

⇒ x = 25 - 6 = 19

or, x < 19

So, the cost of 1 adult ticket is x and is less than $19.

The cost of 1 child ticket is y and is less than $6.

Hence, by above statement, NO the cost of 1 adult ticket can not be $ 20 and the cost of 1 children ticket is not $ 8.