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?

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.

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p