<h3>
No, the cost of 1 adult ticket cannot be $ 20 and the cost of 1 children ticket is not $ 8.</h3>
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.
= 68%