Each child ticket for a ride costs $3, while each adult ticket costs $5. If the ride collected a total of $115, and 33 tickets were sold, how many of each type of ticket were sold?

A. 8 child and 25 adult
B. 17 child and 16 adult
C. 156 child and 41 adult
D. 25 child and 8 adult

Easiest way, SAT tip of the day... Is to plug each answer in and see which works.

so 3c + 5a=115

and a+c=33
so if we plug in then 3*8 is 24 and 5*25 is 125 so that doent work so i just go on from their!
Thanks

To solve this problem, we can set up a system of equations based on the information given. Let's assume that the number of child tickets sold is represented by "C", and the number of adult tickets sold is represented by "A".

The first equation we can create is based on the total number of tickets sold: C + A = 33. This equation represents the fact that the total number of child tickets sold (C) plus the total number of adult tickets sold (A) equals 33.

The second equation we can create is based on the total amount collected: 3C + 5A = 115. This equation represents the fact that the total cost of the child tickets (3C) plus the total cost of the adult tickets (5A) equals $115.

To solve this system of equations, we can use either substitution or elimination. Let's use substitution:

From the first equation, we can express C in terms of A: C = 33 - A.

Substituting this expression for C into the second equation, we get: 3(33 - A) + 5A = 115.

Simplifying this equation, we get: 99 - 3A + 5A = 115.

Combining like terms, we get: 2A = 16.

Dividing both sides of the equation by 2, we get: A = 8.

Substituting this value back into the first equation, we can solve for C: C + 8 = 33.

Subtracting 8 from both sides of the equation, we get: C = 25.

Therefore, the answer is D. 25 child tickets and 8 adult tickets were sold.

I'll be glad to check your answer.