Question

Chelsea has $45 to spend at the fair. she spends $20 on admission and $15 on snacks. she wants to play a game that costs $0.65 per game. write an inequality to find the minimum number of times, x, Chelsea can play the game. using this inequality, determine the maximum number of times she can play the game

Answer 1

Chelsea already spent $20 and $15:

$45-20-15=10$

So now she has $10 left to spend on the game she wants to play. If the number of times she plays this game is x and the game costs $0.65, you can model this with the following inequation:

$0.65\cdot x\leq10$

If we clear the x from here, we can find the maximum number of times that Chelsea can play this game:

$x\leq\frac{10}{0.65}$$x\leq15.3846$

Since the number of times she can play must be a whole number (you can't play 15.3 times) and this number must be under 15.3846, the maximum number of times Chealsea can play this game is 15.

Bonus: if you substract 15 from the previous result and multiply by how much the game costs, you can get how much money Chealsea has left:

$(15.3846-15)\cdot0.65\cong0.25$

So she leaves the fair with $0.25 after playing the maximum number of times on the game.