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Math question! HELP ME!! ...please? (minimum math level: Algebra 1)?
A local water park has two types of season passes. Plan A costs a one-time fee of $130 for admission, plus $10 for parking every trip. Plan B costs a one-time fee of $49 for parking, plus $26 for admission every trip. How many visits must a person make for plan A and plan B to equal in value?
Can you tell me how you got the answer too? I need to know that too! Can't be a decimal!
3 Answers
- Anonymous1 decade agoFavorite Answer
if there are no decimals allowed, this question doesn't have an answer. there are no multiples of 26 that end in 1, thus it could not be added with 49 to make a number ending in 0 for plan B that would match up with plan A which will always end in 0.
- 1 decade ago
6 visits!
Plan A as an equation is 130+10x.
Plan B as an equation is 49+26x.
You try to find where they meet, so you set up 130+10x=49+26x.
Subtract 49 from both sides, you get 81+30x=26x.
Subtract 10x from both sides, you get 81=16x.
81 divided by 16 = 5.0625
Round UP to the nearest whole number, it is 6.
- 1 decade ago
first make an equation for each pass.
the first one would be 10x+130=y because you pay $10 for every visit plus the original $130
the second one would be 26x+49=y because you pay $26 for every visit plus the original $49
if you want to find out when these are equal, just set them equal to each other and solve for x:
10x+130=26x+49
81=16x
x=5.0625
in most instances you would round this number down, but you would not break even in just 5 visits, so in this case you have to round up. therefore you would have to make 6 visits for the passes to be equal. hope this helps!
