The member pays $9.50 to take a boat out, plus $105 just to be a member.
Let's call ' R ' the number of rentals
Member Cost = (9.50 times R) + 105
Non-member cost = (14.75 times R) .
You're interested in when their costs are equal, so at that point, we can write ...
Member cost = non-member cost
9.5 R + 105 = 14.75 R
Subtract 9.5 R from each side: 105 = 5.25 R
Divide each side by 5.25 : 20 = R
If you're going to rent a boat less than 20 times during the Summer season, rent them at the non-member rate.
More than 20 times in 1 Summer, you'll save money by being a member.
This sounds a bit extreme to me. If the Summer season means ALL of May, June, July, August, and September, then you would need to average more than one rental every 7.65 days ... a hair more than one a week ... in order to save any money by being a member. I love sailboat rental, and I live just 2 miles from Lake Michigan, but even at these prices (cheap), I would never average a rental every week for 5 months. So for me, there would be no benefit in a membership ... at least not in the cost of boat rental.
Another way to do it, with more brain but less algebra:
Each time a member rents a boat, he pays (14.75 - 9.50) = 5.25 LESS than a non-member would pay to rent the same boat. But in order to get that deal, he had to pay $105 "up front", at the beginning of the season, before he ever rented anything.
How many times does he have to 'save' $5.25 before he makes up for the the $105 ?
Set up a problem: 105+9.50b=14.75b. Combine like terms: subtract 9.50b from both sides: 105=5.25b. Get unknown by itself: divide both sides by 5.25: 20=b. The answer is 20. They would have to rent a boat 20 times to pay the same amount. 🙂
The member pays $9.50 to take a boat out,
plus $105 just to be a member.
Let's call ' R ' the number of rentals
Member Cost = (9.50 times R) + 105
Non-member cost = (14.75 times R) .
You're interested in when their costs are equal,
so at that point, we can write ...
Member cost = non-member cost
9.5 R + 105 = 14.75 R
Subtract 9.5 R from each side: 105 = 5.25 R
Divide each side by 5.25 : 20 = R
If you're going to rent a boat less than 20 times during the
Summer season, rent them at the non-member rate.
More than 20 times in 1 Summer, you'll save money by being
a member.
This sounds a bit extreme to me. If the Summer season means
ALL of May, June, July, August, and September, then you would
need to average more than one rental every 7.65 days ... a hair
more than one a week ... in order to save any money by being a
member. I love sailboat rental, and I live just 2 miles from Lake
Michigan, but even at these prices (cheap), I would never average
a rental every week for 5 months. So for me, there would be no
benefit in a membership ... at least not in the cost of boat rental.
Another way to do it, with more brain but less algebra:
Each time a member rents a boat, he pays (14.75 - 9.50) = 5.25
LESS than a non-member would pay to rent the same boat.
But in order to get that deal, he had to pay $105 "up front", at the
beginning of the season, before he ever rented anything.
How many times does he have to 'save' $5.25 before he makes up
for the the $105 ?
($105) / ($5.25) = 20 times .
In order to pay the same amount each member would have to rent a boat out 20 times.
20
Step-by-step explanation:
Let x represent the number of rentals. Then the cost will be the same when ...
105 +9.50x = 14.75x
105 = 5.25x
105/5.25 = x = 20
Each would have to rent a boat 20 times in order to pay the same amount.
The difference between 9.5 and 14.75 is 5.25.
The fee per season 105 divided by the difference (5.25) is 20.
14.75 x 20 = 295
(9.5 x 20) + 105 = 295
190 + 105 = 295
The number of times members and non-members will have to rent a boat in order to pay the same amount=20
Step-by-step explanation:
We can have two expressions to show the total cost paid by a member and non-member;
Total cost by member=Cost per summer season+cost per number of times they rent a boat×number of times they rent a boat
where;
Cost per summer season=$105
Cost per number of times they rent a boat=$9.50
Number of times they rent a boat=n
Replacing;
Total cost by a member=105+(9.5×n)=9.5 n+105equation 1
Total cost by a non-member=Cost per number of times they rent a boat×number of times they rent a boat
where;
Cost per number of times they rent a boat=$14.75
Number of times they rent a boat=n
Replacing;
Total cost by a non-member=(14.75×n)=14.75 nequation 2
To calculate the number of times they would have to rent a boat in order to pay the same amount, we equate equation 1 to equation 2
9.5 n+105=14.75 n
14.75 n-9.5 n=105
5.25 n=105
n=105/5.25
n=20
The number of times members and non-members will have to rent a boat in order to pay the same amount=20
To answer this problem, we need to make a system of equations.
C= cost (in place of y in typical equation)
t= number of times renting boat (in place of x in typical equation)
Members- C=9.5t+105
Non-Members- C=14.75t
You can set them equal to each other because of transitive property.
9.5t+105=14.75t
Subtract 9.5t both sides
105=5.25t
Divide 5.25 on both sides
t=20
So it takes 20 times to rent a boat and have the same cost.
🙂
20 times they have to rent a boat
If you are looking for the equation it would be y=105+9.50x because they pay 105 no matter what and each time they visit they pay an additional 9.50
$105+$24.25=$129.25 that is the answer
Set up a problem: 105+9.50b=14.75b. Combine like terms: subtract 9.50b from both sides: 105=5.25b. Get unknown by itself: divide both sides by 5.25: 20=b. The answer is 20. They would have to rent a boat 20 times to pay the same amount. 🙂