There are two plans for renting a car. Plan A charges $10 per day plus a $50 one-time fee. Plan B charges $15 per day plus a $30 one-time fee. How many days must you rent the car for Plan A to be a better deal than Plan B?
we need
50+10d < 30+15d
so, ...
50 + 10d < 30 + 15d
-10d < - 10d
50 < 30 + 5d
-30 -30
20 < 5d
/5 < 5
4 < d
Please tell me if this is the right answer.
Looks good to me.
If more than 4 days of rental, A is the better deal.
At d=4, A=B=90
After that, A grows more slowly than B.
To determine how many days you must rent the car for Plan A to be a better deal than Plan B, we need to compare the total cost of each plan for a given number of days.
Let's denote the number of days as "n". For Plan A, the total cost will be (10 * n) + 50, since there is a $10 charge per day plus a $50 one-time fee. For Plan B, the total cost will be (15 * n) + 30, as there is a $15 charge per day plus a $30 one-time fee.
Now, we want to find the point where Plan A becomes a better deal than Plan B. In other words, we want to find the value of "n" when the total cost of Plan A is less than the total cost of Plan B.
Setting up an inequality, we have:
(10 * n) + 50 < (15 * n) + 30
Simplifying this inequality, we get:
10n + 50 < 15n + 30
Subtracting 10n from both sides, we have:
50 < 5n + 30
Subtracting 30 from both sides, we get:
20 < 5n
Dividing both sides by 5, we obtain:
4 < n
So, to make Plan A a better deal than Plan B, you must rent the car for at least 4 days.