Question 1188931: A restaurant offered an eat-all-you-can menu last November. They charged 250 pesos for kids below 3 feet and 350 pesos for adults or kids beyond 3 feet. If the restaurant earned 12,450 on the first day, what is the minimum number of kids below 3 feet, if there are more customers who paid 350 pesos? *
Answer by ikleyn(52817) (Show Source):
You can put this solution on YOUR website! .
A restaurant offered an eat-all-you-can menu last November.
They charged 250 pesos for kids below 3 feet and 350 pesos for adults or kids beyond 3 feet.
If the restaurant earned 12,450 on the first day, what is the minimum number of kids below 3 feet,
if there are more customers who paid 350 pesos? *
~~~~~~~~~~~~~~~
In this problem, we have only one equation for two unknowns, which are "a" and "k",
the number of adults and kids, respectively
a*350 + k*250 = 12450. (1)
So, it seems, that there are many solutions.
But we have two additional restrictions.
First, the solutions "a" and "k" must be integer numbers.
Second, we are looking for the solution in integer numbers, where "k" should be minimal.
It provides the UNIQUENESS of the solution.
So, we write from equation (1)
k = ,
and we are looking for the first positive integer k, which we can obtain from this formula, running a = 1, 2, 3, . . .
Fortunately, we do not need run very far, because just the value of k = 5 provides positive integer a = 32.
So, the ANSWER to the problem's question is:
+------------------------------------------------------------+
| the minimum number of kids under given condition is 5. |
+------------------------------------------------------------+
Solved.
------------------
The way to facilitate the required calculations is to use Excel software, if you have it in your computer.
Then you can make your calculations pleasant, instead having them torturous.
|
|
|