A convention manager finds that she has $ 1410, made up of twenties and fifties. She has a total of 48 bills. How many fifty-dollar bills does the manager have?
Let x = 50s and y = 20s
x + y = 48
x = 48 - y
50x + 20y = 1410
Substitute 48 - y for x in the second equation and solve for x. Put that value in the first equation to find y. Check by putting both values into the second equation.
To solve this problem, we can set up a system of equations.
Let's say the number of twenty-dollar bills is x, and the number of fifty-dollar bills is y.
From the given information, we can create two equations:
Equation 1: 20x + 50y = 1410 (This equation represents the total value of the money)
Equation 2: x + y = 48 (This equation represents the total number of bills)
Now, we can solve this system of equations to find the values of x and y.
We'll start by rearranging Equation 2 to solve for x: x = 48 - y.
Substituting this value of x into Equation 1, we get:
20(48 - y) + 50y = 1410
Expanding and simplifying the equation gives:
960 - 20y + 50y = 1410
Combine like terms:
30y = 1410 - 960
30y = 450
Divide both sides by 30 to isolate y:
y = 450 / 30 = 15
So, the manager has 15 fifty-dollar bills.