The cash in the school canteen contains x quarters and (30-x) dimes. If the total value of the coins is $5.85, how many of each kind of coins are there?
just add up the values of the coins:
25x + 10(30-x) = 585
Now find x, and go from there.
X=8
To solve this problem, we can set up an equation based on the given information.
Let's start by representing the number of quarters as "x" and the number of dimes as "30 - x".
The value of x quarters is 0.25x dollars, and the value of (30 - x) dimes is 0.10(30 - x) dollars.
Since the total value of the coins is $5.85, we can create the equation:
0.25x + 0.10(30 - x) = 5.85
Now, let's solve for x:
0.25x + 0.10(30 - x) = 5.85
0.25x + 3 - 0.10x = 5.85
Combine like terms:
0.25x - 0.10x + 3 = 5.85
0.15x + 3 = 5.85
Subtract 3 from both sides:
0.15x = 2.85
Divide both sides by 0.15:
x = 2.85 / 0.15
x ≈ 19
Therefore, there are approximately 19 quarters (x) and 30 - x = 30 - 19 = 11 dimes in the school canteen.