The cash register in the school canteen contains x quarters and (30-x) dimes. If the total value of the coins is $5.86, how many of each kind of coin are there?
To solve this problem, we can set up two equations based on the given information.
Let's assume that the value of one quarter is $0.25 and the value of one dime is $0.10.
First, we can set up an equation based on the number of coins:
x quarters + (30 - x) dimes = total number of coins
Now, let's set up an equation based on the total value of the coins:
0.25x + 0.10(30 - x) = 5.86
Multiplying the equation by 100 to eliminate decimals, we get:
25x + 10(30 - x) = 586
Simplifying the equation, we have:
25x + 300 - 10x = 586
Combining like terms, we get:
15x + 300 = 586
Subtracting 300 from both sides, we have:
15x = 286
Dividing both sides by 15, we get:
x = 19
Therefore, there are 19 quarters and (30 - 19) = 11 dimes in the cash register.
To find out how many of each kind of coin are in the cash register, we can set up a system of equations based on the information given.
Let's use the variables:
q = number of quarters
d = number of dimes
We are given two pieces of information:
1) There are x quarters, so q = x.
2) The total value of the coins in the cash register is $5.86, which can be expressed as:
0.25q + 0.10d = 5.86
Substituting q = x into the equation, we get:
0.25x + 0.10d = 5.86
We also know that the number of dimes is (30 - x):
d = 30 - x
Now we can substitute this value for d in the previous equation:
0.25x + 0.10(30 - x) = 5.86
Expanding and simplifying the equation gives:
0.25x + 3 - 0.10x = 5.86
0.15x + 3 = 5.86
Subtracting 3 from both sides:
0.15x = 2.86
Dividing both sides by 0.15:
x ≈ 19.07
Since x represents the number of quarters, we cannot have a fraction of a coin. Therefore, we round down the value of x to the nearest whole number:
x = 19
Using the equation for d, we can find the number of dimes:
d = 30 - 19
d = 11
So, there are 19 quarters and 11 dimes in the cash register.
Form the "value" equation:
25x + 10(30-x) = 586
solve for x