Khadija has $0.90 worth of nickels and dimes. She has a total of 10 nickels and dimes altogether. By following the steps below, determine the number of nickels, x and the number of dimes y, that Khadija has.
Determine three ways to have a total of 10 coins and = 90 cents
Method 1:
Let x be the number of nickels and y be the number of dimes.
We know that the total number of coins is 10, so we have the equation:
x + y = 10
The value of x nickels is 5x cents, and the value of y dimes is 10y cents.
We also know that the total value of the coins is $0.90, so we have the equation:
5x + 10y = 90
Simplify the first equation by subtracting x from both sides:
y = 10 - x
Now substitute this value into the second equation:
5x + 10(10 - x) = 90
5x + 100 - 10x = 90
-5x + 100 = 90
-5x = -10
x = 2
Substitute this value of x back into the first equation to solve for y:
2 + y = 10
y = 8
Therefore, Khadija has 2 nickels and 8 dimes.
Method 2:
Another way to solve this problem is to use a trial and error method.
Since we have a total of 10 coins, we can start by assuming that we have 9 dimes and 1 nickel:
0.10 * 9 + 0.05 * 1 = 0.90
This does not add up to $0.90, so let's try with 8 dimes and 2 nickels:
0.10 * 8 + 0.05 * 2 = 0.80 + 0.10 = 0.90
This adds up correctly, so Khadija has 2 nickels and 8 dimes.
Method 3:
We can also solve this problem by using algebraic equations without substituting for one variable.
Let x be the number of nickels and y be the number of dimes.
We have the following equations:
x + y = 10
0.05x + 0.10y = 0.90
To eliminate decimals, we multiply the second equation by 100:
5x + 10y = 90
Now we can solve the two equations simultaneously:
x + y = 10
5x + 10y = 90
By multiplying the first equation by -5, we can eliminate the x variable:
-5(x + y) = -5 * 10
-5x - 5y = -50
Add this equation to the second equation:
-5x - 5y + 5x + 10y = -50 + 90
5y = 40
Divide both sides by 5:
y = 8
Substitute this value of y into the first equation to solve for x:
x + 8 = 10
x = 10 - 8
x = 2
Therefore, Khadija has 2 nickels and 8 dimes.