nicole has $2.20 in nickels and quarters. she has 16 coins all together. how many of each does she have?
chris has $3.20 in dimes and quarters. he has 17 all together. how many of each does he have?
Your last four money questions are the same type
If will do one, then you try the others.
let the number of nickels be x
then the number of quarters is 16-x
5x + 25(16-x) = 220
5x + 400 - 25x = 220
-20x = -180
x = 9
so 9 nickels and 7 quarters
check: 9 nickels and 7 quarters = 16 coins
value: 9x5 = 7x25 = 220
It's not necessary to change names on this forum. It's obvious that these four questions were posted by one person.
yeah i know i didn't think you would answer all of thes questions. can you help me w/ at least 2 please. i need to know the steps then i can do the other problems
where is the 25 and 5x from?
25 = 25 cents = 1 quarter
5 = 5 cents = 1 nickel
PAM AND PAT HAVE $20 ALTOGETHER PAM HAS $4 MORE THAN PAT HOW MUCH MONEY DOES PAM HAS $
To solve these types of problems, we can set up a system of equations to represent the given information.
Let's start with Nicole's situation:
Let n represent the number of nickels Nicole has.
Let q represent the number of quarters Nicole has.
We know that Nicole has a total of 16 coins, so we can write the equation:
n + q = 16 (Equation 1)
We also know that the value of the nickels ($0.05) plus the value of the quarters ($0.25) equals $2.20. We can write the equation for the total value:
0.05n + 0.25q = 2.20 (Equation 2)
Now, let's move on to Chris's situation:
Let d represent the number of dimes Chris has.
Let q represent the number of quarters Chris has.
We know that Chris has a total of 17 coins, so we can write the equation:
d + q = 17 (Equation 3)
We also know that the value of the dimes ($0.10) plus the value of the quarters ($0.25) equals $3.20. We can write the equation for the total value:
0.10d + 0.25q = 3.20 (Equation 4)
Now we have a system of equations, and we can solve them using various methods like substitution or elimination. Let's use the elimination method to solve this system.
First, let's multiply Equation 1 by -0.10 and Equation 3 by 0.05 to eliminate d and n from the equations:
-0.10n - 0.10q = -1.60 (Equation 1 multiplied by -0.10)
0.05d + 0.05q = 0.85 (Equation 3 multiplied by 0.05)
Adding these two equations together, we have:
-0.10n +0.05d -0.10q +0.05q = -1.60 + 0.85
-0.10n +0.05d = -0.75
Now, we have eliminated q from the equations. Let's call this new equation Equation 5.
Next, let's multiply Equation 2 by 2 and Equation 4 by 0.40 to eliminate decimals:
0.10n + 0.50q = 4.40 (Equation 2 multiplied by 2)
0.40d + 0.40q = 6.40 (Equation 4 multiplied by 0.40)
Adding these two equations together, we have:
0.10n +0.40d +0.50q +0.40q = 4.40 + 6.40
0.10n +0.40d +0.90q = 10.80
Now, we have eliminated q from these equations. Let's call this new equation Equation 6.
Let's rearrange Equation 5 and Equation 6:
-0.10n +0.05d = -0.75 (Equation 5)
0.10n +0.40d +0.90q = 10.80 (Equation 6)
Now, we can solve this system of equations using any method. We can use substitution or elimination again to find the values of n, d, and q, which represent the number of nickels, dimes, and quarters.