Need help setting up equation so I can solve it.
Jose has $6.70 in an assortment of coins. If he has 3 more quarters than pennies,11 more dimes than pennies, and 2.5 times as many nickels as pennies, how many of each type of coin does he have?
Just write down what they have told you.
q = p+3
d = p+11
n = 5/2 p
p+5n+10d+25q = 670
Now just solve for p,n,d,q
If that looks hard, solve for p first. That gives
p + 5(5/2 p) + 10(p+11) + 25(p+3) = 670
Once you have p, then you can easily figure the others...
To set up the equation, let's assign variables to the number of each type of coin Jose has.
Let:
P = number of pennies
Q = number of quarters
D = number of dimes
N = number of nickels
We are given the following information:
1. Jose has $6.70 in total. We need to convert this into cents since all the given quantities are in terms of coins. Therefore, $6.70 becomes 670 cents.
2. Jose has 3 more quarters than pennies. This can be written as Q = P + 3.
3. Jose has 11 more dimes than pennies. This can be written as D = P + 11.
4. Jose has 2.5 times as many nickels as pennies. This can be written as N = 2.5P.
Now, let's set up the equation for the total value of the coins:
The value of each penny is 1 cent, so the value of all the pennies will be P cents.
The value of each quarter is 25 cents, so the value of all the quarters will be 25Q cents.
The value of each dime is 10 cents, so the value of all the dimes will be 10D cents.
The value of each nickel is 5 cents, so the value of all the nickels will be 5N cents.
The sum of the values of all the coins is $6.70 or 670 cents:
P + 25Q + 10D + 5N = 670
Now, substitute the expressions for Q, D, and N from the given information into the equation:
P + 25(P + 3) + 10(P + 11) + 5(2.5P) = 670
Simplifying the equation will give you a quadratic equation that you can solve to find the value of P (number of pennies), and then substitute that back into the given information to find the values of Q, D, and N.