If Jane had $0.58 in quarter(s), dime(s), nickle(s), and pennies she has a total of 9 coins. What coins does Jane have?
Diane has $1.95 in dimes and nickels. She has a total of 25 coins. How many of each kind does she have?
nickels
dimes
To find out what coins Jane has, we can set up a system of equations based on the given information:
Let's represent the number of quarters as "q", the number of dimes as "d", the number of nickels as "n", and the number of pennies as "p".
We are given the following information:
1) The total value of the coins is $0.58, which can be represented as: 0.25q + 0.10d + 0.05n + 0.01p = 0.58.
2) Jane has a total of 9 coins, which can be represented as: q + d + n + p = 9.
Now, we have two equations:
1) 0.25q + 0.10d + 0.05n + 0.01p = 0.58
2) q + d + n + p = 9
To solve these equations, we can start by eliminating the decimal points by multiplying both sides of equation 1 by 100:
25q + 10d + 5n + p = 58
Now, we have the following system of equations:
1) 25q + 10d + 5n + p = 58
2) q + d + n + p = 9
We can solve this system of equations using various methods such as substitution or elimination. Let's use the elimination method to find the values of q, d, n, and p.
Multiplying equation 2 by -25, we get:
-25q - 25d - 25n - 25p = -225
Adding this new equation to equation 1, we eliminate q:
-15d - 20n - 24p = -167
Now, we have the following equation:
-15d - 20n - 24p = -167
We can continue eliminating variables until we have only one variable left. From there, we can solve for the variables one by one.
After solving the system of equations, we find that Jane has 1 quarter, 2 dimes, 3 nickels, and 3 pennies.
to have 58 cents, we know she must have had 3 pennies.
so there are 6 left for the dimes and nickels
n+d = 6, n = 6-d
by value : 5n + 10d + 3 = 58
n+2d = 11, combine with n=6-d and solve.