a purse contains 26 coins, some of which are dimes and the rest are nickels. Altogether, the coins are worth more than $2.10. At least how many dimes are in the purse?
Let x = # dimes and y = # of nickels.
x + y = 26
y = 26-x
10x + 5y = 210
Substitute 26-x for y in second equation and solve for x. Insert that value into the first equation and solve for y. Check by inserting both values into the second equation.
at least 16
at least 17
To solve this problem, we'll have to set up some equations and use a bit of algebra. Let's assume the number of dimes in the purse is "d" and the number of nickels is "n".
We're given two pieces of information:
1. The purse contains 26 coins, which means the sum of dimes and nickels must equal 26.
d + n = 26 -- Equation 1
2. The total value of the coins is more than $2.10. Since a dime is worth 10 cents and a nickel is worth 5 cents, the value equation can be written as:
10d + 5n > 210 cents -- Equation 2
Now let's solve this system of equations to determine the minimum number of dimes in the purse.
First, we can rewrite Equation 1 in terms of "n" to get:
n = 26 - d -- Equation 3
Now substitute Equation 3 into Equation 2:
10d + 5(26 - d) > 210
10d + 130 - 5d > 210
5d > 80
d > 16
Since we're looking for the minimum number of dimes, we round up the value of "d" to the next whole number, which is 17.
Therefore, there are at least 17 dimes in the purse.