Answer:
Explanation:
To start, let's use algebra to set up some equations based on the information we have. We know that Trip has a total of 15 coins and their total value is 95 cents. Let's call the number of nickels Trip has "n" and the number of dimes "d". Then we can write:
n + d = 15 (since Trip has a total of 15 coins)
0.05n + 0.1d = 0.95 (since the total value of the coins is 95 cents)
Next, we know that four of the coins are each worth twice as much as the rest. Let's call the value of the "rest" coins "x". Then the value of the four coins that are worth twice as much is 2x. We can set up an equation based on the total value of the coins using these values:
4(2x) + 11x = 95
Simplifying this equation, we get:
19x = 95
x = 5
Now we know that each of the "rest" coins is worth 5 cents. We can substitute this value into the equation for the total value of the coins to get:
4(2*5) + 11(5) = 95
Simplifying this equation, we get:
40 + 55 = 95
So our values for n and d are correct: Trip has 11 nickels and 4 dimes.
To justify this argument, we can use a proof by contradiction. Suppose that Trip had a different number of nickels or dimes. If Trip had more than 11 nickels, then the total value of the nickels alone would be greater than 55 cents, which is impossible since the total value of all the coins is only 95 cents. Similarly, if Trip had more than 4 dimes, then the total value of the dimes alone would be greater than 40 cents, which is also impossible. Therefore, Trip must have exactly 11 nickels and 4 dimes, and our argument is justified.