By the end of the 1-day drive the youth had collected 48.15 If there were four times as many dimes as nickels how many of each type of coin was collected?

Answer 1

-47.15

Answer 2

If we assume that the youth only has dimes and nickles, this problem can be solved with a system of equations. Let d = number of dimes and n = number of nickels. first, to make things easier, convert dollars to cents. 48.15 dollars = 4815 cents. Here is the first equation: 10d + 5n = 4815 This makes sense, because a dime is worth 10 cents and a nickel is worth 5 cents. Therefore, you take the number of dimes and multiply it by 10, and add this to the number of nickels multiplied by 5, and you get your total amount of cents. The second equation: d = 4n This makes sense because you have four times as many dimes as nickels. Next, some substitution. 10(4n) + 5n = 4815. We can substitute 4n for d because our second equation shows that they are equal. 40n + 5n = 4815 45n = 4815 n = 107 Now we go back and solve for d. I would use the second equation. d = 4(107) d = 428 Now we go back and check using the first equation. 10 (428) + 5 (107) = 4815 4280 + 535 = 4815 4815 = 4815 cents Therefore, you have 428 dimes and 107 nickels.