A purse contains 5-cent coins and 10-cent coins worth a total of $1.05. If the 5-cent coins were replaced with 10-cent coins and the 10-cent coins were replaced with 5-cent coins, the coins would be worth a total of $2.15. How many coins are in the purse? HELP PLEASE WITH WORK
Accepted Solution
A:
Let f and t represent the numbers of 5- and 10-cent coins in the purse. (We expect these numbers to be positive integers.) .. 5f +10t = 105 . . . . . . current value .. 10f +5t = 215 . . . . . . value if numbers are swapped
Multiply the first equation by -2 and add to the second. .. -2(5f +10t) +(10f +5t) = -2(105) +215 .. -15t = 5 .. t = -1/3
No such coin arrangement is possible.
_____ If you simply add the two original equations, you get .. 15(t + f) = 320 .. t + f = 21 1/3 . . . . . There are 21 1/3 coins in the purse.
Taking that at face value, we find the average coin value is .. 1.05/21.3333... = 4.921875¢ . . . . less than the value of a nickel
There is no combination of nickels and dimes that will make the average value of a single coin be less than 5¢. (The average value must be between 5¢ and 10¢.)