MATH SOLVE

5 months ago

Q:
# 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¢.)

.. 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¢.)