SOLUTION: A cashier has 25 bills consisting of twice as many ones as tens, two fewer fives than ones, and the rest twenties. If the total value is $140, write a system of equations to model

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: A cashier has 25 bills consisting of twice as many ones as tens, two fewer fives than ones, and the rest twenties. If the total value is $140, write a system of equations to model       Log On


   

Question 1095284: A cashier has 25 bills consisting of twice as many ones as tens, two fewer fives than ones, and the rest twenties. If the total value is $140, write a system of equations to model the situation. (do not solve)
Found 2 solutions by CubeyThePenguin, ikleyn:
Answer by CubeyThePenguin(3113) About Me  (Show Source):
You can put this solution on YOUR website!
w = number of 1s
x = number of 5s
y = number of 10s
z = number of 20s

w + x + y + z = 25
w = 2y
x = w - 2
w + 5x + 10y + 20z = 140


Answer by ikleyn(52855) About Me  (Show Source):
You can put this solution on YOUR website!
A cashier has 25 bills consisting of twice as many ones as tens,
two fewer fives than ones, and the rest twenties.
If the total value is $140, write a system of equations to model the situation. (do not solve)
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            It is very stupid way/instruction to solve this problem using a system of equations.

            It can be solved using  ONE  SINGLE  equation in  ONE  SINGLE  unknown. 


Let x be the number of ten-dollar bills.


Then the number of one-dollar bills is 2x; the number of five-dollar bills is (2x-2)

and the twenty-dollar bills are the rest  25-x - 2x - (2x-2) = 27-5x.


Having it, we write the total money equation


    10x + 1*(2x) + 5*(2x-2) + 25*(27-5x) = 140  dollars.


The next step is to simplify and solve it for x, and then restore all needed values.

It is how this problem  SHOULD  BE  SOLVED  and is  EXPECTED  to  be  SOLVED.