SOLUTION: Mr Abbey bought 8 plates and 12 drinking cups from a shop . A plate cost him 5$ more than a drinking cup . If he spent 940$ altogether how much did a plate and a drinking cup cost

Algebra ->  Finance -> SOLUTION: Mr Abbey bought 8 plates and 12 drinking cups from a shop . A plate cost him 5$ more than a drinking cup . If he spent 940$ altogether how much did a plate and a drinking cup cost      Log On


   

Question 1208453: Mr Abbey bought 8 plates and 12 drinking cups from a shop . A plate cost him 5$ more than a drinking cup . If he spent 940$ altogether how much did a plate and a drinking cup cost
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
system%288p%2B12c=940%2Cp-c=5%29

system%282p%2B3c=235%2Cp-c=5%29

system%282p%2B3c=235%2C3p-3c=15%29

add: 5p=250

p=50

c=p-5
c=50-5
c=45

---
The plate and cup, one each together cost $95.

Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
Mr Abbey bought 8 plates and 12 drinking cups from a shop.
A plate cost him 5$ more than a drinking cup. If he spent 940$ altogether
how much did a plate and a drinking cup cost
~~~~~~~~~~~~~~~~~ 

Let "c" be the cost for a cup, in dollars..
Then the cost for a plate is (c+5) dollars.


Write an equation for total cost

    12c + 8*(c+5) = 940  dollars.


Simplify, solve and find "c"

    12c + 8c + 40 = 940,

    12c + 8c = 940 - 40,

       20c   =    900

         c   =    900/20 = 90/2 = 45.


ANSWER.  One cup costs $45.  One plate costs  $45 + $5 = $50.


CHECK.   12*45 + 8*50 = 940  dollars,  total cost.   ! correct !

Solved.