SOLUTION: Kim bought a total of $2.65 worth of postage stamps in four denominations. If she bought an equal number of 5-cent and 25-cent stamps and twice as many 10-cent stamps as 5-cent

Algebra ->  Expressions-with-variables -> SOLUTION: Kim bought a total of $2.65 worth of postage stamps in four denominations. If she bought an equal number of 5-cent and 25-cent stamps and twice as many 10-cent stamps as 5-cent      Log On


   

Question 339528:
Kim bought a total of $2.65 worth of postage stamps in four denominations. If she bought an equal number of 5-cent and 25-cent stamps and twice as many 10-cent stamps as 5-cent stamps, what is the least number of 1-cent stamps she could have bought ?
(A) 5(B) 10(C) 15(D) 20(E) 25
Could you explain the above problem Thanks

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

Let x = the number of 1-cent stamps
Let y = the number of 5-cent stamps and the number of 25-cent stamps

Then 2y = the number of 10-cent stamps

1x + 5y + 25y + 10(2y) = 265
         x + 30y + 20y = 265
               x + 50y = 265
                     x = 265 - 50y

x will be the least possible when what we must subtract from 265
namely 50y is as large as possible. That will be when y is as large
as possible.  Since x must not be negative:

                    x > 0 
            265 - 50y > 0
            265 - 50y > 0
                 -50y > -265  
                    y < %28-265%29%2F%28-50%29
                                                      
                    y <53%2F10 
                    y <5%263%2F10 

So y is the largest possible when y = 5

So x is smallest possible when x = 265 - 50(5) = 265 - 250 = 15

Answer: 15 1-cent stamps, choice C.

(5 5-cent stamps, 5 25-cent stamps and 10 10-cent stamps.)

Edwin