I should know this but it is frusterating me: story problem

• Jan 4th 2008, 11:05 AM
Audriella
I should know this but it is frusterating me: story problem
I dont know what it is with me and story problems but i can never seem to figure them out no matter how long i try.

so for your help i will be eternally grateful. :)

How many pounds of hamburger worth \$1.05 per pound must be mixed with 60 pounds of hamburger worth \$0.90 per poundto produce hamburger worth\$1.00 per pound?
• Jan 4th 2008, 11:38 AM
colby2152
Quote:

Originally Posted by Audriella
I dont know what it is with me and story problems but i can never seem to figure them out no matter how long i try.

so for your help i will be eternally grateful. :)

How many pounds of hamburger worth \$1.05 per pound must be mixed with 60 pounds of hamburger worth \$0.90 per poundto produce hamburger worth\$1.00 per pound?

Let \$\displaystyle x\$ be the pounds of missing hamburger!

\$\displaystyle 1.05x+.9*60=(60+x)*1\$

\$\displaystyle 1.05x+54=60+x\$

\$\displaystyle 0.05x=6\$

\$\displaystyle x=120\$ pounds
• Jan 4th 2008, 11:43 AM
Audriella
Thanks
and i know i should have known that
but thank you.
• Jan 4th 2008, 11:44 AM
Jhevon
Colby is correct, but just to explain what happened here:

Let \$\displaystyle x\$ be the amount we require

Then at the end we will have \$\displaystyle x + 60\$ pounds of hamburger. We want this to cost \$1 per pound. So the cost will be \$\displaystyle 1(x + 60) = x + 60\$

Now the cost of the 60-pound hamburger is: \$\displaystyle 60(0.90) = 54\$
And the cost of burger \$\displaystyle x\$ is: \$\displaystyle 1.05x\$

We want the total cost to be the same as \$\displaystyle x + 60\$, hence we have:

\$\displaystyle 54 + 1.05x = x + 60\$

Now solve for \$\displaystyle x\$, and you will get Colby's answer