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Mathematics 21 Online
OpenStudy (moazzam07):

Craig wants to use an elevator to carry identical packages having the same weight. Each package weighs 4 pounds and Craig weighs 90 pounds. If the elevator can carry a maximum of 330 pounds at a time, which inequality shows the maximum number of packages, n, that Craig can carry with himself in the elevator if he is the only passenger? n ≥ 60 n ≤ 60 n ≤ 236 n ≥ 236

OpenStudy (johnweldon1993):

Okay so lets find an expression for the total weight in the elevator We have his weight (90) and the weight of each package (p) which is 4 pounds each...so (4p) So the total weight in the elevator at any given time is \[\large 90 + 4p\] Now...we know the elevator cannot handle more than 330 pounds...so we need the total weight to always be less than or equal to that So \[\large 90 + 4p \le 330\] And we just need to simplify that a bit and solve for 'p'

OpenStudy (moazzam07):

-4 from both sides?

OpenStudy (johnweldon1993):

Dont be afraid to tag me back here :) So if we have \[\large 90 + 4p \le 330\] our goal is to isolate the 'p' since that is what we are solving for So first...we see that 90 is being added to the 4p...so lets subtract 90 from both sides of our equation to cancel that out \[\large \cancel{90 - 90} + 4p \le 330 - 90\] Giving us \[\large 4p \le 240\] Now..we have a 4 being multiplied to the ''...so to cancel that out...we divide both sides of the equation by 4 \[\large \frac{\cancel{4}p}{\cancel{4}} \le \frac{240}{4}\] And we end up with?

OpenStudy (moazzam07):

p < 240

OpenStudy (moazzam07):

or equal to

OpenStudy (johnweldon1993):

Not quite...remember up there we needed to divide the 240 by 4

OpenStudy (moazzam07):

yes

OpenStudy (moazzam07):

60?

OpenStudy (johnweldon1993):

There we go So we have \(\large p \le 60\)

OpenStudy (johnweldon1993):

Or in your case 'n' ...sorry should have kept the variables the same

OpenStudy (moazzam07):

ah ok

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