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Mathematics 8 Online
OpenStudy (anonymous):

This one is a bit of a challenge to me; it confuses my brain :P Rachel has an envelope that contains only pennies, nickels, dimes, and quarters. There is more than $.85 and less than $1.60 in the envelope. There is at least one of each coin in the envelope. Using the information above, and the following clues, solve the problem: There are 3 times as many nickels as dimes. There are twice as many dimes as quarters. There are an equal number of pennies as quarters. ~~~ medal to anyone who can figure this out :P

OpenStudy (kinggeorge):

Well, since you have at least one of each, \(.25+.1+.05+.01 = .41\) so you're really looking for coins that add to more than \(.85-.41=.44\) and less than \(1.60-.41=1.19\). After that is where it gets more complex.

OpenStudy (anonymous):

thats confusing :O

OpenStudy (anonymous):

\[85<p+5n+10d+25q<160\]\[n=3d=6q\]\[d=2q\]\[p=q\]

OpenStudy (anonymous):

what?? D:

OpenStudy (anonymous):

\[85<76q<160\rightarrow q=2\]

OpenStudy (kinggeorge):

That's much better way to do it than I was trying.

OpenStudy (anonymous):

the choices areee: $1.02 $1.22 $1.51 $1.52

OpenStudy (anonymous):

So there are 2 pennies, 12 nickels, 4 dimes, and 2 quarters.

OpenStudy (anonymous):

ohhh okk :P thanks, ill just go ahead and give yall both medals

OpenStudy (anonymous):

I think that's a dollar fifty-two.

OpenStudy (anonymous):

lol :3 ok thanks

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