Greg,
You will need to use simultaneous equations as follows:
Let number of Dimes = D and number of Nickels = N
Then equation 1
10D + 5N = 505
Equation 2
D+N = 63
Multiply Equation 2 by 10 giving equation 3
10D + 10N = 630
Now subtract equation 1 from equation 3 giving equation 4
5N = 125
N = 25 ie 25 Nickels
Now substitute in equation 2
D + 25 = 63
D = 38 ie 38 Dimes
Answer is 38 Dimes and 25 Nickels
Regards,
Brian Kliner
Business Manager
Swift LG - GIS Division
> -----Original Message-----
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
> Sent: 03 October 2002 07:48
> To: [EMAIL PROTECTED]
> Subject: MI-L MI - Algebra question
>
>
> Hi All,
> How do I transcribe this into an algebraic equation and solution?
>
> dimes and nickels, 63 coins in all, totaling $5.05 - how many
> nickels? how
> may dimes?
>
> Thanks,
> Greg
>
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