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by manelgirona » Sun Mar 07, 2010 10:24 am
Fred has a jar full of nickels, dimes and quarters, in the ratio 2:5:10, respectively. If the total value of these coins is 15,50, how many dimes are in Fred's jar?

a) 17
b) 25
c) 50
d) 85
e) 155


OA: 25
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by thephoenix » Sun Mar 07, 2010 10:42 am
manelgirona wrote:Fred has a jar full of nickels, dimes and quarters, in the ratio 2:5:10, respectively. If the total value of these coins is 15,50, how many dimes are in Fred's jar?

a) 17
b) 25
c) 50
d) 85
e) 155


OA: 25

nickel=.05$
dime=.1$
q=.25$

so total value is 2x(.05)+5x(.1$)+10x(0.25$)=15.50$

solving x=5
so # of dimes=5*5=25
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by manelgirona » Sun Mar 07, 2010 11:11 am
How do you arrive at these values?

nickel=.05$
dime=.1$
q=.25$
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