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January 30, 2015

January 30, 2015

Posted by **april** on Friday, August 28, 2009 at 6:34pm.

- math word problem -
**Reiny**, Friday, August 28, 2009 at 7:36pmlet the number of pennies be x

"chris has twice as many 1 dollar bills as pennies"

----> $1 -- 2x

"he has twice as many pennies as he has dimes"

-----> dimes -- x/2

"he has twice as many dimes as he has 10 dollar bills"

----> $10 -- x/4

value of his money in pennies:

x + 10(x/2) + 100(2x) + 1000(x/4)

= x + 5x + 200x + 250x

= 456x pennies

There is no unique answer, but we can argue as follows:

for the variables x, 2x, x/2 and x/4 we must have positive whole numbers, so the smallest value of x possible is 4

so he could have

one $10 bill

2 dimes

8 $1 bills and

4 pennies for a total of $18.24

or x = 8, then he would have $36.48 etc.

- math word problem -
**MathMate**, Friday, August 28, 2009 at 7:36pmThere is not enough information to give a numerical answer.

Let t=number of ten dollar bills.

"he has twice as many dimes as he has 10 dollar bills", so the amount A is

A=10*t+0.1*(2t)

"he has twice as many pennies as he has dimes"

A=10*t+0.1*(2t)+0.01*4t

"chris has twice as many 1 dollar bills as pennies"

A=10*(t)+0.1*(2t)+0.01*(4t)+1*(8t)

=18.24t

Therefore the total amount Chris has is 18.24 times the number of $10 bills he possesses.

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