Elementary Algebra Question

Algebra
Tutor: None Selected Time limit: 1 Day

There are 21 quarters and dimes in a can whose value adds to $3.60. How many of each coin are in the can?

May 16th, 2015

There are 21 quarters and dimes in a can whose value adds to $3.60. How many of each coin are in the can?

N=number of nickels; D=number of dimes

N+D=21
D=21-N Use this to substitute for D

$0.05N+$0.10D=$31.60 Substitute for D. (This converts number of coins to value)
$0.05N+$0.10(21-N)=$3.60
$0.05N+$2.10-$0.10N=$3.60 Subtract $2.10 from each side.
-$0.05N -0.10N= $1.50

You take it up from there 

May 16th, 2015

There are 21 quarters and dimes in a can whose value adds to $3.60. How many of each coin are in the can?

Q=number of quarters; D=number of dimes

Q+D=21
D=21-q Use this to substitute for D

$0.25Q+$0.10D=$3.60 Substitute for D. (This converts number of coins to value)
$0.25Q+$0.10(21-Q)=$3.60
$0.25Q+$2.10-$0.10Q=$3.60 Subtract $2.10 from each side.
$0.25-0.10Q= $1.50

-0.15Q = $1.50

Q= 10 

So Number of dimes = 21-Q 

So number of dimes = 11

May 16th, 2015

The second one is correct answer Sorry There was a confusion

May 16th, 2015

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