##### 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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May 16th, 2015
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May 16th, 2015
Dec 7th, 2016
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