Warred has 40 coins (all nickels, dimes, and quarters) worth $4.05. He has 7 more nickels than dimes. How many quarters does Warren have?

Question

This is a algebra one word problem about coins

 

Answer 1

N + D + Q = 40  <--- 40 coins in all

 

5N + 10D + 25Q = 405 <--- amount in pennies to avoid the decimals

 

N = D + 7 <---- 7 more nickels than dimes

 

3 unknowns, 3 equations...

 

step 1

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Substitutes the third equation into the first and second equation.

 

D + 7 + D + Q = 40

 

5(D+7) + 10D + 25Q = 405

 

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step 2: cleans it up

 

2D + Q + 7 = 40

 

5D + 35 + 10D + 25Q = 405 <---- distributes the 5

 

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step 3: combines like terms and writes in standard form

 

2D + Q = 33

15D + 25Q  = 370

 

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step 4: solves the first equation for Q and substitutes it into the second equation:

 

Q = 33 - 2D

 

So,

15D + 25(33 - 2D) = 370

 

15D + 825 - 50D = 370

 

825 - 35D = 370

 

825 - 370 = 35D

 

455 = 35D

 

D = 455/35 = 13

 

Q = 33 - 2D = 33 - 2(13) = 33 - 26 = 7

 

That means N = 40 - 13 - 7 = 20 nickels.

 

7 quarters

13 dimes

20 nickles

 

7 x 25 + 13 x 10 + 20 x 5 = 175 + 130 + 100 = 405

and there are in fact 7 more nickles than dimes.