SOLUTION: There are 27 nickels, dimes, and quarters in the drawer with a value of $3. How many coins of each type are there if there are three times as many nickels as there are dimes?

Question 625312: There are 27 nickels, dimes, and quarters in the drawer with a value of $3. How many coins of each type are there if there are three times as many nickels as there are dimes?
Found 2 solutions by jankiz, MathTherapy:
Answer by jankiz(13) (Show Source): You can put this solution on YOUR website!
If you have 27 nickels, and that is 3 times more than dimes, you divide 27 by three. Then you get 9 dimes. Then, multiply 27 by 5 to get the value of nickels, wich is $1.35. Then 9*10 to get the value of dimes, which is 90 cent. Then you add 1.35 to .90 cent, and get 2.25. So you need 75 cents more, which equals 3 quarters. In conclusion, you need 27 nickels, 9 dimes, and 3 quarters!
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!

There are 27 nickels, dimes, and quarters in the drawer with a value of $3. How many coins of each type are there if there are three times as many nickels as there are dimes?

Let amount of nickels, dimes, and quarters be N, D, and Q, respectively

Then: N + D + Q = 27

Also, N = 3D, since there are 3 times as many nickels as dimes

Substituting 3D for N, we now have: 3D + D + Q = 27, or 4D + Q = 27 ----- eq (i)

Also, .05(3D) + .1(D) + .25(Q) = 3 ---- .15D + .1D + .25Q = 3 ---- .25D + .25Q = 3 ---- eq (ii)

4D + Q = 27 ----- eq (i)
- 4D - 4Q = - 48 ---- Multiplying eq (ii) by - 16 ----- eq (iii)
- 3Q = - 21 ------ Adding eqs (iii) & (i)

Q, or amount of quarters = , or

4D + 7 = 27 ------ Substituting 7 for Q in eq (i)

4D = 20

D, or amount of dimes = , or

Since there are 5 dimes, then there are 3 * 5, or nickels

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