SOLUTION: oe has a collection of nickels and dimes that is worth $7.65. If the number of dimes was doubled and the number of nickels was increased by 35, the value of the coins would be $14.

Question 530447: oe has a collection of nickels and dimes that is worth $7.65. If the number of dimes was doubled and the number of nickels was increased by 35, the value of the coins would be $14.30. How many nickels and dimes does he have?
Found 2 solutions by oberobic, MathTherapy:
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
With coin problems you have to keep track of the count of each coin and the value.
n = number of nickels
5n = value of the nickels in cents
d = number of dimes
10d = value of the times in cents
.
5n + 10d = 765 cents
.
2(10d) + 5(n+35) = 1430 cents
20d + 5n + 165 = 1430
20d + 5n = 1265
.
With two equations and two unknowns, you can solve the system of equations.
.
5n + 10d = 765
5n + 20d = 1265
----------------
-10d = -500
.
d = 50
.
Substitute d = 50 to find n.
.
5n + 10d = 765
5n + 10(50) = 765
5n = 265
n = 53
.
Check the values to make sure the above solution is the correct answer.
.
5(53) = 265
10(50) = 500
265 + 500 = 765 cents
.
Check the other equation, too.
.
2(10(50)) + 5(53+35) = 1430 ??
2(500) + 265 + 175 = 1430
Correct.
.
Answer: Joe has 53 nickels and 50 dimes.
.
Done.
Answer by MathTherapy(10858) (Show Source): You can put this solution on YOUR website!

Joe has a collection of nickels and dimes that is worth $7.65. If the number of dimes was doubled and the number of nickels was increased by 35, the value of the coins would be $14.30. How many nickels and dimes does he have?

Let amount of dimes be D, and amount of nickels, N

Then: . 1(D) + .05(N) = 7.65 ------ 10D + 5N = 765 ----- Multiplying by 100 to clear decimals ----- eq (i)

Also, .1(2D) + .05(N + 35) = 14.3 ----- .2D + .05N + 1.75 = 14.3 ------- .2D + .05N = 12.55 ------- 20D + 5N = 1,255 ----- Multiplying by 100 to clear decimals ----- eq (ii)

10D + 5N = 765 ------- eq (i)
20D + 5N = 1,255 ----- eq (ii)
---------------------------------
- 10D = - 490 ----- Subtracting eq (ii) from (i)

D, or amount of dimes = , or , or

10(49) + 5N = 765 ------- Substituting 49 for D in eq (i)
490 + 5N = 765
5N = 765 � 490
5N = 275

N, or amount of nickels = , or

I'll leave the check up to you to perform.

Send comments and �thank-yous� to �D� at MathMadEzy@aol.com
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