SOLUTION: 40 coins =$4.05 Have 7 more nickles than dimes. how many quarters do yo have?

Algebra ->  Customizable Word Problem Solvers  -> Coins -> SOLUTION: 40 coins =$4.05 Have 7 more nickles than dimes. how many quarters do yo have?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 518958: 40 coins =$4.05 Have 7 more nickles than dimes. how many quarters do yo have?
Found 2 solutions by Maths68, MathTherapy:
Answer by Maths68(1474) About Me  (Show Source):
You can put this solution on YOUR website!
Dimes = x
Nickles = x+7
Quarters = 40-(2x+7)=40-2x-7=33-2x
Total Amount = $4.05*100=405 cents
5x+10(x+7)+25(33-2x)=405
5x+10x+70+825-50x=405
-35x+895=405
-35x+895=405-895
-35x=-490
-35x/-35=-490/-35
x=14


Dimes = x = 14
Nickles = x+7 = 14+7 = 21
Quarters = 33-2x = 33-2(14) = 33-28=5


Check
=====
5*14+10*21+25*5=405
70+210+125=405
405=405

Answer by MathTherapy(10553) About Me  (Show Source):
You can put this solution on YOUR website!

40 coins =$4.05 Have 7 more nickles than dimes. how many quarters do yo have?

Let the amount of quarters be Q, and dimes D
Then the amount of nickels = D + 7

We can then say that: D + D + 7 + Q = 40 ----- 2D + Q = 33 -------- eq (i)

Also, .1D + .05(D + 7) + .25Q = 4.05 ----- .1D + .05D + .35 + .25Q = 4.05 ---- .15D + .25Q = 3.7 -------- eq (ii)

2D + Q = 33 -------- eq (i)
.15D + .25Q = 3.7 -------- eq (ii)
- 6D – 3Q = - 99 ------- Mutiplying eq (i) by – 3 -------- eq (iii)
6D + 10Q = 148 ------- Mutiplying eq (ii) by 40 --------- eq (iv)
7Q = 49 --------- Adding eqs (iii) & (iv)

Q, or amount of quarters = 49%2F7, or highlight_green%287%29

2D + 7 = 33 ------ Substituting 7 for Q in eq (i)
2D = 26

D, or amount of dimes = 26%2F2, or highlight_green%2813%29

Therefore, amount of nickels = highlight_green%2820%29 (13 + 7)

------
Check
------

13 dimes = $1.30
20 nickels = $1.00
7 quarters = $1.75

$1.30 + 1.00 + 1.75 = $4.05

$4.05 = $4.05 (TRUE)

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