SOLUTION: Warren has 40 coins (all nickels, dimes, and quarters) worth $4.05. He has 7 more nickels than dimes. How many quarters does Warren have? I have been stumped for over an hour wi

Algebra ->  Finance -> SOLUTION: Warren has 40 coins (all nickels, dimes, and quarters) worth $4.05. He has 7 more nickels than dimes. How many quarters does Warren have? I have been stumped for over an hour wi      Log On


   



Question 219366: Warren has 40 coins (all nickels, dimes, and quarters) worth $4.05. He has 7 more nickels than dimes. How many quarters does Warren have? I have been stumped for over an hour with this question. Can you help?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Warren has 40 coins (all nickels, dimes, and quarters) worth $4.05. He has 7 more nickels than dimes. How many quarters does Warren have?
:
Write an equation for each statement:
:
"Warren has 40 coins"
n + d + q = 40
:
"(all nickels, dimes, and quarters) worth $4.05."
.05n + .10d + .25q = 4.05
:
"He has 7 more nickels than dimes."
n = d + 7
:
How many quarters does Warren have?
:
Substitute (d+7) for n in both equations, simplify, results:
(d+7) + d + q = 40
2d + q = 40 - 7
2d + q = 33
q = (33-2d); can use this for substitution
and
.05(d+7) + .10d + .25q = 4.05
.05d + .35 + .10d + .25q = 4.05
.15d + .25q = 4.05 - .35
.15d + .25d = 3.70
:
Substitute (33-2d) for q in the above equation
.15d + .25(33-2d) = 3.70
.15d + 8.25 - .5d = 3.70
.15d -.5d = 3.70 - 8.25
-.35d = -4.55
d = %28-4.55%29%2F%28-.35%29
d = +13 dimes
:
Find n using n = d+7
n = 13 + 7
n = 20 nickels
:
Find q using q = (33-2d)
q = 33 - 2(13)
q = 33 - 26
q = 7 quarters, which is what they asked for
;
:
Check solution in the total$ equation
.05(20) + .10(13) + .25q(7) =
1.00 + 1.30 + 1.75 = 4.05