SOLUTION: Jess has 3 more nickels than dimes for a total of $1.50. How many of each coin did he have?

Question 209214This question is from textbook Pre-Algebra
: Jess has 3 more nickels than dimes for a total of $1.50. How many of each coin did he have?
This question is from textbook Pre-Algebra

Found 3 solutions by stanbon, josmiceli, MathTherapy:
Answer by stanbon(48558)   (Show Source):
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Jess has 3 more nickels than dimes for a total of $1.50. How many of each coin did he have?
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Quantity Eq: n = d + 3
Value Eq: 5n + 10d = 150 cents
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Substitute to solve for "d":
5(d+3)+10d = 150
15d + 15 = 150
15d = 135
d = 9 (# of dimes)
n = d+3 = 9+3 = 12 (# of nickels)
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Cheers,
Stan H.
Reply to stanbon@comcast.net

Answer by josmiceli(6786)   (Show Source):
You can put this solution on YOUR website!
Suppose he had all nickels. That would be
nickels
Suppose I change of those nickels
for dimes. I have
or dimes
and
or nickels
Now I'll change nickels into dimes:
-----------------------------------
18 nickels
6 dimes
----------
16 nickels
7 dimes
----------
14 nickels
8 dimes
----------
12 nickels
9 dimes
That looks like the answer- 3 more nickels than dimes
check:

OK

Answer by MathTherapy(639)   (Show Source):
You can put this solution on YOUR website!
Jess has 3 more nickels than dimes for a total of $1.50. How many of each coin did he have?

Let the amount of dimes be D

Since he has 3 more nickels than dimes, then he has (D + 3) nickels

Since the coins total $1.50, then we have:
.1D + .05(D + 3) = 1.5

.1D + .05D + .15 = 1.5

.15D = 1.35

dimes

Since he has 3 more nickels than dimes, then he has 9 + 3, or nickels