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Sharu has 2.35 in nickels and dimes. If he has a total of thirty-two coins how many of each coin does he have?

2007-12-31 07:44:56 · 7 answers · asked by anth 1 in Science & Mathematics Mathematics

7 answers

We need 2 equations here [more coming]

Total amount of money = 2.35 dollars
Each nickel is worth 0.05 cents (let X denote nickels)
Each dime is worth 0.10 cents (let Y denote dimes)
Total amount of coins is 32

So we get the following:
0.05x + 0.10y = 2.35
x + y = 32 --> y=32-x (plug this into the 1st equation)

0.05x + 0.10y = 2.35
0.05x + 0.10(32-x) = 2.35
0.05x + 3.2 - 0.10x = 2.35
-0.05x = -0.85
x= 17

Since we know that x + y = 32
(17) + y = 32 --> y =15

Therefore, we have 17 nickels and 15 dimes.

2007-12-31 07:54:52 · answer #1 · answered by ¿ /\/ 馬 ? 7 · 0 1

n = nickels
d = dimes

total value = value of nickels + value of dimes
2.35 = .05n + .10d

Total coins = # of nickels + # of dimes
32 = n + d

We want to substitute one equation into the other to get it down to 1 variable.

32 = n + d
32 - d = n
2.35 = .05(32 - d) + .10d
2.35 = 1.6 - .05d + .10d
.75 = .05d
15 = d

15 dimes

15 + n = 32
n = 32 - 15
n = 17

17 nickels

2007-12-31 15:59:09 · answer #2 · answered by Anonymous · 0 1

5n + 10d = 235
n + d = 32

so half of knowing where to start - is expressing our problem

so let's isolate one of the coins from the second equation
by subtracting n from both sides

n-n+d = 32-n
therefore
d = 32 - n (the number of dimes equals the total number minus the number of nickels) so easy - yeah!

now substitute into the first equation

5n + 10(32-n) = 235

5n + 320 - 10n = 235

320 - 5n = 235

subtract 235 both sides

320- 235 - 5n = 235 - 235

85 - 5n = 0

add 5n to both sides

85 = 5n

and you have it from here

once u get n =

substitute back into the second equation to see how many dimes you have

have a great 2008

2007-12-31 15:57:53 · answer #3 · answered by tom4bucs 7 · 0 3

Set up a system of equations.
n=nickels
d=dimes
I will be using cents.

1) 235 = 5 * n + 10 * d
2) 32 = n + d

There are many ways to solve this: substitution, matrices, and elimination. I will be using substitution.

n = 32 - d ...(from equation 2)
235 = 5 * (32 - d) + 10 * d ... (substitute for n into equaiton 1)
235 = 160 - 5d + 10d ...(simplify)
75 = 5d
d=15
n=17 (put d into equation 2)

2007-12-31 15:57:35 · answer #4 · answered by Goddard 5 · 0 1

Nickels = n, Dimes = d

We have two equations:
1) .05n + .1d = 2.35
2) n + d = 32

Manipulate these two equations, eliminate one variable, solve for the remaining variable, then substitute the value of the variable into either equation and find the value of the other variable!

2007-12-31 15:56:40 · answer #5 · answered by =) 1 · 0 3

32-n=d
5n+10d=235
n=17
d=15

2007-12-31 15:56:11 · answer #6 · answered by Mark L 3 · 0 3

N + D = 32
5N + 10D = 245
divide both sides by 5
N + 2D = 49
subtract the first equation
D = 17
plug into the first equation
N + 17 = 32
N = 15

check .75 + 1.70 = $2.45

2007-12-31 15:56:09 · answer #7 · answered by holdm 7 · 0 2

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