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If tommy has 175 dollars in 1, 5, 10 and 25 cent coins, how many coins of each would he have if he has 3576 coins total.

 May 6, 2016

Best Answer 

 #4
avatar+561 
+5

Just take away 70 one cent coins and add 2 ten cents and 2 twenty-five cents then.

 May 7, 2016
 #1
avatar+561 
+5

Let w = number of 1 cent coins, x = number of 5 cent coins, y = number of 10 cent coins, and z = number of 25 cent coins.

 

We know two things.

 

\(0.01w+0.05x+0.1y+0.25z=175\)

 

\(w+x+y+z=3576\)

 

Let's eliminate w because we can.

 

\(0.01(3576-x-y-z)+0.05x+0.1y+0.25z=175\)

 

\(0.04x+0.09y+0.24z=139.24\)

 

There will be more than one solution for this problem. Let's settle for finding just one of them by saying z=0 and y=0.

 

\(0.04x=139.24\)

 

\(x=3481\)

 

\(w+3481=3576\)

 

\(w=95\)

 

One solution to this problem would be 95 one cent coins, 3481 five cent coins 0 ten cent coins, 0 twenty-five cent coins.

 May 6, 2016
 #2
avatar+36915 
+5

Don't really think that answered the question, Will ...it says there ARE 10 and 25 cent coins.   We need more info for a discreet answer to this question....there are 4 variables and only two equations relating them, so there are possibly multiple solutions.

 May 6, 2016
 #4
avatar+561 
+5
Best Answer

Just take away 70 one cent coins and add 2 ten cents and 2 twenty-five cents then.

Will85237  May 7, 2016
 #3
avatar
+5

There are MANY solutions to this problem. We can, however, proceed as follows:

 

Set quarters to =$50 x 4=200 quarters
Set dimes to      =$50 x 10=500 dimes
Amount remaining in nickels and pennies
=$75 for a total of 2,876 nickels and pennies
n + p =2876
.05n +.01p=7500, solve for n, p
p=1,720 x .01 =$17.20
n=1,156 x .05 =$57.80
d=500 x .10  =$50.00
q=200 x .25  =$50.00
Totals: 3,576 coins for a total of $175.00

 May 7, 2016
 #5
avatar+118587 
0
 May 7, 2016

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