SOLUTION: Dana inserted eight coins, consisting of dimes and nickels, into a vending machine to purchase a Snickers bar for 55 cents. How many coins of each type did she use? Book answer: 3

Algebra ->  Customizable Word Problem Solvers  -> Coins -> SOLUTION: Dana inserted eight coins, consisting of dimes and nickels, into a vending machine to purchase a Snickers bar for 55 cents. How many coins of each type did she use? Book answer: 3      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 50641This question is from textbook College Algebra
: Dana inserted eight coins, consisting of dimes and nickels, into a vending machine to purchase a Snickers bar for 55 cents. How many coins of each type did she use?
Book answer: 3 dimes, 5 nickels. This question is from textbook College Algebra

Found 2 solutions by checkley71, AnlytcPhil:
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
D+N=8 & 10D+5N=55 OR D=8-N THEN 10(8-N)+5N=55 OR 80-10N+5N=55 OR 80-55=5N OR
25=5N OR N=5 THEN D+5=8 OR D=8-5 OR D=3 THEN THERE WERE 5 NICKELS & 3 DIMES

Answer by AnlytcPhil(1807) About Me  (Show Source):
You can put this solution on YOUR website!
Dana inserted eight coins, consisting of dimes and nickels, into a vending
machine to purchase a Snickers bar for 55 cents. How many coins of each type
did she use?
Book answer: 3 dimes, 5 nickels.

You can do this with one unknown or with two.

With one unknown:P

Let x be the number of dimes.

Then use the principle:  

number in 2nd part = total number - number in 1st part  

                number of nickels = total number of coins - number of dimes

                number of nickels = 8 - x        


Make this chart:

               Worth of each coin   Number of coins       Value          
DIMES                  
NICKELS                 

Now we know that a dime has worth 10¢ and a nickel has worth 5¢

               Worth of each coin   Number of coins       Value          
DIMES                  10 ¢                
NICKELS                 5 ¢            

Next we fill in the number of coins.  We have x dimes and 8-x nickels. So
fill these in:


               Worth of each coin   Number of coins       Value            
DIMES                  10 ¢                x                
NICKELS                 5 ¢              8 - x          


Next we fill in the value of all the dimes by multiplying 10¢ by x dimes,
getting 10x ¢, so fill in 10x ¢ under value of the dimes:


               Worth of each coin   Number of coins       Value          
DIMES                  10 ¢                x                10x ¢
NICKELS                 5 ¢              8 - x         

Next we fill in the value of all the nickels by multiplying 5¢ by 8 - x
nickels, getting 5(8 - x) ¢, so fill that in under value of the nickels:


               Worth of each coin   Number of coins       Value          
DIMES                  10 ¢                x                10x ¢
NICKELS                 5 ¢              8 - x         5(8 - x) ¢ 

Now we can get our equation by the fact that the two expressions under
"value" must total 55¢.

So the equation is  10x + 5(8 - x) = 55

Solve that and get x = 3

So there are 3 dimes

Since there are 8 - x nickels, this is 8 - 3 or 5 nickels.

That's the way using one unknown

-----------------------------------

Or, if you want to use two unknowns:

>>...Dana inserted eight coins, consisting of dimes and nickels...<<

Number of dimes + Number of nickels = Number of coins

Translation:                  D + N = 8

>>...into a vending machine to purchase a Snickers bar for 55 cents...<<

Money in the D dimes + Money in the N nickels = Money dropped in machine

                           10D + 5N = 55

So we have the system of 2 equations in one unknown            

                              D + N = 8
                           10D + 5N = 55

Solve that system of equations and get D = 3, N = 5

-----------------------------

Either method is perfectly correct

Edwin