SOLUTION: Rachelle has $ 4.45 in dimes and nickles in her car. The number of nickles is twenty more than the number of dimes. How many of each type of coin does she have?

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Question 1204507: Rachelle has
$
4.45
in dimes and nickles in her car. The number of nickles is twenty more than the number of dimes. How many of each type of coin does she have?
Found 4 solutions by josgarithmetic, MathLover1, greenestamps, ikleyn:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
How many nickels is d+20, if number of dimes is d.



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Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!

let dime be and nickel
=$
=$
if Rachelle has $ in dimes and nickels, we have
.....eq.1
if the number of nickels is twenty more than the number of dime, we have
...eq.2
go to
.....eq.1, substitute
.....solve for








go to

...eq.2, substitute


Rachelle has dimes and nickels.




Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


Using formal algebra....

x = # of dimes
x+20 = # of nickels

The total value is $4.45, or 445 cents:

10(x)+5(x+20) = 445
10x+5x+100 = 445
15x+100 = 445
15x = 345
x = 345/15 = 23

ANSWER: x=23 dimes and x+20=43 nickels

CHECK: 23(10)+43(5)=230+215=445

You can get good mental exercise by solving the problem informally, which uses essentially the same calculations as the formal algebraic solution.

The "extra" 20 nickels have a value of 5(20) = 100 cents; the remaining coins then have a total value of 445-100 = 345 cents.

The remaining coins are equal numbers of dimes and nickels. The value of one dime and one nickel is 15 cents. So the number of nickels and dimes remaining is 345/15 = 23.

And of course again we find there are 23 dimes and 23+20 = 43 nickels.
Answer by ikleyn(52786)   (Show Source): You can put this solution on YOUR website!
.
Rachelle has $4.45 in dimes and nickles in her car. The number of nickles is twenty more
than the number of dimes. How many of each type of coin does she have?
~~~~~~~~~~~~~~~~~~


        Good problem to solve it mentally and to train your mind. 


Put twenty extra nickels aside, for a minute.


Then the remaining collection consists of equal number of nickels and dimes
and is worth 445 - 5*20 = 345 cents.


One nickel PLUS one dime are worth 5+10 = 15 cents;
so, there are 345/15 = 23 groups consisting of one nickel and one dime.


From it, we conclude that the initial collection is  23 dimes  and  23 + 20 = 43 nickels.


ANSWER.  23 dimes and 43 nickels.


CHECK.  23*10 + 43*5 = 445 cents = $4.45.    ! correct !

Solved.



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