Jayden has a jar containing 80 coins all of which are nickels and quarters. The total value of the coins is $14.60. How many of each coin does Jayden have

Question

Using elimination or subsistution 

Answer 1

We will address this word problem step by step while explaining every step. We will use substitution method.

Let n be the number of nickels.

Let q be the number of quarters.

We are told: q + n = 80 ,call this equation1

Also we know the dollar values: 0.25q + 0.05n = 14.60 ,call this equation2

Let's simplify equation1 and get, n = 80 - q

Let's plug n into equation2 and get, 0.25q + 0.05(80-q) = 14.6

Let's expand and get, 0.25q + 0.05*80 - 0.05*q =14.6

This simplifies to: 0.25q + 4 - 0.05q =14.6

Bring like terms together and get: 0.25q - 0.05q =14.6 - 4

This simplifies to: 0.20q = 10. 6

Solve for q and get, q = 10.6/0.20 = 53

From equation1 we know: q + n =80, Let's plug in q=53 and get, 53 + n = 80

Let's solve for n and get, n = 80 - 53 = 27

Answer: So we have 27 nickels and 53 quarters.

Extra step: Let's check our answer by using equation2:

0.25*53 +0.05*27 = 13.25+1.35 = 14.60

Our calculated 14.6 = 14.6 from equation2.

So we are correct!

Answer 2

Choose appropriate variables (either x and y, or N and Q, or other).  Let:

     N = number of nickels

     Q = number of quarters

 

Now, realize that the value of the nickels is 5N and the value of the quarters is 25Q.

 

Translate:

"containing 80 coins "          means     N+Q = 80    [eq1]

"total value of the coins is $14.60"     means    5N + 25Q = 1460    [eq2]

     [note: It is easier to use cents than to keep the decimal point]

 

The math:

(1) elimination  -   the coefficient of one of the variables must be the same (or opposite) to subtract (or add) the equations.

 

     5N    +    5Q     =     400              [5*eq1]

     5N    +   25Q    =    1460             [eq2]

  ----------------------------------     [elimination; subtract equations]

                 -20Q    =   -1060

                      Q    =     53             [divide both sides by (-20)]

      N     +      Q     =     80           [eq1 again]

  -------------------------------       [elimination; subtract equations]

     -N                    =    -27

       N                   =     27          [divide both sides by (-1)]

 

Check:

  Is 5*27 + 25*53 = 1460   ?

      135  +  1325  =  1460   ?

            1460 = 1460   ?yes

 

(2) substitution -  isolate (solve) for one variable in terms of the other in one equation, then substitute that into the other equation

 

       N = 80 - Q             [from eq1]

 

     5(80-Q) + 25Q = 1460             [substitute for N in eq2]

      400 - 5Q + 25Q = 1460

        400 + 20Q = 1460

         20Q = 1060

           Q = 53

 

    Now, substitute for Q in either equation:

         N + 53 = 80          [eq1 is pretty easy]

          N = 27

 

[note: same check applies]