in a purse misty has $1.91.

Question

In a purse misty has $1.91, including one more nickel than quarters. She also has twice as many dimes as nickles. But, she also has eight more pennies than dimes. How many of each of the four coins does she have?

Answer 1

Let N = numbers of nickels

     D = number of dimes

     Q = number of quarters

     P = number of pennies

 

Need 4 independent equations to solve for the 4 unknowns, N, D, Q, and P:

 

1)   $0.05*N + $0.10*D + $0.25*Q + $0.01*P = $1.91     (divide by $1 all terms)

 

   0.05*N + 0.10*D + 0.25*Q + 0.01*P = 1.91     (multiply all terms by 100 to eliminate decimals)

     

    5*N + 10*D + 25*Q + P = 191

 

2)   N = Q + 1     or Q = N -1

3)   D = 2*N

4)   P = D + 8

 

Substitute Equation 3) into Equation 4):

 

P = 2*N + 8

 

Now substitute this equation and Equation 3) into Equation 1):

 

5*N + 10*(2*N) + 25*(N - 1) + (2*N + 8) = 191  (combine like terms and solve for N)

 

5*N + 20*N + 25*N - 25 + 2*N + 8 = 191

 

52*N = 191 + 25 - 8 = 208

 

N = 208/52 = 4 nickels

 

D = 2*N = 2*4 = 8 dimes

 

P = 2*N + 8 = 2*4 + 8 = 16 pennies

 

Q = N -1 = 4 - 1 = 3 quarters

 

checking results,

 

$0.05*(4) + $0.10*(8) + $0.25*(3) + $0.01*(16) = $0.20 + $0.80 + $0.75 + $0.16 = $1.91   checks out