Question

3. Gwen Chester has $41.70 in her piggybank She has one more than threetimes as many dimes as she has nickels, and she has five times as manyquarters as nickels. How many ofeach coin does she have?00

Answer 1

Given data:

Total: $41.70

She has one more than three times as many dimes (d) as she has nickels (n):

`1+3d=n`

she has five times as many quarters(q) as nickels(n):

`5q=n`

1 d is equivalient to $0.10

1 n is equivalent to $0.05

1 q is equivalent to $0.25

`0.10d+0.05n+0.25q=41.70`

Use the next equations to solve the problem

`\begin{gathered} 0.10d+0.05n+0.25q=41.70 \\ 1+3d=n \\ 5q=n \end{gathered}`

1. Solve d in the second equation:

`\begin{gathered} 1+3d=n \\ 3d=n-1 \\ d=(n)/(3)-(1)/(3) \end{gathered}`

2. Solve q in the third equation:

`\begin{gathered} 5q=n \\ q=(n)/(5) \end{gathered}`

3. Substitute the d and q in the first equation by the values you get in steps 1 and 2:

`0.10((n)/(3)-(1)/(3))+0.05n+0.25((n)/(5))=41.70`

4. Solve n:

`\begin{gathered} (0.10n)/(3)-(0.10)/(3)+0.05n+(0.25n)/(5)=41.70 \\ \\ \text{add (0.10/3) in both sides of the equation:} \\ (0.10n)/(3)-(0.10)/(3)+(0.10)/(3)+0.05n+(0.25n)/(5)=41.70+(0.10)/(3) \\ \\ (0.10n)/(3)+0.05n+(0.25n)/(5)=(125.1+0.10)/(3) \\ \\ (0.10n)/(3)+0.05n+(0.25n)/(5)=(125.2)/(3) \\ \\ (0.10n+0.15n)/(3)+(0.25n)/(5)=(125.2)/(3) \\ \\ (0.25n)/(3)+(0.25n)/(5)=(125.2)/(3) \\ \\ (1.25n+0.75n)/(15)=(125.2)/(3) \\ \\ (2n)/(15)=(125.2)/(3) \\ \\ \text{Multiply both sides of the equation by 15} \\ 15((2n)/(15))=15((125.2)/(3)) \\ \\ 2n=626 \\ \\ \text{Divide both sides of the equation into 2:} \\ (2)/(2)n=(626)/(2) \\ \\ n=313 \end{gathered}`

5. Use the value of n=313 to solve d and q:

`\begin{gathered} d=(n)/(3)-(1)/(3) \\ \\ d=(313)/(3)-(1)/(3) \\ \\ d=(312)/(3) \\ \\ d=104 \end{gathered}`