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anonymous

  • one year ago

Lupe, who works at a fast food restaurant received $6.10 in tips one afternoon, all in quarters, dimes, and nickels. There were five less dimes than quarters and seven less nickels than dimes. How many of each kind of coin was there?

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  1. mathstudent55
    • one year ago
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    Let's choose variables. Let n = number of nickels Let d = number of dimes Let q = number of quarters Since a nickel is worth $0.05, a dime $0.10 and a quarter $0.25, the total value of the coins is 0.05n + 0.10d + 0.25q = 6.10

  2. mathstudent55
    • one year ago
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    Now we can get two other equations using the information we are given. "There were five less dimes than quarters" means d = q - 5 "seven less nickels than dimes" means n = d - 7

  3. mathstudent55
    • one year ago
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    Now we have a system of equations in three variables that we can solve.

  4. anonymous
    • one year ago
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    Okay, I get confused after this. I'm confused on how to solve because I have two variables.

  5. mathstudent55
    • one year ago
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    \(\begin{cases} 0.05n + 0.10d + 0.25q = 6.10 \\ d = q - 5 \\ n = d - 7 \end{cases}\) Rewrite the second and third equations so all variables line up below the first equation: \(\begin{cases} 5n + 10d + 25q = 610 ~~~~~~~Eq.1\\ ~~~~~~~~~~~~~~d - ~~~q = - 5 ~~~~~~~~Eq. 2\\ ~~n ~~~~~-d ~~~~~~~~~~= - 7 ~~~~~~~~Eq. 3\end{cases}\) \(~5n + 35d = 485\) 25 * Eq. 2 + Eq. 1 \(35n - 35d = -245\) 35 * Eq. 3 \(40n = 240\) Add last two equations above. \(n = 6\) Substitute n = 6 in Eq. 3 to get \(6 - d = -7\) \(-d = -13\) \(d = 13\) Substitute d = 13 in Eq. 2 to get \(13 - q = -5\) \(-q = -18\) \(q = 18\) Answer: 6 nickels, 13 dimes, 18 quarters Check: 6 * $0.05 + 13 * $0.10 + 18 * $0.25 = $0.30 + $1.30 + $4.50 = $6.10

  6. anonymous
    • one year ago
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    Thank you so much!

  7. mathstudent55
    • one year ago
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    You're welcome.

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