Please help me. I'm a 38 year old trying to go back to school and I need help. Please show work. A vending machine accepts nickels, dimes and quarters. At the end of the day, there were three more dimes than nickels and four fewer quarters than nickels. If the total value of the coins was $4.10 find the number of each type of coins.
Word problems with coins
- Thread starter DORIS
- Start date
Hello, DORIS!
Okay, it seems that some baby-talk is in order.
(Don't be offended; this is my normal method of explanation.)
\(\displaystyle \text{Let: }\:N \:=\: \text{number of nickels}\)
\(\displaystyle \text{The total value of the nickels is: }\,5N\text{ cents.}\)
\(\displaystyle \text{"There were 3 more dimes than nickels": }\
\:=\:N+3\)
\(\displaystyle \text{The total value of the dimes is: }\,10(N+3)\text{ cents.}\)
\(\displaystyle \text{"There were 4 fewer quarters than nickels": }\:Q \:=\:N - 4\)
\(\displaystyle \text{The total value of the quarters is: }\,25(N-4)\text{ cents.}\)
\(\displaystyle \text{"The total value is 410 cents."}\)
. . \(\displaystyle \underbrace{5N}_{\text{nickels}} + \underbrace{10(N + 3)}_{\text{dimes}} + \underbrace{25(N-4)}_{\text{quarters}} \;=\;410\)
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I assume you can finish the problem, but just in case . . .
\(\displaystyle 5N + 10N + 30 + 25N - 100 \;=\;410\)
. . \(\displaystyle 40N \:=\:480 \quad\Rightarrow\quad N \:=\:12\)
\(\displaystyle \text{Then: }\:\begin{array}{cccccc}D &=& N+3 &=& 15 \\ Q &=& N-4 &=& 8\end{array}\)
\(\displaystyle \text{There were: \;12 nickels, 15 dimes, and 8 quarters.}\)
Okay, it seems that some baby-talk is in order.
(Don't be offended; this is my normal method of explanation.)
\(\displaystyle \text{Let: }\:N \:=\: \text{number of nickels}\)
\(\displaystyle \text{The total value of the nickels is: }\,5N\text{ cents.}\)
\(\displaystyle \text{"There were 3 more dimes than nickels": }\
\:=\:N+3\)\(\displaystyle \text{The total value of the dimes is: }\,10(N+3)\text{ cents.}\)
\(\displaystyle \text{"There were 4 fewer quarters than nickels": }\:Q \:=\:N - 4\)
\(\displaystyle \text{The total value of the quarters is: }\,25(N-4)\text{ cents.}\)
\(\displaystyle \text{"The total value is 410 cents."}\)
. . \(\displaystyle \underbrace{5N}_{\text{nickels}} + \underbrace{10(N + 3)}_{\text{dimes}} + \underbrace{25(N-4)}_{\text{quarters}} \;=\;410\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I assume you can finish the problem, but just in case . . .
\(\displaystyle 5N + 10N + 30 + 25N - 100 \;=\;410\)
. . \(\displaystyle 40N \:=\:480 \quad\Rightarrow\quad N \:=\:12\)
\(\displaystyle \text{Then: }\:\begin{array}{cccccc}D &=& N+3 &=& 15 \\ Q &=& N-4 &=& 8\end{array}\)
\(\displaystyle \text{There were: \;12 nickels, 15 dimes, and 8 quarters.}\)