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Solving Coin and Ticket Problems: Nickels, Dimes, and Ticket Sales

This guide presents solutions to three real-world problems involving coins and ticket sales. The first problem identifies how many nickels and dimes are among 42 coins that total $3.00. The second calculates how many adult tickets must be sold by a drama club to reach $1005, given a total capacity of 250 seats. Lastly, it determines the cost of bacon alone when combined with eggs. Each problem leverages simple algebraic equations to arrive at clear answers.

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Solving Coin and Ticket Problems: Nickels, Dimes, and Ticket Sales

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  1. 6.2 System Applications Alg Con

  2. 1. Your grandmother gives your little brother 42 coins. Some are dimes and some are nickels, but they total to $3.00. Find how many nickels and how many dimes your little brother has. D = Dimes N = Nickels D + N = 42 0.10D + 0.05N = 3.00 D = 42 - N 0.10(42 - N) + 0.05N = 3.00 D = 42 - N 4.2 - .1N + 0.05N = 3.00 D = 42 - 24 4.2 - 0.05N = 3.00 D = 18 - 0.05N = -1.2 N = 24 There are 24 nickels and 18 dimes that were given by Grandmother

  3. 2. The drama club is selling tickets to its play. The cost of an adult ticket is $4.50, and the cost of a student ticket is $2.50. The theatre can hold 250 people and they want to make $1005. How many adult tickets must they sell? A = Adult S = Student A + S = 250 4.5A + 2.5S = 1005 A = 250 - S 4.5(250 - S) + 2.5 S = 1005 A = 250 - S 1125 – 4.5S + 2.5 S = 1005 A = 250 - 60 1125 – 2S = 1005 A = 190 - 2S = -120 S = 60 The Drama Club must sell 190 Adult Tickets

  4. 3. At Penny’s Diner, two eggs with bacon cost $3.15. One egg with bacon costs $2.20. At these rates, how much would bacon alone cost? E = Eggs B = Bacon 2E + B = 3.15 E + B = 2.20 E = 2.2 - B 2(2.2 – B) + B = 3.15 4.4 – 2B + B = 3.15 4.4 – B = 3.15 -B = -1.25 B = 1.25 Bacon alone would cost $1.25

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