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  ### Math Word Problems - GED, PSAT, SAT, ACT, GRE Preparation6.4 Coins Word Problems - 2
  
   Example: Tim has 11 coins in his pocket that have a total value of \$1. If these coins consists of nickels and dimes only, how many of each kind are there? Solution: Let 'n' be the number of nickels and 'd' be the number of dimes. The word problem can be translated into equation n+d=11 There are 5 cents in a nickel and 'n' nickels equals an amount of 5n. There are 10 cents in a dime and 'd' dimes equals to an amount of 10d. The total value is \$1 or 100 cents. Another equation that can be written as 5n+10d=100 Solve these two equations for n and d to find the number of nickels n and dimes d n+d=11 implies n = 11 - d 5n+10d=100 substituting n = 11 - d in the equation 5n+10d=100 5(11 - d)+10d=100 55 - 5d + 10d = 100 55 + 5d = 100 5d = 100 - 55 5d = 45 d = 9 n = 11 - d = 11 - 9 = 2 Tim has 9 dimes and 2 nickels Verification: substitute value of n and d in the equation n + d = 11 which is 9 + 2 = 11 or substitute value of n and d in the equation 5n+10d=100 which is 5n+10d= 5x2 + 10x9 = 10 + 90 = 100 Example: Kim has 10 coins that total \$2.10. The coins are nickels and quarters only. How many of each kind are there? Solution: Let 'n' be the number of nickels and 'q' be the number of quarters. The word problem can be translated into this equation as n+q=10 Another equation that can be written is 5n+25q=210 Solve these two equations for n and q n+q=10 implies that n = 10 - q 5n+25q=210 5(10 - q)+25q=210 50 - 5q + 25q = 210 50 + 20q = 210 20q = 160 q = 8 n = 10 - q = 10 - 8 = 2 There are 8 quarters and 2 nickles Example: Lilly has 21 coins that total \$4.50. The coins are dimes and quarters only. How many of each kind are there? Solution: Let 'd' be the number of dimes and 'q' be the number of quarters. The word problem can be translated into this equation d+q=21 Another equation that can be written is 10d+25q=450 Solve these two equations for d and q d+q=21 10d+25q=450 d+q=21 implies d = 21 - q substituting in the equation 2 we have 10d+25q=450 10(21 - q)+25q=450 210 - 10q + 25q = 450 210 + 15q = 450 15q = 240 q = 16 d = 21 - q = 21 - 16 = 5 Directions: Solve the following word problems.
   Q 1: Ron has 38 coins that total \$6.70. The coins are nickels and quarters only. How many of each kind are there?q = 24 and n = 10q = 21 and n = 14q = 24 and n = 14 Q 2: Kelly has 16 coins that total \$3.20. The coins are nickels and quarters only. How many of each kind are there? q = 12 and n = 4q = 12 and n = 6q = 10 and n = 6 Q 3: Peter has 31 coins in his pocket that have a total value of \$2.80. If these coins consists of nickels and dimes only, how many of each kind are there? d = 25 and n = 6d = 20 and n = 6d = 21 and n = 8 Q 4: Kim has 15 coins that total \$2.35. The coins are nickels and quarters only. How many of each kind are there?q = 5 and n = 10q = 8 and n = 7q = 10 and n = 5 Q 5: Dan has 22 coins that total \$2.50. The coins are dimes and quarters only. How many of each kind are there?d = 2 and q = 20d = 18 and q = 4d = 20 and q = 2 Q 6: Eric has 29 coins that total \$6.50. The coins are dimes and quarters only. How many of each kind are there?q = 24 and d = 5q = 22 and d = 7q = 24 and d = 8 Q 7: Dan has 8 coins that total \$1.70. The coins are dimes and quarters only. How many of each kind are there?q = 6 and d = 2q = 2 and d = 6q = 3 and d = 3 Q 8: Mary has 20 coins in his pocket that have a total value of \$1.75. If these coins consists of nickels and dimes only, how many of each kind are there? d = 12 and n = 8d = 16 and n = 6d = 15 and n = 5 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!