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### Math Word Problems - GED, PSAT, SAT, ACT, GRE Preparation6.7 Coin Word Problems - 5

 Example: Sue has \$5.30 in nickels, dimes, and quarters. If there are 2 more quarters than nickels, and 5 times as many dimes as nickels, how many of each coin is there? Solution: Let 'n' be the number of nickels, 'd' be the number of dimes, and 'q' be the number of quarters. The word problem can be translated into this equation as: 5n+10d+25q=530 Another equation that can be written is q=n+2 Another equation that can be written is d=5n we have 3 equation and we can solve for three values q, d and n 5n+10d+25q=530 q=n+2 d=5n 5n+10(5n)+25(n+2)=530 5n+50n+25n+50=530 80n = 480 n = 6 substituting n = 6 in the equation q = n + 2 and d = 5n we have q = n + 2 = 6+ 2 = 8 d = 5n = 5 x 6 = 30 Therefore n = 6 d = 30 and q = 8 Example: It costs \$4 for adult ticket and \$3 for child ticket. How many children and 4 adult take if they spent \$22. Solution: Let 'a' be the price for the adult ticket and 'c' be the price for child ticket We can translate the word problem into equation as 4a + 3c = 22 Also we know that there are 4 adults therefore a = 4 substituting a = 4 in the equation 4a + 3c = 22 we have 4a + 3c = 22 4(4) + 3c = 22 16 + 3c = 22 3c = 22 - 16 3c = 6 c = 2 Example: The total money collected for a concert were \$54. The total number of tickets sold were 17. They sold regular seats for \$2 each, box seats for \$3 each, and balcony seats for \$4 each. If there were 2 times as many balcony seats as box seats sold, how many of each kind were sold? Solution: Let 'a' be the number of regular seats, 'b' be the number of box seats, and 'c' be the number of balcony seats. The word problem can be translated into this equation a+b+c=17 Another equation that can be written is 2a+3b+4c=54 Another equation that can be written is c=2b Solving the three unknowns: a,b, and c with three equations a+b+c=17 2a+3b+4c=54 c=2b substituting value of c in equation 1 and 2 we have a+b+c=17 2a+3b+4c=54 a+3b=17 2a+11b=54 multiplying be 2 for equation 1 2a+6b=34 2a+11b=54 2a+6b=34 -2a-11b=-54 -------------------------- -5b=-20 b = 4 z = 2b = 8 a+b+c = 17 a = 17 - 8 - 4 = 5 They sold 5 regular seats, 4 box seats and 8 balcony seats. Example: The total money collected for a football game were \$30. General admission tickets cost \$1 each, reserved seats \$2 each, and box seats \$3 each. If there were 4 times as many general admission as reserved seat tickets, and 2 times as many reserved as box seat tickets sold, how many of each kind were sold? Solution: Let 'a' be the number of general admission tickets, 'b' be the number of reserved tickets, and 'c' be the number of box seats sold. The word problem can be translated into this equation 1a+2b+3c=30 Another equation that can be written is a=4b Another equation that can be written is b=2c There are three unknowns (x,y, and z) and three equations. 1a+2b+3c=30 a=4b b=2c substituting a = 4b and c =b/2 in the equation 1a+2b+3c=30 1a+2b+3c=30 4b+2b+3(b/2)=30 multiplying both sides by 2 8b+4b+3b=60 15b = 60 b = 4 a = 4b = 4 x 4 = 16 c = b/2 = 4/2 = 2 Directions: Solve the following word problems.