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Coin Problems

Coin Problems. 11/22/11. Solving Coin Problems. To solve coin problems, organize the facts in the problem using verbal sentences or a table. Number of coins × Value of each = Total value of the coin in cents coins in cents.

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Coin Problems

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  1. Coin Problems 11/22/11

  2. Solving Coin Problems To solve coin problems, organize the facts in the problem using verbal sentences or a table. Number of coins × Value of each = Total value of the coin in cents coins in cents

  3. Helen has $1.35 in her bank in nickels and dimes. There are 9 more nickels than dimes. Find the number she has of each kind. Who is the “x man” in the story? = ˟ value quantity total nickels .05 x + 9 .05(x + 9) dimes .10 x .10x $1.35

  4. = 6 = 15 Let x = the number of dimes x + 9 = the number of nickels .10x + .05(x + 9) = 1.35 .10x + .05x + .45 = 1.35 .15x + .45 = 1.35 -.45 -.45 .15x = .90 .15 .15 x = 6 Check: 6 dimes = 6(.10) = $0.60 15 nickels = 15(.05) = $0.75 Total: $0.60 + 0.75 = $1.35 $1.35 = $1.35 √

  5. A boy has $3.20 in his bank made up of nickels, dimes, and quarters. There are three times as many quarters as nickels, and 5 more dimes than nickels. How many coins of each kind are there? × = value quantity total nickels .05 x .05x dimes .10 x + 5 .10(x + 5) .25 .25(3x) 3x quarters $3.20

  6. = 3 = 8 = 9 Let x = number of nickels x + 5 = number of dimes 3x = number of quarters .05x + .10(x + 5) + .25(3x) = 3.20 .05x + .10x + .50 + .75x = 3.20 .90x +.50 = 3.20 -.50 -.50 .90x = 2.70 .90 .90 x = 3 Check: 3 nickels = 3(.05) = $0.15 8 dimes = 8(.10) = $0.80 9 quarters = 9(.25) = $2.25 Total: $0.15 + 0.80 + 2.25 = $3.20 $3.20 = $3.20 √

  7. Vikihas $2.80 in quarters and dimes. The number of dimes is 7 less than the number of quarters. Find the number she has of each kind. = ˟ value quantity total dimes .10 x - 7 .10(x – 7) quarters .25 x .25x 2.80

  8. = 10 = 3 Let x = number of quarters x – 7 = number of dimes .25x + .10(x – 7) = 2.80 .25x + .10x - .70 = 2.80 .35x - .70 = 2.80 + .70 + .70 .35x = 3.50 .35 .35 x = 10 Check: 10 quarters = 10(.25) = $2.50 3 dimes = 3(.10) = $0.30 Total: $2.50 + 0.30 = $2.80 $2.80 = $2.80 √

  9. Total - 11/29/11

  10. A purse contains $1.35 in nickels and dimes. In all there are 15 coins. How many coins of each kind are there? = value X quantity total .05(15 – x) nickels .05 15 - x x dimes .10 .10x For example: If the sum of two number is 15 and one of the numbers is 6, then the other number is 15 – 6 or 9. 15 $1.35

  11. = 12 = 3 Let x = the number of dimes 15 – x = the number of nickels .10x + .05(15 – x) = 1.35 .10x + .75 - .05x = 1.35 .05x + .75 = 1.35 -.75 -.75 .05x = .60 .05 .05 x = 12 Check: 12 dimes = .10(12) = $1.20 3 nickels = .05(3) = $0.15 Total: $1.20 + 0.15 = $1.35 $1.35 = $1.35√

  12. A purse that contains $3.20 in quarters and dimes has, in all, 20 coins. Find the number of each kind of coin. × = value quantity total dimes .10 x .10x quarters .25 20 - x .25(20 - x) 20 $3.20

  13. = 12 = 8 Let x = # of dimes 20 – x = # of quarters .10x + .25(20 – x) = $3.20 .10x + 5.00 - .25x = 3.20 5.00 - .15x = 3.20 -5.00 -5.00 -.15x = -1.80 -.15 -.15 x = 12 Check: 12 dimes: .10(12) = $1.20 8 quarters: .25(8) = $2.00 Total: $1.20 + 2.00 = $3.20 $3.20 = $3.20 √

  14. Mildred bought 1-cent stamps, 33-cent stamps, and 34-cent stamps for $22.45. The number of 1-cent stamps exceeded the number of 34-cent stamps by 50. The number of 33-cent stamps was 10 less than twice the number of 34-cent stamps. How many of each stamp did she buy? × = value quantity total 1-cent .01 x + 50 .01(x + 50) 33-cent .33 2x - 10 .33(2x – 10) x .34x 34-cent .34 $22.45

  15. = 25 = 40 = 75 Let x = # of 34-cent stamps 2x – 10 = # of 33-cent stamps x + 50 = # of 1-cent stamps .01(x + 50) + .33(2x – 10) + .34x = $22.45 .01x + .50 + .66x – 3.30 + .34x = 22.45 1.01x – 2.80 = 22.45 + 2.80 +2.80 1.01x = 25.25 1.01 1.01 x = 25 Check: (25) 34-cent stamps: .34(25) =$8.50 (40) 33-cent stamps: .33(40) = $13.20 (75) 1- cent stamps: .01(75) = $0.75 Total: $8.50 + 13.20 + 0.75 = $22.45 $22.45 = $22.45 √

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