8-6 Digit and Coins Problems Warm-up Problems 1.If a car travels at a constant speed of 30 miles per hour, how long will it take to travel 96 miles. 2.Zeb.

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8-6 Digit and Coins Problems Warm-up Problems 1.If a car travels at a constant speed of 30 miles per hour, how long will it take to travel 96 miles. 2.Zeb is 5 years older than Yolanda. The sum of their ages is 41. How old are they?

Chapter Digit and Coins Problems

Example 1 The sum of the digits of a two-digit number is 14. If the digits are reversed, the number is 36 greater than the original number. Find the original number. Let x = the tens digit and y = the ones digit. There are two statements in this problem. The sum of the digits is 14 x + y = 14 The new # is 36 greater than the original #. 10y + x = x + y 9y – 9x = 36 We have a system of equations. x + y = 14 x + y = 14 9y – 9x = 36 –9x + 9y = 36

9(x + y) = 9(14) 9x + 9y = 126 –9x + 9y = 36 18y = 162 y = 9 x + y = 14 x + 9 = 14 x = 5 The solution is (5,9). The original number is 5(10) + 9 or 59.

Try This a.The sum of a two-digit number is 5. If the digits are reversed, the new number is 27 more than the original number. Find the original number.

Example 2 Kami has some nickels and dimes. The value of the coins is $1.65. There are 12 more nickels than dimes. How many of each kind of coin does Kami have? let d = the number of dimes and n = the number of nickels. d + 12 = n The value of nickels in cents is 5n and the value of dimes is 10n. Since we have the values of the coins in cents, we must use 165 for the total value. 5n + 10d = 165 We can now solve using the substitution method. d + 12 = n 5n + 10d = 165

d + 12 = n 5n + 10d = 165 5(d + 12) + 10d = 165 5d d = d + 60 = d = 105 d = 7 d + 12 = n = n 19 = n Kami has 7 dimes and 19 nickels.

Try This b.On a table there are 20 coins, some quarters and some dimes. Their value is $3.05. How many of each are there?

Example 3 There were 411 people at a play. Admission was $5 for adults and $3.75 for children. The receipts were $ How many adults and how many children attended? let a = the number of adults and c = the number of children. Since the total number of people is 411. a + c = 411 The receipts from the adults and from children equal the total receipts. 5.00a c = NumberAdmission Price ($) Receipts ($) Adult Child Total NumberAdmission Price ($) Receipts ($) Adulta a Childc c Total

a + c = a + 500c = a c = a + 375c = –500a + 500c = –500a – 375c = – c = 7625 c = 61 a + c = 411 a + 61 = 411 a = 350 There were 350 adults and 61 children at the play.

Try This c.There were 166 paid admissions to a game. The price was $2 for adults and $0.75 for children. The amount taken in was $ How many adults and how many children attended?