Word Expression Algebraic Expression Example. Let z first odd integer Then z 2 second consecutive odd integer z 4 third consecutive odd integer

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Hector Ward
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1 3.6 Applications REVIEW from Section 1.6 Five Step Word Problem Method 1) Identify a variable. 2) Write an equation. 4) State your answer. 5) Check your answer. Consecutive Integers Word Expression Algebraic Expression Example Three consecutive integers Three consecutive even integers Let x first integer Then x 1 second consecutive integer 2 third consecutive integer x x 1 x 2 x 4 1 Let y first even integer Then y 2 second consecutive even integer y 4 third consecutive even integer y y 2 y Three consecutive odd integers Let z first odd integer Then z 2 second consecutive odd integer z 4 third consecutive odd integer z z 2 z Note: Notice the expressions used for odd and even integers are the same, y first even integer, y 2 second even integer, y 4 third even integer and z first odd integer, z 2 second odd integer, z 4 third odd integer. This is because if you start with an even number to get to the next even number you will need to add 2. Likewise, if you start with an odd number to get to the next odd number you will need to add 2. So, what makes your consecutive integers odd or even, depends on which number you start with, not the expression used. 1

2 Example 1: The three sides of a triangle are consecutive integers and the perimeter is 72 inches. Find the measure of each side of the triangle. Solution: k k 1 k 2 1) Identify a variable. k = the length of the first side of the triangle 2) Write an equation. If the sides are consecutive integers and k = the length of the first side of the triangle, then the lengths of the other two sides should be k 1and k 2. The perimeter of a triangle is the sum of the lengths of the three sides. So we get, k k 1 k 2 72 length of first side length of second side length of third side k k 1 k 2 72 k k 1 k k Perimeter 3k k 23 4) State your answer. We still need to plug the k 23 into k 1and k 2 to find the lengths of the other two sides. k 23 k k So, the lengths of the sides of the triangle are 23 inches, 24 inches, and 25 inches. You Try It 1: The three sides of a triangle are consecutive integers and the perimeter is 57 centimeters. Find the measure of each side of the triangle. 2

3 Example 2: The length and width of a rectangle are consecutive odd integers. The perimeter of the rectangle is 168 centimeters. Find the length and width of the rectangle. Solution: W W 2 1) Identify a variable. W = the width of the rectangle 2) Write an equation. If the length and width of a rectangle are consecutive odd integers and W is the width of the rectangle, then W 2 = L, the length of the rectangle. The perimeter of a rectangle is given by the formula, P 2L 2W. So we get, 2 W 2 2 W 168 length width W W Perimeter W W W 4 2W 168 4W W W 41 4) State your answer. We still need to plug the W 41into W 2 to find the length of the rectangle. W 41 W So, width and length of the rectangle are 41 centimeters and 43 centimeters, respectively. You Try It 2: The length and width of a rectangle are consecutive odd integers. The perimeter of the rectangle is 120 meters. Find the length and width of the rectangle. 3

4 Example 3: Ray inherits $15,000 and decides to invest in two different types of accounts, a savings account paying 2% interest, and a certificate of deposit paying 4% interest. He decides to invest $3,000 more in the certificate of deposit than in savings. Find the amount invested in each account. Solution: 1) Identify a variable. s = the amount invested in the savings account From the information given, we know the amount invested in the certificate of deposit is $3000 more than what is invested in the savings account amount invested in the certificate of deposit So, the amount invested in the certificate of deposit = 3000 s Account Type Amount Invested 3000 Savings Account (2% interest) s Certificate of Deposit (4% interest) 3000 s Total s A table can be useful to organize the information given. 2) Write an equation. Using the table above, we can see that if we add the amounts invested in each account, we should get the total amount invested, $15,000. Hence we get, s 3000 s Since there is no coefficient in front of the parentheses (other than the invisible 1), we can remove them. s 3000 s s s s ) State your answer. We still need to plug the s 6000 into 3000 s to find the amount invested in the certificate of deposit. s s So, Ray invested $6,000 in the savings account and $9,000 in the certificate of deposit. You Try It 3: Dylan invests a total of $2,750 in two accounts, a savings account paying 3% interest, and a mutual fund paying 5% interest. He invests $250 less in the mutual fund than in savings. Find the amount invested in each account. 4

5 Example 4: Jose cracks open his piggy bank and finds that he has $3.45, all in nickels and dimes. He has 12 more dimes than nickels. How many dimes and nickels does Jose have? Solution: 1) Identify a variable. n = the number of nickels From the information given, we know the number of dimes is 12 more than the number of nickels the number of dimes So, the number of dimes = 12 n Type of Value Number Total Value Coin of Coin of Coins Nickel 5 n 5n Dime n n 12 n Note: We will write the values using cents so we do not need to worry about decimals. Total Amount 345 To complete the table, we multiply the value of the coin with the number of coins to get the total value of each coin. 2) Write an equation. Using the table above, we can see that if we add the total values of each coin, we should get the total amount, 345. Hence we get, 5n10 12 n 345 5n10 12 n 345 5n n n n n 15 4) State your answer. We still need to plug the n 15into 12 n to find the number of dimes. n n So, Jose has 15 nickels and 27 dimes. You Try It 4: David keeps his change in a bowl made by his granddaughter. There is $1.95 in change in the bowl, all in dimes and quarters. There are two fewer quarters than dimes. How many dimes and quarters does he have in the bowl? 5

6 Example 5: A large children s organization purchases tickets to the circus. The organization has a strict rule that every five children must be accompanied by one adult. Hence, the organization orders five times as many child tickets as it does adult tickets. Child tickets are $3 each and adult tickets are $6 each. If the total costs of tickets is $420, how many child and adult tickets were purchased? Solution: 1) Identify a variable. A = the number of Adult tickets purchased From the information given, we know the number of Child tickets is 5 times the number of Adult tickets the number of Child tickets 5 So, the Child tickets purchased = 5A Type of Price per Number of Tickets Cost Ticket Ticket Purchased Adult 6 A 6A 3 5A 15A Child 3 5A Total Cost 420 To complete the table, we multiply the price per ticket with the number of tickets purchased to get the cost. 2) Write an equation. Using the table above, we can see that if we add the cost of each type of ticket, we should get the total cost, 420. Hence we get, 6A15A 420 6A15A420 21A A 20 4) State your answer. We still need to plug the A 20 into 5A to find the number of Child tickets purchased. A 20 5A So, the organization purchased 20 Adult tickets and 100 Child tickets. A You Try It 5: Emily purchased tickets to the IMAX theater for her family. An adult ticket costs $12 and a child ticket costs $4. She buys two more child tickets than adult tickets and the total cost is $136. How many adult and child tickets did she buy? 6

Only to be used for arranged hours, Will count as two activites. Math 31 Activity # 5 Word Problems

Only to be used for arranged hours, Will count as two activites. Math 31 Activity # 5 Word Problems Math 31 Activity # 5 Word Problems Your Name: USING MATH TO SOLVE REAL LIFE PROBLEMS 1. Read the question carefully till you understand it, then assign well- defined variable(s) to the unknown in complete