Slide 1

Word problems with Linear and Quadratic Equations Prepared by Doron Shahar Slide 2 Warm-up Write each description as a mathematical statement. Page 9 #1, 3 more than twice a number Page 9 #2, The sum of a number and 16 is three times the number. 1.2.1 The sum of three consecutive integers is 78. What is the smallest of the three integers? 1.5.1 The sum of the square of a number and the square of 7 more than the number is 169. What is the number? Prepared by Doron Shahar Slide 3 Page 9 #1 Write 3 more than twice a number as a mathematical statement. Variable: Let x be the number. Expression: Prepared by Doron Shahar Slide 4 Page 9 #2 Write the sum of a number and 16 is three times the number as a mathematical statement. Variable: Let x be the number. Equation: Prepared by Doron Shahar Slide 5 1.2.1 The sum of three consecutive integers is 78. What is the smallest of the three integers? Variable: Let x be the smallest of the three integers. Equation: Want: The smallest of the three integers. Examples of consecutive integers: 1, 2, 3 25, 26, 27 17, 18,19 x, x+1, x+2 Prepared by Doron Shahar Slide 6 1.5.1 The sum of the square of a number and the square of 7 more than the number is 169. What is the number? Variable: Let x be the number. Equation: Want: The number. Prepared by Doron Shahar Slide 7 Approach to word problems Solving word problems involves two key steps: 1. Constructing the equations you are to solve. 2. Solving the equations you found. The word problems in sections 1.2 and 1.5 will lead to linear and quadratic equations, which we have just learned how to solve. Therefore, we shall not solve any of the word problems in class. Rather we will focus on constructing the equations. That is, we will translate the word problems into equations. Prepared by Doron Shahar Slide 8 1.5.2 Projectile Problem A stone is thrown downward from a height of 274.4 meters. The stone will travel a distance of s meters in t seconds, where s=4.9t 2 +49t. How long will it take the stone to hit the ground? Variable: Let T seconds be the time it will take the stone to hit the ground. Equation: To find the equation, lets try using a picture. Want: The time it will take the stone to hit the ground. Prepared by Doron Shahar Slide 9 1.5.2 Equation Initial Height T is the time it will take until the stone hits the ground. Distance traveled in T seconds 274.4 meters 4.9T 2 +49T meters Equation: Prepared by Doron Shahar Slide 10 1.5.2 Summary Variables: Let T seconds be the time until the stone hits the ground. Equation: Want: The time it will take the stone to hit the ground. Solutions to equation: Note that T=14 cannot be the answer to the word problem, because T denotes the time it will take the stone to hit the ground. So T>0. Solution to word problem: It will take 4 seconds for the stone to hit the ground. Prepared by Doron Shahar Slide 11 Extra Projectile Problem A cannonball is launched from a cannon. The height, h, in feet of the cannonball t seconds after it leaves the cannon is given by the equation h=16t 2 +63t+4. When will the cannon ball hit the ground? Variable: Let T seconds be the time from when the cannonball is launched until it hits the ground. Equation: To find the equation, lets try using a picture. Want: The time when the cannonball hits the ground. Prepared by Doron Shahar Slide 12 Extra problem: Equation T is the time when the cannonball hits the ground. Equation: After T seconds, the cannonball is on the ground, and has a height of 0 feet. The height, h, in feet of the cannonball t seconds after it leaves the cannon is given by the equation h=16t 2 +63t+4. Prepared by Doron Shahar Slide 13 Extra problem: Summary Variables: Let T seconds be the time from the cannonball is launched until it hits the ground. Equation: Want: The time when the cannonball hits the ground. Solutions to equation: Note that T=0.0625 cannot be the answer to the word problem, because T denotes the time until the cannonball hits the ground. So T>0. Solution to word problem: The cannonball will hit the ground about 4 seconds after it was launched. Prepared by Doron Shahar Slide 14 1.5.4 Geometry Problem The area of a square is numerically 60 more than the perimeter. Determine the length of the side of the square. Want: Variables: Length of the side of the square Let S be the length of the side of the square. Let A be the area of the square. Let P be the perimeter of the square. Equations: Facts from geometry Information given in the problem. Prepared by Doron Shahar Slide 15 1.5.4 Equations Equations: Want: Key Variable: Length of the side of the square Let S be the length of the side of the square. What method can we use to solve for S given these equations? Substitution Now solve for S. Prepared by Doron Shahar Slide 16 1.5.4 Summary Key Equation: Want: Key Variable: Length of the side of the square Let S be the length of the side of the square. Solutions to equation: Note that S=6 cannot be the answer to the word problem, because S denotes the length of the side of the square. So S>0. Solution to word problem: The side of the square is 10 units long. Prepared by Doron Shahar Slide 17 1.5.5 Geometry Problem The length of a rectangle is 7 centimeters longer than the width. If the diagonal of the rectangle is 17 centimeters, determine the length and width. Want: Variables: Let W cm be the width of the rectangle. Let D cm be the perimeter of the square. Equations: Let L cm be the length of the rectangle. Length and Width of the rectangle Information given in the problem. Fact from geometry Prepared by Doron Shahar Slide 18 1.5.5 Equations Equations: Want: Key Variables: Length and width of rectangle What method can we use to solve for L and W given these equations? Substitution Now solve for W. Then solve for L. Let W cm be the width of the rectangle. Let L cm be the length of the rectangle. Prepared by Doron Shahar Slide 19 1.5.5 Summary Key Equations: Want: Key Variable: Solutions to first equation: Note that W=15 cannot be the answer to the word problem, because W denotes the widths of the rectangle. So W>0. Solution to word problem: The width of the rectangle is 8 cm. Length and width of rectangle Let W cm be the width of the rectangle. Let L cm be the length of the rectangle. The length of the rectangle is 15 cm. Prepared by Doron Shahar Slide 20 Key Idea for many word problems In the previous problems, the equations we are trying to find are almost given, or else come from basic facts about geometry. Next we will have to find the equations. Prepared by Doron Shahar Slide 21 Key Idea for many word problems Once we have an equation of the form, We may need to use formulas to replace the amount when dealing with values and costs. Prepared by Doron Shahar Slide 22 1.2.2 General problem Tina has $6.30 in nickels and quarters in her coin purse. She has a total of 54 coins. How many of each coin does she have? Want: Variables: The number of nickels and the number of quarters. Let N be the number of nickels. Let Q be the number of quarters. Well, it depends on how many coins she has that are not in her coin purse. Prepared by Doron Shahar Slide 23 1.2.2 Equations Number of nickels Number of quarters Total Number of coins Dollar Value of nickels Dollar Value of quarters Dollar Value of all coins What is the amount being added? Number of coins Dollar Value Prepared by Doron Shahar Slide 24 1.2.2 Summary Tina has $6.30 in nickels and quarters in her coin purse. She has a total of 54 coins. How many of each coin does she have? Want: Variables: The number of nickels and the number of quarters. Let N be the number of nickels. Let Q be the number of quarters. Equations: What method might we use to solve this system of equations? Elimination Prepared by Doron Shahar Slide 25 1.2.3 General problem A company produces a pair of skates for $43.53 and sells a pair for $89.95. If the fixed costs are $742.72, how may pairs must the company produce and sell in order to break even? Want: Variables: The number of pairs of skates that must be produced to break even. Let S be the number of pairs of skates that must be produced to break even. Prepared by Doron Shahar Slide 26 1.2.3 Equation Fixed Cost Production Cost Total Cost What is the amount being added? Cost Fixed costsBreak even Rent Money Earned = Total Cost Prepared by Doron Shahar Slide 27 1.2.3 Summary A company produces a pair of skates for $43.53 and sells a pair for $89.95. If the fixed costs are $742.72, how may pairs must the company produce and sell in order to break even? Want: Variables: The number of pairs of skates that must be produced to break even. Let S be the number of pairs of skates that must be produced to break even. Equation: Prepared by Doron Shahar Slide 28 Warm-up Page 9, How much pure salt is in 5 gallons of a 20% salt solution? Page 20 #3, An alloy contains 40% gold. Represent the number of grams of gold present in G grams of the alloy. Page 9, What is the equation involving distance, rate, and time? Page 20 #1, A car is travelling M mph for H hours. Represent the number of miles traveled. Prepared by Doron Shahar Slide 29 Key Idea for many word problems In the previous problems, the equations we are trying to find are almost given, or else come from basic facts about geometry. Next we will have to find the equations. Prepared by Doron Shahar Slide 30 Key Idea for many word problems Once we have an equation of the form, We may need to use formulas to replace the amount when dealing with percents and rates. Prepared by Doron Shahar Slide 31 Key Idea for mixture problems Below is the formula for percents. Page 9, How much pure salt is in 5 gallons of a 20% salt solution? Prepared by Doron Shahar Slide 32 Key Idea for mixture problems Below is the formula for percents. Page 20 #3, An alloy contains 40% gold. Represent the number of grams of gold present in G grams of the alloy. Prepared by Doron Shahar Slide 33 1.2.4 Mixture problem A premium mix of nuts costs $12.99 per pound, while almonds cost $6.99 per pound. A shop owner adds almonds into the premium mix to get 90 pounds of nuts that cost $10.99 per pound. How many pounds of almonds did she add? Want: Variables: The number of pounds of almonds she added. Let A be the number of pounds of almonds she added. Let P be the number of pounds of the premium mix of nuts. Prepared by Doron Shahar Slide 34 1.2.4 Equations Pounds of almonds Pounds of premium mix Total Pounds in mixture Cost of almonds Cost of premium mix Total cost of mixture What is the amount being added? Mass (Pounds) Cost Prepared by Doron Shahar Slide 35 1.2.4 Summary A premium mix of nuts costs $12.99 per pound, while almonds cost $6.99 per pound. A shop owner adds almonds into the premium mix to get 90 pounds of nuts that cost $10.99 per pound. How many pounds of almonds did she add? Want: Key Variable: The number of pounds of almonds she added. Let A be the number of pounds of almonds. Equations: What method might we use to solve this system of equations? Elimination Prepared by Doron Shahar Slide 36 1.2.5 Mixture problem A chemist wants to strengthen her 40L stock of 10% acid solution to 20%. How much 24% solution does she have to add to the 40L of 10% solution in order to obtain a mixture that is 20% acid? Want: Variable: The volume of the 24% solution that the chemist needs to add to obtain a mixture that is 20% acid. Let V liters be the volume of the 24% solution that needs to be added. Prepared by Doron Shahar Slide 37 1.2.5 Equation Volume of acid in beaker 1 Volume of acid in beaker 2 Total volume of acid in beaker 3 What is the amount being added? Volume of acid 10% acid 24% acid 20% acid Beaker 1 Beaker 2 Beaker 3 40 liters of solution V liters of solution (40+V) liters of solution Prepared by Doron Shahar Slide 38 1.2.5 Summary A chemist wants to strengthen her 40L stock of 10% acid solution to 20%. How much 24% solution does she have to add to the 40L of 10% solution in order to obtain a mixture that is 20% acid? Want: Variable: The volume of the 24% solution that the chemist needs to add to obtain a mixture that is 20% acid. Let V liters be the volume of the 24% solution that needs to be added. Equation: Prepared by Doron Shahar Slide 39 Key Idea for rate problems Below is the formula for rates. Page 20 #1, A car is travelling M mph for H hours. Represent the number of miles traveled. Prepared by Doron Shahar Slide 40 1.2.7 Distance-Rate-Time Two motorcycles travel towards each other from Chicago and Indianapolis (350km apart). One is travelling 110 km/hr, the other 90 km/hr. If they started at the same time, when will they meet? Want: Variable: The time when they will meet. Let T hours be the time from when they started until they meet. Prepared by Doron Shahar Slide 41 1.2.7 Equation Distance traveled by motorcycle 1 Distance traveled by motorcycle 2 Total distance traveled by both What is the amount being added? Distance Motorcycle 1Motorcycle 2 110 km/hr 90 km/hr Chicago Indianapolis 350 km Distance traveled by Motorcycle 1 Distance traveled by Motorcycle 2 Where they meet Prepared by Doron Shahar Slide 42 1.2.7 Summary Two motorcycles travel towards each other from Chicago and Indianapolis (about 350km apart). One is travelling 110 km/hr, the other 90 km/hr. If they started at the same time, when will they meet? Want: Variable: The time when they will meet. Let T hours be the time from when they started until they meet. Equation: Prepared by Doron Shahar Slide 43 1.5.3 Distance-Rate-Time Amy travels 450 miles in her car at a certain speed. If the car had gone 15 mph faster, the trip would have taken 1 hour less. Determine the speed of Amys car. Want: Variables: The speed of Amys car. Let A mph be the speed of Amys car. Let T hours by the time it takes Amy to drive 450 miles. Prepared by Doron Shahar Slide 44 1.5.3 Equations Amys car A mph 450 miles for T hours Amys car (A+15) mph 450 miles for (T-1) hours Prepared by Doron Shahar Slide 45 1.5.3 Summary Amy travels 450 miles in her car at a certain speed. If the car had gone 15 mph faster, the trip would have taken 1 hour less. Determine the speed of Amys car. Want: Key Variable: The speed of Amys car. Let A mph be the speed of Amys car. Equations: What method might we use to solve this system of equations? Substitution Prepared by Doron Shahar Slide 46 1.2.9 Shared Work Problem Suppose a journeyman and apprentice are working on making cabinets. The journeyman is twice as fast as his apprentice. If they complete one cabinet in 14 hours, how many hours does it take for the journeyman working alone to make one cabinet? Want: Variables: The time it takes the journeyman working alone to make one cabinet. Let J hours be the time it takes the journeyman to make one cabinet working alone. Let A hours be the time it takes the apprentice to make one cabinet working alone. Prepared by Doron Shahar Slide 47 1.2.9 Equations The journeyman is twice as fast as his apprentice. Journeymans rate 2 Apprentices rate Multiply both sides of the equation by AJ Prepared by Doron Shahar Slide 48 1.2.9 Equations Amount of the cabinet built by the journeyman in 14 hours Amount of the cabinet built by the apprentice in 14 hours Amount of the cabinet built by both in 14 hours What is the amount being added? Amount of the cabinet built The answer is NOT time!! Prepared by Doron Shahar Slide 49 1.2.9 Summary Want: Key Variable: The time it takes the journeyman working alone to make one cabinet. Let J be the time it takes the journeyman to make one cabinet working along. Equations: What method might we use to solve this system of equations? Substitution Prepared by Doron Shahar Slide 50 1.5.6 Shared Work Problem It take Julia 16 minutes longer to chop vegetables than it takes Bob. Working together, they are able to chop the vegetables in 15 minutes. How long will it take each of them if they work by themselves? Want: Variables: The time it takes each of them working alone to chop the vegetables. Let J minutes be the time it takes Julia to chop the vegetables working alone. Let B minutes be the time it takes Bob to chop the vegetables working alone. Prepared by Doron Shahar Slide 51 1.5.6 Equations Amount of the vegetables chopped by Julia in 15 minutes Amount of the vegetables chopped by Bob in 15 minutes Amount of the vegetables chopped by both in 15 minutes What is the amount being added? Amount of vegetables chopped The answer is NOT time!! It take Julia 16 minutes longer to chop the vegetables than it takes Bob. Prepared by Doron Shahar Slide 52 1.5.6 Summary Want: Variables: The time it takes each of them working alone to chop the vegetables. Let J minutes be the time it takes Julia to chop the vegetables working alone. Let B minutes be the time it takes Bob to chop the vegetables working alone. Equations: What method might we use to solve this system of equations? Substitution Prepared by Doron Shahar Slide 53 1.2.8 Shared Work Problem Suppose it takes Mike 3 hours to grade one set of homework and it takes Jenny 2 hours to grade one set of homework. If they grade together, how long will it take to grade one set of homework? Want: Variables: The time it will take them to grade one set of homework if they work together. Let T hours be the time it will take them to grade one set of homework if they work together. Prepared by Doron Shahar Slide 54 1.2.8 Equations Amount of the homework set graded by Mike in T hours Amount of the homework set graded by Jenny in T hours Amount of the homework set graded by both in T hours What is the amount being added? Amount of the homework set graded The answer is NOT time!! Prepared by Doron Shahar Slide 55 1.2.8 Summary Suppose it takes Mike 3 hours to grade one set of homework and it takes Jenny 2 hours to grade one set of homework. If they grade together, how long will it take to grade one set of homework? Want: Variables: The time it will take them to grade one set of homework if they work together. Let T be the time it will take them to grade one set of homework if they work together. Equation: Prepared by Doron Shahar Slide 56 Key Idea for relative motion problems The boats speed Direction of Current Directions of Boats B mph C mph DownstreamUpstream going downstream isgoing upstream is Prepared by Doron Shahar Slide 57 1.2.6 Relative Motion Problem A boat travels down a river with a current. Travelling with the current, a trip of 66 miles takes 3 hours while the return trip travelling against the current takes 4 hours. How fast is the current? Want: Variables: The speed of the current. Let C mph be the speed of the current. Let B mph be the speed of the boat in still water. Prepared by Doron Shahar Slide 58 1.2.6 Equations B mph C mph B mph C mph For 3 hours For 4 hours 66 miles Prepared by Doron Shahar Slide 59 1.2.6 Summary A boat travels down a river with a current. Travelling with the current, a trip of 66 miles takes 3 hours while the return trip travelling against the current takes 4 hours. How fast is the current? Want: Key Variable: The speed of the current. Let C mph be the speed of the current. Equations: What method might we use to solve this system of equations? Elimination Prepared by Doron Shahar Slide 60 1.5.7 Relative Motion Maria traveled upstream along a river in a boat a distance of 39 miles and the came right back. If the speed of the current was 1.3 mph and the total trip took 16 hours, determine the speed of the boat relative to the water. Want: Variables: The speed of the boat relative to the water. Let B mph be the speed of the boat relative to the water. Let T hours be the time of the trip upstream. Prepared by Doron Shahar Slide 61 1.2.6 Equations B mph 1.3 mph B mph 1.3 mph For T hours For 16T hours 39 miles Prepared by Doron Shahar Slide 62 1.5.7 Summary Maria traveled upstream along a river in a boat a distance of 39 miles and the came right back. If the speed of the current was 1.3 mph and the total trip took 16 hours, determine the speed of the boat relative to the water. Want: Key Variable: The speed of the boat relative to the water. Let B mph be the relative speed of the boat. Equations: What method might we use to solve this system of equations? Substitution or Elimination Prepared by Doron Shahar Slide 63 Review of Word Problems Dont forget units Sanity checks: Check that your answer makes sense Examples: Lengths, times, and speeds should not be negative. If Michael and Rachel can each build a bookcase working alone in under a day, it should not take them more than a day to build a bookcase working together. If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is.John Louis von Neumann Prepared by Doron Shahar