Lesson 7 M Exit Ticket Sample Solutions Bob can paint a fence in hours, and working with Jen, the two of them painted a fence in hours. How long would it have taken Jen to paint the fence alone? Let x represent the time it would take Jen to paint the fence alone. Then, Bob can paint the entire fence in hours; therefore, in one hour he can paint of the fence. Similarly, Jen can paint of the fence in one hour. We know that it x took them two hours to complete the job, and together they can paint of the fence in one hour. We then have to solve the equation: + x x 0x + 0 0x x 0x x + 0 x x 0 3. Thus, it would have taken Jen 3 hours and 0 minutes to paint the fence alone. Problem Set Sample Solutions. If two inlet pipes can fill a pool in one hour and 30 minutes, and one pipe can fill the pool in two hours and 30 minutes on its own, how long would the other pipe take to fill the pool on its own?. + x. We find that x 3. 7; therefore, it takes 3 hours and 4 minutes for the second pipe to fill the pool by itself.. If one inlet pipe can fill the pool in hours with the outlet drain closed, and the same inlet pipe can fill the pool in. hours with the drain open, how long does it take the drain to empty the pool if there is no water entering the pool? x. We find that x 0; therefore, it takes 0 hours for the drain to empty the pool by itself. 3. It takes 3 minutes less time to travel 0 miles by car at night than by day because the lack of traffic allows the average speed at night to be 0 miles per hour faster than in the daytime. Find the average speed in the daytime. 0 t 3 0 + t We find that t 80. The time it takes to travel 0 miles by car at night is 80 minutes, which is 3 hours. Since 0 40, the average speed in the daytime is 40 miles per hour. 3 Lesson 7: 308 This work is derived from Eureka Math and licensed by Great Minds. 0 Great Minds. eureka-math.org This file derived from ALG II-M-TE-.3.0-07.0
Lesson 7 M 4. The difference in the average speed of two trains is miles per hour. The slower train takes hours longer to travel 70 miles than the faster train takes to travel 0 miles. Find the speed of the slower train. 0 70 t t + We find that t 3, so it takes 3 hours for the faster train to travel 0 miles, and it takes hours for the slower train to travel 70 miles. The average speed of the slower train is 34 miles per hour.. A school library spends $80 a month on magazines. The average price for magazines bought in January was 70 cents more than the average price in December. Because of the price increase, the school library was forced to subscribe to 7 fewer magazines. How many magazines did the school library subscribe to in December? 80 x + 0. 70 80 x 7 The solution to this equation is. 0, so the average price in December is $. 0. Thus the school subscribed to 3 magazines in December.. An investor bought a number of shares of stock for $, 00. After the price dropped by $0 per share, the investor sold all but 4 of her shares for $, 0. How many shares did she originally buy? 00 0 x x 4 + 0 This equation has two solutions: 3 and 0. Thus, the investor bought either 3 or 0 shares of stock. 7. Newton s law of universal gravitation, F Gm m r, measures the force of gravity between two masses m and m, where r is the distance between the centers of the masses, and G is the universal gravitational constant. Solve this equation for G. G Fr m m 8. Suppose that x+y xy. a. Show that when x a a and y, the value of t does not depend on the value of a. a+ When simplified, we find that t ; therefore, the value of t does not depend on the value of a. b. For which values of a do these relationships have no meaning? If a is 0, then x has no meaning. If a, then y has no meaning. 9. Consider the rational equation R x + y. a. Find the value of R when x and y 3 4. So R 3. R + 3 4 Lesson 7: 309 This work is derived from Eureka Math and licensed by Great Minds. 0 Great Minds. eureka-math.org This file derived from ALG II-M-TE-.3.0-07.0
Lesson 7 M b. Solve this equation for R, and write R as a single rational expression in lowest terms. There are two approaches to solve this equation for R. The first way is to perform the addition on the right: R x + y y xy + x xy x + y xy. The second way is to take reciprocals of both sides and then simplify: R x + y y xy + x xy (x + y) xy. In either case, we find that R xy x+y. 0. Consider an ecosystem of rabbits in a park that starts with 0 rabbits and can sustain up to 0 rabbits. An equation that roughly models this scenario is 0 P +, t + where P represents the rabbit population in year t of the study. a. What is the rabbit population in year 0? Round your answer to the nearest whole rabbit. If t 0, then P 4. ; therefore, there are 4 rabbits in the park. b. Solve this equation for t. Describe what this equation represents in the context of this problem. 0 P t P 0 This equation represents the relationship between the number of rabbits, P, and the year, t. If we know how many rabbits we have, 0 < P < 0, we will know how long it took for the rabbit population to grow that much. If the population is 0, then this equation says we are in year 0 of the study, which fits with the given scenario. c. At what time does the population reach 0 rabbits? If P 0, then t 0 300 0 0 40 4; therefore, the rabbit population is 0 in year 4 of the study. 0 Lesson 7: 30 This work is derived from Eureka Math and licensed by Great Minds. 0 Great Minds. eureka-math.org This file derived from ALG II-M-TE-.3.0-07.0
Lesson 7 M Extension:. Suppose that Huck Finn can paint a fence in hours. If Tom Sawyer helps him paint the fence, they can do it in 3 hours. How long would it take for Tom to paint the fence by himself? Huck paints the fence in hours, so his rate of fence painting is fence per hour. Let T denote the percentage of the fence Tom can paint in an hour. Then fence (( + T) fence per hour) (3 hours). 3 + T + T 3( + T) T + T T So, Tom can paint of the fence in an hour. Thus, Tom would take 7. hours to paint the fence by himself.. Huck Finn can paint a fence in hours. After some practice, Tom Sawyer can now paint the fence in hours. a. How long would it take Huck and Tom to paint the fence together? The amount of fence that Huck can paint per hour is, and the amount that Tom can paint per hour is. So, together they can paint + of the fence per hour. Suppose the entire job of painting the fence takes h hours. Then, the amount of the fence that is painted is h ( + ). Since the entire fence is painted, we need to solve the equation h ( + ). h ( + ) h + 30 + 30 So, together they can paint the fence in 30 hours, which is hours and 44 minutes. Lesson 7: 3 This work is derived from Eureka Math and licensed by Great Minds. 0 Great Minds. eureka-math.org This file derived from ALG II-M-TE-.3.0-07.0
Lesson 7 M b. Tom demands a half-hour break while Huck continues to paint, and they finish the job together. How long does it take them to paint the fence? Suppose the entire job of painting the fence takes h hours. Then, Huck paints at a rate of of the fence per hour for h hours, so he paints h of the fence. Tom paints at a rate of of the fence per hour for h hour, so he paints (h ) of the fence. Together, they paint the whole fence; so, we need to solve the following equation for h: h + (h ) h + h h + 3 h 0 ( h + 3 h) 0 h + 0h h. Thus, it takes hours, which is hours 7 minutes, to paint the fence with Tom taking a hour break. c. Suppose that they have to finish the fence in 3 hours. What s the longest break that Tom can take? Suppose the entire job of painting the fence takes 7 hours, and Tom stops painting for b hours for his break. Then, Huck paints at a rate of of the fence per hour for 7 7 hours, so he paints of the fence. Tom paints at 0 a rate of of the fence per hour for (7 b) hours, so he paints (7 b) of the fence. Together, they paint the whole fence; so, we need to solve the following equation for b: 7 0 + (7 b) 7 0 + 7 b 0 ( 7 0 + 7 b ) 0 4 + 3 0b 0 4 + 3 0 0b b 7 0. Thus, if Tom takes a break for 7 0 hours, which is hour and 4 minutes, the fence will be painted in 3 hours. Lesson 7: 3 This work is derived from Eureka Math and licensed by Great Minds. 0 Great Minds. eureka-math.org This file derived from ALG II-M-TE-.3.0-07.0