#$&* course MTH 158 015. `* 15
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Given Solution: * * Good. The details: If x is the amount lent at 16%, then the amount lent at 19% is 1,000,000 - x. Interest on x at 16% is .16 x, and interest on 1,000,000 - x at 19% is .19 (1,000,000 - x). This is to be equivalent to a single rate of 18%. 18% of 1,000,000 is 180,000 so the total interest is 1,000,000. So the total interest is .16 x + .19(1,000,000 - x), and also 180,000. Setting the two equal gives us the equation .16 x + .19(1,000,000 - x) = 180,000. Multiplying both sides by 100 to avoid decimal-place errors we have 16 x + 19 ( 1,000,000 - x) = 18,000,000. Using the distributive law on the right-hand side we get 16 x + 19,000,000 - 19 x = 18,000,000. Combining the x terms and subtracting 19,000,000 from both sides we have -3 x = 18,000,00 - 19,000,000 so that -3 x = -1,000,000 and x = -1,000,000 / (-3) = 333,333 1/3. ** STUDENT QUESTION I did the table based on the example in the book, but I just can’t seem to get a grasp on how to set word problems up in a formula. Is there anything I could study to help me get a better understanding of this? INSTRUCTOR RESPONSE The problem states that there is $1 million to lend. Some can be lent at 16%, some at 19%. The goal is to average an 18% return. We will use the following procedure: State in words each quantity we wish to find. Pick a quantity to represent with the variable (typically the variable is x, but if it's helpful we can use any other symbol for the variable), and state what the quantity is and what the variable is. Write every quantity relevant to the problem in terms of x. State what is equal to what (look for two different ways to express the same quantity). Write an equation to represent the equality. Solve the equation for the unknown. Find the value of every quantity which depends on the unknown. Make sense of your results. 1. Stating the quantities in words: We want to find the amount lent at 16%, and the amount lent at 19%. 2. Picking the quantity equal to the variable: We could let the variable x represent either of these quantities. Let's choose to let x represent the amount lent at 16%. 3. Listing relevant quantities: The quantities relevant to the problem are the $1 million we have to lend, the unknown amount lent at 16%, the unknown amount lent at 19%, and the return We have already decided to let the amount lent at 16% be represented by x. The entire $1 million is lent at one of the two rates, so the amount lent at 19% will be $1 million - x. The return on the amount x lent at 16% is 16% of that amount, i.e., 16% of x, or .16 x. The return on the amount x lent at 19% is 19% of that amount, i.e., 19% of ($1 million - x), or .19 ( $1 million - x). The total return is therefore .16 x + .19 ( $l million - x) The total return is to be 18% of the original $1 million, so the total return is $180 000. 4. What is equal to what? total return.= return on amount lent at 16% + return on amount lent at 19% 5. Write an equation We have two expressions for the total return. One expression is $180 000. The other is .16 x + .19 ( $1 million - x). Setting these equal gives us our equation: .16 x + .19 ( $1 million - x) = 180 000 6. The equation has already been solved in the given solution. We get x = $333 333 1/3. 7. The amount invested at 19% is $1 million - x = $1 million - $333 333 1/3 = $666 666 2/3. 8. We make sense of these numbers as follows: 16% * $333 333 1/3 = $53 333 1/3. 19% * $666 666 2/3 = $126 666 2/3. So the return is $53 333 1/3 + $126 666 2/3 = $180 000. $180 000 is 18% of $1 million, which verifies our solution. STUDENT QUESTION I do not understand this problem. I don’t understand why the 18% is coming into play. INSTRUCTOR RESPONSE If you lend all the money at 19%, you get a 19% return (assuming you are repaid as planned). If all the money is lent at 16%, you get a 16% return. If you lend part at 19% and part at 16%, your return will be between 16% and 19%. Is wasn't stated, but presumably the 16% loan is less risky for you that the 19% loan. So you wouldn't automatically lend all your money at 19%. Now, presumably you need an 18% return. You can get this by lending part of the money at 16%, and part at 19%. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: * 1.7.26 \ 36 (was 1.2.36). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) 3 mph current, upstream takes 5 hr, downstream 2.5 hr. Speed of boat? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I couldn’t figure this one out. I read the close examples and worked them with the aid of the book, but they answered the problem for the speed of the boat not the speed of the current. I know it is likely going to be one of those problems where you simply set up a formula for the speed of the current. That seems to have always been my weak point in math. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * STUDENT SOLUTION: Speed of the boat is 9 mph, I used the equation 5(x - 3) = 2.5(x + 3) Reasoning is that it took 5 hours for the boat to travel against the 3mph current, and then traveled the same distance with the 3mph current in 2.5 hours. INSTRUCTOR COMMENT: Good. The details: If we let x be the water speed of the boat then its actual speed upstream is x - 3, and downstream is x + 3. Traveling for 5 hours upstream, at speed x - 3, we travel distance 5 ( x - 3). Traveling for 2.5 hours downstream, at speed x + 3, we travel distance 2.5 ( x + 3). The two distance must be the same so we get 5 ( x - 3) = 2.5 ( x + 3) or 5 x - 15 = 2.5 x + 7.5. Adding -2.5 x + 15 to both sides we get 2.5 x = 22.5 so that x = 22.5 / 2.5 = 9. So the water speed is 9 mph. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This problem is actually a lot easier than the one in the book. The solution makes sense to me but I don’t know if I would’ve ever thought to do it this way. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: * 1.7.36 \ 32 (was 1.2.42). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) pool enclosed by deck 3 ft wide; fence around deck 100 ft. Pond dimensions if pond square, if rectangular 3/1 ratio l/w, circular; which pond has most area? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I drew diagrams for this problem and got that the decks dimensions would be 25’ x 25’ giving us a 100’ perimeter for a 100’ long deck. Subtracting 3’ from all sides of the square for the deck I got a 19’ x 19’ pool which is 361’ For the circle I plugged 100’ into the circumference formula and reverse engineered the problem to find the diameter to be 31.85’. 31.85 -6 = 25.85 so the diameter of the pool is 25.85’. A=πr^2 so A = π12.925^2 A = π167.055625 which is roughly 524’. I couldn’t figure out the rectangular pool. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * If the deck is circular then its circumference is C = 2 pi R and its radius is r = C / (2 pi). C is the 100 ft length of the fence so we have R = 100 ft / ( 2 pi ) = 50 ft / pi. The radius of the circle is 3 ft less, due to the width of the deck. So the pool radius is r = 50 ft / pi - 3 ft. This gives us pool area A = pi r^2 = pi ( 50 / pi - 3)^2 = pi ( 2500 / pi^2 - 300 / pi + 9) = 524, approx.. If the pool is square then the dimensions around the deck are 25 x 25. The dimensions of the pool will be 6 ft less on each edge, since each edge spans two widths of the deck. So the area would be A = 19 * 19 = 361. The perimeter of the rectangular pool spans four deck widths, or 12 ft. The perimeter of a rectangular pool is therefore 12 ft less than that of the fence, or 100 ft - 12 ft = 88 ft. If the pool is rectangular with length 3 times width then we first have for the 2 l + 2 w = 88 or 2 (3 w) + 2 w = 88 or 8 w = 88, giving us w = 11. The width of the pool will be 11 and the length 3 times this, or 33. The area of the pool is therefore 11 * 33 = 363. The circular pool has the greatest area, the rectangular pool the least. ** STUDENT QUESTIONS WITH INSTRUCTOR RESPONSES: a) if it is square, what are the dimensions (I don't understand what dimensions it is looking for. Is it asking for the pond dimensions? or the deck dimensions? And either one, I would not even begin to know how to solve. I have drawn a pond on my paper and put a 3 ft diameter around it with a fence that is 100 ft long. Is it 100ft all the way around? or is it 100ft on one side? I just don't understand word problems.) The fence around the deck is 100 ft long. If the pond is square, then the 3-ft-wide deck around it is also square, and so therefore is the fence. What are the dimensions of a square with a perimeter of 100 ft? What would be the dimensions of the pond? b) if it is rectangular, and the length of the pond is to be three times its width, what are its dimensions (Same thing here, I don't even know what it is asking to even set up the problem to solve it) Sketch a picture of a rectangle which is 3 times as long as wide, labeling the width w and the length 3w. Sketch a 3 ft wide deck around the rectangle. In terms of w, what are the length and width of this rectangle? In terms of w, what is the perimeter of this rectangle? Set your expression for the perimeter equal to 100 ft and solve for w. What therefore are the length, width and area of the pond? c) if pond is circular, what is its diameter (I know to find the diameter of a circle, it is 2 times the radius, but I don't know what the radius of this pond is or how to find it.) You should know that the circumference of a circle is 2 pi * radius or pi * diameter. Whichever you prefer to use, set it equal to 100 ft and solve for either radius or diameter, as appropriate. Having found the radius or diameter of the fence, what is the radius or diameter of the pond, taking into consideration the 3 ft deck? What therefore is the area of the pond? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I see how to do the rectangular pool now. I started to do it that way but something was telling me that it was incorrect. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: * 1.7.68 \ 44 (was 1.2.54). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) 20 lb bag 25% cement 75% sand; how much cement to produce 40% concentration? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I first put 5lbs/25% as this is the percentage of cement in the bag. 5lbs/.25 + xlbs/.15 would yield 40% concrete. So 6.67lbs of concrete would equal 40% concrete. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * If x stands for the amount of cement added then we have the following: Original amount of cement in bag is 25% of 20 lb, or 5 lb. Original amount of sand in bag is 75% of 20 lb, or 15 lb. The final amount of cement will therefore be 5 lb + x, the final amount of sand will be 15 lb and the final weight of the mixture will be 20 lb + x. The mix has to be 40%, so (amt of cement) / (total amt of mixture) = .40. This gives us the equation (5 + x) / (20 + x) = .40. Multiplying both sides by 20 + x we have 5 + x = .40 ( 20 + x ). After the distributive law we have 5 + x = 80 + .40 x. Multiplying by 100 we get 500 + 100 x = 800 + 40 x. Adding -40 x - 500 to both sides we have 60 x = 300 so that x = 300 / 60 = 5. We should add 5 lbs of cement to the bag. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I worked that problem totally wrong. I tried to think about it too much rather than put it into a word problem. Again, word problems are my weakest point and when I come to a situation where I have to have one I try to reason it out instead. ------------------------------------------------ Self-critique Rating:2
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Given Solution: * * Solution from Previous Student and Instructor Comment: It's not possible, adding a 25% solution to a 48% solution is only going to dilute it, I don't really know how to prove that algebraically, but logically that's what I think. (This is much like the last problem, that I don't really understand). INSTRUCTOR COMMENT: Right but the 48% solution is being added to the 25% solution. Correct statement, mostly in your words Adding a 48% solution to a 25% solution will never give you a 58% solution. Both concentrations are less than the desired concentration. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): What would be the equation for this problem? ------------------------------------------------ Self-critique Rating:1