course Mth 158 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I have to admit I had help setting this up. I did the table based on the example in the book, but I had to ask my husband how I would set up the formula. 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?
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. ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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: Boat heads upstream with current of 3 mph upstream trip 5 hours downstream trip 2.5 hours x = speed of boat 5(x-3) / 2.5( x+3) = if it took the boat 5 hours to travel upstream and the current was 3mph you would subtract 3 from the upstream and add 3 to the downstream time. 5x – 15 / 2.5x + 7.50 5x – 2.5x / 7.50 + 15 2.5x / 22.50 divide both sides by 2.5 x = 9 mph confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I have no idea how I came up with the answer 9 when I started out the problem with a division. I guess it should have been = since division at that point really doesn’t make sense.
Your expression
5(x-3) / 2.5( x+3) should have been the equation 5(x-3) = 2.5( x+3). Expressed in this form the steps of your solution would make sense. ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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: A pond is enclosed by a deck 3ft wide. the fence surrounding the deck is 100 ft long. a) if it is square, what are the dimensions b) if it is rectangular, and the length of the pond is to be three times its width, what are its dimensions c) if pond is circular, what is its diameter d) which pond has the most area I have tried for about an hour to search the book to help me understand how to solve these problems, but I have so much trouble understand them. I don’t know how to answer these. I will see my tutor this week, but it will be after I take the test. In the meantime, I will try to learn how to solve these types of problems on my own. confidence rating: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Even after reading the given solution, I don’t think I could solve the problem. I will continue to work on this for this week.
A detailed self-critique of your thinking on each line the given solution might be helpful. Feel free to submit one.
------------------------------------------------ Self-critique Rating: 0 ********************************************* 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: x = cement added to produce 40% concentration in 20lb bag 25 + x = 40/100 25% + x would give you 40% of 100% 100(25+x) = 100 (40) multiply both sides by 100 2500 + 100x = 4000 add -2500 to both sides 100x = 4000 – 2500 100x = 1500 divide both sides by 100 x = 15% 15% added to 25% = 40% cement confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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 have no idea if I did it right. I see how the given solution would work. I just gave it my best try to see if I could come up with the right mixture.
A detailed, line-by-line self-critique would be useful here.
------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 1.7.57 \ 52 (was 1.2.60). 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): without solving what's wrong with prob how many liters 48% soln added to 20 liters of 25% soln to get 58% soln? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: How many liters of 25% ethanol should be added to 20 liters of 48% ethanol to obtain a solution of 58% ethanol? .25 + .48 = 73% which is more than 58% confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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. **
&#Good responses. See my notes and let me know if you have questions. &#