Math 20 Take Home Quiz 12 Sections 17 And 18 Due Wednesday S
Math 20take Home Quiz 212sections 17 And 18due Wednesday Septemb
Math 20 Take-home Quiz Sections: 1.7 and 1.8 Due Wednesday, September 24th Correct answers without corresponding work will be given no credit. You may use a scientific calculator, but its use is unnecessary. Collaboration is permitted, but solutions must be your own.
1. Solve two of the following three problems. [21 2 points each]
- A jar of coins consists of only nickels, dimes, and quarters. The total value of the coins is $5.40. There are the same amount of nickels as quarters, and three more dimes than nickels. How many coins of each type are in the jar?
- A carpenter saws a board that is 21 feet long into three pieces. The center piece is 1 foot shorter than twice the length of the piece on the left, and the piece on the right is three feet longer than the center piece. Find the length of each piece.
- A couple is planning their wedding, and they have narrowed their section to two locations. One location costs $2,000 plus $35 per guest, and the other location is $1,350 plus $40 per guest. How many guests would have to be invited for both venues to be equally costly?
2. Solve two of the following three problems. [21 2 points each]
- Nine fluid ounces of a 15% alcohol solution are mixed with 11 fluid ounces of an 8% alcohol solution. What percent alcohol is the mixture?
- How many liters of a 1% glucose solution should a pharmacist mix with 0.5 liters of a 5% glucose solution to obtain a 2% glucose solution?
- Two marathon runners leave the starting gate, one running 12 mph and the other 10 mph. If they maintain the pace, how long will it take for them to be one-quarter of a mile apart?
Paper For Above instruction
In this paper, I will address the selected problems from the provided math quiz, solving two from each set and demonstrating the process and reasoning behind each solution. The problems involve a mix of algebra, geometry, and word problems requiring translation into algebraic equations. I will present each problem, outline the steps taken to solve it, and conclude with the answer, ensuring clarity and logical progression throughout.
Problem Set 1, Problem 1: Coins in the Jar
Given that the jar contains only nickels, dimes, and quarters, with the total value being $5.40, and the conditions that the number of nickels equals the number of quarters and there are three more dimes than nickels, we set up variables and equations accordingly.
Let N represent the number of nickels. Then, the number of quarters is also N, and the number of dimes is N + 3. The total value can be expressed as:
Value of nickels = 5 cents × N,
Value of quarters = 25 cents × N,
Value of dimes = 10 cents × (N + 3).
Summing these gives:
5N + 25N + 10(N + 3) = 540 cents.
Simplifying:
5N + 25N + 10N + 30 = 540,
which combines to:
40N + 30 = 540.
Subtracting 30 from both sides:
40N = 510,
Dividing both sides by 40:
N = 12.75.
Since the number of coins must be an integer, and this result indicates a fractional value, there is an inconsistency. Reevaluating the calculations, I realize that the correct total value in cents is 540, and the equations need to ensure integer solutions. Rechecking the calculations shows that the mistake was in the placement of values.
Actually, the total value in cents is 540, so summing the coin values: 5N + 25N + 10(N + 3) = 540.
Calculating step-by-step:
5N + 25N + 10N + 30 = 540, thus:
(5 + 25 + 10)N + 30 = 540,
which simplifies to:
40N + 30 = 540,
Subtract 30: 40N = 510,
Divide: N = 12.75.
Again, non-integer; perhaps the initial assumption needs rechecking. Alternatively, perhaps the total value was meant to be $5.40, which is 540 cents, confirming that calculation.
Considering that, N must be an integer, and thus, the original problem might be designed for an approximate solution or requires a re-examination. After careful analysis, the best estimate is that N=12.75, which isn't practical in real coins. Therefore, the solution suggests that the total coins are approximately 13 nickels and quarters, with 16 dimes (since 13 + 3 = 16). The total value then:\n
(13 × 5) + (13 × 25) + (16 × 10) = 65 + 325 + 160 = 550 cents or $5.50, slightly more than $5.40, indicating a small discrepancy likely due to rounding or the problem's constraints.
Problem Set 1, Problem 2: Sawn Board Lengths
The total length of the board is 21 feet, divided into three pieces: left (L), center (C), and right (R).
From the problem: C = 2L - 1 (the center is one foot shorter than twice the left), and R = C + 3 (the right is three feet longer than the center).
Additionally, the sum of the three segments equals 21 feet:
L + C + R = 21.
Substituting R and C in terms of L:
C = 2L - 1,
R = (2L - 1) + 3 = 2L + 2.
Adding all three:
L + (2L - 1) + (2L + 2) = 21.
Simplify:
L + 2L - 1 + 2L + 2 = 21,
Combine like terms:
(L + 2L + 2L) + ( -1 + 2) = 21,
5L + 1 = 21.
Subtract 1:
5L = 20,
Divide both sides by 5:
L = 4.
Now, find C and R:
C = 2(4) - 1 = 8 - 1 = 7 feet,
R = 2(4) + 2 = 8 + 2 = 10 feet.
The lengths are:
- Left piece: 4 feet
- Center piece: 7 feet
- Right piece: 10 feet
This sums to 21 feet, confirming the solution.
Problem Set 2, Problem 1: Alcohol Solution Mixture
Two solutions with different volumes and alcohol concentrations are mixed:
- 9 oz of 15% alcohol solution
- 11 oz of 8% alcohol solution
Calculate the total amount of alcohol in each solution:
Alcohol in first solution: 0.15 × 9 = 1.35 oz,
Alcohol in second solution: 0.08 × 11 = 0.88 oz.
Combined total alcohol: 1.35 + 0.88 = 2.23 oz.
Total volume after mixing: 9 + 11 = 20 oz.
The concentration of alcohol in the mixture is:
Percentage = (total alcohol / total volume) × 100 = (2.23 / 20) × 100 = 11.15%.
Thus, the mixture contains approximately 11.15% alcohol.
Problem Set 2, Problem 2: Glucose Solution Mixing
The pharmacist needs to mix a certain volume of 1% glucose solution with 0.5 liters of 5% glucose solution to get a 2% solution.
Let V represent the volume in liters of the 1% solution to be added.
The total amount of glucose in the 1% solution is 0.01 V liters, and in the 0.5 liters of 5% solution, it is 0.05 × 0.5 = 0.025 liters.
The total glucose after mixing: 0.01 V + 0.025 liters.
The total volume of the mixture: V + 0.5 liters.
We want the final concentration to be 2%, so:
Final glucose = 0.02 × (V + 0.5).
Set up the equation:
0.01 V + 0.025 = 0.02(V + 0.5),
0.01 V + 0.025 = 0.02 V + 0.01.
Subtract 0.01 V from both sides:
0.025 = 0.01 V + 0.01.
Subtract 0.01:
0.015 = 0.01 V.
Divide both sides by 0.01:
V = 1.5 liters.
Therefore, the pharmacist should mix 1.5 liters of the 1% glucose solution with 0.5 liters of the 5% glucose solution to obtain a 2% glucose solution.
Problem Set 2, Problem 3: Marathon Runners
One runner runs at 12 mph, the other at 10 mph. The problem asks for the time when they are one-quarter mile apart.
Let T be the time in hours when they are 0.25 miles apart.
The distance traveled by each runner after T hours is:
- Runner 1: 12 T miles
- Runner 2: 10 T miles
The difference in their distances is:
|12 T - 10 T| = 2 T miles.
Set this equal to 0.25 miles:
2 T = 0.25,
T = 0.125 hours.
Convert hours to minutes:
0.125 hours × 60 minutes/hour = 7.5 minutes.
Thus, it takes approximately 7.5 minutes for the runners to be a quarter mile apart.
Conclusion
Through solving these problems, we see the essential role of translating word problems into algebraic expressions, setting up equations based on the given conditions, and solving systematically. These problems also highlight how different areas of mathematics, such as algebra and geometry, play vital roles in solving real-world problems. Understanding how to construct and manipulate equations enables critical thinking and problem-solving skills applicable across various contexts.
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