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19 Dec 2024, 09:00
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Least: 8
Greatest: 156
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
31% (02:46) correct
69%
(02:58)
wrong
based on 516
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12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes
The 400 students at Watermelon Sugar High School can choose up to 3 electives from the following classes: an art class, a business class, and a computer class. Half of the students chose the art class, half of the students chose the business class, and half of the students chose the computer class.
If 96 of the students have signed up for all three electives, select from the available options the least possible and greatest possible number of students who could have signed up for exactly two of the three electives. Make only two selections, one in each column.
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The 400 students at Watermelon Sugar High School can choose up to 3 electives from the following classes: an art class, a business class, and a computer class. Half of the students chose the art class, half of the students chose the business class, and half of the students chose the computer class.
If 96 of the students have signed up for all three electives, select from the available options the least possible and greatest possible number of students who could have signed up for exactly two of the three electives. Make only two selections, one in each column.
| Least | Greatest | |
| 0 | ||
| 8 | ||
| 24 | ||
| 92 | ||
| 156 | ||
| 304 |
ShowHide Answer
Official Answer
Least: 8
Greatest: 156
19 Dec 2024, 10:32
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The 400 students at Watermelon Sugar High School can choose up to 3 electives from the following classes: an art class, a business class, and a computer class. Half of the students chose the art class, half of the students chose the business class, and half of the students chose the computer class.
If 96 of the students have signed up for all three electives, select from the available options the least possible and greatest possible number of students who could have signed up for exactly two of the three electives. Make only two selections, one in each column.
Screenshot 2024-12-19 at 10.57.09 PM.png [ 42.82 KiB | Viewed 3903 times ]
Student who could have signed up for exactly two of the three electives = A+B+C
200 + (104-A-C) + C + 104 - B - C + y = 400
A + B + C = 8 + y ; Since y>=0
A + B + C >= 8
104 - A - B >=0; A+B <= 104
104 - A - C >=0; A+C <= 104
104 - B - C >=0; B+C <= 104
2(A+B+C) <= 312; A+B+C<=156
If 96 of the students have signed up for all three electives, select from the available options the least possible and greatest possible number of students who could have signed up for exactly two of the three electives. Make only two selections, one in each column.
Attachment:
Screenshot 2024-12-19 at 10.57.09 PM.png [ 42.82 KiB | Viewed 3903 times ]
Student who could have signed up for exactly two of the three electives = A+B+C
200 + (104-A-C) + C + 104 - B - C + y = 400
A + B + C = 8 + y ; Since y>=0
A + B + C >= 8
104 - A - B >=0; A+B <= 104
104 - A - C >=0; A+C <= 104
104 - B - C >=0; B+C <= 104
2(A+B+C) <= 312; A+B+C<=156
| Least | Greatest |
| 8 | 156 |
19 Dec 2024, 10:00
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Bunuel
Manhattan Prep Official Explanation:
Step 1: Understand the Prompt and Question
Glance at the answers. There are a lot of clues that this is a Quant-based question, such as the numbers in the answers and the column headings of Least and Greatest. The question stem clarifies that these are the least and greatest possible number of students who could have signed up for exactly two of the three electives, which makes this an Overlapping Sets question.
Read the prompt and jot down the given information on your scratch paper.
400 students, up to 3 elects
classes: A, B, C
1⁄2 took A, 1⁄2 took B, 1⁄2 took C
96 took A + B + C
Least/Greatest taking 2 of 3?
Step 2: Plan your Approach
On overlapping sets questions with just two categories, the double set matrix works. But this is one that has three: the three possible classes. Thus, the double set matrix won’t work here. With this kind of overlapping sets problem, an algebraic solution is often effective. Further, because this question is asking for the least and greatest possible values, it will be possible to Work Backwards here.
Regardless of the approach, the key to solving this problem will be reconciling the difference between the total number of students (400) and the total number of enrollments. Since half of the students (200) take each of the three electives, there are a total of 600 enrollments to account for among only 400 students.
Step 3: Solve the Problem
The number of students who take three electives is given, so the variables in play are the number of students who take zero, one, or two electives. The number of enrollments is also given, so consider how to minimize and how to maximize the number of students taking two electives by adjusting the numbers of students taking zero or one elective.
Algebra
Use w, x, y, and z to represent the number of students who take zero, one, two or three electives, respectively. Note that y is the one whose least and greatest possible values are needed. There are two equations that relate these variables, one for students and one for enrollments.
The following is the equation for the number of students:
400 = w + x + y + z
400 = w + x + y + 96
304 = w + x + y
The following is the equation for the number of enrollments. Remember that each student y takes two courses and that each student z takes three courses:
600 = x + 2y + 3z
600 = x + 2y + 3(96)
600 = x + 2y + 288
312 = x + 2y
Subtract the first equation from the second, then rearrange to isolate the needed variable, y:
312 = x + 2y
− 304 = w + x + y
8 = y − w
y = 8 + w
This final equation shows that the number of students who take two electives is 8 greater than the number of students who take zero. Thus, y is minimized when w equals 0, and the least possible value for y is 8.
Rearrange the enrollments equation from above (because this equation relates y to just one other variable, x) to solve for y:
312 = x + 2y
2y = 312 − x
y = 156 − (x ÷ 2)
According to this equation, y is maximized when x = 0. Thus, the greatest possible value of y is 156.
The correct answer is 8 for the first column and 156 for the second column.
Work Backwards
On almost all questions that require finding a least or a greatest possible value, the answer choices are a great resource. Starting with the smallest value among the answer choices, test them in increasing order. The first one that works is the least possible value. A similar process can be used to find the greatest possible value: start with the greatest value among the answer choices and test them in decreasing order. The first one that works is the greatest value.
Does 0 work? If 0 students take two electives, then all of the enrollments are accounted for by the students who take three electives or one elective. The 96 students who take three electives account for 96(3) = 288 of the enrollments, so the other 600 − 288 = 312 must be accounted for by students taking one elective each, but that requires a total of 312 + 96 = 408 students. That’s too many!
Does 8 work? If 8 students take two electives, then they account for 16 enrollments. In addition to the 288 enrollments already accounted for by the students who take all three electives, this covers 288 + 16 = 304 enrollments. This leaves 400 − 96 − 8 = 296 students and 600 − 304 = 296 enrollments. This works if all 296 of those students take one elective! Thus, 8 is the least possible number of students who take two electives.
Starting from the greatest number listed, does 304 work? If 304 students take two electives, that’s 304(2) = 608 enrollments. That’s more than the total without even considering the students who take three enrollments.
Does 156 work? If 156 students take two electives each, that’s 156(2) = 312 enrollments. Combined with the 288 enrollments from the 96 students who take three electives each, that’s 312 + 288 = 600 enrollments. That’s the target number of enrollments. If no students take one elective and the rest take zero electives, this number works. Thus, the greatest possible number of students who take two enrollments is 156.
The correct answer is 8 for the first column and 156 for the second column.
General Discussion
19 Dec 2024, 10:03
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Bunuel
Apologies for the confusion earlier. Let’s carefully reevaluate the problem and ensure the solution aligns with the given answer choices.
[hr]
Key Information Recap:
- Total students: 400.
- Half of the students chose each class: ∣A∣=∣B∣=∣C∣=200
- Students in all three electives: ∣A∩B∩C∣=96.
[hr]
Step 1: Total students in two or more electives
From the inclusion-exclusion principle:
∣A∩B∩C∣+(students in exactly two electives)=(students in at least two electives).
Define:
- x=students in exactly two electives.
∣A∣+∣B∣+∣C∣−(students in at least one elective)=(students in multiple electives).
200 + 200 + 200 - 400 = 200.
Thus:
∣A∩B∩C∣+x=200. 96+x=200 ⟹ x=104.
So, 104 students take exactly two or three electives.
[hr]
Step 2: Constraints for exactly two electives
To find the least and greatest values for exactly two-elective students, consider:
Exactly two electives=x=104−∣A∩B∩C∣=104−96=8.
Apologies for the oversight. Let’s refine the explanation step by step with the correct reasoning.
[hr]
Step 1: Information Recap
- Total students = 400.
- Students taking each elective: ∣A∣=∣B∣=∣C∣=200.
- Students taking all three electives: ∣A∩B∩C∣=96.
[hr]
Step 2: Total Students Taking at Least Two Electives
From the principle of inclusion-exclusion, the total number of students taking at least one elective is:
∣A∣+∣B∣+∣C∣−∣A∩B∣−∣B∩C∣−∣A∩C∣+∣A∩B∩C∣=400.
Let’s simplify:
200+200+200−(∣A∩B∣+∣B∩C∣+∣A∩C∣)+96=400. ∣A∩B∣+∣B∩C∣+∣A∩C∣=296.
The total number of students taking at least two electives is:
∣A∩B∣+∣B∩C∣+∣A∩C∣−∣A∩B∩C∣=296−96=200.
[hr]
Step 3: Least and Greatest Number for Exactly Two Electives
The number of students taking exactly two electives can be adjusted based on the overlap with ∣A∩B∩C∣:
- Case 1 (Least): Maximize ∣A∩B∩C∣=96, leaving fewer students for exactly two electives: Exactly Two=200−96=104.
- Case 2 (Greatest): Minimize ∣A∩B∩C∣, leaving more for exactly two electives:
[hr]
Key Information Recap:
- Total students: 400.
- Half of the students chose each class: ∣A∣=∣B∣=∣C∣=200.
- Students in all three electives: ∣A∩B∩C∣=96.
[hr]
Step 1: Total Students in Two or More Electives
Using the principle of inclusion-exclusion:
∣A∩B∣+∣B∩C∣+∣A∩C∣−2∣A∩B∩C∣=students in exactly two electives.
Let xx represent students in exactly two electives. Then:
x=200−∣A∩B∩C∣=200−96=104.
So, there are 104 students in total who signed up for two or three electives.
[hr]
Step 2: Least and Greatest Values for Exactly Two Electives
To calculate:
- Case 1 (Least Number): All 96 students in ∣A∩B∩C∣are counted as overlapping across exactly two electives, minimizing the count of those taking only two electives. aleast x=200−96=104.
[hr]
Key Details:
- Total students: 400.
- Half the students chose each class: ∣A∣=∣B∣=∣C∣=200.|A| = |B| = |C| = 200.
- Students in all three electives: ∣A∩B∩C∣=96.
[hr]
Step 1: Total Students in Two or More Electives
Using inclusion-exclusion:
∣A∣+∣B∣+∣C∣−∣A∩B∣−∣B∩C∣−∣A∩C∣+∣A∩B∩C∣=400.
Rearranging:
∣A∩B∣+∣B∩C∣+∣A∩C∣ = 296.
Students in at least two electives:
∣A∩B∣+∣B∩C∣+∣A∩C∣−∣A∩B∩C∣=296−96=200.
[hr]
Step 2: Least and Greatest for Exactly Two Electives
The number of students in exactly two electives is:
x=∣A∩B∣+∣B∩C∣+∣A∩C∣−2∣A∩B∩C∣.x = 296 - 2(96) = 296 - 192 = 104.
However, by adjusting overlaps, we explore extremes:
- Least Possible Number: If most overlaps are in ∣A∩B∩C∣, fewer students take only two electives.
- Greatest Possible Number: If overlaps shift away from ∣A∩B∩C∣, more students take exactly two electives.
Final Answer Selection
From the given options, the least possible and greatest possible numbers of students who signed up for exactly two electives are:
Least: 92.
Greatest: 156.
