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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 19 Dec 2024, 13:55
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Now we are told Total students are 400
and given the equation T = G1 + G2 +G3 - (G1G2 + G1G3 + G2G3) + None


Maximum number of students from 2 electives can be 400 - 96 = 304, assuming the rest of the students are at least in two electives at a time.
On the other side, the minimum no of students from 2 electives can be 400 = 200 + 200 + 200 -2*96 -x + 0 -> x = 8 students are on 2 electives at a time.

Worth noting that, as we are told they are students, there can't be members from no groups, thus none = 0
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 19 Dec 2024, 14:09
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Quote:
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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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 19 Dec 2024, 14:45
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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.
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600 = Art + Business + Computer - (Art + Business) - (Art + Computer) - (Business + Computer) - 2 (Art + Business + Computer)

400 = 200 + 200 + 200 - (Art + Business) - (Art + Computer) - (Business + Computer) - 2 * 96

(Art + Business) - (Art + Computer) - (Business + Computer) = 200 - 192

(Art + Business) - (Art + Computer) - (Business + Computer) = 8

This is the least possible value

600 = One Course + 2 * (Two Course) + 3 * (Three courses)

For maximum overlap, One Course should be zero.

600 - 3(96)/2 = Two Course

300 - 144 = Two Course

Two Course = 156

This is the maximum Overlap

Least = 8
Greatest = 156
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 19 Dec 2024, 18:45
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x=sum of EXACTLY 2−group overlaps
n=Neither

Total=A+B+C−(sum of EXACTLY 2−group overlaps)−2∗(all three)+Neither

400 = 200 + 200 + 200 - x - 2*96 + n

x = 8 + n

So x must be 8 at most.
¿Can it be 0 at least as one of the answers says?.
No, because if it were 0, then onlyA=104, onlyB=104 and onlyC=104. And the students would be 104+104+104+96=408, not 400 as it is said.

Least=8 and Greatest=8
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 19 Dec 2024, 20:22
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400 students = (# art) + (# business) + (# computer) - 2 x (# students choosing all 3) - (# students choosing only 2)

400 = 200 + 200 + 200 - 2 x 96 - (# students choosing only 2)
(# students choosing only 2) = 8

Answer: 8 for both greatest and least
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 19 Dec 2024, 23:12
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Explanation:

We need to find the minimum and maximum number of students who signed up for exactly two of the three electives (Art, Business, Computer).

Given:
  • Total Students: 400
  • Students in each elective: 200 in Art, 200 in Business, 200 in Computer
  • Students in all three electives: 96
Finding Exactly Two Electives:
  1. Total Selections:
    • Each elective has 200 students.
    • Some students are counted in multiple electives.
  2. Using the Principle of Inclusion-Exclusion:
    Total = ∣A∣+∣B∣+∣C∣− (Exactly Two) − 2 × (All Three)
    400 = 200 + 200 + 200 − (Exactly Two) − 2 × 96
    400 = 600 − (Exactly Two) − 192
    Exactly Two = 600 − 192 − 400 = 8
    • Minimum Number: 8 students
  3. Maximum Number of Exactly Two Electives:
    • To maximize, and minimize the number of students taking only one elective or none.
    • The highest possible number of students taking exactly two electives is 156.
Final Selections:
  • Least: 8
  • Greatest: 156
Answer:
  • Least: 8
  • Greatest: 156
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 20 Dec 2024, 00:15
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Least and greatest are both 8

We know there are 400 students in total at the school.
We know half are in art (A), business (B), and computer (C) = 200 students each.
If 96 are in A, B, and C, then there are 104 in A, B, and C who can be in 1 or 2 classes.
The combos of 2 classes are A&B, A&C, and B&C.

To solve for those in 2 classes: 96 + x - 2x + 3*104 = 400 --> 408 - x = 400 --> x = 8

Or, let's consider the options:

If 0 students are in 2 classes, then we have 96 + 3*104 = 96 + 312 = 408 unique students which doesn't equal 400 total students.

If 8 students are in 2 classes, let's assume all 8 are in the same 2 classes, such as A&B. Then, we have 96 + 8 + (2*(104 - 8)+104) = 96 + 8 + 2*96 + 104 = 400 unique students which equals 400 total students. Let's now assume there are 4 in A&B and 4 in A&C. Then, we have 96 + 24 + ((104 - 8)+2*(104-4)) = 96 + 8 + 96 + 2*100 = 400 unique students again. This is the solution.

If 24 students are in 2 classes, let's assume all 24 are in the same 2 classes, such as A&B. Then, we have 96 + 24 + (2*(104 - 24)+104) = 96 + 24 + 2*80 + 104 = 384 unique students which doesn't equal 400 total students. Let's now assume there are 8 in each of the 2 combos of classes. Then, we have 96 + 24 + (3*(104 - 16)) = 96 + 24 + 3*88 = 384 unique students again which doesn't equal 400 total students.

If 92 students are in 2 classes, let's assume all 96 are in the same 2 classes, such as A&B. Then, we have 96 + 92 + (2*(104 - 92)+104) = 96 + 8 + 2*80 + 104 = 368 unique students which doesn't equal 400 total students.

If 156 students are in 2 classes, they must be in all 3 combos are 2 classes, so let's assume they're split evenly. Then, every student who is not in 3 classes is in 2 classes, so we have 96 + 156 + (3*(104-104)) = 96 + 156 = 252 unique students which doesn't equal 400 total students.

It is not possible for 304 students to be in 2 classes because we already know 156 is the maximum number of unique students in 2 classes.
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 20 Dec 2024, 00:23
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Three circles intersect: Then there are students attending one (O), two(T), three(Th) or none courses. Overall: 400.
Art: O+T+Th=200
Business: O+T+Th=200
Computer: O+T+Th=200
None:N
Add all: O1+O2+O3+T1+T2+T3+3Ths+N=600
T+2Ths+N=200 (as 600-400)
Ths=96
T+N=200-162
T+N=8.
Greatest T=8, least T=0
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 20 Dec 2024, 00:23
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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.
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Let A = number of student taking art class
Let B = number of student taking business class
Let C = number of student taking computer class
Let A∩B= students taking both art and business.
Let A∩C= students taking both art and computer.
Let B∩C= students taking both business and computer
Let x = number of student choosing exactly two electives = A∩B+A∩C+B∩C
Let A∩B∩C= students taking art, business and computer class

A = B = C = 400/2 = 200

For at least two of the three electives
400 = A+B+C-x-A∩B∩C
400 = 200+200+200-2x+x-96*3+96
400 = 600-288+96-x
400 = 408-x
x = 8

Therefore at least 8 students have exactly two of the three electives
------------------------------------------------------------------------
For at most two of the three electives
Let y = number of student choosing only 1 elective

400 = y+A∩B∩C
400 = y+96
y = 304
Therefore at most 304 students have exactly two of the three electives
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 20 Dec 2024, 00:34
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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

Minimise:
When we put 104= A only , 96= exact 3, 96=B only, 96= C only, we would need min 8 = B and C only for total to be 400

Maximise:
We can assume none also to be there
In venn, we first put 96 for exact 3, then we have 304 remaining and 104 for each class. We try to overlap the remaining circles such that 52= B and A only , 52= A and C only. Now we have 400-104-96= 200 people remaining. We overlap remaining people such that B and C only =52. Thus exact 2 together =52*3=156

Ans 8, 156
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 20 Dec 2024, 00:51
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We can draw a simple venn diagram for all three groups of students (as attached). However, we need to remember that there's some X number of students that (potentially) selected none of the suggested courses, as the task doesn't say choosing one was mandatory.

Therefore, the total number of students can be defined through a simple 3-circle venn formula but with the addition of X:
\(400 = a+b+c-(ab+bc+ac)-2abc +x = (200+200+200) -(ab+bc+ac) - 2*96 + x = 600 - 192 -(ab+bc+ac) +x = 408 -(ab+bc+ac) + x\)
Therefore, what we're looking for is equal to:
\((ab+bc+ac)=408-400+x = x+8\)

  • So, if none of the students went without electives, \( x=0 \) and \((ab+bc+ac)=8\)

However, the maximum number of students not selecting anything is:
\(96+x+ (ab+bc+ac)<=400,\) so \(96 + 2x+8<=400\) and \(2x<=296\)
So \(x<=148\), as we know for a fact that 96 people chose 3 subjects and a certain amount of people chose only 2.
  • Therefore, max X value is 14, and max \((ab+bc+ac) = 148+8=156\)

The answers are: 8 and 156.
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 20 Dec 2024, 00:53
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 20 Dec 2024, 02:52
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We can visualize this problem as a Venn like diagram. If we assume the distinct subsets as:
A-Only Art
B-Only Business
C-Only Computer
D-Art & Business Only
E-Business & Computer Only
F- Art & Computer Only
G-All Three

Define the equations as follows given the conditions:
A+B+C+D+E+F+G=400 - Statement 1
A+D+F+G=200 - Statement 2 (Art has half the students)
B+D+E+G=200 - Statement 3 (Business has half the students)
C+E+F+G=200 - Statement 4 (Computer has half the students)

Add statements 2,3 & 4 together & subtract statement 1 from the combined sum, we get
D+E+F+2*G=200
G is equal to 96 as 96 people joined all three classes (Fixed)
Hence the highest value for D+E+F is 8 (200-192) & lowest value is 0 as the remaining 8 are or can not be a part of any of the two clubs only .

Highest Value=8
Lowest Value=0
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 20 Dec 2024, 03:36
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Hi everyone :)

This is an Ovelap question with Venndiagram.
We will use the equation (also you can illustrate the Venndiagram).

Total = A+B+C - D - 2E + None
400 = 200+200+200 - D - 2*96 + N
400 - 408 = -D+N
D = 8 + N

Max of D: (Working Backward)
if N = 296 D = 304
Then (we can take one of the classes for example):
Art class (200) = 96 + 101 + 101. (101 it's approx 304/3: i divided the number into 3: 101,101,102)
and it's impossible because then Art become negative.
if N = 148 D = 156
Then: Art class (200) = 96 + 52 +52 => Making Only who study in Art class as 0 students (52 it's 156/3)
Thus: the number of Max D is 156

Min of D:
if N = 0 D = 8
Art class (200) = 96 + 3 + 3 = 102 so Only A is 98.
Computer class (200) = 96 + 2 + 3 = 101 so Only C is 99.
Business class (200) = 96 + 2 + 3 = 101 so Only B is 99.
All together is 400, and None is 0.
Least D is 8.
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 20 Dec 2024, 03:55
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Let ONLY Art class students be a, ONLY Business class students be b, and ONLY computer class students be c.
Let the students common to Art and Business classes be d.
Let the students common to Art and Computer classes be e.
Let the students common to Computer and Business classes be f.

a+d+e+96 = 200
b+d+f+96 = 200
c+e+f+96 = 200

adding all three equations,

(a+b+c) + 2(d+e+f) = 312

Now, (d+e+f) is what we're looking for, hence, we can replace this with x.

(a+b+c) + 2x = 312

Now, we can put in options in place of x to get our answers quickly.

For least, if we put in x=0, a+b+c = 312
if a+b+c = 312 and common to all are 96, this total exceeds 400.
Hence, 0 is not the least.

if we put in x=8
a+b+c = 312 -16 = 296

If a+b+c = 296, and common to all is 96,

296+96 = 392
we put in x = 8, then,
392+8 = 400.

Therefore, least = 8.

For greatest, let's put x=304, then,

(a+b+c) + 2*304 = 312
(a+b+c) = 312 - 608 = -296

(a+b+c) can't be negative, hence, x cannot be 304.

Put x = 156,
(a+b+c) = 312 - 2*156
a+b+c = 0

156 +96 = 252, but total students are 400. therefore, x cannot be 156.

Put in x = 92,
(a+b+c) = 312 - 2*92 = 128

128+96+92 = 316, therefore, x cannot be 92.

Put x = 24,
(a+b+c) = 312 - 2*24 = 264

264 +96+24 = 384 , therefore, x cannot be 24.

We know that x = 8 satisfies the condition, hence,

x=8 is the greatest as well.

Final Answer - 8,8




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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.
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 20 Dec 2024, 04:19
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Let Art class be A, Business class be B, and Computer class be C

The 400 students can choose up to 3 electives. Since the question does not mention anything about any student not taking up an elective,
Students who have chosen neither electives are 0.

Half of the students chose the art class --> A = 200
Half of the students chose the business class --> B = 200
Half of the students chose the computer class --> C = 200

Students signed up for all 3 electives (A n B n C) = 96

Using formula,
Total = A+B+C - (Sum of exactly 2 group overlaps) - 2*(A n B n C) + Neither
400 = 200+200+200 - x - 2*96 + 0, we get
x = 8

This means that a maximum of 8 students can choose exactly 2 electives out of 3.
Minimum number of students who will opt for exactly 2 electives out of 3 will be 0.

Hence the answer is:
Minimum = 0
Maximum = 8
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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.
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 20 Dec 2024, 04:19
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Using the formula:

total = art + business + computer - exactlytwo - 2*three + neither

400 = 200 + 200 + 200 - exactlytwo - 2*96 + neither

exactlytwo = 8 + neither

If neither=8, exactlytwo=0, students in only art=104 (to be 200 adding 96), students in only business=104 (to be 200 adding 96) and students in only computer=104 (to be 200 adding 96). But then the number of students=416(104+104+104+96+8), what is impossible. So, among the answer options, only 8 in both answers works.

Answers Least=8, Greatest=8
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 20 Dec 2024, 04:41
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Let the students who have taken only art be a, only business be b & only computer be c
Students that took art & business but not computer be d
Students that took business & computer but not art be e
Students that took art & computer but not business be f
Students who took all the subjects be g

So, a + d + f + g = 200
b + d + e + g = 200
c + e + f + g = 200
g = 96

we need, e + g + f (min & max)

i) minimum can be 0 as a + b + c could equal 304

ii) maximum -
400 = a + b + c + d + e + f + g
400 - 96 = a + b + c + d + e + f = 304.... (a)

We know that : a + b + c + 2 (e + f + g) + 3g = 600
i.e. a + b + c + 2 (e + f + g) = 600 - 3*96= 600 - 288 = 312.... (b)

Subtracting a & b, we get -
e + f + g = 312 - 304 = 8

Hence, 0 & 8

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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.
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 20 Dec 2024, 05:44
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In TPA questions its always better to form a strategy first before solving. Well this is a detail bulky question.

The stem tells us that;

  • 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.
  • 96 of the students have signed up for all three electives

And we have to select;

  • greatest possible number of students who could have signed up for exactly two of the three electives
  • least possible number of students who could have signed up for exactly two of the three electives

Look at the venn diagram below to substantiate your understanding;

Using the basic formula for solving venn diagram;

(200+200+200)-(a+b+c ) -2(96)=400
or
a+b+c = 8 (least value)

For finding the maximum value we'll have to assume that none of the students are enrolled in a single class only.
I.e. the vacant region of the venn diagram = 0

(0+0+0)+(a+b+c)+96=400
or
a+b+c=304

Hence we can fill the respective options here.
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 20 Dec 2024, 06:17
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Given : total students registered for any of the 3 courses is 200.

Students taking 3 electives : 200
Let's assume all students have signed up for atleast 1 class.

Students signed for 1 class + students signed for 2 classes + students signed for 3 classes = 400.

Let's assume : x+ y + z = 400.
Given z = 96. Hence x + y = 304.

Total seats = 600. If 96 students have registered for all 3 courses. Remaining seats = 600-(96*3) = 312.
Assuming no student is registered for a single course, divide 312/2 , giving 156 students choosing 2 classes.

Hence, greatest possible no. of students signed up for 2 electives is 156.

Now for the least, assume that all students have chosen single elective. That gives 104 students per elective (since 96 have chosen all 3 and 200-96 = 104).
Adding for 3 electives, total becomes 104*3 = 312, but x+y = 304. The values cannot exceed 304. Hence we can say that 312-304 students have to choose 2 classes , which is 8.

Hence, lowest possible no. of students signed up for 2 electives is 8.

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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.
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