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805+ Level|   Math Related|         
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__Poisonivy__
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 20 Dec 2024, 06:19
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Let,
A= Students taking art = 200
B= Students taking business =200
C= Students taking comp = 200
AB=Students taking art and business but not computer
BC=Students taking business and computer but not art
CA=Students taking art and computer but not Business.
ABC= Students taking all 3= 96.

A+B+C- (AB+ BC+ CA+ 2ABC)= 400
=> 3x 200 - (AB+ BC+ CA+ 2x 96)= 400
=> 600- 192 -(AB+ BC+ CA)= 400
=> AB+ BC+ CA= 8. least possible value

to maximize AB+ BC+ CA,

AB+ BC+ CA= 400-92= 304
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 20 Dec 2024, 06:44
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Art(A) = 200
Business(B) = 200
Computer(C) = 200

Total number of students = 400

Total number of students in different electives = 200 + 200 + 200 = No of students chosen 1 elective only(One_E) + 2 * No of students have chosen two electives (Two_E) + 3 * No of students have chosen three electives(Three_E)

Given Three_E = 96

Total number of students = One_E + Two_E + Three_E

600 = 3 * 96 + 2* Two_E + One_E ---- (1)
400 = Two_E + One_E + 96 -----(2)

One on value for Two_E satifies these equations where Two_E = 8
This is the greatest and least possible value for Two_E



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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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aarati17
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 20 Dec 2024, 07:15
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Ans: Least possible value = 8 and greatest possible number = 8

We know,
Total number of students = 400
Total number of students who signed up for all electives = 96
Total number of students in each elective = total number of students/2 = 400/2 = 200

Since the number of students in all three electives is give and we cannot change that, there will be only one value for total number of students in 2 electives i.e., the maximum and the minimum value will be the same

Let,
Total number of students who selected only 1 elective = a
Total number of students who selected 2 electives = b
Total number of students who selected all 3 electives = c

We know:
a + b + c = 400

Additionally,
In a Venn diagram of 3 intersecting circles, since the common area intersecting all three circles is counted thrice and the area where only 2 circles meet is counted twice, we can write an equation:
a + 2b + 3c = 200+200+200 =600

We know c= 96. So, substituting the value of c:

a+2b=312 ----------------------Equation(1)
a+b=304 -----------------------Equation (2)

Subtracting equation (2) from equation (1)
b=312-304 = 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, 08:15
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Using the overlapping set formula Total students = Arts+Business+Computer - those who took exactly two - 2 X all the three classes+ Neither

So 400= 600-x-2x196
X=8 Max and Minimum is zero
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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, 08:21
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Bunuel
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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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Wow great question,

We know that total(t) = 400.

Lets call the 3 sets with A,B,C.

y is the #students who didn't choose any of A,B,C.

n(AUBUC) = n(A)+n(B)+n(C) - n(A^B) -n(B^C) -n(A^C) +n(A^B^C)
and
n(AUBUC)= t - Y => 400-Y


What are we looking for number of students who could have signed up for exactly two of the three electives (Lets call this x)
x=> n(A^B)+n(A^C)+n(B^C) - 3n(A^B^C).
So n(A^B)+n(A^C)+n(B^C) = x+3*96

So 400 - y = n(A)+n(B)+n(C) - n(A^B) -n(B^C) -n(A^C) +n(A^B^C)
400 - y = 600 - (x+3*96)+96
=> x = y+8.

So y can be 0, Hence minimum x is 8.

To get x maximum.
y maximum is 400 - (96+52*3) = 148.

So x Maximum is 156.

IMO 8,156.
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 20 Dec 2024, 09:00
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Total students = 400
Students who signed up for all three electives = 96


A = students who chose art = 400/2 = 200
B = students who chose business = 400/2 = 200
C = students who chose computer = 400/2 = 200
X = number of students who chose exactly two electives
Y = number of students who chose exactly one elective
Z = number of students who chose all three electives = 96

Total number of students who signed up for at least one elective = A+B+C−X−2Z

Now for least case:
Total Students Signed Up for At Least One Elective = A+B+C−X−2Z
=> A+B+C−X−2Z=400
=> 200+200+200−X−2(96)=400
=> X = 8


Now for greatest case:
=> 400=Y+X+Z
=> 400=Y+X+96
=> Y+X = 304
=> X = 304 (Y = 0 since minimized for the greatest condition)

LEast possible = 8
Greatest = 304
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12 Days of Christmas GMAT Competition - Day 4: The 400 students at 25 Mar 2025, 04:47
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@[color=#683d3d]KarishmaB[/color] How to solve this question using the max min approach explained in your youtube https://youtu.be/oLKbIyb1ZrI

I tried applying the same concepts, but unable to get the right answer. Thank you!
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