Venn diagram Of the 25 students in a certain class, 12

I drew up a venn diagram.

Assume x students enrolled in all three courses. A number of students enrolled in only spanish and speech, B only in speech and music and C only in music and spanish.

Total is 25.

Hence 25= 12-(A+B+X) +15 -(B+C+X) +18-(C+A+X) +A +B +C+X
or 25=38-A-B-C -2X
or 13=A+B+C +2X
from here X could be 0.

Nitin 38 shd be 45...
why x shd be zero ? Can u elaborate....

Anyone ..?

Logically speaking wihtout the aids of any diagram, the least is 0. There is nothing in the question that says there must be students taking all three, so the least must be 0.

Of the 25 students in a certain class, 12 enrolled in spanish course, 15 enrolled in a speech course, and 18 enrolled in a music
course. What is the least possible number of the students who enrolled in all the three courses?

a) 0
b) 2
c) 3
d) 5
e) 8

since it is asking for least values, i would start evaluating with choice a. It is not possible as even if we assume all 15 of speech course have enrolled for music and none of them have enrolled for spanish then also total number of people is 27 !=25
choice b is possible as if 2 people are enrolled for all courses and remaining 13 of speach course student are common with music. 3 of the other music course members has enrolled for spanish. it leaves 7 people who have enrolled only for spanish.
so it gives total no. of students as 2+3+13+7 =25

so i would choose B

Guys, I am not getting 0 by any combination... my least is coming to 2, which is B.

Can someone explain how you got 0, if the OA is A.

Explanation:

for any question which ask for minimum number, we should look for maximizing other numbers. In this case it asks for the least common value. So lets say all the 15 people, who have taken speech course have taken music courses as well. and the remaining 3 people of music course have taken spanish. so we have 15 + 3 + 9 = 27, which is two more than the universe of 25. so atleast 2 people should have taken all the three courses.

regards,

Amardeep

If there were more than 50 enrollments then there would be a certain amount of students taking 3 courses...since if 25 + 25 = 2 courses each is exceeded, then the remainder is the amount of people taking 3 courses. If the number is over 75, then we know that someone is taking a course twice.

Here is my thinking :

If we assume 2 people have not taken any courses then number of the students who enrolled in all the three courses can be 0.

Let all 15 speech also be enrolled for music ..
Then remaining 3 of music be also enrolled for spanish
So those who like two only is =15+3=18
Students doing only spanish =12-3=9
Total =15+3+9=27....
if we assume this extra 2 people have not taken any courses then number of the students who enrolled in all the three courses can be 0.

Or
by formulating eqns ..we will get x+y+z+2R=20
x,y,z --> region overlapping ONLY two..R-all three region
x+y+z can take value = 20
e.g 10 take both speech +music..8 take music + spanish ...4 take spanish+speech ...then Only two can go upto 22...(total will become 23 ...the remaining two can be not taking any)
so R will be zero

Hi suithink,

correct me if I am wrong: If there are 25 students in the class then they should be taking atleast one of the courses, otherwise what will make them to be part of the class. so by this logic, every student have to take atleast one course.

Do you have OE as well.

regards,

Amardeep

I would also say= 0

Here is how it might work in an example:

Group1=only Spanish
Group2=only Speech
Group3=only Music
Group4=combined course Music and Speech
Group5=combined course Music and Spanish
Group6=combined course Spanish and Speech
Group7= all courses

G1=0 students
G2=0 students
G3=5 students
G4=8 students
G5=5 students
G6=7 students
G7=0 students

My answer is A.

By using a venn diagram:

Spn = 12
Spc = 15
Mus = 18

x = studnets in all three courses
a-x = sts in Spn + Spc
b-x = sts in Spn + Mus
c-x = sts in Spc + Muc

only Spn = 12 - (a-x + b-x + x) = 12-a-b+x
only Spc = 15 - (a-x + c-x + x) = 15-a-c+x
only Mus = 18 - (b-x + c-x + x) = 18-b-c+x

(12-a-b+x) + (15-a-c+x) + (18-b-c+x) = 25

2a+2b+2c - 3x = 20

2(a+b+c) - 3x = 20

2(a+b+c) - 20 = 3x OR 3x = 2(a+b+c) - 20

If x = 1 it means 3 then 2(a+b+c) must be equal to 23 which is not possible.

Taking x = 0: 3x = 0 then 2(a+b+c) must be equal to 20 which can be possible.

So the answer is A