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Venn diagram problem


aravindm

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How do I go about solving the problem below? Can any one help please?

 

In a group of 30 students, 25 are taking mathematics, 22 English, and 19 history. Every

student is taking at least one of the courses. The greatest number of students who COULD be

taking all three courses is x. The least number of students who COULD be taking all three

courses is y. What is the value of x + y?

a. 17

b. 19

c. 22

d. 23

e. 24

 

Thank you.

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  • 4 weeks later...

Total number of registrations that came in is 25+22+19=66, hence 36 duplicate registrations.

 

Say "x" number of students take all 3 subjects, "a" number take maths and eng, "b" take maths and history and "c" take history and maths.

 

Duplicate entries are for a+b+c+x.

 

so, 25-(a+b+x) + 19-(b+c+x) + 22-(a+c+x) + a+b+c+x = 30

or, 25 -a -b -x + 19 - b -c -x + 22 -a -c -x + a+b+c+x = 30

or, 25 +19 -b -x + 22 -a -c -x = 30

or 66 -(a+b+c+2x) = 30

or a+b+c+2x = 36 => duplicate registrations.... How do we get range of values for x ? We need to know number of students who took 2 subjects

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How do I go about solving the problem below? Can any one help please?

 

In a group of 30 students, 25 are taking mathematics, 22 English, and 19 history. Every

student is taking at least one of the courses. The greatest number of students who COULD be

taking all three courses is x. The least number of students who COULD be taking all three

courses is y. What is the value of x + y?

a. 17

b. 19

c. 22

d. 23

e. 24 = 25 not 24

 

Thank you.

 

Clearly x = 19. To find y, assume that the Math and English students overlap as little as possible. Since there are 5 students who are not taking Math, the smallest overlap is 22- 5 = 17. Then if the History students overlap the 8 math students not taking English and the 5 English students not taking math, there must be 19 - 5 - 8 = 6 history students also taking both Math and English. Thus y = 6 and x + y = 25

hence the final answer choice is E

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  • 2 months later...
Clearly x = 19. To find y, assume that the Math and English students overlap as little as possible. Since there are 5 students who are not taking Math, the smallest overlap is 22- 5 = 17. Then if the History students overlap the 8 math students not taking English and the 5 English students not taking math, there must be 19 - 5 - 8 = 6 history students also taking both Math and English. Thus y = 6 and x + y = 25

hence the final answer choice is E

 

I belive you are wrong my friend..

x is not = 19.

if so, it means that there are only 27 students total.

can someone please verfay the real answer?

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  • 1 month later...

It's clearly that y = 6. Because 25 + 22 + 19 = 66. Assume that if every student took 2 courses so the number would be 60 for 30 students. The number 66 means that at least 6 students have taken 3 courses.

 

Now let's find x. In my views, to find x, we find the greatest number of students WHO took 2 courses and then comparing with the number of students took the 3rd one.

 

First, let find the greatest number of students who takes Math and English among 30 students. We have 25 for Math and 22 for English, thus the number taking both Math and English is 25 + 22 - 30 = 17. And we have 19 History students. Comparing 17 and 19, the greatest number of students that take 3 courses is 17.

 

Repeat with (1) Math and History comparing with English and (2) History and English comparing with Math, then we have the greatest number would be 17.

 

Then we have x + y = 17 + 6 = 23.

 

It is my explanation. Hope that it is correct :D.

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  • 3 weeks later...
  • 4 weeks later...

Here is my take on the problem.

 

I came up with 17 for x, applying the same thinking as yuppievn.

 

To determine y, I started with the smallest subject (history)

30 -19 = 11 (number of student not yet assigned to a class)

assigning these student to English

22-11= 11 (number of student taking both english and history)

Of these 11 student only 6 need to take math --> 25+11-y = 30 -->(solve for y)--> y=6

 

I think the correct answer is 23

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  • 2 weeks later...

I believe a4nsd is correct about finding y, but not quite accurate for x.

 

I can't draw a Venn diagram here, so instead: a=number of students studying only math, b= only English, c= only history, ab=math and English, etc.

 

If x were 19, then abc=19, a is at most 6, b is at most 3, c is zero, adding up to only 28. (Having students in ab only lowers the total.) But there are 30 students, so this fails.

 

The way to find x is to assume that no students are taking two classes. There are 30-25=5 students NOT taking math -- all of these are either only-English or only-history, going on this assumption. From here I used a little trial and error. If all the 5 students are taking only history, then abc is 19-5=14. If all 5 are taking only English, then abc is 22-5=17. But if abc=17, then there are two extra history students who aren't in abc. We can make abc=18 by having 4 students take only English and one take only history.

 

Double checking: abc=18, b=4, c=1, a=7, ab=ac=bc=0. This adds up to 30. x=18.

 

Answer: E, 24.

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  • 4 weeks later...

Finding x was easier, and x is 18 as mentioned in previous posts. (Check their calculations).

x=19 only leaves 28 students in total.

 

 

Now, y can only have 4 values: 1, 4, 5, 6 (Since x+y has to be greater than 18)

In this problem, Venn diagrams helped me a lot. I could only figure out hit& trial method in such a scenario.

 

 

 

 

H= History students

E=English Students

M=Maths Students

 

 

HUEUM= H+E+M-H&E-H&M-M&E+H&M&E

30=19+22+25-H&E-H&M-M&E+H&M&E

H&M&E= H&E+H&M+M&E-36......................(1)

 

 

 

 

y is equal to least possible value of H&M&E. We need to find the value of H&M&E such that it is postive yet closest to zero.

 

 

 

 

 

 

You have to draw a Venn Diagram to understand this calculation:

 

 

Assume y = 1

The biggest value of H&E+H&M+M&E can be (1)+(3+1)+(21+1) = 27

According to eq (1), it should be 37

 

 

Using Venn's diagram:

Consider H&M&E=4

The biggest value of H&E+H&M+M&E can be (4)+(3+4)+(18+4) = 33

According to eq (1), it should be 40

 

 

Consider H&M&E=5

The biggest value of H&E+H&M+M&E can be (5)+(3+5)+(17+5) = 35

According to eq (1), it should be 41

 

 

Consider H&M&E=6

The biggest value of H&E+H&M+M&E can be (6)+(3+6)+(16+6) = 37

According to eq (1), it should be 42

 

 

Hence, none of the solutions is correct (I checked multiple times), I kept on doing this calculation, because I was almost thr to find the answer.

 

 

So on..Continue this trend, which is now clearly repetitive

 

 

And you will find the value of y = 11, which produces H&E+H&M+M&E = 47

 

 

x+y = 11+18 = 29

 

 

 

 

Let me clarify my point with the exact scenario

 

 

H=19

E=22

M=25

H&E=11

H&M=14

M&E=22

H&M&E=11

This is a case, where Students with only maths are 0, students with only history are 0, students with only history are 5.

 

 

Total Number of students = 19+22+25-11-14-22+11 = 30

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  • 6 months later...
Please correct me and provide a better solution to this problem

 

This is how I did it and got the answer as 24 (18 + 6). Hope this helps.

 

First lets find the greatest number of students taking all three subjects. Since 19 is the smallest number of number of people taking a particular subject, let's assume that x=19. Plugging that into the Venn diagram where either a student takes only a single subject or all three, we get that the total number of students is (25-19)+(22-19)+(19-19)+19 = 6 + 3 + 0 + 19 = 28. So this doesn't work.

Next, let's try x=18. Which works because if 18 students take all three, 7 students take only M, 4 students take only E, and 1 student takes only H. Therefore, 18+7+4+1 = 30. ( and it works for other combinations as well since 25 = 18 + 7 ; 22 = 18 + 4 and 19 = 18 + 1). Hence x = 18.

 

Now for the least number of students taking all three, it's really easy to find. We know that total number of students = 30. Total number of registrations = 25 + 22 + 19 = 66. Now let's assume that each student takes at the most two subjects, and no one takes three. then the max number of registrations = 30 * 2 = 60. Which leaves us with 66 - 60 = 6. Therefore, our assumption wasn't correct and 6 of those 30 students have to have taken at least 3 subjects. Hence, y is 6 at least.

We can check this result,

If we have only M = 0, only E = 0, and only H = 0, and only (ME) = 11; only (EH) = 5 and only (MH) = 8 with MHE being 6, we find that.

Total M = 11 + 8 + 6 = 25

Total E = 5 + 11 + 6 = 22

Total H = 5 + 8 + 6 = 19

and total number of students = 11 + 5 + 8 + 6 = 30.

Hence y = 6.

 

So the grand sum is x + y = 18 + 6 = 24.

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