In a state school. a student secures an 'overall distinction

Anyone have a solution for why the min is 2?

Any quick way to derive the numbers of two subjects in common?

I try to avoid over formula-ing whenever possible. My approach to this problem was to identify the pattern.

So for the min, I started with a 0 in the middle and saw how the totals in each course and overall changed as I incremented upwards.

For the Max, I started with a 23 and incremented downwards.

Fortunately with this question, I found the answer very quickly. However, if the answer was further away, I would still have been able to guess it once I established the pattern of how the other course numbers had to change when I changed the middle number.

Does anyone know if there is a better way of solving this problem

Besides how did you arrive at the final answer using the venn diagram?

For min, you seemed to have considered values for AB, BC and AC, whereas for max you ve used values for A,B,C

Please clarify?

Its easy to calculate that the minimum is 2. Obviously Venn diagrams make it simpler .. but if solving without it.

so above 75%

23 in math
28 in physics
31 in chemistry


to get at minimum, we have to try minimize the number of students who scored more than 75% in more than one subject.

So if 23 scored over 75% in math in a class of 40, then 17 students did not. Since we are trying to minimize the number of students who got 75% in more than one subject, lets take the case where all those who failed to get 75% in math got 75% in physics.. thats 17 students who got 75% in physics, but we know 28 scored above 75% in physics. So the remaining 11 should be people who scored 75% + in maths as well.


So there are minimum 11 people who scored above 75% in maths and physics.. --> 29 who have not scored 75% in math and physics. If we assume that all of these 29 scored 75% + in chemistry, 31 people scored 75% in chemistry.. so there are at least 2 people who scored 75 % in all subjects.

Hope that is clear.

The method i used was
Total no of students P union M union C (40) = P+M+C (82) - students with distinction in only 2 subjects(not 3) -2* students with overall distinction in all 3 (P intersection M intersection C)
This got me to
P inter M inter C = 21 - ½ students with distinction in 2 subjects.

So for max possible number tried with 0 students with 2 sub distinction, and it worked as i could reach sum total of 40 in venn diagram.
For min i could not keep the 2 subject students as 42 students, i could not keep it 40, as then the 3 subject intersection would be 1, making it 41 students total. So tried 38 , which worked and set in the venn diagram too, as 38 students with 2 subjects and 2 with all 3 and none with only 1 subject.

Hi thanks a lot

Based on what you've said I've calculated what you've said, theoretically.

n(A U B U C) = A+B+C - n( Students who secured 75+ in exactly 1 subject) -2(ABC)
40= 23+28+31 - (T) - 2ABC
2ABC= 82-40 - T = 42-T
ABC = (42/2) -T/2 = 21-T/2

Now consider the minimum in the options given : 0,1,2

Plugging in ABC= 0
21- (T/2) = 0

T=42.However this is not possible because this exceeds the total number of students 40. (T is the a subset of 40)

Plugging 1
ABC=1 = (42-T)/2
42-T= 2
T= 40 . It is possible.

So then the Min number of Overall distinction= 1

Similarly for max we can plug in 22,21 and 23 from the answer choices.

(42-T)/2= 21 ; T= 0 (Possible)

(42-T)/2= 22 ; T= -ve (Not possible)

(42-T)/2= 23 ; T= -ve (Not possible)

So that leaves me with Max number= 21 and Minimum= 1

Now someone please tell me where I am going wrong

Hi
I understood how you got the 9, 12 and 17 in the 2nd venn diagram

But can you explain the first venn dia

Thanks.. Got it now..

Why can't the max be 23? Can anyone explain how this venn diagram arrived at 21 ?

Assume
A = Number of Students who scored 75% in only 1 subject
B = Number of Students who scored 75% in 2 subjects
C = Number of Students who scored 75% in 3 subjects

A+B+C = 40 (Total number of students)
A+2B+3C = 23+28+31 = 82
Solving these 2 :
B+2C = 42 & C-A = 2
To find minimum students: Apply A=0 => C=2
To find Maximum: Apply B=0 => C=42/2=21

Let

A = no of people who have exactly 1

B = no of people who have exactly 2

C = no of people who have exactly 3

We can set up 2 equations:


A + B + C = 40

1(A) + 2(B) + 3(C) = (23 + 28 + 31) = 82


Maximize C

Step 1: solve by subtracting the 2equations

A + 2B + 3C = 82
- ( A + B + C = 40)
____________
B + 2C = 42

“surplus” of 42——-each person who gets two will account for +1 of this surplus and each person who gets all three (and a commendation) will account for +2 of this surplus

To maximize the no. of people who are part of all 3 sets, we should minimize the no. of people part of 2 sets (i.e., minimize B)

Let B = 0 ———> 0 + 2C = 42 ——-> C = 21

If C = 21, can both of the original equations be satisfied?

A + B + 21 = 40
A + 2B + (3)(21) = 82
————————
A + B = 19
A + 2B = 82 - 63 = 19

We can have 19 people get exactly 1 pass and have 21 get all 3 passes and the commendation

21 is the Maximum Value of C


Minimum

A + B + C = 40

To minimize the no. of ppl who pass all three (C) ———> we should Maximize the no. of ppl who pass exactly two (B)———-> for which we should Minimize the no. of ppl who pass exactly one (A)

Case 1: let’s try A = 0

0 + B + C = 40
0 + 2B + 3C = 82
———————

To solve for both equations through combination, multiply the 1st equation by *(-2) on each side and then Add the two equations

2B + 3C = 82
+ (-2B - 2C = 80)
_____________

C = 2 = Minimum no. of people part of all three sets

Does C = 2 and A = 0 satisfy our equations?

40 = 0 + B + 2 ————> B = 38 ppl who pass Exactly Two

82 = 0 + 2B + (3)(2)
76 = 2B ———> B = 38 ppl who pass exactly two

Works!

Answer:

Max = 21
Min = 2

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