SNAP Venn Diagrams Questions

Let the number of students who took exactly one exam be ‘x’.
The number of students who took exactly two exams be ‘y’ and the number of students who took exactly three exams be ‘z’.
So, x + y + z = 200
x + 2y + 3z = 112 + 160 + 128 = 400
From these two equations, we get, y + 2z = 200
To minimize ‘z’, we have to maximize ‘y’. If z = 0, y = 200. In this case, x = 0
So, the minimum number of students who took all the three tests is 0.

Given e = 8

Number of students shortlisted for A and B is 14

 e + b = 14

       b = 6

Number of students shortlisted for A and C is 12

e + d = 12

      d = 4

Number of students shortlisted for B and C is 15

e + f = 15

      f = 7

Given 

Number of students shortlisted for B is 26

b + f + e + c = 26

6 + 7 + 8 + c = 26

                  c = 5

Number of students shortlisted for C is 29

d + e + f + g = 29

                 g = 10

Number of students shortlisted for A is 21

b + e + d + a = 21

                  a = 3

Number of students shortlisted for only B = c = 5

Number of students shortlisted for only C = g = 10

Ratio = 5:10 = 1:2

Answer is option B

Let total number of students=100x

10x=students who cleared the cut off in all the sections

5x=students who  cleared none of the sections.

Remaining = 85x

Out of these 85x , 30% cleared only 1st section , 20% cleared only 2nd section, 30% cleared only section 3 

Together they constitute 60% of 85x

Remaining= 40% of 85x =1020

On solving x=30

Total students=100x= 100*30=3000

Exactly 1 subject = a+b+c ---> Represented by X

Exactly 2 subjects=d+e+f -----> Represented by Y

Exactly 3 subjects= g -----> Represented by Z

So X + Y+ Z + none = total --------------> (I)

German + French + Spanish = (a+b+c) +2(d+e+f) + 3(g) = X+ 2Y + 3Z -----------> (II)

So X+2Y+3Z= 30+28+32=90 

Given X = 54

So 2Y +3Z = 36--------> (1)

Given ,

French and German = 6 => d+g = 6

German and Spanish = e+g = 8

French and Spanish = f + g = 10

adding all the three (d+e+f) + 3g = 24

Y + 3Z = 24 ------> (2)

solving 1 and 2 you get Y=12 and Z=4

Therefore Total = X+Y+Z+None = 54+12+4=70

Since P(B) and P(A) are mutually exclusive and exhaustive , thus, P(B)+P(A)=1

But we know that P(B)=3 P(A) , thus substituting in the above equation we get P(B)=3/4 and P(A)=1/4

Now they have asked P(B_) i.e. 1-P(B)=1-(3/4)=1/4 

With the given information, we can draw the following Venn diagram:

Let the number of students who cleared the cut-off in exactly one, two and three subjects be a, b and c, respectively, and the number of students who cleared the cut-off in both VA and DI be y.

Thus, c = 40

a + 2b + 3c = 80 + 70 + 60 = 210

a + 2(5 + 10 + y) + 3(40) = 210

a + 2y = 210 - 120 - 30

a + 2y = 60

We have to find the minimum value of a.

'a' is at least 5 as the students in only QA = 5.

At a = 5, y = $$\frac{55}{2}$$. Since y needs to be an integer, a cannot be 5.

At a = 10, y = 25. But in this case, the total number of students in VA = 75. Thus, a cannot be 10.

At a = 20, y = 20. This case satisfies all the conditions.

Hence, the minimum number of students who did not clear cut off in exactly two subjects = Minimum number of students who cleared only one subject = 20.

The answer is option C.

Exactly two consecutive months include both July-August and August-September. We cannot include July-September, as these months are not consecutive.
             

N - none of the three months

Number of people who read in July and August only = 7

Number of people who read in August and September only = 2

Therefore, the number of surveyed people who have read exactly for two consecutive months = 7+2 = 9

Answer is option B.

Given Readership of newspaper X = 8.7

only X + 2.5 + 0.5 + 1 = 8.7

only X = 4.7

Given Readership of newspaper Y = 9.1

only Y + 2.5 + 1.5 + 0.5 = 9.1

only Y = 4.6

Given Readership of newspaper Z = 5.6

only Z + 0.5 + 1 + 1.5 = 5.6

only Z = 2.6

Number of people who read only one newspaper = 4.7 + 4.6 + 2.6 = 11.9 lakhs