CAT 2018 Question Paper Slot 2

Let 'x' be the the number of Old visitors. Then, the number of middle-aged visitors = 2x.

Also, the number of Young visitors = 2*2x = 4x

$$\Rightarrow$$ x+2x+4x = 140

$$\Rightarrow$$ x = 20

It is given that total of 55 Economy tickets were sold out. 

It is given that Young visitors half the total number of Platinum tickets that were sold.

Let 'Y' be the number of Platinum tickets bought by the Young visitors. 

Then,the number of Platinum tickets sold = 2Y.

Consequently, we can say that the number of Gold tickets sold = 140 - 55 - 2Y = 85 - 2Y.

Let us assume that 'Z' is the number of Economy tickets bought by the Old visitors. It is given that the number of Gold tickets bought by Old visitors was equal to the number of Economy tickets bought by Old visitors.

It is given that the number of Old visitors buying Gold tickets was strictly greater than the number of Young visitors buying Gold tickets.

Z > 42 - Y

$$\Rightarrow$$ Z + Y > 42 ... (1)

The number of Middle-aged visitors buying Gold tickets = 43 - (Y+Z)

Since (Y+Z) > 42, then We can say that (Y+Z)$$_{min}$$ = 43. 

Hence, the number of Middle-aged visitors buying Gold tickets = 43 - 43 = 0

Let 'x' be the the number of Old visitors. Then, the number of middle-aged visitors = 2x.

Also, the number of Young visitors = 2*2x = 4x

$$\Rightarrow$$ x+2x+4x = 140

$$\Rightarrow$$ x = 20

It is given that total of 55 Economy tickets were sold out. 

It is given that Young visitors half the total number of Platinum tickets that were sold.

Let 'Y' be the number of Platinum tickets bought by the Young visitors. 

Then,the number of Platinum tickets sold = 2Y.

Consequently, we can say that the number of Gold tickets sold = 140 - 55 - 2Y = 85 - 2Y.

Let us assume that 'Z' is the number of Economy tickets bought by the Old visitors. It is given that the number of Gold tickets bought by Old visitors was equal to the number of Economy tickets bought by Old visitors.

Let us check with the help of options. 

Option (A): The numbers of Gold and Platinum tickets bought by Young visitors were equal.

Y = 42 - Y

$$\Rightarrow$$ Y = 21. Hence, this statement can be true. 

Option (B): The numbers of Middle-aged and Young visitors buying Gold tickets were equal

43 - (Y+Z) = 42 - Y

$$\Rightarrow$$ Z = 1. Hence, this statement can be true. 

Option (C): The numbers of Old and Middle-aged visitors buying Platinum tickets were equal

20 - 2Z = (Y+2Z) - 20

$$\Rightarrow$$ Y+4Z = 40. Hence, this statement can be true. 

Option (D): The numbers of Old and Middle-aged visitors buying Economy tickets were equal

Z = 17 - Z

$$\Rightarrow$$ Z = 8.5. This is not possible as Z has to be an integer. Hence, we can say that this statement is false.

It is given that:  High Q > Best Ed > Cosmopolitan  and Education Aid > A-one

We can say that High Q > Cosmopolitan

We can see that both High Q and Cosmopolitan got same points in reputation (R) and placement quality (P). High Q received more points in infrastructure (I) than Cosmopolitan whereas Cosmopolitan received more points in faculty Quality (F) than High Q.

Hence, we can say that Infrastructure's weight should be greater than Faculty quality. i.e.   I > F

Similarly, We can see that both Best Ed and Cosmopolitan got same points in faculty Quality (F) and placement quality (P). Best Ed received more points in reputation (R) than Cosmopolitan whereas Cosmopolitan received more points in infrastructure (I) than Best Ed.

Hence, we can say that reputation's weight should be greater than infrastructure. i.e.   R > I

Similarly, We can see that both Education Aid and A-one got same points in faculty Quality (F) and reputation (R). Education Aid received more points in infrastructure (I) than A-one whereas A-one received more points in placement quality (P) than Education Aid.

Hence, we can say that reputation's weight should be greater than infrastructure. i.e.   I > P

So basically there are two possible cases: R > I > P > F or  R > I > F > P

Case 1:   Order of weights assigned = R > I > P > F 

R = 0.4, I = 0.3. P = 0.2, F = 0.1

In this case overall score received by Best Ed = 0.1*40+0.4*30+0.2*20+0.3*20 = 26

In this case overall score received by High Q = 0.1*30+0.4*20+0.2*20+0.3*40 = 27

We can see that High Q's overall score is higher than Best Ed. Hence, this is a possible case. 

Case 2:   Order of weights assigned = R > I > F > P

R = 0.4, I = 0.3. P = 0.1, F = 0.2

In this case overall score received by Best Ed = 0.2*40+0.4*30+0.1*20+0.3*20 = 28

In this case overall score received by High Q = 0.2*30+0.4*20+0.1*20+0.3*40 = 28

We can see that High Q's overall score is not greater than the overall score received Best Ed. Hence, this case is not possible.

Now that we know the weight of each parameter, we can calculate the overall score and accreditation received by each college.

We can see that weight of the faculty quality parameter = 0.1. Hence,option D is the correct answer. 

It is given that:  High Q > Best Ed > Cosmopolitan  and Education Aid > A-one

We can say that High Q > Cosmopolitan

We can see that both High Q and Cosmopolitan got same points in reputation (R) and placement quality (P). High Q received more points in infrastructure (I) than Cosmopolitan whereas Cosmopolitan received more points in faculty Quality (F) than High Q.

Hence, we can say that Infrastructure's weight should be greater than Faculty quality. i.e.   I > F

Similarly, We can see that both Best Ed and Cosmopolitan got same points in faculty Quality (F) and placement quality (P). Best Ed received more points in reputation (R) than Cosmopolitan whereas Cosmopolitan received more points in infrastructure (I) than Best Ed.

Hence, we can say that reputation's weight should be greater than infrastructure. i.e.   R > I

Similarly, We can see that both Education Aid and A-one got same points in faculty Quality (F) and reputation (R). Education Aid received more points in infrastructure (I) than A-one whereas A-one received more points in placement quality (P) than Education Aid.

Hence, we can say that reputation's weight should be greater than infrastructure. i.e.   I > P

So basically there are two possible cases: R > I > P > F or  R > I > F > P

Case 1:   Order of weights assigned = R > I > P > F 

R = 0.4, I = 0.3. P = 0.2, F = 0.1

In this case overall score received by Best Ed = 0.1*40+0.4*30+0.2*20+0.3*20 = 26

In this case overall score received by High Q = 0.1*30+0.4*20+0.2*20+0.3*40 = 27

We can see that High Q's overall score is higher than Best Ed. Hence, this is a possible case. 

Case 2:   Order of weights assigned = R > I > F > P

R = 0.4, I = 0.3. P = 0.1, F = 0.2

In this case overall score received by Best Ed = 0.2*40+0.4*30+0.1*20+0.3*20 = 28

In this case overall score received by High Q = 0.2*30+0.4*20+0.1*20+0.3*40 = 28

We can see that High Q's overall score is not greater than the overall score received Best Ed. Hence, this case is not possible.

Now that we know the weight of each parameter, we can calculate the overall score and accreditation received by each college.

From the table, we can see that three received the accreditation of AAA.

It is given that:  High Q > Best Ed > Cosmopolitan  and Education Aid > A-one

We can say that High Q > Cosmopolitan

We can see that both High Q and Cosmopolitan got same points in reputation (R) and placement quality (P). High Q received more points in infrastructure (I) than Cosmopolitan whereas Cosmopolitan received more points in faculty Quality (F) than High Q.

Hence, we can say that Infrastructure's weight should be greater than Faculty quality. i.e.   I > F

Similarly, We can see that both Best Ed and Cosmopolitan got same points in faculty Quality (F) and placement quality (P). Best Ed received more points in reputation (R) than Cosmopolitan whereas Cosmopolitan received more points in infrastructure (I) than Best Ed.

Hence, we can say that reputation's weight should be greater than infrastructure. i.e.   R > I

Similarly, We can see that both Education Aid and A-one got same points in faculty Quality (F) and reputation (R). Education Aid received more points in infrastructure (I) than A-one whereas A-one received more points in placement quality (P) than Education Aid.

Hence, we can say that reputation's weight should be greater than infrastructure. i.e.   I > P

So basically there are two possible cases: R > I > P > F or  R > I > F > P

Case 1:   Order of weights assigned = R > I > P > F 

R = 0.4, I = 0.3. P = 0.2, F = 0.1

In this case overall score received by Best Ed = 0.1*40+0.4*30+0.2*20+0.3*20 = 26

In this case overall score received by High Q = 0.1*30+0.4*20+0.2*20+0.3*40 = 27

We can see that High Q's overall score is higher than Best Ed. Hence, this is a possible case. 

Case 2:   Order of weights assigned = R > I > F > P

R = 0.4, I = 0.3. P = 0.1, F = 0.2

In this case overall score received by Best Ed = 0.2*40+0.4*30+0.1*20+0.3*20 = 28

In this case overall score received by High Q = 0.2*30+0.4*20+0.1*20+0.3*40 = 28

We can see that High Q's overall score is not greater than the overall score received Best Ed. Hence, this case is not possible.

Now that we know the weight of each parameter, we can calculate the overall score and accreditation received by each college.

From the table we can see that Education Aid scored the highest overall score = 48. 

It is given that:  High Q > Best Ed > Cosmopolitan  and Education Aid > A-one

We can say that High Q > Cosmopolitan

We can see that both High Q and Cosmopolitan got same points in reputation (R) and placement quality (P). High Q received more points in infrastructure (I) than Cosmopolitan whereas Cosmopolitan received more points in faculty Quality (F) than High Q.

Hence, we can say that Infrastructure's weight should be greater than Faculty quality. i.e.   I > F

Similarly, We can see that both Best Ed and Cosmopolitan got same points in faculty Quality (F) and placement quality (P). Best Ed received more points in reputation (R) than Cosmopolitan whereas Cosmopolitan received more points in infrastructure (I) than Best Ed.

Hence, we can say that reputation's weight should be greater than infrastructure. i.e.   R > I

Similarly, We can see that both Education Aid and A-one got same points in faculty Quality (F) and reputation (R). Education Aid received more points in infrastructure (I) than A-one whereas A-one received more points in placement quality (P) than Education Aid.

Hence, we can say that reputation's weight should be greater than infrastructure. i.e.   I > P

So basically there are two possible cases: R > I > P > F or  R > I > F > P

Case 1:   Order of weights assigned = R > I > P > F 

R = 0.4, I = 0.3. P = 0.2, F = 0.1

In this case overall score received by Best Ed = 0.1*40+0.4*30+0.2*20+0.3*20 = 26

In this case overall score received by High Q = 0.1*30+0.4*20+0.2*20+0.3*40 = 27

We can see that High Q's overall score is higher than Best Ed. Hence, this is a possible case. 

Case 2:   Order of weights assigned = R > I > F > P

R = 0.4, I = 0.3. P = 0.1, F = 0.2

In this case overall score received by Best Ed = 0.2*40+0.4*30+0.1*20+0.3*20 = 28

In this case overall score received by High Q = 0.2*30+0.4*20+0.1*20+0.3*40 = 28

We can see that High Q's overall score is not greater than the overall score received Best Ed. Hence, this case is not possible.

Now that we know the weight of each parameter, we can calculate the overall score and accreditation received by each college.

From the table we can see that none of the mentioned college received an overall scores between 31 and 40, both inclusive. Hence, option A is the correct answer. 

Let 'x' be the number of students enrolled in all three sports. Then the number of students enrolled only in L = 2x

It is given that there are a total of 17 students enrolled in G. Also, ten students enrolled in G are also enrolled in at least one more sport. Hence, the number of students enrolled in only G = 17 - 10 = 7

The number of students enrolled only in G is one less than the number of students enrolled only in L. Hence, the number of students enrolled only in L = 7+1 

$$\Rightarrow$$ 2x = 8 

$$\Rightarrow$$ x = 4

Let us assume that 'y' students are enrolled in K and L but not G. Then, the number of students enrolled only in K = y + 4

Let us assume that 'z' be the the number of students enrolled in G and K but not L. Then, the number of students enrolled G and L bot not K = 10 - 4 - z = 6 - z

It is given that a total of 39 students in the sports. 

7 + z + 4 + 6 - z + 8 + y + y + 4 = 39 

$$\Rightarrow$$ y = 5 

Number of students enrolled in G = 17 

Number of students enrolled in K = 9 + 4 + 5 + z = 18 + z

Number of students enrolled in L = 6 - z + 4 + 5 + 8 = 23 - z

It is given that the maximum student enrollment is in L.

$$\Rightarrow$$ 23 - z > 18 + z

$$\Rightarrow$$ 2z < 5

$$\Rightarrow$$ z < 2.5 

Therefore, we can say that z can take three values = {0, 1, 2}

The number of students  enrolled in both G and L but not in K = 6 - z. This number will be minimum when 'z' is maximum. z_{max} = 2

Therefore, the minimum number of students enrolled in both G and L but not in K = 6 - 2 = 4

Let 'x' be the number of students enrolled in all three sports. Then the number of students enrolled only in L = 2x

It is given that there are a total of 17 students enrolled in G. Also, ten students enrolled in G are also enrolled in at least one more sport. Hence, the number of students enrolled in only G = 17 - 10 = 7

The number of students enrolled only in G is one less than the number of students enrolled only in L. Hence, the number of students enrolled only in L = 7+1 

$$\Rightarrow$$ 2x = 8 

$$\Rightarrow$$ x = 4 

Let us assume that 'y' students are enrolled in K and L but not G. Then, the number of students enrolled only in K = y + 4

Let us assume that 'z' be the the number of students enrolled in G and K but not L. Then, the number of students enrolled G and L bot not K = 10 - 4 - z = 6 - z

It is given that a total of 39 students in the sports. 

7 + z + 4 + 6 - z + 8 + y + y + 4 = 39 

$$\Rightarrow$$ y = 5 

Number of students enrolled in G = 17 

Number of students enrolled in K = 9 + 4 + 5 + z = 18 + z

Number of students enrolled in L = 6 - z + 4 + 5 + 8 = 23 - z

It is given that the maximum student enrollment is in L.

$$\Rightarrow$$ 23 - z > 18 + z

$$\Rightarrow$$ 2z < 5

$$\Rightarrow$$ z < 2.5 

Therefore, we can say that z can take three values = {0, 1, 2}

It is given that the numbers of students enrolled in K and L are in the ratio 19:22.

$$\dfrac{18+z}{23-z} = \dfrac{19}{22}$$

z = 1 which is a possible solution as well. 

In this case the number of students enrolled in L = 23 - z = 23 - 1 = 22. Hence, option C is the correct answer.

Let 'x' be the number of students enrolled in all three sports. Then the number of students enrolled only in L = 2x

It is given that there are a total of 17 students enrolled in G. Also, ten students enrolled in G are also enrolled in at least one more sport. Hence, the number of students enrolled in only G = 17 - 10 = 7

The number of students enrolled only in G is one less than the number of students enrolled only in L. Hence, the number of students enrolled only in L = 7+1 

$$\Rightarrow$$ 2x = 8 

$$\Rightarrow$$ x = 4 

Let us assume that 'y' students are enrolled in K and L but not G. Then, the number of students enrolled only in K = y + 4

Let us assume that 'z' be the the number of students enrolled in G and K but not L. Then, the number of students enrolled G and L bot not K = 10 - 4 - z = 6 - z

It is given that a total of 39 students in the sports. 

7 + z + 4 + 6 - z + 8 + y + y + 4 = 39 

$$\Rightarrow$$ y = 5 

Number of students enrolled in G = 17 

Number of students enrolled in K = 9 + 4 + 5 + z = 18 + z

Number of students enrolled in L = 6 - z + 4 + 5 + 8 = 23 - z

It is given that the maximum student enrollment is in L.

$$\Rightarrow$$ 23 - z > 18 + z

$$\Rightarrow$$ 2z < 5

$$\Rightarrow$$ z < 2.5 

Therefore, we can say that z can take three values = {0, 1, 2}

Hence, the number of students enrolled in K = 18 + z = {18, 19, 20} 

It is given that after withdrawal the number of students enrolled in K went down by one. This one student must have left sports K. Hence we can say that the remaining 3 students must have left either G or L.

Before withdraw there were a total of 24 students were enrolled in exactly 1 sports, 11 students were enrolled in exactly 2 courses and 4 students were enrolled in all three courses. 

The students which were enrolled in all three sports, withdrew from one of the sports. Hence, we can say that now the number of students who were enrolled in exactly 2 courses = 11 + 4 = 15.

It is given that the number of students enrolled in G was six less than the number of students enrolled in L. Let 'a' be the number of students who were enrolled in G and K but not L. Then, the number of students who were enrolled in L and K but not G = a + 5 

Consequently, we can say that the number of students enrolled in G and L but not K = 15 - (2a + 5) = 10 - 2a

Number of students enrolled in this case = a + a+5 + 9 = 14 + 2a. We can see that '14+2a' is an even number. It is given that the number of students enrolled in K went down by one. Therefore, we can say that the number of students enrolled in K earlier was an odd number. 

Hence, the number of students enrolled in K = 18 + z = {18, 19, 20} 

We can see that only '19' is an odd number. Hence, we can say that the number of students enrolled in K after withdrawal = 18

$$\Rightarrow$$ 14 + 2a = 18

$$\Rightarrow$$ a = 2

From the diagram we can see that the number of students enrolled in both G and K = 2.

Let 'x' be the number of students enrolled in all three sports. Then the number of students enrolled only in L = 2x

It is given that there are a total of 17 students enrolled in G. Also, ten students enrolled in G are also enrolled in at least one more sport. Hence, the number of students enrolled in only G = 17 - 10 = 7

The number of students enrolled only in G is one less than the number of students enrolled only in L. Hence, the number of students enrolled only in L = 7+1 

$$\Rightarrow$$ 2x = 8 

$$\Rightarrow$$ x = 4 

Let us assume that 'y' students are enrolled in K and L but not G. Then, the number of students enrolled only in K = y + 4

Let us assume that 'z' be the the number of students enrolled in G and K but not L. Then, the number of students enrolled G and L bot not K = 10 - 4 - z = 6 - z

It is given that a total of 39 students in the sports. 

7 + z + 4 + 6 - z + 8 + y + y + 4 = 39 

$$\Rightarrow$$ y = 5 

Number of students enrolled in G = 17 

Number of students enrolled in K = 9 + 4 + 5 + z = 18 + z

Number of students enrolled in L = 6 - z + 4 + 5 + 8 = 23 - z

It is given that the maximum student enrollment is in L.

$$\Rightarrow$$ 23 - z > 18 + z

$$\Rightarrow$$ 2z < 5

$$\Rightarrow$$ z < 2.5 

Therefore, we can say that z can take three values = {0, 1, 2}

Hence, the number of students enrolled in K = 18 + z = {18, 19, 20} 

It is given that after withdrawal the number of students enrolled in K went down by one. This one student must have left sports K. Hence we can say that the remaining 3 students must have left either G or L.

Before withdraw there were a total of 24 students were enrolled in exactly 1 sports, 11 students were enrolled in exactly 2 courses and 4 students were enrolled in all three courses. 

The students which were enrolled in all three sports, withdrew from one of the sports. Hence, we can say that now the number of students who were enrolled in exactly 2 courses = 11 + 4 = 15.

It is given that the number of students enrolled in G was six less than the number of students enrolled in L. Let 'a' be the number of students who were enrolled in G and K but not L. Then, the number of students who were enrolled in L and K but not G = a + 5 

Consequently, we can say that the number of students enrolled in G and L but not K = 15 - (2a + 5) = 10 - 2a

Number of students enrolled in this case = a + a+5 + 9 = 14 + 2a. We can see that '14+2a' is an even number. It is given that the number of students enrolled in K went down by one. Therefore, we can say that the number of students enrolled in K earlier was an odd number. 

Hence, the number of students enrolled in K = 18 + z = {18, 19, 20} 

We can see that only '19' is an odd number. Hence, we can say that the number of students enrolled in K after withdrawal = 18

$$\Rightarrow$$ 14 + 2a = 18

$$\Rightarrow$$ a = 2

From the diagram we can see that the number of students enrolled in both G and L = 6. Hence, option A is the correct answer.