CAT 2018 LRDI Slot 2 | Previous Year CAT Paper

Answer: Option A

Explanation :

​​​​​​​

From point 2, Best Ed is better than Cosmopolitan. The grades of the two colleges on F and P are the same. If Best Ed is better than Cosmopolitan, the weightage assigned to R must be higher than that assigned for I.

From point 3, Education Aid is better than A-one. The grades of the two colleges on F and R are the same. If Education Aid is better than A-one, the weightage assigned to I must be higher than that assigned for P.

Therefore we have, R > I > P.

From point 1, High Q is better than Best Ed. High Q has worse grades than Best Ed on F and R and has same grade on P and a better grade only on I. We have already figured out that R > I > P. If a better grade only on I reverses the effect of worse grades on F and R, I must be assigned higher weightage than that assigned for F.

Therefore, we have the following two possibilities:

R > I > P > F or R > I > F > P.

That means R = 0.4, I = 0.3, P = 0.2 and F= 0.1 or R = 0.4, I = 0.3, F = 0.2 and P = 0.1 are the two possibilities.

Using the two possibilities, let us calculate the scores of High Q and Best Ed.

Possibility 1: R= 0.4, I = 0.3, P =0.2, F = 0.1

​​​​​​​

Possibility 2: R= 0.4, I = 0.3, F =0.2, P = 0.1

​​​​​​​

Since High Q is better than Best Ed, possibility 1 is valid and possibility 2 is ruled out.

Therefore, we have R = 0.4, I = 0.3, P = 0.2 and F = 0.1

Now all the questions can be answered.

Hence, option (a).

Answer: 3

Explanation :

The weight assigned to parameter R is highest and is at 0.4.

A college that receives 50 points on all the four parameters scores 50 out of 50.

A college that receives 40 on parameter R loses 4 points on R and can score maximum 46.

Further, a college that receives 30 or below on parameter R cannot score 45.

Therefore we need to consider only three colleges, namely A-one, Education Aid and Fancy.

​​​​​​​

A-one lost 10 × 0.3 = 3 points on I.

Education Aid lost 10 × 0.2 = 2points on P.

Fancy lost 10 × 0.2 + 10 × 0.3 = 5points on P and I.

Therefore, these three colleges receive the accreditation AAA.

Answer: 3

Answer: 48

Explanation :

The weight assigned to parameter R is highest and is at 0.4.

A college that receives 50 points on all the four parameters scores 50 out of 50.

A college that receives 40 on parameter R loses 4 points on R and can score maximum 46.

Further, a college that receives 30 or below on parameter R cannot score 45.

Therefore we need to consider only three colleges, namely A-one, Education Aid and Fancy.

​​​​​​​

A-one lost 10 × 0.3 = 3 points on I.

Education Aid lost 10 × 0.2 = 2points on P.

Fancy lost 10 × 0.2 + 10 × 0.3 = 5points on P and I.

it can be seen that the highest overall score among the eight colleges is 50 - 2 = 48. Therefore the required answer is 48.

Answer: 48

Answer: Option A

Explanation :

We have already figured out that three colleges have overall score of 45 or above. Therefore we need to check for only the remaining five colleges.

The college that earned 0 on R (Global) lost 50 × 0.4 = 20 points on R. Therefore it can score maximum 30 points. We need not calculate the overall score of that college further.

There are three colleges that scored 20 on R (Cosmopolitan, Dominance and High Q).

They lost 30 × 0.4 = 12 points on R.

However, Cosmopolitan and Dominance colleges lost 30 × 0.3 = 9 more points on I. Therefore those colleges cannot score more than 30 points.

We have already figured out that High Q and Best Ed had an overall scores of 27 and 26 respectively.

Therefore no college scored between 31 and 40 points.

Hence, option (a).

Answer: Option C

Explanation :

Using the information given in the diagram, we get the following: 

​​​​​​​

 Note: The numbers in the table represent the area of the boxes in square units.

From statement 1, Alfa and Bravo have two products each in Blockbuster category.

From statement 6, Charlie had higher revenue than Bravo from products in Blockbuster category. Therefore the product with revenue 9 belongs to Charlie and the product with revenue 4 belongs to Bravo.

From statement 2, Alfa, Charlie and Bravo had 3, 2 and 1 products respectively in No-hope category.

From statement 7, Bravo and Charlie had same revenue from products in No-hope category. Therefore Charlie had revenue (1 + 3 = 4) and the other product with revenue 1 belongs to Alfa.

From statement 4, there were 4 products of Alfa and 3 products of Bravo in the Doubtful category.

From statement 5, the product with revenue 9 belonged to Bravo and the products with revenue 1 and 6 belonged to Alfa in the Doubtful category.

So far we have the following:

​​​​​​​

From statement 3, each company had one product each in the Promising category.

From statement 8, the product with revenue 2 belonged to Alfa, the product with revenue 9 belonged to Charlie and hence the product with revenue 3 belonged to Bravo.

Now we have the following:

​​​​​​​

Now all the questions can be answered.

Hence, option (c).

Answer: Option C

Explanation :

Using the information given in the diagram, we get the following: 

​​​​​​​

 Note: The numbers in the table represent the area of the boxes in square units.

From statement 1, Alfa and Bravo have two products each in Blockbuster category.

From statement 6, Charlie had higher revenue than Bravo from products in Blockbuster category. Therefore the product with revenue 9 belongs to Charlie and the product with revenue 4 belongs to Bravo.

From statement 2, Alfa, Charlie and Bravo had 3, 2 and 1 products respectively in No-hope category.

From statement 7, Bravo and Charlie had same revenue from products in No-hope category. Therefore Charlie had revenue (1 + 3 = 4) and the other product with revenue 1 belongs to Alfa.

From statement 4, there were 4 products of Alfa and 3 products of Bravo in the Doubtful category.

From statement 5, the product with revenue 9 belonged to Bravo and the products with revenue 1 and 6 belonged to Alfa in the Doubtful category.

So far we have the following:

​​​​​​​

From statement 3, each company had one product each in the Promising category.

From statement 8, the product with revenue 2 belonged to Alfa, the product with revenue 9 belonged to Charlie and hence the product with revenue 3 belonged to Bravo.

Now we have the following:

​​​​​​​

Number of products that Bravo had in the required categories is 1, 3, 1 and 2 respectively.

Hence, option (c).

Answer: Option D

Explanation :

Using the information given in the diagram, we get the following: 

​​​​​​​

 Note: The numbers in the table represent the area of the boxes in square units.

From statement 1, Alfa and Bravo have two products each in Blockbuster category.

From statement 6, Charlie had higher revenue than Bravo from products in Blockbuster category. Therefore the product with revenue 9 belongs to Charlie and the product with revenue 4 belongs to Bravo.

From statement 2, Alfa, Charlie and Bravo had 3, 2 and 1 products respectively in No-hope category.

From statement 7, Bravo and Charlie had same revenue from products in No-hope category. Therefore Charlie had revenue (1 + 3 = 4) and the other product with revenue 1 belongs to Alfa.

From statement 4, there were 4 products of Alfa and 3 products of Bravo in the Doubtful category.

From statement 5, the product with revenue 9 belonged to Bravo and the products with revenue 1 and 6 belonged to Alfa in the Doubtful category.

So far we have the following:

​​​​​​​

From statement 3, each company had one product each in the Promising category.

From statement 8, the product with revenue 2 belonged to Alfa, the product with revenue 9 belonged to Charlie and hence the product with revenue 3 belonged to Bravo.

Now we have the following:

​​​​​​​

Statement 4 is false while all the other statements are correct.

Hence, option (d).

Answer: Option C

Explanation :

Using the information given in the diagram, we get the following: 

​​​​​​​

 Note: The numbers in the table represent the area of the boxes in square units.

From statement 1, Alfa and Bravo have two products each in Blockbuster category.

From statement 6, Charlie had higher revenue than Bravo from products in Blockbuster category. Therefore the product with revenue 9 belongs to Charlie and the product with revenue 4 belongs to Bravo.

From statement 2, Alfa, Charlie and Bravo had 3, 2 and 1 products respectively in No-hope category.

From statement 7, Bravo and Charlie had same revenue from products in No-hope category. Therefore Charlie had revenue (1 + 3 = 4) and the other product with revenue 1 belongs to Alfa.

From statement 4, there were 4 products of Alfa and 3 products of Bravo in the Doubtful category.

From statement 5, the product with revenue 9 belonged to Bravo and the products with revenue 1 and 6 belonged to Alfa in the Doubtful category.

So far we have the following:

​​​​​​​

From statement 3, each company had one product each in the Promising category.

From statement 8, the product with revenue 2 belonged to Alfa, the product with revenue 9 belonged to Charlie and hence the product with revenue 3 belonged to Bravo.

Now we have the following:

​​​​​​​

Bravo’s total revenue = Rs. 34 crore.

Hence, option (c).

Answer: Option A

Explanation :

Suppose the total sales of the four models was 100 in 2016. Accordingly, we can fill the following table:

​​​​​​​

Hence, option (a).

Answer: Option C

Explanation :

Suppose the total sales of the four models was 100 in 2016. Accordingly, we can fill the following table:

​​​​​​​

Cxqi had the highest profit in 2016.

Hence, option (c).

Answer: Option B

Explanation :

Suppose the total sales of the four models was 100 in 2016. Accordingly, we can fill the following table:

​​​​​​​

Bysi had the highest profit in 2017.

Hence, option (b).

Answer: Option D

Explanation :

Suppose the total sales of the four models was 100 in 2016. Accordingly, we can fill the following table:

​​​​​​​

The profit of all brands, except Cxqi, went up in 2017 from 2016.

Hence, option (d).

Answer: 4

Explanation :

Since 10 students who play G enrolled in at least one other sport, the number of students who enrolled only in G = 17 - 10 = 7. Therefore x = 7.

Also from statement 1, the number of students = who enrolled in

all the three sports $\frac{7+1}{2}$ = 4

As 17 students enrolled in G, the number of students who did not enrol in G = 39 - 17 = 22.

Therefore, x + 1 + z + y + z = 22. We know that x = 7 and y = 4. Therefore we have the following: 8 + 4 + 2z = 22 or z = 5.

From statement 5, 6-w>wor w=0 or 1 or 2.

Now first two questions can be answered.

The required answer is 6 - 2 = 4.

Answer: 4

Answer: Option C

Explanation :

If the ratio of the numbers of students enrolled in K and L are in the ratio 19:22,

$\frac{18+w}{23-w}=\frac{19}{22}$

Therefore w = 1.

Therefore total enrollment in L

= 23 − 1 = 22.

Hence, option (c).

Answer: 2

Explanation :

Out of 4 students who are enrolled in all the three, suppose ‘a’ students dropped out of L and ‘b’ students dropped out of K. Therefore the number of students who dropped out of G = 4 - a - b.
Therefore we have the following:

If the number of students enrolled in K reduced by 1 that means out of the 4 students who had enrolled in all the three, one student dropped out of K i.e. b = 1.

Now, if the number of students enrolled in G was 6 less than the number of students enrolled in L, we have the following:

(7 + w + a + 6 − w + b) + 6

= 6 − w + b + 9 − a − b + 8

∴19 + a + b = 23 − w − a

∴2a + b + w = 4

Since b = 1, the only solution for the equation 2a + b + w = 4 is a = 1, b = 1 and w = 1.

Now both the questions can be answered.

The required number of students = w + a = 2.

Answer: 2

Answer: Option D

Explanation :

The required number of students

= 6 − w + b = 6 − 1 + 1 = 6.

Hence, option (d).

Answer: Option A

Explanation :

is ≡ 35 and as ≡ 56 ⇒ s ≡ 5

∴ a ≡ 6 and i ≡ 3

Letters ‘i’ and ‘d’ are common in words ‘bird’ and ‘india’. Numbers ‘1’ and ‘3’ are common in there codes. We know that i ≡ 3. Therefore, d ≡ 1. Also, br = 34

India ≡ 13366, d ≡ 1 a ≡ 6 and i ≡ 3 ⇒ n = 6

of ≡ 79 means o = 7 or 9

As peacock ≡ 5688999, code for ‘o’ must be ‘9’. Therefore, ‘f ≡ 7’.

national ≡ 13666689, a ≡ 6, i ≡ 3 , o ≡ 9 and n = 6 ⇒ tl ≡ 19

the ≡ 458 and tl ≡ 18 ⇒ t ≡ 8 and hence, l ≡ 1

Consider designated ≡ 1135556678.

As d ≡ 1 a ≡ 6, i ≡ 3, n ≡ 6, s ≡ 5, t ≡ 8; eeg ≡ 557. Therefore, e ≡ 5 and g ≡ 7

Now in peacock ≡ 5688999 we know codes for letters e, a, and o.

Therefore, pcck ≡ 8899 i.e., c = 8 or 9

If c ≡ 8, 9 codes both ‘p’ and ‘k’. As ‘9’ codes two letters and one of them in ‘o’, it can not be code for both ‘p’ and ‘k’. Hence, ‘c’ must be coded as 9. And 8 must be the code for both ‘p’ and ‘k’.

Thus, we have

br = 34 and

1 ≡ d, l

3 ≡ i

5 ≡ s, e

6 ≡ a, n

7 ≡ f, g

8 ≡ t, p, k

9 ≡ o, c

Now both the questions can be answered.

The code for the letter L = 1.

Hence, option (a).

Answer: Option C

Explanation :

is ≡ 35 and as ≡ 56 ⇒ s ≡ 5

∴ a ≡ 6 and i ≡ 3

Letters ‘i’ and ‘d’ are common in words ‘bird’ and ‘india’. Numbers ‘1’ and ‘3’ are common in there codes. We know that i ≡ 3. Therefore, d ≡ 1. Also, br = 34

India ≡ 13366, d ≡ 1 a ≡ 6 and i ≡ 3 ⇒ n = 6

of ≡ 79 means o = 7 or 9

As peacock ≡ 5688999, code for ‘o’ must be ‘9’. Therefore, ‘f ≡ 7’.

national ≡ 13666689, a ≡ 6, i ≡ 3 , o ≡ 9 and n = 6 ⇒ tl ≡ 19

the ≡ 458 and tl ≡ 18 ⇒ t ≡ 8 and hence, l ≡ 1

Consider designated ≡ 1135556678.

As d ≡ 1 a ≡ 6, i ≡ 3, n ≡ 6, s ≡ 5, t ≡ 8; eeg ≡ 557. Therefore, e ≡ 5 and g ≡ 7

Now in peacock ≡ 5688999 we know codes for letters e, a, and o.

Therefore, pcck ≡ 8899 i.e., c = 8 or 9

If c ≡ 8, 9 codes both ‘p’ and ‘k’. As ‘9’ codes two letters and one of them in ‘o’, it can not be code for both ‘p’ and ‘k’. Hence, ‘c’ must be coded as 9. And 8 must be the code for both ‘p’ and ‘k’.

Thus, we have

br = 34 and

1 ≡ d, l

3 ≡ i

5 ≡ s, e

6 ≡ a, n

7 ≡ f, g

8 ≡ t, p, k

9 ≡ o, c

The code for the letter B = 3 or 4.

Hence, option (c).

Answer: Option C

Explanation :

is ≡ 35 and as ≡ 56 ⇒ s ≡ 5

∴ a ≡ 6 and i ≡ 3

Letters ‘i’ and ‘d’ are common in words ‘bird’ and ‘india’. Numbers ‘1’ and ‘3’ are common in there codes. We know that i ≡ 3. Therefore, d ≡ 1. Also, br = 34

India ≡ 13366, d ≡ 1 a ≡ 6 and i ≡ 3 ⇒ n = 6

of ≡ 79 means o = 7 or 9

As peacock ≡ 5688999, code for ‘o’ must be ‘9’. Therefore, ‘f ≡ 7’.

national ≡ 13666689, a ≡ 6, i ≡ 3 , o ≡ 9 and n = 6 ⇒ tl ≡ 19

the ≡ 458 and tl ≡ 18 ⇒ t ≡ 8 and hence, l ≡ 1

Consider designated ≡ 1135556678.

As d ≡ 1 a ≡ 6, i ≡ 3, n ≡ 6, s ≡ 5, t ≡ 8; eeg ≡ 557. Therefore, e ≡ 5 and g ≡ 7

Now in peacock ≡ 5688999 we know codes for letters e, a, and o.

Therefore, pcck ≡ 8899 i.e., c = 8 or 9

If c ≡ 8, 9 codes both ‘p’ and ‘k’. As ‘9’ codes two letters and one of them in ‘o’, it can not be code for both ‘p’ and ‘k’. Hence, ‘c’ must be coded as 9. And 8 must be the code for both ‘p’ and ‘k’.

Thus, we have

br = 34 and

1 ≡ d, l

3 ≡ i

5 ≡ s, e

6 ≡ a, n

7 ≡ f, g

8 ≡ t, p, k

9 ≡ o, c

Only for 8 and 9, the complete list of letters associated is identified.

Hence, option (c).

Answer: Option B

Explanation :

is ≡ 35 and as ≡ 56 ⇒ s ≡ 5

∴ a ≡ 6 and i ≡ 3

Letters ‘i’ and ‘d’ are common in words ‘bird’ and ‘india’. Numbers ‘1’ and ‘3’ are common in there codes. We know that i ≡ 3. Therefore, d ≡ 1. Also, br = 34

India ≡ 13366, d ≡ 1 a ≡ 6 and i ≡ 3 ⇒ n = 6

of ≡ 79 means o = 7 or 9

As peacock ≡ 5688999, code for ‘o’ must be ‘9’. Therefore, ‘f ≡ 7’.

national ≡ 13666689, a ≡ 6, i ≡ 3 , o ≡ 9 and n = 6 ⇒ tl ≡ 19

the ≡ 458 and tl ≡ 18 ⇒ t ≡ 8 and hence, l ≡ 1

Consider designated ≡ 1135556678.

As d ≡ 1 a ≡ 6, i ≡ 3, n ≡ 6, s ≡ 5, t ≡ 8; eeg ≡ 557. Therefore, e ≡ 5 and g ≡ 7

Now in peacock ≡ 5688999 we know codes for letters e, a, and o.

Therefore, pcck ≡ 8899 i.e., c = 8 or 9

If c ≡ 8, 9 codes both ‘p’ and ‘k’. As ‘9’ codes two letters and one of them in ‘o’, it can not be code for both ‘p’ and ‘k’. Hence, ‘c’ must be coded as 9. And 8 must be the code for both ‘p’ and ‘k’.

Thus, we have

br = 34 and

1 ≡ d, l

3 ≡ i

5 ≡ s, e

6 ≡ a, n

7 ≡ f, g

8 ≡ t, p, k

9 ≡ o, c

(X, Y, Z) can be coded with the same digit. (I, B, M) can have code as ‘3’. (S, E, Z) can be coded with number ‘5’. As ‘S’ and ‘E’ are coded 5, only one more letter has code 5. Thus, (S, U, V) cannot be coded with the same digit.

Hence, option (b).

Answer: 1200

Explanation :

Suppose the base exchange rates of A, B and C w.r.t L are 100m, 120m and m respectively.

Therefore we have the following:

​​​​​​​​​​​​​​

Given: The outlet received 88000 units of L by selling A. Therefore the number of units of A sold 

$=\frac{88000}{110 m}=\frac{800}{m}$

From point 3, the number of units of L received by selling B = $\frac{9}{5}$ × 88000 = 158400

Therefore the number of units of B sold = $\frac{158400}{132 m}=\frac{1200}{m}$

From points 5 and 6, the number of units of A bought = 800 + $\frac{800}{m}$

Therefore, the number of units of L paid to buy A

= 95m $\left(800+\frac{800}{m}\right)$ = 76000 (1 + m)

From point 2, the number of units of L paid to buy B = 

$\frac{3}{5}$ × 76000 (1 + m) = 45600 (1 + m)

Therefore the number of units of B bought = $\frac{45600(1+m)}{114 m}=400+\frac{400}{m}$

From points 5 and 6,

$\frac{1200}{m}=400+\frac{400}{m}.$ Solving for m, m = 2

Therefore we have the following

​​​​​​​​​​​​​​

If the number of units of C sold = x, from points 5 and 6, the number of units of C bought = x + 3000.

Therefore, we get, 1.9(x+3000)=2.2x or solving for x, x = 19000.

Using the value of m, we get the following table for the number of units of A, B and C bought and sold.

​​​​​​​​​​​​​​

Answer: 1200

Answer: Option C

Explanation :

Suppose the base exchange rates of A, B and C w.r.t L are 100m, 120m and m respectively.

Therefore we have the following:

​​​​​​​​​​​​​​

Given: The outlet received 88000 units of L by selling A. Therefore the number of units of A sold 

$=\frac{88000}{110 m}=\frac{800}{m}$

From point 3, the number of units of L received by selling B = $\frac{9}{5}$ × 88000 = 158400

Therefore the number of units of B sold = $\frac{158400}{132 m}=\frac{1200}{m}$

From points 5 and 6, the number of units of A bought = 800 + $\frac{800}{m}$

Therefore, the number of units of L paid to buy A

= 95m $\left(800+\frac{800}{m}\right)$ = 76000 (1 + m)

From point 2, the number of units of L paid to buy B = 

$\frac{3}{5}$ × 76000 (1 + m) = 45600 (1 + m)

Therefore the number of units of B bought = $\frac{45600(1+m)}{114 m}=400+\frac{400}{m}$

From points 5 and 6,

$\frac{1200}{m}=400+\frac{400}{m}.$ Solving for m, m = 2

Therefore we have the following

​​​​​​​​​​​​​​

If the number of units of C sold = x, from points 5 and 6, the number of units of C bought = x + 3000.

Therefore, we get, 1.9(x+3000)=2.2x or solving for x, x = 19000.

Using the value of m, we get the following table for the number of units of A, B and C bought and sold.

​​​​​​​​​​​​​​

The outlet sold 19000 units of currency C on that day.

Hence, option (c).

Answer: 240

Explanation :

Suppose the base exchange rates of A, B and C w.r.t L are 100m, 120m and m respectively.

Therefore we have the following:

​​​​​​​​​​​​​​

Given: The outlet received 88000 units of L by selling A. Therefore the number of units of A sold 

$=\frac{88000}{110 m}=\frac{800}{m}$

From point 3, the number of units of L received by selling B = $\frac{9}{5}$ × 88000 = 158400

Therefore the number of units of B sold = $\frac{158400}{132 m}=\frac{1200}{m}$

From points 5 and 6, the number of units of A bought = 800 + $\frac{800}{m}$

Therefore, the number of units of L paid to buy A

= 95m $\left(800+\frac{800}{m}\right)$ = 76000 (1 + m)

From point 2, the number of units of L paid to buy B = 

$\frac{3}{5}$ × 76000 (1 + m) = 45600 (1 + m)

Therefore the number of units of B bought = $\frac{45600(1+m)}{114 m}=400+\frac{400}{m}$

From points 5 and 6,

$\frac{1200}{m}=400+\frac{400}{m}.$ Solving for m, m = 2

Therefore we have the following

​​​​​​​​​​​​​​

If the number of units of C sold = x, from points 5 and 6, the number of units of C bought = x + 3000.

Therefore, we get, 1.9(x+3000)=2.2x or solving for x, x = 19000.

Using the value of m, we get the following table for the number of units of A, B and C bought and sold.

​​​​​​​​​​​​​​

The base exchange rate of B w.r.t. L on that day was 240

Therefore the required answer is 240.

Answer: 240

Answer: Option C

Explanation :

Suppose the base exchange rates of A, B and C w.r.t L are 100m, 120m and m respectively.

Therefore we have the following:

​​​​​​​​​​​​​​

Given: The outlet received 88000 units of L by selling A. Therefore the number of units of A sold 

$=\frac{88000}{110 m}=\frac{800}{m}$

From point 3, the number of units of L received by selling B = $\frac{9}{5}$ × 88000 = 158400

Therefore the number of units of B sold = $\frac{158400}{132 m}=\frac{1200}{m}$

From points 5 and 6, the number of units of A bought = 800 + $\frac{800}{m}$

Therefore, the number of units of L paid to buy A

= 95m $\left(800+\frac{800}{m}\right)$ = 76000 (1 + m)

From point 2, the number of units of L paid to buy B = 

$\frac{3}{5}$ × 76000 (1 + m) = 45600 (1 + m)

Therefore the number of units of B bought = $\frac{45600(1+m)}{114 m}=400+\frac{400}{m}$

From points 5 and 6,

$\frac{1200}{m}=400+\frac{400}{m}.$ Solving for m, m = 2

Therefore we have the following

​​​​​​​​​​​​​​

If the number of units of C sold = x, from points 5 and 6, the number of units of C bought = x + 3000.

Therefore, we get, 1.9(x+3000)=2.2x or solving for x, x = 19000.

Using the value of m, we get the following table for the number of units of A, B and C bought and sold.

​​​​​​​​​​​​​​

The buying exchange rate of C w.r.t. L on that day was 1.90

Hence, option (c).

Answer: Option A

Explanation :

From Ganeshan’s statement, there were two candidates in room 102. From Erina’s statement, she was the only candidate in her room.

However, from Balram’s statement, there were at least three candidates in the room 101. Therefore, there were 4 candidates in room 101, 2 candidates in room 102 and 1 candidate in room 103.

Using the given statements, we can generate the following:

​​​​​​​​​​​​​​

Now all the questions can be answered.

Hence, option (a).

Answer: Option D

Explanation :

From Ganeshan’s statement, there were two candidates in room 102. From Erina’s statement, she was the only candidate in her room.

However, from Balram’s statement, there were at least three candidates in the room 101. Therefore, there were 4 candidates in room 101, 2 candidates in room 102 and 1 candidate in room 103.

Using the given statements, we can generate the following:

​​​​​​​​​​​​​​

Noone was there in the room when Ganeshan entered.

Hence, option (c).

Answer: Option D

Explanation :

From Ganeshan’s statement, there were two candidates in room 102. From Erina’s statement, she was the only candidate in her room.

However, from Balram’s statement, there were at least three candidates in the room 101. Therefore, there were 4 candidates in room 101, 2 candidates in room 102 and 1 candidate in room 103.

Using the given statements, we can generate the following:

​​​​​​​​​​​​​​

Erina reached the venue at 7:45 am

Hence, option (c).

Answer: Option C

Explanation :

From Ganeshan’s statement, there were two candidates in room 102. From Erina’s statement, she was the only candidate in her room.

However, from Balram’s statement, there were at least three candidates in the room 101. Therefore, there were 4 candidates in room 101, 2 candidates in room 102 and 1 candidate in room 103.

Using the given statements, we can generate the following:

​​​​​​​​​​​​​​

If Ganeshan entered before Divya, Balaram entered at 7.25 am.

Hence, option (c).

Answer: Option A

Explanation :

From points 1 and 2, the number of tickets bought by Young people = 80, the number of tickets bought by the middle aged people = 40 and the number of tickets bought by the old people = 20.

Using points 3 & 4, we get the following

​​​​​​​​​​​​​​

Now all the questions can be answered.

We have, 20 - 2y = x + 2y - 20. Therefore, x + 4y = 40.

Therefore, x = 40 - 4y = 4(10 - y)

or 2x = 8(10 - y). Therefore, x is a multiple of 8. Out of the given options, 32 is a possible answer.

Hence, option (a).

Answer: 3

Explanation :

We have, 20 − 2y = 17 − y or y = 3. Therefore the number of old visitors buying Gold tickets = 3.

Therefore the required answer is 3.

Answer: 3

Answer: 0

Explanation :

We have, y > 42 − x or x + y > 42. Now, the number of middle-aged visitors buying Gold tickets

= 43 − (x + y). Since x + y > 42, the number of middle-aged visitors buying Gold tickets = 0.

Therefore the required answer is 0.

Answer: 0

Answer: Option B

Explanation :

Option 1: The numbers of Middle-aged and Young visitors buying Gold tickets were equal i.e. 43 − x − y = 42 − x or y = 1. No condition precludes the possibility of y having a value of 1.

Option 2: The numbers of Old and Middle-aged visitors buying Economy tickets were equal i.e. y = 17 − y. This is not possible because in that case y will not be a natural number.
 
Option 3: The numbers of Gold and Platinum tickets bought by Young visitors were equal i.e. x = 42 − x i.e. x = 21. This is possible.

Option 4: The numbers of Old and Middle-aged visitors buying Platinum tickets were equal i.e. 20 - 2y = x + 2y - 20 or x + 4y = 40. This is possible.

Hence, option (b).