CAT 2020 Question Paper | CAT LR/DI
CAT Question Paper | CAT Previous Year Paper

From the information given

Thus, in all 6 musicians are expert in tabla, 5 musicians are expert in mridangam and 2 musicians are expert in ghatam.

From (2), D is expert in ghatam. From (4) and (5), I is expert in ghatam. i.e., Ghatam: D and I

Now, as H is not expert in mridangam and ghatam, he is expert in tabla only. Also J is neither expert in tabla and ghatam. Therefore, he must be expert in mridangam only.

A, B and J are expert in mridangam. Also either F or G (but not both) is expert in mridangam. So one more expert in mridangam is either E or C.

F, G, D and H are experts in tabla. One among A and B is also expert in tabla. So one more expert in tabla is either C or E.

Only tabla: (F/G), (C/E), H

Only mridangam: (A/B), (E/C), J

Only ghatam: I

Tabla and mridangam (but not ghatam): (A/B), (F/G)

Tabla and ghatam (but not mridangam): D

We know that Only tabla: (F/G), (C/E), H

As C is expert in mridangam, E is expert in tabla. F is not expert in mridangam, so F is expert in tabla.

Thus, E, F and H are the three musicians who are experts in tabla but not in either mridangam or ghatam.

 Hence, option 4.

From the information given

Thus, in all 6 musicians are expert in tabla, 5 musicians are expert in mridangam and 2 musicians are expert in ghatam.

From (2), D is expert in ghatam. From (4) and (5), I is expert in ghatam. i.e., Ghatam: D and I

Now, as H is not expert in mridangam and ghatam, he is expert in tabla only. Also J is neither expert in tabla and ghatam. Therefore, he must be expert in mridangam only.

A, B and J are expert in mridangam. Also either F or G (but not both) is expert in mridangam. So one more expert in mridangam is either E or C.

F, G, D and H are experts in tabla. One among A and B is also expert in tabla. So one more expert in tabla is either C or E.

Only tabla: (F/G), (C/E), H

Only mridangam: (A/B), (E/C), J

Only ghatam: I

Tabla and mridangam (but not ghatam): (A/B), (F/G)

Tabla and ghatam (but not mridangam): D

C and E CANNOT have any musician who is an expert in both table and mridangam but not in ghatam.

 Hence, option 2.

Thus, in all 6 musicians are expert in tabla, 5 musicians are expert in mridangam and 2 musicians are expert in ghatam.

From (2), D is expert in ghatam. From (4) and (5), I is expert in ghatam. i.e., Ghatam: D and I

Now, as H is not expert in mridangam and ghatam, he is expert in tabla only. Also J is neither expert in tabla and ghatam. Therefore, he must be expert in mridangam only.

A, B and J are expert in mridangam. Also either F or G (but not both) is expert in mridangam. So one more expert in mridangam is either E or C.
 
F, G, D and H are experts in tabla. One among A and B is also expert in tabla. So one more expert in tabla is either C or E.

Only tabla: (F/G), (C/E), H

Only mridangam: (A/B), (E/C), J

Only ghatam: I

Tabla and mridangam (but not ghatam): (A/B), (F/G)

Tabla and ghatam (but not mridangam): D

J is DEFINITELY an expert in mridangam but not in either tabla or ghatam.

 Hence, option 1.

Thus, in all 6 musicians are expert in tabla, 5 musicians are expert in mridangam and 2 musicians are expert in ghatam.

From (2), D is expert in ghatam. From (4) and (5), I is expert in ghatam. i.e., Ghatam: D and I 

Now, as H is not expert in mridangam and ghatam, he is expert in tabla only. Also J is neither expert in tabla and ghatam. Therefore, he must be expert in mridangam only.

A, B and J are expert in mridangam. Also either F or G (but not both) is expert in mridangam. So one more expert in mridangam is either E or C.

F, G, D and H are experts in tabla. One among A and B is also expert in tabla. So one more expert in tabla is either C or E.

Only tabla: (F/G), (C/E), H

Only mridangam: (A/B), (E/C), J

Only ghatam: I

Tabla and mridangam (but not ghatam): (A/B), (F/G)

Tabla and ghatam (but not mridangam): D

H is DEFINITELY an expert in tabla but not in either mridangam or ghatam.

 Hence, option 4.

As patients were first randomly assigned into two groups of equal size, exactly 500 patients were given at least one of the four medicines. We can draw following diagram using the given information.

treated with type B and type C medicines only, ‘z’ be the number of patients treated with medicines of types B, C and D, but not with medicine of type A, ‘y’ be the number of patients treated with medicines of type B and D only, ‘x’ be the number of patients treated with medicines of types A, B and D, but not with medicine of type C.

Using (a), 25 + 20 + 30 + 40 + 20 + 50 + 35 + x = 250 ⇒ x = 30

Using (d), 40 + 20 + x + z = 100 ⇒ 40 + 20 + 30 + z = 100 ⇒ z = 10

Using (a), 30 + 40 + 20 + 20 + 50 + 20 + w + z = 210 ⇒ 30 + 40 + 20 + 20 + 50 + 20 + w + 10 = 210 ⇒ w = 20

Using (f), 25 + m + 20 + 10 = 75 ⇒ m = 20

Now 500 = 250 + m + w + z + y + 20 + 20 + 10 ⇒ 500 = 250 + 20 + 20 + 10 + y + 20 + 20 + 10 ⇒ y

= 150

Therefore,

The number of patients who were treated with medicine type D = 20 + 50 + 10 + 20 + 35 + 30 + 150 + 10 = 325

Therefore, the required answer is 325.

 Answer: 325.

As patients were first randomly assigned into two groups of equal size, exactly 500 patients were given at least one of the four medicines. We can draw following diagram using the given information.

where ‘m’ be the number of patients treated with type B medicine only, ‘w’ be the number of patients treated with type B and type C medicines only, ‘z’ be the number of patients treated with medicines of types B, C and D, but not with medicine of type A, ‘y’ be the number of patients treated with medicines of type B and D only, ‘x’ be the number of patients treated with medicines of types A, B and D, but not with medicine of type C.

Using (a), 25 + 20 + 30 + 40 + 20 + 50 + 35 + x = 250 ⇒ x = 30

Using (d), 40 + 20 + x + z = 100 ⇒ 40 + 20 + 30 + z = 100 ⇒ z = 10

Using (a), 30 + 40 + 20 + 20 + 50 + 20 + w + z = 210 ⇒ 30 + 40 + 20 + 20 + 50 + 20 + w + 10 = 210 ⇒ w = 20

Using (f), 25 + m + 20 + 10 = 75 ⇒ m = 20

Now 500 = 250 + m + w + z + y + 20 + 20 + 10 ⇒ 500 = 250 + 20 + 20 + 10 + y + 20 + 20 + 10 ⇒ y 

= 150

Therefore,

The number of patients treated with medicine types B and D only = 150

Therefore, the required answer is 150.

 Answer: 150.

As patients were first randomly assigned into two groups of equal size, exactly 500 patients were given at least one of the four medicines. We can draw following diagram using the given information.

where ‘m’ be the number of patients treated with type B medicine only, ‘w’ be the number of patients treated with type B and type C medicines only, ‘z’ be the number of patients treated with medicines of types B, C and D, but not with medicine of type A, ‘y’ be the number of patients treated with medicines of type B and D only, ‘x’ be the number of patients treated with medicines of types A, B and D, but not with medicine of type C.

Using (a), 25 + 20 + 30 + 40 + 20 + 50 + 35 + x = 250 ⇒ x = 30

Using (d), 40 + 20 + x + z = 100 ⇒ 40 + 20 + 30 + z = 100 ⇒ z = 10

Using (a), 30 + 40 + 20 + 20 + 50 + 20 + w + z = 210 ⇒ 30 + 40 + 20 + 20 + 50 + 20 + w + 10 = 210 ⇒ w = 20

Using (f), 25 + m + 20 + 10 = 75 ⇒ m = 20

Now 500 = 250 + m + w + z + y + 20 + 20 + 10 ⇒ 500 = 250 + 20 + 20 + 10 + y + 20 + 20 + 10 ⇒ y

= 150

Therefore,

The number of patients who were treated with medicine types B, C and D, but not type A was 10.

Therefore, the required answer is 10.

Answer: 10.

As patients were first randomly assigned into two groups of equal size, exactly 500 patients were given at least one of the four medicines. We can draw following diagram using the given information.

where ‘m’ be the number of patients treated with type B medicine only, ‘w’ be the number of patients treated with type B and type C medicines only, ‘z’ be the number of patients treated with medicines of types B, C and D, but not with medicine of type A, ‘y’ be the number of patients treated with medicines of type B and D only, ‘x’ be the number of patients treated with medicines of types A, B and D, but not with medicine of type C.

25 + 30 + 20 + 35 + Number of patients treated with type B medicine + 20 + 20 + 10 = 500

Number of patients treated with type B medicine + 160 = 500

Number of patients treated with type B medicine = 500 – 160 = 340

Therefore, the required answer is 340.

 Answer: 340.

Given:

It is known that over the decade, the institutes each got into two contracts with two of these vendors, and each vendor got into two contracts with two of these institutes.

Therefore, considering point (V), A had one 3-year contract and one single-year contract. B had one 7- year contract and one single-year contract. C had one 3-year contract and one single-year contract. D had one 4-year contract and one single-year contract.

Using point (III), we can conclude that Y and W have only one-year contracts. Thus, 4 one-year contracts were with Y and W.

Using points (III) and (IV), W had two single-year contracts in 2012. Y did not have any contract in 2012. So, among Z and X, either Z has 2 contracts or X has two contract in 2012.

Using point (II) and (III), X, Y and W did not have 7-year contract. Therefore using point (I) we can conclude that Z must have had 7-year contract with Institute B. Also, another contract Z has must be a 3-year or a 4-year contract.

Case (i): Z has a 4-year contract.

Then X will have two 3-year contracts. One from 2010-2012 and other from 2013-2015.

So Z has to have two contracts in 2012. The 4-year contract must have been from 2010-2013 and the 7- year contract from 2013-2019 OR the 7-year contract must have been from 2010-2016 and the 4-year contract from 2016-2019

But in that case, there will be only 4 contracts in 2012. Therefore this case is invalid.

Thus, Z had a 3-year contract either with institute A or C. Therefore, X had one 4-year with D and one 3- year contract. As D did not have contract in 2010 and C did not have contract in 2011, A had contract in 2010-2012 with Vendor X and D had contract from 2012-2015 with vendor X. Again as institute C did not have a contact in 2011, B had a 7-year contract (2010-2016) with vendor Z and C had a 3-year contract (2017-2019) with vendor Z.

As institutes B and D each had exactly one contract in 2012, Remaining contracts must be with A and C. 

Therefore, both contracts of W must be single-year contracts with institutes A and C. As D did not have any contract in 2010, Y must have had a single-year contract with B in 2010 and with D in 2019.

Thus, we have

Institutes A, B and vendors W, X had more than one contracts in any year.

 Hence, option 1.

Given:

It is known that over the decade, the institutes each got into two contracts with two of these vendors, and each vendor got into two contracts with two of these institutes.

Therefore, considering point (V), A had one 3-year contract and one single-year contract. B had one 7- year contract and one single-year contract. C had one 3-year contract and one single-year contract. D had one 4-year contract and one single-year contract.

Using point (III), we can conclude that Y and W have only one-year contracts. Thus, 4 one-year contracts were with Y and W.

Using points (III) and (IV), W had two single-year contracts in 2012. Y did not have any contract in 2012. So, among Z and X, either Z has 2 contracts or X has two contract in 2012.

Using point (II) and (III), X, Y and W did not have 7-year contract. Therefore using point (I) we can conclude that Z must have had 7-year contract with Institute B. Also, another contract Z has must be a 3-year or a 4-year contract.

Case (i): Z has a 4-year contract.

Then X will have two 3-year contracts. One from 2010-2012 and other from 2013-2015.

So Z has to have two contracts in 2012. The 4-year contract must have been from 2010-2013 and the 7- year contract from 2013-2019 OR the 7-year contract must have been from 2010-2016 and the 4-year contract from 2016-2019

But in that case, there will be only 4 contracts in 2012. Therefore this case is invalid.

Thus, Z had a 3-year contract either with institute A or C. Therefore, X had one 4-year with D and one 3- year contract. As D did not have contract in 2010 and C did not have contract in 2011, A had contract in 2010-2012 with Vendor X and D had contract from 2012-2015 with vendor X. Again as institute C did not have a contact in 2011, B had a 7-year contract (2010-2016) with vendor Z and C had a 3-year contract (2017-2019) with vendor Z.

As institutes B and D each had exactly one contract in 2012, Remaining contracts must be with A and C.

Therefore, both contracts of W must be single-year contracts with institutes A and C. As D did not have any contract in 2010, Y must have had a single-year contract with B in 2010 and with D in 2019.

Thus, we have

Institutes A and B had multiple contracts during the same year.

 Hence, option 4.

Given:

It is known that over the decade, the institutes each got into two contracts with two of these vendors, and each vendor got into two contracts with two of these institutes.

Therefore, considering point (V), A had one 3-year contract and one single-year contract. B had one 7- year contract and one single-year contract. C had one 3-year contract and one single-year contract. D had one 4-year contract and one single-year contract.

Using point (III), we can conclude that Y and W have only one-year contracts. Thus, 4 one-year contracts were with Y and W.

Using points (III) and (IV), W had two single-year contracts in 2012. Y did not have any contract in 2012. So, among Z and X, either Z has 2 contracts or X has two contract in 2012.

Using point (II) and (III), X, Y and W did not have 7-year contract. Therefore using point (I) we can conclude that Z must have had 7-year contract with Institute B. Also, another contract Z has must be a 3-year or a 4-year contract.

Case (i): Z has a 4-year contract.

Then X will have two 3-year contracts. One from 2010-2012 and other from 2013-2015.

So Z has to have two contracts in 2012. The 4-year contract must have been from 2010-2013 and the 7- year contract from 2013-2019 OR the 7-year contract must have been from 2010-2016 and the 4-year contract from 2016-2019

But in that case, there will be only 4 contracts in 2012. Therefore this case is invalid.

Thus, Z had a 3-year contract either with institute A or C. Therefore, X had one 4-year with D and one 3- year contract. As D did not have contract in 2010 and C did not have contract in 2011, A had contract in 2010-2012 with Vendor X and D had contract from 2012-2015 with vendor X. Again as institute C did not have a contact in 2011, B had a 7-year contract (2010-2016) with vendor Z and C had a 3-year contract (2017-2019) with vendor Z.

As institutes B and D each had exactly one contract in 2012, Remaining contracts must be with A and C.

Therefore, both contracts of W must be single-year contracts with institutes A and C. As D did not have any contract in 2010, Y must have had a single-year contract with B in 2010 and with D in 2019.

Thus, we have

There were exactly 3 contracts in 2010.

Hence, option 1.

Given:

It is known that over the decade, the institutes each got into two contracts with two of these vendors, and each vendor got into two contracts with two of these institutes.

Therefore, considering point (V), A had one 3-year contract and one single-year contract. B had one 7- year contract and one single-year contract. C had one 3-year contract and one single-year contract. D had one 4-year contract and one single-year contract.

Using point (III), we can conclude that Y and W have only one-year contracts. Thus, 4 one-year contracts were with Y and W.

Using points (III) and (IV), W had two single-year contracts in 2012. Y did not have any contract in 2012. So, among Z and X, either Z has 2 contracts or X has two contract in 2012.

Using point (II) and (III), X, Y and W did not have 7-year contract. Therefore using point (I) we can conclude that Z must have had 7-year contract with Institute B. Also, another contract Z has must be a 3-year or a 4-year contract.

Case (i): Z has a 4-year contract.

Then X will have two 3-year contracts. One from 2010-2012 and other from 2013-2015. 

So Z has to have two contracts in 2012. The 4-year contract must have been from 2010-2013 and the 7- year contract from 2013-2019 OR the 7-year contract must have been from 2010-2016 and the 4-year contract from 2016-2019

But in that case, there will be only 4 contracts in 2012. Therefore this case is invalid.

Thus, Z had a 3-year contract either with institute A or C. Therefore, X had one 4-year with D and one 3- year contract. As D did not have contract in 2010 and C did not have contract in 2011, A had contract in 2010-2012 with Vendor X and D had contract from 2012-2015 with vendor X. Again as institute C did not have a contact in 2011, B had a 7-year contract (2010-2016) with vendor Z and C had a 3-year contract (2017-2019) with vendor Z.

As institutes B and D each had exactly one contract in 2012, Remaining contracts must be with A and C.

Therefore, both contracts of W must be single-year contracts with institutes A and C. As D did not have any contract in 2010, Y must have had a single-year contract with B in 2010 and with D in 2019.

Thus, we have

During this period there was only one contract in exactly 3 years.

Hence, option 3.

Given:

It is known that over the decade, the institutes each got into two contracts with two of these vendors, and each vendor got into two contracts with two of these institutes.

Therefore, considering point (V), A had one 3-year contract and one single-year contract. B had one 7- year contract and one single-year contract. C had one 3-year contract and one single-year contract. D had one 4-year contract and one single-year contract.

Using point (III), we can conclude that Y and W have only one-year contracts. Thus, 4 one-year contracts were with Y and W.

Using points (III) and (IV), W had two single-year contracts in 2012. Y did not have any contract in 2012. So, among Z and X, either Z has 2 contracts or X has two contract in 2012.

Using point (II) and (III), X, Y and W did not have 7-year contract. Therefore using point (I) we can conclude that Z must have had 7-year contract with Institute B. Also, another contract Z has must be a 3-year or a 4-year contract.

Case (i): Z has a 4-year contract.

Then X will have two 3-year contracts. One from 2010-2012 and other from 2013-2015.

So Z has to have two contracts in 2012. The 4-year contract must have been from 2010-2013 and the 7- year contract from 2013-2019 OR the 7-year contract must have been from 2010-2016 and the 4-year contract from 2016-2019

But in that case, there will be only 4 contracts in 2012. Therefore this case is invalid.

Thus, Z had a 3-year contract either with institute A or C. Therefore, X had one 4-year with D and one 3- year contract. As D did not have contract in 2010 and C did not have contract in 2011, A had contract in 2010-2012 with Vendor X and D had contract from 2012-2015 with vendor X. Again as institute C did not have a contact in 2011, B had a 7-year contract (2010-2016) with vendor Z and C had a 3-year contract (2017-2019) with vendor Z.

As institutes B and D each had exactly one contract in 2012, Remaining contracts must be with A and C.

Therefore, both contracts of W must be single-year contracts with institutes A and C. As D did not have any contract in 2010, Y must have had a single-year contract with B in 2010 and with D in 2019.

Thus, we have

D had a contract with Y in 2019.

 Hence, option 4.

Given:

It is known that over the decade, the institutes each got into two contracts with two of these vendors, and each vendor got into two contracts with two of these institutes.

Therefore, considering point (V), A had one 3-year contract and one single-year contract. B had one 7- year contract and one single-year contract. C had one 3-year contract and one single-year contract. D had one 4-year contract and one single-year contract.

Using point (III), we can conclude that Y and W have only one-year contracts. Thus, 4 one-year contracts were with Y and W.

Using points (III) and (IV), W had two single-year contracts in 2012. Y did not have any contract in 2012. So, among Z and X, either Z has 2 contracts or X has two contract in 2012.

Using point (II) and (III), X, Y and W did not have 7-year contract. Therefore using point (I) we can conclude that Z must have had 7-year contract with Institute B. Also, another contract Z has must be a 3-year or a 4-year contract.

Case (i): Z has a 4-year contract.

Then X will have two 3-year contracts. One from 2010-2012 and other from 2013-2015.

So Z has to have two contracts in 2012. The 4-year contract must have been from 2010-2013 and the 7- year contract from 2013-2019 OR the 7-year contract must have been from 2010-2016 and the 4-year contract from 2016-2019

But in that case, there will be only 4 contracts in 2012. Therefore this case is invalid.

Thus, Z had a 3-year contract either with institute A or C. Therefore, X had one 4-year with D and one 3- year contract. As D did not have contract in 2010 and C did not have contract in 2011, A had contract in 2010-2012 with Vendor X and D had contract from 2012-2015 with vendor X. Again as institute C did not have a contact in 2011, B had a 7-year contract (2010-2016) with vendor Z and C had a 3-year contract (2017-2019) with vendor Z.

As institutes B and D each had exactly one contract in 2012, Remaining contracts must be with A and C.

Therefore, both contracts of W must be single-year contracts with institutes A and C. As D did not have any contract in 2010, Y must have had a single-year contract with B in 2010 and with D in 2019.

Thus, we have

Among the given options, there were two or more contracts in 2015.

 Hence, option 3.

We can try to give maximum possible final rating (or maximum sum of ratings) to each driver and then calculate the monthly payment of each driver by adding his fixed payment to his bonus value. Let’s start one by one.

For Arun – he has exactly one rating of 5 and at least one rating of 4. Maximum possible ratings that we can give him on other three parameters are 4, 3 and 3.

Monthly payment = Fixed + Bonus = 1000 + 950 = Rs. 1950.

For Barun – he has exactly one rating of 5, and at least one rating of 3. Maximum possible ratings on other three parameters will be 4, 4 and 3.

Monthly payment = Fixed + Bonus = 1200 + 760 = Rs. 1960.

For Chandan – he has got two 5’s, and at least one 2. On other two parameters, we can give him maximum ratings of 4 and 4.

Bonus = 100 × 4 = 400.

Monthly payment = Fixed + Bonus = 1400 + 400 = Rs. 1800.

For Damodaran – he has at least one rating of 3. Maximum possible ratings that we can give him on other four parameters are 5, 5, 4 and 4.

Monthly payment = Fixed + Bonus = 1300 + 630 = Rs. 1930.

For Eman – he has at least one rating of 2. Other four ratings that we can give him maximum would be 5, 5, 4 and 4; in order to ensure that he gets a bonus.

Bonus = 200 × 4 = 800.

Monthly payment = Fixed + Bonus = 1100 + 800 = Rs. 1900.

So the maximum monthly payment comes out to be Rs. 1960, which is for driver Barun.

Hence, option 3.

We can try to give minimum possible final rating (or minimum sum of ratings) to each driver and then calculate the monthly payment of each driver by adding his fixed payment to his bonus value. Let’s start one by one.

For Arun – he has exactly one rating of 5 and at least one rating of 4. Minimum possible ratings that we can give him in other three parameters will be 2, 2 and 3.

Monthly payment = Fixed + Bonus = 1000 + 800 = Rs. 1800.

For Barun – he has exactly one rating of 5, and at least one rating of 3. Minimum possible ratings in other three parameters will be 2, 2, 3.

Thus bonus = 200 × 3 = 600 and monthly payment = Fixed + Bonus = 1200 + 600 = Rs. 1800.

For Chandan – he has got two 5’s, and at least one 2. In other two parameters, we can give him minimum ratings of 2 and 3.

And monthly payment = Fixed + Bonus = 1400 + 340 = Rs. 1740.

For Damodaran – he has at least one rating of 3. In order to ensure him a bonus, minimum possible ratings that we can give him in other four parameters are 2, 2, 3 and 5.

So minimum final rating of Damodaran

Bonus = 150 × 3 = 450.

Thus monthly payment = Fixed + Bonus = 1300 + 450 = Rs. 1750.

For Eman – he has at least one rating of 2. Other four ratings that we can give him minimum would be 2, 3, 3 and 5; in order to ensure that he gets a bonus.

Bonus = 200 × 3 = 600.

Monthly payment would be = Fixed + Bonus = 1100 + 600 = Rs. 1700.

As we can see, out of all five drivers, Eman will receive a minimum monthly payment which is equal to Rs. 1700.

Hence, option 1.

We have to ensure bonus to Eman with minimum possible sum of ratings. And he has already got one 2.

We can give him one more 2, two 3’s, and one 5. This gives him a total rating of 2 + 2 + 3 + 3 + 5 = 15, and this will be the minimum possible.

In this case his minimum possible final rating will be average of total of ratings

= 15/5 = 3

Hence, option 1.

In this case, we have to ensure that Damodaran gets maximum possible rating points without getting any bonus. He already has one ‘3’, we can give him two 5’s, one 4 and one 1, since he should not get bonus. 

In this case, his total rating = 3 + 5 + 5 + 4 + 1 = 18. Or we can give him one more 3, two 4’s and a 2.

But in this case his total rating = 3 + 3 + 4 + 4 + 2 = 16.

We can see that maximum sum of ratings that Damodaran can receive is ‘18’. So his maximum possible overall rating = average of total ratings

= 18/5 = 3.6

Hence, option 1.

Given: Each of them missed either one or two examinations.

Therefore, every student appeared in either 4 or 3 examinations.

Given: If a student missed only one examination, then the marks awarded in that examination was the average of the best three among the four scores in the examinations they appeared for.

The student who missed exactly one examination can have different marks in all subjects. Also we can conclude that if a student has different marks in all the subjects, he must have missed exactly one examination.

Given: If a student missed two examinations, then the marks awarded in each of these examinations was the average of the best two among the three scores in the examinations they appeared for.

Therefore, these students must have equal marks in two subjects.

Alva: Equal marks in Hindi and Science

Bithi: Equal marks in Science and Social Science

Carl: Equal marks in Mathematics and Social Science

Deep: Equal marks in Hindi and Science

Esha: Different marks in all subjects

Foni: Equal marks in English and Social Science

As Esha scored different marks in all subjects, she must have missed exactly one examination. Carl have scored equal marks in Mathematics and Social Science. But there is no student who missed these two examinations. So, Carl must have missed one examination. Both Alva and Deep have scored equal marks in Hindi as well as Science. But there is only one such student. So, among Alva and Deep, one missed both Hindi and Science examinations and the other missed exactly one examination.

Students missing exactly one examination: Esha, Carl, Alva/Deep Student missing exactly two examinations: Bithi(Science & Social Science), Foni(English & Social Science)
and Deep/Alva(Hindi & Science)

We can also conclude that the students who missed exactly one examination, missed Mathematics, Hindi and English.

There is one student who missed Mathematics only and no other exam. As per the given information, marks in Mathematics for that student must be average of the best three of the other four subjects’ exams. We can test this one by one. If Alva is that student, then her score in Mathematics (70) must be average of 80, 75 and 75. But that’s not true. So Alva cannot be that student. Deep and Esha got their highest scores (100 and 95 respectively) in Mathematics so they definitely cannot be that student as they got less than Mathematics in each of the other subjects. We can check that Carl got 90 in Mathematics, and apart from Mathematics, his highest three scores are 100, 90, 80 – the average of which is 90. Thus Carl missed only Mathematics examination.

Now we need to find the student who missed only English among Alva, Deep and Esha. We can eliminate Alva as Alva scored maximum marks in English as compared to remaining subjects. This student cannot be Deep as average of (100, 90, 90) ≠ 70. So, this student has to be Esha. Note that average of (85, 95 and 60) is 80. So, Esha missed only English examination.

Now either Alva or Deep missed only Hindi exam and another student missed both Hindi and science exams. The one who missed both Hindi and Science exams should get same marks in both subjects, and those should be average marks of best two subjects from other three. We can see from the given table that both Alva and Deep got same marks in Hindi and science.

Alva got 75 which is the average of her best two marks – 80 and 75. Also Deep got 90 which is also the average of his best two marks – 100 and 80. So one of these is the student who missed Hindi and Science. Also, the one who missed only Hindi exam can also be one of Alva or Deep. Alva got 75 in Hindi, which is average of her best three marks – 80, 75 and 70. Deep got 90 in Hindi which is average of his best three marks – 100, 90 and 80.

Hence, on the basis of above points, we can finally make two cases regarding misses (M) and appearances (A) of these six students as follows:

Case 1

Case 2

In each of the two cases, we are sure about the examinations missed by Bithi (Science and Social Science), Carl (Mathematics), Esha (English) and Foni (English and Social Science).
But we cannot be sure about which examinations have been missed by Alva and Deep, as one of them has missed only
Hindi and other one has missed both Hindi and Science.

Answer: 4

Given: Each of them missed either one or two examinations.

Therefore, every student appeared in either 4 or 3 examinations.

Given: If a student missed only one examination, then the marks awarded in that examination was the average of the best three among the four scores in the examinations they appeared for.
The student who missed exactly one examination can have different marks in all subjects. Also we can conclude that if a student has different marks in all the subjects, he must have missed exactly one examination.

Given: If a student missed two examinations, then the marks awarded in each of these examinations was the average of the best two among the three scores in the examinations they appeared for.

Therefore, these students must have equal marks in two subjects. 

Alva: Equal marks in Hindi and Science

Bithi: Equal marks in Science and Social Science

Carl: Equal marks in Mathematics and Social Science

Deep: Equal marks in Hindi and Science

Esha: Different marks in all subjects

Foni: Equal marks in English and Social Science

As Esha scored different marks in all subjects, she must have missed exactly one examination. Carl have scored equal marks in Mathematics and Social Science. But there is no student who missed these two examinations. So, Carl must have missed one examination. Both Alva and Deep have scored equal marks in Hindi as well as Science. But there is only one such student. So, among Alva and Deep, one missed both Hindi and Science examinations and the other missed exactly one examination.

Students missing exactly one examination: Esha, Carl, Alva/Deep Student missing exactly two examinations: Bithi(Science & Social Science), Foni(English & Social Science) and Deep/Alva(Hindi & Science)

Answer: 3.

Given: Each of them missed either one or two examinations.

Therefore, every student appeared in either 4 or 3 examinations.

Given: If a student missed only one examination, then the marks awarded in that examination was the average of the best three among the four scores in the examinations they appeared for.

The student who missed exactly one examination can have different marks in all subjects. Also we can conclude that if a student has different marks in all the subjects, he must have missed exactly one examination.

Given: If a student missed two examinations, then the marks awarded in each of these examinations was the average of the best two among the three scores in the examinations they appeared for.

Therefore, these students must have equal marks in two subjects.

Alva: Equal marks in Hindi and Science

Bithi: Equal marks in Science and Social Science

Carl: Equal marks in Mathematics and Social Science

Deep: Equal marks in Hindi and Science

Esha: Different marks in all subjects

Foni: Equal marks in English and Social Science 

As Esha scored different marks in all subjects, she must have missed exactly one examination. Carl have scored equal marks in Mathematics and Social Science. But there is no student who missed these two examinations. So, Carl must have missed one examination. Both Alva and Deep have scored equal marks in Hindi as well as Science. But there is only one such student. So, among Alva and Deep, one missed both Hindi and Science examinations and the other missed exactly one examination.

Students missing exactly one examination: Esha, Carl, Alva/Deep

Student missing exactly two examinations: Bithi(Science & Social Science), Foni(English & Social Science) and Deep/Alva(Hindi & Science)

We can also conclude that the students who missed exactly one examination, missed Mathematics, Hindi and English.

There is one student who missed Mathematics only and no other exam. As per the given information, marks in Mathematics for that student must be average of the best three of the other four subjects’ exams. We can test this one by one. If Alva is that student, then her score in Mathematics (70) must be average of 80, 75 and 75. But that’s not true. So Alva cannot be that student. Deep and Esha got their highest scores (100 and 95 respectively) in Mathematics so they definitely cannot be that student as they
got less than Mathematics in each of the other subjects. We can check that Carl got 90 in Mathematics, and apart from Mathematics, his highest three scores are 100, 90, 80 – the average of which is 90. Thus Carl missed only Mathematics examination.

Now we need to find the student who missed only English among Alva, Deep and Esha. We can eliminate Alva as Alva scored maximum marks in English as compared to remaining subjects. This student cannot be Deep as average of (100, 90, 90) ≠ 70. So, this student has to be Esha. Note that average of (85, 95 and 60) is 80. So, Esha missed only English examination.

Now either Alva or Deep missed only Hindi exam and another student missed both Hindi and science exams. The one who missed both Hindi and Science exams should get same marks in both subjects, and those should be average marks of best two subjects from other three. We can see from the given table that both Alva and Deep got same marks in Hindi and science. Alva got 75 which is the average of her best two marks – 80 and 75. Also Deep got 90 which is also the average of his best two marks – 100 and
80. So one of these is the student who missed Hindi and Science. Also, the one who missed only Hindi exam can also be one of Alva or Deep. Alva got 75 in Hindi, which is average of her best three marks – 80, 75 and 70. Deep got 90 in Hindi which is average of his best three marks – 100, 90 and 80.

Hence, on the basis of above points, we can finally make two cases regarding misses (M) and appearances (A) of these six students as follows:

Case 1

Case 2

One student out of Alva and Deep has missed Science examination (along with Hindi). Other student who has missed Science examination is Bithi. 

Hence, option 1.

Given: Each of them missed either one or two examinations.

Therefore, every student appeared in either 4 or 3 examinations.

Given: If a student missed only one examination, then the marks awarded in that examination was the average of the best three among the four scores in the examinations they appeared for.

The student who missed exactly one examination can have different marks in all subjects. Also we can conclude that if a student has different marks in all the subjects, he must have missed exactly one examination.

Given: If a student missed two examinations, then the marks awarded in each of these examinations was the average of the best two among the three scores in the examinations they appeared for.

Therefore, these students must have equal marks in two subjects.

Alva: Equal marks in Hindi and Science

Bithi: Equal marks in Science and Social Science
Carl: Equal marks in Mathematics and Social Science

Deep: Equal marks in Hindi and Science

Esha: Different marks in all subjects

Foni: Equal marks in English and Social Science

As Esha scored different marks in all subjects, she must have missed exactly one examination. Carl have scored equal marks in Mathematics and Social Science. But there is no student who missed these two examinations. So, Carl must have missed one examination. Both Alva and Deep have scored equal marks in Hindi as well as Science. But there is only one such student. So, among Alva and Deep, one missed both Hindi and Science examinations and the other missed exactly one examination.

Students missing exactly one examination: Esha, Carl, Alva/Deep Student missing exactly two examinations: Bithi(Science & Social Science), Foni(English & Social Science) and Deep/Alva(Hindi & Science) We can also conclude that the students who missed exactly one examination, missed Mathematics, Hindi and English.

There is one student who missed Mathematics only and no other exam. As per the given information, marks in Mathematics for that student must be average of the best three of the other four subjects’ exams. We can test this one by one. If Alva is that student, then her score in Mathematics (70) must be average of 80, 75 and 75. But that’s not true. So Alva cannot be that student. Deep and Esha got their highest scores (100 and 95 respectively) in Mathematics so they definitely cannot be that student as they
got less than Mathematics in each of the other subjects. We can check that Carl got 90 in Mathematics, and apart from Mathematics, his highest three scores are 100, 90, 80 – the average of which is 90. Thus Carl missed only Mathematics examination.

Now we need to find the student who missed only English among Alva, Deep and Esha. We can eliminate Alva as Alva scored maximum marks in English as compared to remaining subjects. This student cannot be Deep as average of (100, 90, 90) ≠ 70. So, this student has to be Esha. Note that average of (85, 95 and 60) is 80. So, Esha missed only English examination.

Now either Alva or Deep missed only Hindi exam and another student missed both Hindi and science exams. The one who missed both Hindi and Science exams should get same marks in both subjects, and those should be average marks of best two subjects from other three. We can see from the given table that both Alva and Deep got same marks in Hindi and science. Alva got 75 which is the average of her best two marks – 80 and 75. Also Deep got 90 which is also the average of his best two marks – 100 and
80. So one of these is the student who missed Hindi and Science. Also, the one who missed only Hindi exam can also be one of Alva or Deep. Alva got 75 in Hindi, which is average of her best three marks – 80, 75 and 70. Deep got 90 in Hindi which is average of his best three marks – 100, 90 and 80.

Alva and Deep are the two students who did not appear for Hindi examination. 

Hence, option 4.

Given: Each of them missed either one or two examinations.

Therefore, every student appeared in either 4 or 3 examinations.

Given: If a student missed only one examination, then the marks awarded in that examination was the average of the best three among the four scores in the examinations they appeared for.

The student who missed exactly one examination can have different marks in all subjects. Also we can conclude that if a student has different marks in all the subjects, he must have missed exactly one examination.

Given: If a student missed two examinations, then the marks awarded in each of these examinations was the average of the best two among the three scores in the examinations they appeared for.

Therefore, these students must have equal marks in two subjects.

Alva: Equal marks in Hindi and Science

Bithi: Equal marks in Science and Social Science

Carl: Equal marks in Mathematics and Social Science

Deep: Equal marks in Hindi and Science

Esha: Different marks in all subjects

Foni: Equal marks in English and Social Science

As Esha scored different marks in all subjects, she must have missed exactly one examination. Carl have scored equal marks in Mathematics and Social Science. But there is no student who missed these two examinations. So, Carl must have missed one examination. Both Alva and Deep have scored equal marks in Hindi as well as Science. But there is only one such student. So, among Alva and Deep, one missed both Hindi and Science examinations and the other missed exactly one examination.
Students missing exactly one examination: Esha, Carl, Alva/Deep Student missing exactly two examinations: Bithi(Science & Social Science), Foni(English & Social Science) and Deep/Alva(Hindi & Science) We can also conclude that the students who missed exactly one examination, missed Mathematics, Hindi and English.

There is one student who missed Mathematics only and no other exam. As per the given information, marks in Mathematics for that student must be average of the best three of the other four subjects’ exams. We can test this one by one. If Alva is that student, then her score in Mathematics (70) must be average of 80, 75 and 75. But that’s not true. So Alva cannot be that student. Deep and Esha got their highest scores (100 and 95 respectively) in Mathematics so they definitely cannot be that student as they
got less than Mathematics in each of the other subjects. We can check that Carl got 90 in Mathematics, and apart from Mathematics, his highest three scores are 100, 90, 80 – the average of which is 90. Thus Carl missed only Mathematics examination.

Now we need to find the student who missed only English among Alva, Deep and Esha. We can eliminate Alva as Alva scored maximum marks in English as compared to remaining subjects. This student cannot be Deep as average of (100, 90, 90) ≠ 70. So, this student has to be Esha. Note that average of (85, 95 and 60) is 80. So, Esha missed only English examination.

Thus, Esha and Foni did not appear for the English Examination.

Hence, option 2.

Given: Each of them missed either one or two examinations.

Therefore, every student appeared in either 4 or 3 examinations.

Given: If a student missed only one examination, then the marks awarded in that examination was the average of the best three among the four scores in the examinations they appeared for.

The student who missed exactly one examination can have different marks in all subjects. Also we can conclude that if a student has different marks in all the subjects, he must have missed exactly one examination.

Given: If a student missed two examinations, then the marks awarded in each of these examinations was the average of the best two among the three scores in the examinations they appeared for.

Therefore, these students must have equal marks in two subjects.

Alva: Equal marks in Hindi and Science

Bithi: Equal marks in Science and Social Science

Carl: Equal marks in Mathematics and Social Science

Deep: Equal marks in Hindi and Science

Esha: Different marks in all subjects

Foni: Equal marks in English and Social Science

As Esha scored different marks in all subjects, she must have missed exactly one examination. Carl have scored equal marks in Mathematics and Social Science. But there is no student who missed these two examinations. So, Carl must have missed one examination. Both Alva and Deep have scored equal marks in Hindi as well as Science. But there is only one such student. So, among Alva and Deep, one missed both Hindi and Science examinations and the other missed exactly one examination.

Students missing exactly one examination: Esha, Carl, Alva/Deep 

Student missing exactly two examinations: Bithi(Science & Social Science), Foni(English & Social Science) and Deep/Alva(Hindi & Science)

We can also conclude that the students who missed exactly one examination, missed Mathematics, Hindi and English.

There is one student who missed Mathematics only and no other exam. As per the given information, marks in Mathematics for that student must be average of the best three of the other four subjects’ exams. We can test this one by one. If Alva is that student, then her score in Mathematics (70) must be average of 80, 75 and 75. But that’s not true. So Alva cannot be that student. Deep and Esha got their highest scores (100 and 95 respectively) in Mathematics so they definitely cannot be that student as they
got less than Mathematics in each of the other subjects. We can check that Carl got 90 in Mathematics, and apart from Mathematics, his highest three scores are 100, 90, 80 – the average of which is 90. Thus Carl missed only Mathematics examination.

Hence, option 4.