CAT DILR Marathon Questions 2026 PDF with Video Solutions

Since the colleagues from Mumbai and Kolkata cannot see each other but sit opposite to each other, we know that they face away from the table. We know that R is from Mumbai, and thus we can see where D will also sit. Now the arrangement will look like this :

A, who is from Kolkata, sits third to the left of C. This means that C will be third to the left of A because both of them will be facing different directions. The person sitting second to the right of R is from Shillong. Now the arrangement will look like this :

The neighbours of A are from Hyderabad and Delhi.

Now, we also know that the persons from Delhi and Chennai are sitting opposite (Metro cities), and we are also told that they are P and Q.

So, C can't be from Delhi and Chennai. And we also know that the persons from Hyderabad and Chandigarh are opposite to each other.

Because one of A's neighbours is from Hyderabad, C is from Chandigarh.

Now we know that D and B are sitting opposite. So, B is from Shillong. And the only left-out city is Jaipur, so D is from Jaipur. And the only left-out person is S. So, S is from Hyderabad.

The question says that D(who is from Jaipur) is a neighbour of P. So, Q is from New Delhi.

Since the colleagues from Mumbai and Kolkata cannot see each other but sit opposite to each other, we know that they face away from the table. We know that R is from Mumbai, and thus we can see where D will also sit. Now the arrangement will look like this :

A, who is from Kolkata, sits third to the left of C. This means that C will be third to the left of A because both of them will be facing different directions. The person sitting second to the right of R is from Shillong. Now the arrangement will look like this :

The neighbours of A are from Hyderabad and Delhi.

Now, we also know that the persons from Delhi and Chennai are sitting opposite (Metro cities), and we are also told that they are P and Q.

So, C can't be from  Delhi and Chennai. And we also know that the persons from Hyderabad and Chandigarh are opposite to each other.

Because one of A's neighbours is from Hyderabad, C is from Chandigarh.

Now we know that D and B are sitting opposite. So, B is from Shillong. And the only left-out city is Jaipur, so D is from Jaipur. And the only left-out person is S. So, S is from Hyderabad.

From the figure, we can see that B is third to the right of S, who is from Hyderabad.

Since the colleagues from Mumbai and Kolkata cannot see each other but sit opposite to each other, we know that they face away from the table. We know that R is from Mumbai, and thus we can see where D will also sit. Now the arrangement will look like this :

A, who is from Kolkata, sits third to the left of C. This means that C will be third to the left of A because both of them will be facing different directions. The person sitting second to the right of R is from Shillong. Now the arrangement will look like this :

The neighbours of A are from Hyderabad and Delhi.

Now, we also know that the persons from Delhi and Chennai are sitting opposite (Metro cities), and we are also told that they are P and Q.

So, C can't be from Delhi and Chennai. And we also know that the persons from Hyderabad and Chandigarh are opposite to each other.

Because one of A's neighbours is from Hyderabad, C is from Chandigarh.

Now we know that D and B are sitting opposite. So, B is from Shillong. And the only left-out city is Jaipur, so D is from Jaipur. And the only left-out person is S. So, S is from Hyderabad.

From the figure, we can see that P/Q sits second to the right of S. But we know for sure that the person sitting there is from New Delhi.

Since the colleagues from Mumbai and Kolkata cannot see each other but sit opposite to each other, we know that they face away from the table. We know that R is from Mumbai, and thus we can see where D will also sit. Now the arrangement will look like this :

A, who is from Kolkata, sits third to the left of C. This means that C will be third to the left of A because both of them will be facing different directions. The person sitting second to the right of R is from Shillong. Now the arrangement will look like this :

The neighbours of A are from Hyderabad and Delhi.

Now, we also know that the persons from Delhi and Chennai are sitting opposite (Metro cities), and we are also told that they are P and Q.

So, C can't be from Delhi and Chennai. And we also know that the persons from Hyderabad and Chandigarh are opposite to each other.

Because one of A's neighbours is from Hyderabad, C is from Chandigarh.

Now we know that D and B are sitting opposite. So, B is from Shillong. And the only left-out city is Jaipur, so D is from Jaipur. And the only left-out person is S. So, S is from Hyderabad.

There are two arrangements in total.

Since the colleagues from Mumbai and Kolkata cannot see each other but sit opposite to each other, we know that they face away from the table. We know that R is from Mumbai, and thus we can see where D will also sit. Now the arrangement will look like this :

A, who is from Kolkata, sits third to the left of C. This means that C will be third to the left of A because both of them will be facing different directions. The person sitting second to the right of R is from Shillong. Now the arrangement will look like this :

The neighbours of A are from Hyderabad and Delhi.

Now, we also know that the persons from Delhi and Chennai are sitting opposite (Metro cities), and we are also told that they are P and Q.

So, C can't be from Delhi and Chennai. And we also know that the persons from Hyderabad and Chandigarh are opposite to each other.

Because one of A's neighbours is from Hyderabad, C is from Chandigarh.

Now we know that D and B are sitting opposite. So, B is from Shillong. And the only left-out city is Jaipur, so D is from Jaipur. And the only left-out person is S. So, S is from Hyderabad.

We know that A and R do not face the table. Thus, from the figure, we can see that D, who is from Jaipur, also does not face the table.

Given that Akash, Bala, and Chirag participate in a competition from Day 1 to Day 4.
Given the scores are from 1 to 9 and point 1 says score of the participants on any day is distinct.
The highest score on any day can be $$9+8+7=24$$.

Use point 2 to know that the total score on Day 2 is 12.
The values for the sum of the total score of any two days are 22, 30, 37, 29, 27, and 32 in any order.

Let $$a,b,$$ and $$c$$ be the total scores on Day 1, Day 3, and Day 4 in any order.

We know whenever we take sum of total score of any two days from Day 1, Day 2, Day 3, and Day 4, every day will be paired with all the other days thrice. For example, Day 1 will be paired with Day 2, Day 3, and Day 4 respectively once.

When we sum all the values of the total score of any two days, every day's sum will occur thrice.

$$3(a+b+c+12)=22+30+37+29+27+32$$
$$a+b+c=47$$  ------> Eq(1)

Given one of the combination of sum of total score of any two days is 37.
If we assume Day 2 is there in that combination, then the total score of the other day present$$=37-12=25$$, which is not possible.
This implies Day 2 is not there in the combination.
Let $$a$$ and $$b$$ be that combination.

$$a+b=37$$  ------> Eq(2)

Use Eq(1) and Eq(2): $$c=10$$.

The lowest total of any combination is 22 and it is made from 10 and 12.
This implies $$a\ge13,\ b\ge13\ \ $$

Use Eq(2): $$(a,b)$$ can be $$(13,24), (14,23), (15,22)$$ and so on but when we put these values for all other combinations, only one possibility of $$(a,b)$$ exists - $$(17,20)$$.
The total scores on all the days are $$10,12,17,20$$ in any order.

Use point (3) to know the order of total score:
Day 1 - 17
Day 2 - 12
Day 3 - 10
Day 4 - 20

Use point (2), point (3) and point (4) to make the following table:

Let the score of Bala on Day 3 be $$y$$.
$$x+y+2x=12$$
$$3x+y=10$$
$$(x=1,y=7)$$, $$(x=2,y=4)$$, and $$(x=3,y=1)$$ are the possible solutions.
If $$x=2,y=4$$, then the score of Chirag(4) is equal to score of Bala(4) on Day 3. So, not possible.
If $$x=3,y=1$$, then Bala's minimum score will be 1, which contradicts point (4). So, not possible.

$$x=1$$ and $$y=7$$ is correct.

Denote each unknown value by a variable:

$$e+i=17-3=14$$

$$(e,i)$$ can be $$(5,9)$$ and $$(6,8)$$ in any order.

$$f+l=12-3=9$$  -----> Eq(3)

Use point(5): $$l>i\ \ $$
$$l>5$$.

Use Eq(3): $$(f,l)$$ can be $$(3,6)$$ or $$(2,7)$$ or $$(1,8)$$.
$$f$$ can't be equal to 1 since Akash scored 1 on Day 3.
$$f$$ can't be equal to 3 since Bala scored 3 on Day 2.

This implies $$f=2$$ and $$l=7$$.
This gives $$i<7$$.
$$i=5$$ or $$i=6$$. Simultaneously, $$e=9$$ or $$e=8$$

Chirag total score is 18.
$$3+7+2+m=18$$
$$m=6$$.

$$g+j+m=20$$
$$g+j=14$$.

$$g$$ or $$j$$ can't be equal to $$m=6$$.
$$(g,j)$$ can be $$(5,9)$$ in any order.

There are three cases possible:

Case 1:

Case 2:

Case 3:

There are total 3 unique combinations possible.
Option C.

Given that Akash, Bala, and Chirag participate in a competition from Day 1 to Day 4.
Given the scores are from 1 to 9 and point 1 says score of the participants on any day is distinct.
The highest score on any day can be $$9+8+7=24$$.

Use point 2 to know that the total score on Day 2 is 12.
The values for the sum of the total score of any two days are 22, 30, 37, 29, 27, and 32 in any order.

Let $$a,b,$$ and $$c$$ be the total scores on Day 1, Day 3, and Day 4 in any order.

We know whenever we take sum of total score of any two days from Day 1, Day 2, Day 3, and Day 4, every day will be paired with all the other days thrice. For example, Day 1 will be paired with Day 2, Day 3, and Day 4 respectively once.

When we sum all the values of the total score of any two days, every day's sum will occur thrice.

$$3(a+b+c+12)=22+30+37+29+27+32$$
$$a+b+c=47$$ ------> Eq(1)

Given one of the combination of sum of total score of any two days is 37.
If we assume Day 2 is there in that combination, then the total score of the other day present$$=37-12=25$$, which is not possible.
This implies Day 2 is not there in the combination.
Let $$a$$ and $$b$$ be that combination.

$$a+b=37$$ ------> Eq(2)

Use Eq(1) and Eq(2): $$c=10$$.

The lowest total of any combination is 22 and it is made from 10 and 12.
This implies $$a\ge13,\ b\ge13\ \ $$

Use Eq(2): $$(a,b)$$ can be $$(13,24), (14,23), (15,22)$$ and so on but when we put these values for all other combinations, only one possibility of $$(a,b)$$ exists - $$(17,20)$$.
The total scores on all the days are $$10,12,17,20$$ in any order.

Use point (3) to know the order of total score:
Day 1 - 17
Day 2 - 12
Day 3 - 10
Day 4 - 20

Use point (2), point (3) and point (4) to make the following table:

Let the score of Bala on Day 3 be $$y$$.
$$x+y+2x=12$$
$$3x+y=10$$
$$(x=1,y=7)$$, $$(x=2,y=4)$$, and $$(x=3,y=1)$$ are the possible solutions.
If $$x=2,y=4$$, then the score of Chirag(4) is equal to score of Bala(4) on Day 3. So, not possible.
If $$x=3,y=1$$, then Bala's minimum score will be 1, which contradicts point (4). So, not possible.

$$x=1$$ and $$y=7$$ is correct.

Denote each unknown value by a variable:

$$e+i=17-3=14$$

$$(e,i)$$ can be $$(5,9)$$ and $$(6,8)$$ in any order.

$$f+l=12-3=9$$ -----> Eq(3)

Use point(5): $$l>i\ \ $$
$$l>5$$.

Use Eq(3): $$(f,l)$$ can be $$(3,6)$$ or $$(2,7)$$ or $$(1,8)$$.
$$f$$ can't be equal to 1 since Akash scored 1 on Day 3.
$$f$$ can't be equal to 3 since Bala scored 3 on Day 2.

This implies $$f=2$$ and $$l=7$$.
This gives $$i<7$$.
$$i=5$$ or $$i=6$$. Simultaneously, $$e=9$$ or $$e=8$$

Chirag total score is 18.
$$3+7+2+m=18$$
$$m=6$$.

$$g+j+m=20$$
$$g+j=14$$.

$$g$$ or $$j$$ can't be equal to $$m=6$$.
$$(g,j)$$ can be $$(5,9)$$ in any order.

There are three cases possible:

Case 1:

Case 2:

Case 3:

Chirag scored 7 on Day 2.

Given that Akash, Bala, and Chirag participate in a competition from Day 1 to Day 4.
Given the scores are from 1 to 9 and point 1 says score of the participants on any day is distinct.
The highest score on any day can be $$9+8+7=24$$.

Use point 2 to know that the total score on Day 2 is 12.
The values for the sum of the total score of any two days are 22, 30, 37, 29, 27, and 32 in any order.

Let $$a,b,$$ and $$c$$ be the total scores on Day 1, Day 3, and Day 4 in any order.

We know whenever we take sum of total score of any two days from Day 1, Day 2, Day 3, and Day 4, every day will be paired with all the other days thrice. For example, Day 1 will be paired with Day 2, Day 3, and Day 4 respectively once.

When we sum all the values of the total score of any two days, every day's sum will occur thrice.

$$3(a+b+c+12)=22+30+37+29+27+32$$
$$a+b+c=47$$ ------> Eq(1)

Given one of the combination of sum of total score of any two days is 37.
If we assume Day 2 is there in that combination, then the total score of the other day present$$=37-12=25$$, which is not possible.
This implies Day 2 is not there in the combination.
Let $$a$$ and $$b$$ be that combination.

$$a+b=37$$ ------> Eq(2)

Use Eq(1) and Eq(2): $$c=10$$.

The lowest total of any combination is 22 and it is made from 10 and 12.
This implies $$a\ge13,\ b\ge13\ \ $$

Use Eq(2): $$(a,b)$$ can be $$(13,24), (14,23), (15,22)$$ and so on but when we put these values for all other combinations, only one possibility of $$(a,b)$$ exists - $$(17,20)$$.
The total scores on all the days are $$10,12,17,20$$ in any order.

Use point (3) to know the order of total score:
Day 1 - 17
Day 2 - 12
Day 3 - 10
Day 4 - 20

Use point (2), point (3) and point (4) to make the following table:

Let the score of Bala on Day 3 be $$y$$.
$$x+y+2x=12$$
$$3x+y=10$$
$$(x=1,y=7)$$, $$(x=2,y=4)$$, and $$(x=3,y=1)$$ are the possible solutions.
If $$x=2,y=4$$, then the score of Chirag(4) is equal to score of Bala(4) on Day 3. So, not possible.
If $$x=3,y=1$$, then Bala's minimum score will be 1, which contradicts point (4). So, not possible.

$$x=1$$ and $$y=7$$ is correct.

Denote each unknown value by a variable:

$$e+i=17-3=14$$

$$(e,i)$$ can be $$(5,9)$$ and $$(6,8)$$ in any order.

$$f+l=12-3=9$$ -----> Eq(3)

Use point(5): $$l>i\ \ $$
$$l>5$$.

Use Eq(3): $$(f,l)$$ can be $$(3,6)$$ or $$(2,7)$$ or $$(1,8)$$.
$$f$$ can't be equal to 1 since Akash scored 1 on Day 3.
$$f$$ can't be equal to 3 since Bala scored 3 on Day 2.

This implies $$f=2$$ and $$l=7$$.
This gives $$i<7$$.
$$i=5$$ or $$i=6$$. Simultaneously, $$e=9$$ or $$e=8$$

Chirag total score is 18.
$$3+7+2+m=18$$
$$m=6$$.

$$g+j+m=20$$
$$g+j=14$$.

$$g$$ or $$j$$ can't be equal to $$m=6$$.
$$(g,j)$$ can be $$(5,9)$$ in any order.

There are three cases possible:

Case 1:

Case 2:

Case 3:

The absolute difference between the marks of Bala and Akash is $$1,9,$$ and $$7$$ in Case 1, Case 2, and Case 3 respectively.
So, the maximum among them is 9.
Option D.

Given that Akash, Bala, and Chirag participate in a competition from Day 1 to Day 4.
Given the scores are from 1 to 9 and point 1 says score of the participants on any day is distinct.
The highest score on any day can be $$9+8+7=24$$.

Use point 2 to know that the total score on Day 2 is 12.
The values for the sum of the total score of any two days are 22, 30, 37, 29, 27, and 32 in any order.

Let $$a,b,$$ and $$c$$ be the total scores on Day 1, Day 3, and Day 4 in any order.

We know whenever we take sum of total score of any two days from Day 1, Day 2, Day 3, and Day 4, every day will be paired with all the other days thrice. For example, Day 1 will be paired with Day 2, Day 3, and Day 4 respectively once.

When we sum all the values of the total score of any two days, every day's sum will occur thrice.

$$3(a+b+c+12)=22+30+37+29+27+32$$
$$a+b+c=47$$ ------> Eq(1)

Given one of the combination of sum of total score of any two days is 37.
If we assume Day 2 is there in that combination, then the total score of the other day present$$=37-12=25$$, which is not possible.
This implies Day 2 is not there in the combination.
Let $$a$$ and $$b$$ be that combination.

$$a+b=37$$ ------> Eq(2)

Use Eq(1) and Eq(2): $$c=10$$.

The lowest total of any combination is 22 and it is made from 10 and 12.
This implies $$a\ge13,\ b\ge13\ \ $$

Use Eq(2): $$(a,b)$$ can be $$(13,24), (14,23), (15,22)$$ and so on but when we put these values for all other combinations, only one possibility of $$(a,b)$$ exists - $$(17,20)$$.
The total scores on all the days are $$10,12,17,20$$ in any order.

Use point (3) to know the order of total score:
Day 1 - 17
Day 2 - 12
Day 3 - 10
Day 4 - 20

Use point (2), point (3) and point (4) to make the following table:

Let the score of Bala on Day 3 be $$y$$.
$$x+y+2x=12$$
$$3x+y=10$$
$$(x=1,y=7)$$, $$(x=2,y=4)$$, and $$(x=3,y=1)$$ are the possible solutions.
If $$x=2,y=4$$, then the score of Chirag(4) is equal to score of Bala(4) on Day 3. So, not possible.
If $$x=3,y=1$$, then Bala's minimum score will be 1, which contradicts point (4). So, not possible.

$$x=1$$ and $$y=7$$ is correct.

Denote each unknown value by a variable:

$$e+i=17-3=14$$

$$(e,i)$$ can be $$(5,9)$$ and $$(6,8)$$ in any order.

$$f+l=12-3=9$$ -----> Eq(3)

Use point(5): $$l>i\ \ $$
$$l>5$$.

Use Eq(3): $$(f,l)$$ can be $$(3,6)$$ or $$(2,7)$$ or $$(1,8)$$.
$$f$$ can't be equal to 1 since Akash scored 1 on Day 3.
$$f$$ can't be equal to 3 since Bala scored 3 on Day 2.

This implies $$f=2$$ and $$l=7$$.
This gives $$i<7$$.
$$i=5$$ or $$i=6$$. Simultaneously, $$e=9$$ or $$e=8$$

Chirag total score is 18.
$$3+7+2+m=18$$
$$m=6$$.

$$g+j+m=20$$
$$g+j=14$$.

$$g$$ or $$j$$ can't be equal to $$m=6$$.
$$(g,j)$$ can be $$(5,9)$$ in any order.

There are three cases possible:

Case 1:

Case 2:

Case 3:

The sum of score on the four days is 59.