We need to check only for the ones given in the options 
Alva – Hindi = 75 
Average of (80+75+70)/3 = 75 
Deep – Hindi = 90 
Average of (100+90+80)/3 = 90 
Esha – Hindi = 85 
This is not the average of best 3 or 2 subjects 
Hence Alva and Deep.

Now we know that Carl only missed maths and Esha only English. 
One student who missed hindi also missed science 
Exams which can be missed 
Alva – Hindi or Science 
Bithi – Social Science and Science 
Carl – Mathematics 
Deep – Hindi or science 
Esha – English 
Foni – English and Social Science 
Hence 3 

Exams which can be missed 
Alva – Hindi or Science 
Bithi – Social Science and Science 
Carl – Mathematics 
Deep – Hindi or science 
Esha – English 
Foni – English and Social Science 
Either Alva or Deep can be the one who missed both Hindi and Science. Rest we are definite 
Hence 4 .

In this we check for the 4 options 
Carl – Mathematics score = 90 
This is average of best three scores 100 + 80 + 90 = 270/3 = 90 
Alva – Mathematics score = 70 
This cannot be the average of best three or best 2 scores 
Esha – Mathematics score = 95 
This is her highest score hence this is not the possible answer 
Foni - – Mathematics score = 78 
This cannot be the average of best three or best 2 scores 
Hence Answer is Carl. 

Alva and Bithi have scored the highest in English, hence they haven’t missed it. 
Carl and deep have the least score in English, hence they haven’t missed it. 
Esha scored 80 which is the average of (95+85+60)/3 = 80 
Foni scored 83 which is the average of (88+83+78)/3 = 83 

We need to check only for the ones given in the options 
Alva – Hindi = 75 
Average of (80+75+70)/3 = 75 
Deep – Hindi = 90 
Average of (100+90+80)/3 = 90 
Esha – Hindi = 85 
This is not the average of best 3 or 2 subjects 
Hence Alva and Deep.

Let’s start by consolidating all information available:
• We know we have 4 Economics, 3 Sociology, and 1 Anthropology student.
• 5 guides: P, R, and T guide 2 students each. Q and S guide one student each.
• Each student has guide belonging to same department.
• 4 slots: 9am, 9.30am, 10am, and 10.30am. Maximum 3 seminars in each slot. 

Seminars guided by same person never in same slot, always in consecutive slots.
Let’s create a table and fill all further information in it:
• From (1), Economics students have seminars in all 4 slots
• From (2), 10am has only one seminar by ‘A’, hence A must be Economics student
• From (3), 10.30 has another seminar of Anthropology. ‘F’ must be the only Anthropology student.
From (4), 9am has another seminar of Sociology.
• From (5), B & G have seminars in same slot. This slot cannot be either of the last 2 slots, since we
need 2 empty slots. Hence, B & G are scheduled at either 9am or 9.30 am.
• From (6), since P is guiding B & C, C must be a Sociology student too, and B & C must have
seminars in consecutive slots. Hence, C must be 9/9.30 depending on B’s slot.
• From (7), since A’s slot is 10, G (along with B) must be scheduled at 9.30am. Consequently, C
must be scheduled at 9am.
• We know identities of 3 Sociology and 1 Anthropology student, hence remaining must be
Economics students.
• Since Q & S guide 1 student each, they must be either of Anthropology or Sociology guides, and
must guide either of F or G. Hence, R & T are Economics guides.
• Since same guides must have consecutive slots, R must have 10 and 10.30 slots while T has 9 &
9.30 slots. 

R & T are guiding Economics students while P, Q, and S are not. 

Let’s start by consolidating all information available:
• We know we have 4 Economics, 3 Sociology, and 1 Anthropology student.
• 5 guides: P, R, and T guide 2 students each. Q and S guide one student each.
• Each student has guide belonging to same department.
• 4 slots: 9am, 9.30am, 10am, and 10.30am. Maximum 3 seminars in each slot. 

Seminars guided by same person never in same slot, always in consecutive slots.
Let’s create a table and fill all further information in it:
• From (1), Economics students have seminars in all 4 slots
• From (2), 10am has only one seminar by ‘A’, hence A must be Economics student
• From (3), 10.30 has another seminar of Anthropology. ‘F’ must be the only Anthropology student.
From (4), 9am has another seminar of Sociology.
• From (5), B & G have seminars in same slot. This slot cannot be either of the last 2 slots, since we
need 2 empty slots. Hence, B & G are scheduled at either 9am or 9.30 am.
• From (6), since P is guiding B & C, C must be a Sociology student too, and B & C must have
seminars in consecutive slots. Hence, C must be 9/9.30 depending on B’s slot.
• From (7), since A’s slot is 10, G (along with B) must be scheduled at 9.30am. Consequently, C
must be scheduled at 9am.
• We know identities of 3 Sociology and 1 Anthropology student, hence remaining must be
Economics students.
• Since Q & S guide 1 student each, they must be either of Anthropology or Sociology guides, and
must guide either of F or G. Hence, R & T are Economics guides.
• Since same guides must have consecutive slots, R must have 10 and 10.30 slots while T has 9 &
9.30 slots. 

• S may be guiding either of ‘F’ or ‘G’, so not necessarily true
• Similarly, Q may be guiding either of ‘F’ or ‘G’, so not necessarily true
• We are sure that ‘H’ is an Economics student, hence TRUE
• ‘B’ is scheduled in the second slot, hence False 

• If ‘D’ is scheduled in a slot later than Q’s, it means that Q must be in the 9.30am slot, guiding ‘G’.
Hence, S is in the 10.30am slot guiding ‘F’.
• Also, it implies that ‘D’ must be scheduled in the 10.30 slot guided by R, and ‘E’ & ‘H’ are in the
9am or 9.30am slot guided by T.
Hence, both statements are true. 

• If Q is scheduled in the 9.30am slot guiding ‘G’, ‘E’ must also be in 9.30am slot guided by T.
• If Q is scheduled in the 10.30am slot guiding ‘F’, ‘E’ must also be in 10.30am slot guided by R.
Hence, no new inferences can be gained from this information.
Since ‘A’ is guided by R, at least one of ‘D’ and ‘H’ will have to be guided by T. 

If Q is in 10.30am slot, ‘D’ must be in 10am slot. But that is not feasible as A is in the 10am slot
already and no 2 Eco students are in the same slot.
Hence, Q must be in the 9.30am slot and ‘D’ in the 9am slot guided by T. We can see from the
solution that ‘E’ may be guided by either R or T. 

Let’s start by consolidating all information available:
• We know we have 4 Economics, 3 Sociology, and 1 Anthropology student.
• 5 guides: P, R, and T guide 2 students each. Q and S guide one student each.
• Each student has guide belonging to same department.
• 4 slots: 9am, 9.30am, 10am, and 10.30am. Maximum 3 seminars in each slot. 

Seminars guided by same person never in same slot, always in consecutive slots.
Let’s create a table and fill all further information in it:
• From (1), Economics students have seminars in all 4 slots
• From (2), 10am has only one seminar by ‘A’, hence A must be Economics student
• From (3), 10.30 has another seminar of Anthropology. ‘F’ must be the only Anthropology student.
From (4), 9am has another seminar of Sociology.
• From (5), B & G have seminars in same slot. This slot cannot be either of the last 2 slots, since we
need 2 empty slots. Hence, B & G are scheduled at either 9am or 9.30 am.
• From (6), since P is guiding B & C, C must be a Sociology student too, and B & C must have
seminars in consecutive slots. Hence, C must be 9/9.30 depending on B’s slot.
• From (7), since A’s slot is 10, G (along with B) must be scheduled at 9.30am. Consequently, C
must be scheduled at 9am.
• We know identities of 3 Sociology and 1 Anthropology student, hence remaining must be
Economics students.
• Since Q & S guide 1 student each, they must be either of Anthropology or Sociology guides, and
must guide either of F or G. Hence, R & T are Economics guides.
• Since same guides must have consecutive slots, R must have 10 and 10.30 slots while T has 9 &
9.30 slots. 

A is clear from the solution, the first slot has 2 seminars. 

Using Condition F, 75 patients were given only 1 medicine 
Hence only B = 75 -25-20-10 = 20 
And Medicine given to A, B, D but not C = 
250-25-20-30-40-20-50-35 = 30 
Exactly 100 people had taken 3 medicines hence Medicine given to B, C, D but not A = 
100-40-20-30 = 10 
And Since C has a total of 210 patients, hence Only B and C = 
210-20-20-30-40-20-50-10 = 20 
Total patients is 500 
Only B and D = 500 – (250+20+20+10+20+20+10) = 150

Using the Question we are able to fill the Venn diagram in the following manner. 
Using Condition F, 75 patients were given only 1 medicine 
Hence only B = 75 -25-20-10 = 20 
And Medicine given to A, B, D but not C = 
250-25-20-30-40-20-50-35 = 30 
Exactly 100 people had taken 3 medicines hence Medicine given to B, C, D but not A = 
100-40-20-30 = 10 
And Since C has a total of 210 patients, hence Only B and C = 
210-20-20-30-40-20-50-10 = 20 
Total patients is 500 
Only B and D = 500 – (250+20+20+10+20+20+10) = 150

Using Condition F, 75 patients were given only 1 medicine 
Hence only B = 75 -25-20-10 = 20 
And Medicine given to A, B, D but not C = 
250-25-20-30-40-20-50-35 = 30 
Exactly 100 people had taken 3 medicines hence Medicine given to B, C, D but not A = 
100-40-20-30 = 10 
And Since C has a total of 210 patients, hence Only B and C = 
210-20-20-30-40-20-50-10 = 20 
Total patients is 500 
Only B and D = 500 – (250+20+20+10+20+20+10) = 150

Using Condition F, 75 patients were given only 1 medicine 
Hence only B = 75 -25-20-10 = 20 
And Medicine given to A, B, D but not C = 
250-25-20-30-40-20-50-35 = 30 
Exactly 100 people had taken 3 medicines hence Medicine given to B, C, D but not A = 
100-40-20-30 = 10 
And Since C has a total of 210 patients, hence Only B and C = 
210-20-20-30-40-20-50-10 = 20 
Total patients is 500 
Only B and D = 500 – (250+20+20+10+20+20+10) = 150

Total number of valid votes in 4 constituencies = 5,00,000 + 3,25,000 + 6,00,030 + 1,75,000
= 16,00,030
We can consolidate the data about the candidates who lost security in the below table: 

Therefore, total number of votes polled by candidates who lost their security deposit:
45,000 + 2,76,250 + 0 + 61,250 = 3,82,500
% = (3,82,500/16,00, 030)* 100 = 23.9% 

We know that candidates who failed to secure more 1/6 th of total valid votes lose their
security deposit.
Let us put the information we have in a table:
• From (1), we know that the 2 nd runner-up in A polled 95000-10000 = 85000 votes
• From (2), we know that everyone in C secured more than (1/ 6)* 600030 = 100005 votes . Also,
the minimum difference between any 2 candidates’ votes was 10,000.
• From (3), let’s assume the total no. of votes polled in D was 100x. Then votes polled by winner =
37,500 + 5x.
Also, we know that candidates who lost their deposit in D together polled 35% votes . 

Total votes polled in A = 5,00,000
Total votes polled by top 3 candidates = 2,75,000 + 95,000 + 85,000 = 4,55,000
Votes polled by bottom 7 candidates = 5,00,000 – 4,55,000 = 45,000
We can see that 45,000 is clearly less than 1/6 th of 5,00,000 (83,333) . Hence, all the bottom 7
candidates lost their deposit.
Total % votes polled by them = \(({45000 \over {500000}})*100\) = 9% 

We know that candidates who failed to secure more 1/6 th of total valid votes lose their
security deposit.
Let us put the information we have in a table:
• From (1), we know that the 2 nd runner-up in A polled 95000-10000 = 85000 votes
• From (2), we know that everyone in C secured more than (1/ 6)* 600030 = 100005 votes . Also,
the minimum difference between any 2 candidates’ votes was 10,000.
• From (3), let’s assume the total no. of votes polled in D was 100x. Then votes polled by winner =
37,500 + 5x.
Also, we know that candidates who lost their deposit in D together polled 35% votes . 

Total votes polled in B = 3,25,000
1/6 th of 325000 = 54166.66 ̴ 54167
We can see that the winning candidate also secured less than 1/6 th of the total valid votes.
Hence, the rest 11 candidates must have secured lesser votes than that.
Hence, 11 candidates lost their security deposit in B. 

We know that candidates who failed to secure more 1/6 th of total valid votes lose their
security deposit.
Let us put the information we have in a table:
• From (1), we know that the 2 nd runner-up in A polled 95000-10000 = 85000 votes
• From (2), we know that everyone in C secured more than (1/ 6)* 600030 = 100005 votes . Also,
the minimum difference between any 2 candidates’ votes was 10,000.
• From (3), let’s assume the total no. of votes polled in D was 100x. Then votes polled by winner =
37,500 + 5x.
Also, we know that candidates who lost their deposit in D together polled 35% votes . 

Total no. of votes polled in C = 6,00,030
Total no. of candidates = 5
We know that everyone polled more than 1,00,005 votes in C, and minimum diff between any 2 is
10,000.
Hence, minimum votes polled by bottom 4 :
( 1,00,006 ) + ( 1,00, 006 + 10 000) + (1,00,006 + 2*10000) + (1,00,006 + 3*10000)
= 4*1,00,006 + 60,000 = 4,60,024 votes
Maximum votes polled by winning candidate = 6,00,030 – 4,60,024 = 1,40,006
Minimum votes polled by 1 st runner-up = 1,00,006 + 30,000 = 1,30,006
Diff between the two is 10,000. Hence, we can conclude that winner polled exactly 1,40,006
votes. 

We know that candidates who failed to secure more 1/6 th of total valid votes lose their
security deposit.
Let us put the information we have in a table:
• From (1), we know that the 2 nd runner-up in A polled 95000-10000 = 85000 votes
• From (2), we know that everyone in C secured more than (1/ 6)* 600030 = 100005 votes . Also,
the minimum difference between any 2 candidates’ votes was 10,000.
• From (3), let’s assume the total no. of votes polled in D was 100x. Then votes polled by winner =
37,500 + 5x.
Also, we know that candidates who lost their deposit in D together polled 35% votes . 

Total votes polled in D = 100x
Votes secured by top 4 = 37,500 + 5x + 37,500 + 30,000 + 10x = 1,05,000 + 15x
We can also see that the 3 rd runner-up secured less than 1/6 th (10%) votes and hence lost
security deposit. So, minimum 5 and maximum 7 candidates lost security deposit.
We can have 3 cases here:
CASE I: 7 candidates (all except winner) lost security deposit. In this case, top candidate will
have remining 65% votes.
37,500 + 5x = 65x
60x = 37,500
x = 625
Total votes polled = 100x = 62 , 500
1/6 th of 62,500 = 10,417
Votes secured by 1 st runner-up = 37,500 > 10,417
1 st runner up couldn’t have lost deposit. Hence, this case is not feasible.

CASE II: 6 candidates (all except winner & 1 st runner-up) lost security deposit. In this case, top 2
candidates will have remining 65% votes.
37,500 + 5x + 37,500 = 65x
75,000 = 60x
x = 1250
Total votes polled = 100x = 1,25,000
1/6 th of 1,25,000 = 20,833 
Once again, 1 st runner-up secured more than 1/6 th votes and couldn’t have lost deposit. This
case is also not feasible.
CASE III: 5 candidates (all except winner, 1 st runner-up & 2 nd runner-up) lost security deposit.
In this case, top 3 candidates will have remining 65% votes.
75,000 + 5x + 30,000 = 65x
60x = 1,05,000
x = 17 50
Total votes polled = 100x = 1,75,000
1/6 th of 1,75,000 = 29,167
Votes secured by 3 rd runner-up = 10x = 10,500 < 29,167
Hence, this is the only feasible case.
Therefore, total number of valid votes polled in D is 1,75,000. 

The confirmed order we know is D < C < A. ‘B’ could be anywhere in between except more than
A.
Hence, B < C < D < A is not possible as we know that margin of C is greater than that of D.

From (4),
Increase in Electronics sales from 2018 to 2019:
(98+102+70+100) – (78+82+90+80) = 4 0 = Increase in Apparels sales from 2018 to 2019
• From (5),
Total sales in Home Décor in 2019 = 100+72+80+54 = 306
Hence, Total sales in Home Décor in 2018 = 306 – 70 = 236
From (6), we know sales of Home Décor in 2018 of Delhi and Bangalore.
80 + Z + 60 + Z = 236
2Z = 96 

Z = 48
• From (7), we know that
Y – X = B – Y
2Y = B + X
B = 2Y – X -----------( i )
Also,
A – Y = 54 – X ---------(ii)
• F rom (8), we know that Y , 54, and B are in A.P.
54 – Y = B – 54
B = 108 – Y ----------(iii)
Therefore, from ( i ) we get
108 – Y = 2Y – X
X = 3Y – 108 ------------(iv)
And (ii) becomes:
A – Y = 54 – (3Y – 108)
A = 54 – 3Y + 108 + Y
A = 162 – 2Y ----------(v)
• From (4), we knew that

( Y +A+B+54) – (X+ Y+Y +X) = 4 0
Substituting values from (iii), (iv), and (v), we get:
(Y+162-2Y+108-Y+54) – (3Y-108 +Y+Y+3Y-108) = 40
324 – 2Y – 8Y + 216 = 40
10Y = 500
Y = 50
Therefore, we get values of A, B, and X by substituting values of Y.
A = 162 – 2Y = 162 – 100
A = 62
B = 108 – y = 108 – 50
B = 58
X = 3Y – 108 = 150 – 108
X = 42 

Therefore, our final table looks like this:

Mumbai’s sales in Apparel department increased by 12 crores (62cr – 50cr)