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:

We see maximum percentage increase in the Home Décor department in Mumbai, increasing from
48 to 72.
Increase = [(72-48)/ 48]* 100 = 50% 

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:

Total sales amount in 2019:
(50 + 62 + 58 + 54) + (98 + 102 + 70 + 100) + (100 + 72 + 80 + 54)
= (224) + (370) + (306)
= 900 crores 

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:

We can see form the table that Delhi had the highest sales amount in Home Décor in both the
years. 

Hence, cars 7 & 2 are parked near car 3. 

From the options, we can understand that initially, cars 1-6 were in the lot, post which 1 & 4 left
and 7 & 8 came in. 

4, 6, 5, V, 3
• This tells us that 1 & 2 have left and 4, 5, and 6 have joined after 3 entered and 1 & 2 left.
• Also, this implies that 1 & 2 leaving has accommodated 4 positions: 4, 5, 6, & V.
• Therefore, 1 & 2 must be SUVs, and 4, 5, & 6 must be compacts. Nothing can be concluded about
Car 3. 

V, 7, 3, 6, 5
Initial sequence could be 1, 2, 3, 4, 5
Now, we know that 4 was not the first car to leave. However, 6 is the next car and is taking the
position of 4.
The possibilities could be:
• 1 leaves , followed by 4, OR
• 2 leaves, followed by 4
In either case, the positions after both cars leaving would be 1, V, 3, V, 5 or V, 2, 3, V, 5.
Now, 6 arrives and takes the position of 4, skipping the initial vacant position in both cases. This
is only possible when the first vacant position is not enough to accommodate Car 6. Hence, Car
6 must be an SUV. 

Rules:
• Adjacent beads of different colours
• At least one green bead between any 2 blue beads
• At least one green & at least one blue bead between any 2 red beads 

It is clear that using Red would require the other 2 colours too. Hence, combinations of Red +
Blue and Red + Green are not possible.
Using Blue + Green combination:
2 arrangements are possible 

Maximum number of Red beads possible in a configuration = 9 

Minimum number of Blue beads possible in a configuration = 6

If we place 2 red beads at R2-C3 and R3-C2, the rest of the red beads can be maximised in the
following way:
A total of 8 red beads can be placed in this configuration. Hence, we can place 6 more red
beads to maximise their number.

IF Vial C tests positive while A, E, H are negative then any patient with vial A, E, H will not have the disease.

Patient 8 and 14 have E hence they are negative. Patient 2 doesn’t have vial C hence it can’t be positive.

Patient 6 has vial C and none of A, E, H. 

We see patients 9 to 16 have the vial A, among them patient 9 & 11 have vial D, 10 & 12 have vial D & G hence they cannot be disease positive. Patient 14 & 16 have vial G hence they also can’t test positive. Patient 13 – A, C, F, H and Patient 15 – A, C, E, H

Since A, C, H is common we must test either E or F to confirm the disease. 

Let’s look at possible positive patients with given conditions option-wise

• Patient 4 can be positive – B, D, E, G
• Patient 14 can be positive – A, C, F, G
• Patient 4 can be positive – B, D, E, H
• This case is not possible 

4 vials will definitely give us a positive result.

Patient 1 & 2 – 3 vials common and one unique each makes it 5 vials which can give us a positive result.

Patient 1 & 16 – 8 different vials which can lead to positive results

Our answer must contain 4, 5 and 8 which is only option B. 

Assume only F1 & F2 = A

We know from condition 8 that F1 and F2 have 162 students

A + x + 40 + x = 162

F2 = 313 = 162 + 26 + 45 + 30 + x

x = 50 and A = 22 

Now F1 & F4 have the same number of schools

We Calculate the values not common to both to balance the figures

F4 = 50 + 45 + 24 + 20 = 139

F1 = 50 + 25 + 22 + x = 139

x = 42

Now to calculate only F1 and F4

520 – 313 – 26 – 24 – 20 – 25 – 42 – 50 = 20

Assume only F1 & F2 = A

We know from condition 8 that F1 and F2 have 162 students

A + x + 40 + x = 162

F2 = 313 = 162 + 26 + 45 + 30 + x

x = 50 and A = 22 

Now F1 & F4 have the same number of schools

We Calculate the values not common to both to balance the figures

F4 = 50 + 45 + 24 + 20 = 139

F1 = 50 + 25 + 22 + x = 139

x = 42

Now to calculate only F1 and F4

520 – 313 – 26 – 24 – 20 – 25 – 42 – 50 = 20

Assume only F1 & F2 = A

We know from condition 8 that F1 and F2 have 162 students

A + x + 40 + x = 162

F2 = 313 = 162 + 26 + 45 + 30 + x

x = 50 and A = 22 

Now F1 & F4 have the same number of schools

We Calculate the values not common to both to balance the figures

F4 = 50 + 45 + 24 + 20 = 139

F1 = 50 + 25 + 22 + x = 139

x = 42

Now to calculate only F1 and F4

520 – 313 – 26 – 24 – 20 – 25 – 42 – 50 = 20

Assume only F1 & F2 = A

We know from condition 8 that F1 and F2 have 162 students

A + x + 40 + x = 162

F2 = 313 = 162 + 26 + 45 + 30 + x

x = 50 and A = 22 

Now F1 & F4 have the same number of schools

We Calculate the values not common to both to balance the figures

F4 = 50 + 45 + 24 + 20 = 139

F1 = 50 + 25 + 22 + x = 139

x = 42

Now to calculate only F1 and F4

520 – 313 – 26 – 24 – 20 – 25 – 42 – 50 = 20

Now how can we uniquely get 29 and 26

For 29: 5+7+8+9 = 29

For 26:
5+6+7+8 = 26
4+5+8+9 = 26
4+6+7+9 = 26

As per statement 4 Bihari gave the highest rating points in Packaging, and Chirag gave the highest rating points in Hygiene.

So Bihari gives Ravi a 9 in packaging which means the highest points in Hygiene can be 8 only as Chirag gives the highest points in Hygiene and he gives the same score in packaging too.

Also per statement 6 Atal’s rank based on Packaging is the same as that based on Hygiene. This is also true for Deepak, this means that Bihari will be second in Hygiene and thus gives 7 points 

Now how can we uniquely get 29 and 26

For 29: 5+7+8+9 = 29

For 26:
5+6+7+8 = 26
4+5+8+9 = 26
4+6+7+9 = 26

As per statement 4 Bihari gave the highest rating points in Packaging, and Chirag gave the highest rating points in Hygiene.

So Bihari gives Ravi a 9 in packaging which means the highest points in Hygiene can be 8 only as Chirag gives the highest points in Hygiene and he gives the same score in packaging too.

Also per statement 6 Atal’s rank based on Packaging is the same as that based on Hygiene. This is also true for Deepak, this means that Bihari will be second in Hygiene and thus gives 7 points