Digits 1 to 8 are mapped to 3 letters each, 9 is mapped to only 2. 
(3 × 8) + 2 = 26 which is the total number of letters in English. 
Each letter is mapped to a unique code, but remember that each number is mapped
to 2323 letters. 
It is best to start from the small words. Let us attack ‘is’ and ‘as’ first. 
IS = 35, AS = 56. S = 5 , I = 3 , A = 6. 
Now, THE is 458. OF = 79. PEACOCK does not have a ‘7’ in in. It has an O but no F, so F should be
7 and O should be 9. 
THE has 458, DESIGNATED has T and E but no H. Is there a number in THE but not in
DESIGNATED? 
H should be 4. T and E should be 5 and 8 in some order. 
Now let us look at INDIA. INDIA = 13366. I = 3 and A = 6. So, N and D should be 1 and 6 in some
order. BIRD has a D, BIRD = 1334. 
What does this mean? This tells us D = 1 and N = 6. 
Let us have another look at T and E. T and E and 5 and 8 in some order. Is there a word that has
only one of T & E but not both? 
NATIONAL has T but not E. NATIONAL has 8 but not 5. BINGO! We know T has to be 8 and E has
to be 5. 
Now let us go word by word. PEACOCK = P56C9CK. So, PCCK should be 8899. 9 is allotted only
to two numbers. 
We know that O = 9. So, what can we say about 9? 
If C takes 8, then both P and K becomes 9 which is not possible as 9 can be assigned to only two letters. 
Therefore C has to take 9 and P , K takes the value 8. 
DESIGNATED = 1553G66851. The missing number should be G. Or, G = 7. 
NATIONAL = 6683966L. The missing number should be L. Or, L should be 1. 
BIRD = B3R1. 3 and 4 are missing. B and R should be 3 and 4 in some order. 
B and R occur only once each in this sequence, so there is no way of resolving this. 

From the above inferences, we can see that B takes either 3 or 4.

Look at the statements given by Erina and Ganeshan. 
Think about the number of candidates in the rooms. 
One room has only one candidate, one room has 2 candidates, so the third room should have 4
candidates. 
So, number of candidates should be distributed as 4, 2, 1. 
But which room has 2 candidates and which one has 4? 
Room number 102 has 2 candidates. G plus one more. Balaram was the third one to enter 101, so
101 should have had at least 3 candidates. 
This implies 103 has only one candidate. 
Fatima is the 4th one into a room. What does this mean? Akhil should also be in the same room as
Fatima.

A, B and F go to room number 101. C should go to 102, E should go to 103. 
Fatima is the 4th one into a room. Chitra is the last to enter her room. 
So, the one at 7:45 should be entering a different room. Who is this?

A, B and F go to room number 101. C should go to 102, E should go to 103. E should be the one
who goes at 7:45 am. 
We have B, D and G remaining. B and D go into 101, G goes to 102. B comes after D as B is the
third to go into 101. 
Write down the different combinations.

7:10, 7:15 and 7:25 could be D, B, G or D, G, B or G, D, B. 
Let us move on to questions. 
Ganeshan was the first to enter room 102. Therefore no one was there before Ganeshan entered
room 102.

Look at the statements given by Erina and Ganeshan. 
Think about the number of candidates in the rooms. 
One room has only one candidate, one room has 2 candidates, so the third room should have 4
candidates. 
So, number of candidates should be distributed as 4, 2, 1. 
But which room has 2 candidates and which one has 4? 
Room number 102 has 2 candidates. G plus one more. Balaram was the third one to enter 101, so
101 should have had at least 3 candidates. 
This implies 103 has only one candidate. 
Fatima is the 4th one into a room. What does this mean? Akhil should also be in the same room as
Fatima.

A, B and F go to room number 101. C should go to 102, E should go to 103. 
Fatima is the 4th one into a room. Chitra is the last to enter her room. 
So, the one at 7:45 should be entering a different room. Who is this?

A, B and F go to room number 101. C should go to 102, E should go to 103. E should be the one
who goes at 7:45 am. 
We have B, D and G remaining. B and D go into 101, G goes to 102. B comes after D as B is the
third to go into 101. 
Write down the different combinations.

7:10, 7:15 and 7:25 could be D, B, G or D, G, B or G, D, B. 
Let us move on to questions. 
From the table, we can see that Erina was the last one to reach i.e. at 7:45 am.

Look at the statements given by Erina and Ganeshan. 
Think about the number of candidates in the rooms. 
One room has only one candidate, one room has 2 candidates, so the third room should have 4
candidates. 
So, number of candidates should be distributed as 4, 2, 1. 
But which room has 2 candidates and which one has 4? 
Room number 102 has 2 candidates. G plus one more. Balaram was the third one to enter 101, so
101 should have had at least 3 candidates. 
This implies 103 has only one candidate. 
Fatima is the 4th one into a room. What does this mean? Akhil should also be in the same room as
Fatima.

A, B and F go to room number 101. C should go to 102, E should go to 103. 
Fatima is the 4th one into a room. Chitra is the last to enter her room. 
So, the one at 7:45 should be entering a different room. Who is this?

A, B and F go to room number 101. C should go to 102, E should go to 103. E should be the one
who goes at 7:45 am. 
We have B, D and G remaining. B and D go into 101, G goes to 102. B comes after D as B is the
third to go into 101. 
Write down the different combinations.

7:10, 7:15 and 7:25 could be D, B, G or D, G, B or G, D, B. 
Let us move on to questions. 
If Ganeshan entered the venue before Divya, we are looking at possibility 3 from the table. 
Therefore Balaram entered the venue at 7:25 am

Look at the statements given by Erina and Ganeshan. 
Think about the number of candidates in the rooms. 
One room has only one candidate, one room has 2 candidates, so the third room should have 4
candidates. 
So, number of candidates should be distributed as 4, 2, 1. 
But which room has 2 candidates and which one has 4? 
Room number 102 has 2 candidates. G plus one more. Balaram was the third one to enter 101, so
101 should have had at least 3 candidates. 
This implies 103 has only one candidate. 
Fatima is the 4th one into a room. What does this mean? Akhil should also be in the same room as
Fatima.

A, B and F go to room number 101. C should go to 102, E should go to 103. 
Fatima is the 4th one into a room. Chitra is the last to enter her room. 
So, the one at 7:45 should be entering a different room. Who is this?

A, B and F go to room number 101. C should go to 102, E should go to 103. E should be the one
who goes at 7:45 am. 
We have B, D and G remaining. B and D go into 101, G goes to 102. B comes after D as B is the
third to go into 101. 
Write down the different combinations.

7:10, 7:15 and 7:25 could be D, B, G or D, G, B or G, D, B. 
Let us move on to questions. 
From the above table, we can see that Divya is definitely in room 101.

From second constraint, 
OLD Visitors = x, MID = 2x, Young = 4x. 
7x = 140 or, x = 20 , 2x = 40 , 4x = 80. 
Let us fill this in. and then go to constraint 3 and 4.

From constraint 3, Total Platinum = 2y, then YNG platinum should be y. 
From constraint 4, Let us fill OLD-GOLD and OLD-ECO has z.

OLD-GOLD > YNG-GOLD. 
Z > 42 – y Or, y + z > 42 
We need to find MID-GOLD. This is equal to 85 – 2y – z – (42 – y) = 43 – (y + z). 
We know that y + z > 42. So, MID-GOLD has to be 0.

From second constraint, 
OLD Visitors = x, MID = 2x, Young = 4x. 
7x = 140 or, x = 20 , 2x = 40 , 4x = 80. 
Let us fill this in. and then go to constraint 3 and 4.

From constraint 3, Total Platinum = 2y, then YNG platinum should be y. 
From constraint 4, Let us fill OLD-GOLD and OLD-ECO has z.

OLD-PT = MID-ECO. MID-ECO = 17 – z or, OLD-PT = 17 – z. 
17 – z = - 20 – 2z. Or, z = 3.

From second constraint, 
OLD Visitors = x, MID = 2x, Young = 4x. 
7x = 140 or, x = 20 , 2x = 40 , 4x = 80. 
Let us fill this in. and then go to constraint 3 and 4.

From constraint 3, Total Platinum = 2y, then YNG platinum should be y. 
From constraint 4, Let us fill OLD-GOLD and OLD-ECO has z.

Let us go statement-by-statement and see if we can find some easy one that HAS to be FALSE. 
(A) The numbers of Gold and Platinum tickets bought by Young visitors were equal. 
YNG-PT = YNG-GOLD. These two should add up to 42. Or, both should be 21. Total PT should be
42. Total GOLD should be 43. 
All of these appear to be possible. Let us look at the other choices and see if we can find an easier
CLEARLY FALSE statement. 
Else, we will revisit this.

(B) The numbers of Middle-aged and Young visitors buying Gold tickets were equal. 
MID-GOLD = YNG-GOLD. MID-GOLD should also be 42 – y. This tells us that z should be 1. This
also appears prima-facie possible. 
Let us look at the others also before we revisit this one and completely fill the grid.

(C) The numbers of Old and Middle-aged visitors buying Economy tickets were equal. 
OLD-ECO = MID-ECO.These two should add up to 17 so these two CANNOT be equal. 
So, statement C is DEFINITELY FALSE. 
(D) The numbers of Old and Middle-aged visitors buying Platinum tickets were equal. 
OLD-PT = MID-PT. Both of these should be y2𝑦2. This also appears to be alright.

From second constraint, 
OLD Visitors = x, MID = 2x, Young = 4x. 
7x = 140 or, x = 20 , 2x = 40 , 4x = 80. 
Let us fill this in. and then go to constraint 3 and 4.

From constraint 3, Total Platinum = 2y, then YNG platinum should be y. 
From constraint 4, Let us fill OLD-GOLD and OLD-ECO has z.

OLD-PT should be equal to MID-PT, OLD-PT and MID-PT add up to y. 
So, each should be y2𝑦2. Now, we know that y2𝑦2 = 20 – 2z. 
Total number of PT seats = 2y = 4 (20 – 2z) = 8 (10 – z). So, this number should be a multiple of 8. 
The only possible answer is 32

The companies classified their products into four categories based on a combination of scores (out
of 20) on the two parameters - Product popularity and Market potential as given below

From condition 1, Alfa and Bravo should have 2 each, Charlie should have 3. 
From condition 2, Bravo = 1, Charlie = 2, Alfa = 3. 
From condition 3, Alfa = Bravo = Charlie = 1. 
From condition 4, Alpha = 4, Bravo = 3, Charlie = 0. 
Let us capture this differently and then take it from there. 

From condition 5, The big free square is Bravo. 
From condition 6, The big free square is Charlie. 
From condition 7, Charlie has rectangle and Square. 
From condition 8, Chances are the big square in Promising is Charlie’s. 
A) Alfa Block Buster revenues = 9 squares. Charlie Promising Revenues = 9 square units. This is
correct. 
B) Bravo Blockbuster = 10 squares, Alfa doubtful = 12 squares.This is not correct. 
C) No-Hope = 15 squares, Doubtful = 29 squares. This is correct. 
D) Given in the question. This is correct.

The companies classified their products into four categories based on a combination of scores (out
of 20) on the two parameters - Product popularity and Market potential as given below

From condition 1, Alfa and Bravo should have 2 each, Charlie should have 3. 
From condition 2, Bravo = 1, Charlie = 2, Alfa = 3. 
From condition 3, Alfa = Bravo = Charlie = 1. 
From condition 4, Alpha = 4, Bravo = 3, Charlie = 0. 
Let us capture this differently and then take it from there. 

From condition 5, The big free square is Bravo. 
From condition 6, The big free square is Charlie. 

From condition 7, Charlie has rectangle and Square. 
From condition 8, Chances are the big square in Promising is Charlie’s. 
Bravo’s revenues are from 7 companies. 
In No-hope, the revenues should be Rs. 4 Crores. 
In Blockbuster, the revenues should be 4 + 6 = Rs. 10 Crores. 
In Promising, the revenues should be Rs. 2 Crores.(this could be 3) 
In Doubtful, the revenues should be 9 + 6 + 2 = Rs. 17 Crores. 
This adds up to Rs. 33 crores. 
The answer choice should be C. 

The companies classified their products into four categories based on a combination of scores (out
of 20) on the two parameters - Product popularity and Market potential as given below

From condition 1, Alfa and Bravo should have 2 each, Charlie should have 3. 
From condition 2, Bravo = 1, Charlie = 2, Alfa = 3. 
From condition 3, Alfa = Bravo = Charlie = 1. 
From condition 4, Alpha = 4, Bravo = 3, Charlie = 0. 
Let us capture this differently and then take it from there. 

From condition 5, The big free square is Bravo. 
From condition 6, The big free square is Charlie. 
From condition 7, Charlie has rectangle and Square. 
From condition 8, Chances are the big square in Promising is Charlie’s. 
It is between Blockbuster and Doubtful. Doubtful has two small squares whereas everything in
Blockbuster is sizable. Therefore Blockbuster had the highest revenue.

The companies classified their products into four categories based on a combination of scores (out
of 20) on the two parameters - Product popularity and Market potential as given below

From condition 1, Alfa and Bravo should have 2 each, Charlie should have 3. 
From condition 2, Bravo = 1, Charlie = 2, Alfa = 3. 
From condition 3, Alfa = Bravo = Charlie = 1. 
From condition 4, Alpha = 4, Bravo = 3, Charlie = 0. 
Let us capture this differently and then take it from there. 

From condition 5, The big free square is Bravo. 
From condition 6, The big free square is Charlie. 
From condition 7, Charlie has rectangle and Square. 
From condition 8, Chances are the big square in Promising is Charlie’s. 
From the table we can see that Choice B 1,3,1,2 is the correct sequence of no. of products.

From condition 3, we get the above diagram

Condition 6 tells us p – 1 = 7. G-total = 17. G and something else = 10. G-only should be 7.

From condition 1, we get the above diagram.

From condition 6, we can get the diagram. 

From condition 4, we get the above diagram. 

From condition 5, 
Number of students in L = 4 + 5 + 8 + 6 – x = 23- x. 
Number of students in K = 4 + 5 + 9 + x = 18 + x. 
We know that 23 – x > 18 + x. 
5 > 2x. x < 5252. 
x could be 0, 1 or 2.

The other three should have opted out of one or the other of G and L. Let us assume m students
left G, 3 – m should have left L. 
Let us rejig the diagram. 
Total number of students in G = 17 – m. 
Total number of students in L = 20 + m – x. 
20 + m – x – (17 – m) = 3 + 2m – x = 6. 
2m – x = 3. x can only take values 0, 1 and 2. 2m = 3 + x. 
Or, x has to be 1. m has to be 2. 
Both G and L = 7 – x = 6.

From condition 3, we get the above diagram

Condition 6 tells us p – 1 = 7. G-total = 17. G and something else = 10. G-only should be 7.

From condition 1, we get the above diagram.

From condition 6, we can get the diagram. 

From condition 4, we get the above diagram. 

From condition 5, 
Number of students in L = 4 + 5 + 8 + 6 – x = 23- x. 
Number of students in K = 4 + 5 + 9 + x = 18 + x. 
We know that 23 – x > 18 + x. 
5 > 2x. x < 5252. 
x could be 0, 1 or 2.

For 6 - x to be minimum, x should be maximum i.e. x should take 2. 
Therefore the minimum number of students enrolled in both G and L but not in K is 4. 

From condition 3, we get the above diagram

Condition 6 tells us p – 1 = 7. G-total = 17. G and something else = 10. G-only should be 7.

From condition 1, we get the above diagram.

From condition 6, we can get the diagram. 

From condition 4, we get the above diagram. 

From condition 5, 
Number of students in L = 4 + 5 + 8 + 6 – x = 23- x. 
Number of students in K = 4 + 5 + 9 + x = 18 + x. 
We know that 23 – x > 18 + x. 
5 > 2x. x < 5252. 
x could be 0, 1 or 2.

Number of students in L = 4 + 5 + 8 + 6 – x = 23- x. 
Number of students in K = 4 + 5 + 9 + x = 18 + x. 
Ratio of K : L = 19 : 22. or, x = 1. 
Or, number of students in L = 22.

From condition 3, we get the above diagram

Condition 6 tells us p – 1 = 7. G-total = 17. G and something else = 10. G-only should be 7.

From condition 1, we get the above diagram.

From condition 6, we can get the diagram. 

From condition 4, we get the above diagram. 

From condition 5, 
Number of students in L = 4 + 5 + 8 + 6 – x = 23- x. 
Number of students in K = 4 + 5 + 9 + x = 18 + x. 
We know that 23 – x > 18 + x. 
5 > 2x. x < 5252. 
x could be 0, 1 or 2.

We know that G ∩ K ∩ L = 0. We know that the 4 that were originally here are distributed among
the three regions currently having x, 6 – x and 5. 
We know that one student leaves K. So, this student should have gone to the region (G and L but
not K). 
Or, (G and L but not K) will now read 7 – x. 
The other three should have opted out of one or the other of G and L. Let us assume m students
left G, 3 – m should have left L. 
Let us rejig the diagram. 
Total number of students in G = 17 – m. 
Total number of students in L = 20 + m – x. 
20 + m – x – (17 – m) = 3 + 2m – x = 6. 
2m – x = 3. x can only take values 0, 1 and 2. 2m = 3 + x. 
Or, x has to be 1. m has to be 2. 
Both G and K = 3 + x – m. 3 + 1 -2 = 2.

From the above table, we can see that the profits of Azra, Bysi, Dipq went up in 2017 from 2016.

From the above table, we can see that Azra had the highest revenue in 2016. 

From the above table, we can see that Cxqi had the highest profit in 2016.

From the above table, we can see that Bysi had the highest profit in 2017. 

Let us look at this one at a time. Statement 1 does not give us clear comparables, let us start with
statement 2.

Ratings in F and P are identical, so we can infer that Weightage for R > Weightage for I. 
R > I. 

Now, let us move to statement 3.

Ratings in F and R are identical, so we can infer that Weightage for I > Weightage for P. 
I > P. 
Now, let us combine these two inferences – we get R > I > P.

Now, let us look at Statement 1. 
Scores in P are identical. Best Ed has gotten better scores in F and R and a worse score in I. 
Further Best Ed has gotten better score by one grade in F and R, but worse score by two grades
in I. 
So, Best Ed’s better score in two grades is more than offset by High Q’s better score in one
parameter. 

High Q is -1 in F and R, but +2 in I. We know that weightage given for R > I. 
So, the -1 in R has a great impact, so the -1 in F should have much lower impact. 
In other words, weightage for F should be small and that for I should be high. 

Let us account for R > I > P first. 
Of the 4 possibilities, we can easily eliminate the last one as F cannot be 0.4. 
F cannot be 0.3 either, so the 3rd one can also be eliminated.

Even if F were 0.2, High Q would not be higher than Best End – the two scores would be equal. 
So, the weightages have to be F = 0.1, R = 0.4, P = 0.2 and I = 0.3. 
So, the weightages are as follows.

From the above table, we can see that the highest overall score among the eight colleges is 48.

Let us look at this one at a time. Statement 1 does not give us clear comparables, let us start with
statement 2.

Ratings in F and P are identical, so we can infer that Weightage for R > Weightage for I. 
R > I. 

Now, let us move to statement 3.

Ratings in F and R are identical, so we can infer that Weightage for I > Weightage for P. 
I > P. 
Now, let us combine these two inferences – we get R > I > P.

Now, let us look at Statement 1. 
Scores in P are identical. Best Ed has gotten better scores in F and R and a worse score in I. 
Further Best Ed has gotten better score by one grade in F and R, but worse score by two grades
in I. 
So, Best Ed’s better score in two grades is more than offset by High Q’s better score in one
parameter. 

High Q is -1 in F and R, but +2 in I. We know that weightage given for R > I. 
So, the -1 in R has a great impact, so the -1 in F should have much lower impact. 
In other words, weightage for F should be small and that for I should be high. 

Let us account for R > I > P first. 
Of the 4 possibilities, we can easily eliminate the last one as F cannot be 0.4. 
F cannot be 0.3 either, so the 3rd one can also be eliminated.

Even if F were 0.2, High Q would not be higher than Best End – the two scores would be equal. 
So, the weightages have to be F = 0.1, R = 0.4, P = 0.2 and I = 0.3. 
So, the weightages are as follows.

From the above table, we can see that the number of colleges which have overall scores between
31 and 40 both inclusive is 0.

Let us look at this one at a time. Statement 1 does not give us clear comparables, let us start with
statement 2.

Ratings in F and P are identical, so we can infer that Weightage for R > Weightage for I. 
R > I. 

Now, let us move to statement 3.

Ratings in F and R are identical, so we can infer that Weightage for I > Weightage for P. 
I > P. 
Now, let us combine these two inferences – we get R > I > P.

Now, let us look at Statement 1. 
Scores in P are identical. Best Ed has gotten better scores in F and R and a worse score in I. 
Further Best Ed has gotten better score by one grade in F and R, but worse score by two grades
in I. 
So, Best Ed’s better score in two grades is more than offset by High Q’s better score in one
parameter. 

High Q is -1 in F and R, but +2 in I. We know that weightage given for R > I. 
So, the -1 in R has a great impact, so the -1 in F should have much lower impact. 
In other words, weightage for F should be small and that for I should be high. 

Let us account for R > I > P first. 
Of the 4 possibilities, we can easily eliminate the last one as F cannot be 0.4. 
F cannot be 0.3 either, so the 3rd one can also be eliminated.

Even if F were 0.2, High Q would not be higher than Best End – the two scores would be equal. 
So, the weightages have to be F = 0.1, R = 0.4, P = 0.2 and I = 0.3. 
So, the weightages are as follows.

From the above table, we can see that the weight of the faculty quality parameter is 0.1.

Let us look at this one at a time. Statement 1 does not give us clear comparables, let us start with
statement 2.

Ratings in F and P are identical, so we can infer that Weightage for R > Weightage for I. 
R > I. 

Now, let us move to statement 3.

Ratings in F and R are identical, so we can infer that Weightage for I > Weightage for P. 
I > P. 
Now, let us combine these two inferences – we get R > I > P.

Now, let us look at Statement 1. 
Scores in P are identical. Best Ed has gotten better scores in F and R and a worse score in I. 
Further Best Ed has gotten better score by one grade in F and R, but worse score by two grades
in I. 
So, Best Ed’s better score in two grades is more than offset by High Q’s better score in one
parameter. 

High Q is -1 in F and R, but +2 in I. We know that weightage given for R > I. 
So, the -1 in R has a great impact, so the -1 in F should have much lower impact. 
In other words, weightage for F should be small and that for I should be high. 

Let us account for R > I > P first. 
Of the 4 possibilities, we can easily eliminate the last one as F cannot be 0.4. 
F cannot be 0.3 either, so the 3rd one can also be eliminated.

Even if F were 0.2, High Q would not be higher than Best End – the two scores would be equal. 
So, the weightages have to be F = 0.1, R = 0.4, P = 0.2 and I = 0.3. 
So, the weightages are as follows.

From the above table, we can see that the number of colleges which received AAA accredditation
is 3. 

Minimum possible number is when there is 1 of type a and 99 of type b which is in accordance to the condition

Maximum possible number is when 1 of a, 2 of b, 4 of c, 8 of d, 16 of e, 32 of f,

Now left boxes would be 100 – (1+2+…32) = 37

Now if one more type is to be added then we need at least 64 which is not available thus maximum possible is 6.

Let’s try to prove the given options possible using easy numbers.

op1:never possible

op2: 1,30,69 is possible

op3: 1,2,4,18.75 is possible

op4: 1,9,30,60 is possible.

There have to be then at least 31 + `1 + 43 = 75 gifts of same type, Thus maximum possible number of boxes = 5 when all types are lest 1,2,4,18,75

The percentage share in Revenue and cost in the different years are as follows

Assume the cost of the store in 2016 to be 100.
As the profit percentage that year was 100, the revenue of the store in that year would be
200 .
Revenue in 2017 would be 400 and cost of the store in 2018 would be 200 . Given that in
2017,30% of 400 = 40% of cost.
120 = 40% of cost or cost in 2017=300
In 2018,50% of 200 = 40% of Revenue
Revenue in 2018 100/0.4= 250
We have the following values for Revenue and cost for the different years

 Profit percentage of the store in 2017 =100 /300 =33.3%
Profit percentage of the store in 2018 = 50/200 = 25.0%
The required difference 33.3- 25.0 8.3

The percentage share in Revenue and cost in the different years are as follows

Assume the cost of the store in 2016 to be 100.
As the profit percentage that year was 100, the revenue of the store in that year would be
200 .
Revenue in 2017 would be 400 and cost of the store in 2018 would be 200 . Given that in
2017,30% of 400 = 40% of cost.
120 = 40% of cost or cost in 2017=300
In 2018,50% of 200 = 40% of Revenue
Revenue in 2018 100/0.4= 250
We have the following values for Revenue and cost for the different years

Percentage profit of the store in 2018= 50/200 = 25%

The percentage share in Revenue and cost in the different years are as follows

Assume the cost of the store in 2016 to be 100.
As the profit percentage that year was 100, the revenue of the store in that year would be
200 .
Revenue in 2017 would be 400 and cost of the store in 2018 would be 200 . Given that in
2017,30% of 400 = 40% of cost.
120 = 40% of cost or cost in 2017=300
In 2018,50% of 200 = 40% of Revenue
Revenue in 2018 100/0.4= 250
We have the following values for Revenue and cost for the different years

 The ratio of the revenues =160 :100 = 8 :5

The percentage share in Revenue and cost in the different years are as follows

Assume the cost of the store in 2016 to be 100.
As the profit percentage that year was 100, the revenue of the store in that year would be
200 .
Revenue in 2017 would be 400 and cost of the store in 2018 would be 200 . Given that in
2017,30% of 400 = 40% of cost.
120 = 40% of cost or cost in 2017=300
In 2018,50% of 200 = 40% of Revenue
Revenue in 2018 100/0.4= 250
We have the following values for Revenue and cost for the different years

 Profit of the store from the Electronics department in 2016 =100 -30 = 70
Total profit =100 The required percentage = 70%.

In the following the names of the faculties are referred by first letter of their names.
Given that each of MT and ET carries 100 marks. Each of MT and ET has at least four
questions of 5 marks (4 x 5 = 20), at least three questions of 10 marks (3 x 10 =30) and at
least two questions of 15 marks (2 x 15 = 30). These together add up to 80 marks. The
remaining 20 marks can be of the following possible combinations. (Four questions of 5
marks) or (two questions of 10 marks) or (one question of 5 marks and 1question of 15
marks)or (two questions of five marks and one question of ten marks). Hence, the total
number of questions in MT or ET can be 11 or 12 or 13. It is given that ET has more number
of questions than in MT. Hence, MT has 11 or 12 questions.
It is given that the number of questions given by any faculty in both MT and ET together is
the same. If MT has 12 questions and ET has 13 questions, or if MT has 11 questions and ET
has 12 questions, this condition cannot be satisfied. Hence, MT has 11 (Five 5 marks, three
10 marks and three 15 marks) questions and ET has 13 questions (eight 5 marks, three ten
marks and two 15 marks). This implies, each faculty has given four questions in MT and ET
together. Since it is given that faculty A has given only one question in MT and each of the
other faculties has given more than one question, each of the faculties B, C, D, E and F has
given two questions in MT. This implies faculty A has given three questions in ET and all
other faculties have given two questions each in ET. From the given data we get the
following.

Except A, every other faculty gave at least two questions for MT and all the questions of a
faculty appeared consecutively. Hence, 2nd question in MT is given by F, 4th by C, 10th by D. B
also has given two questions and both appeared consecutively. Hence, 6th and 7th questions
are given by B and the 9th question is given by E. In each test first all 5 marks questions
appeared followed by 10 marks questions and then 15 marks questions. It can be
understood that MT has five 5 marks, three 10 marks and three 15 marks. Hence, in MT
questions 1 to 5 carry 5 marks each, 6 to 8 carry 10 marks each and 9 to 11 carry 15 marks
each.
It can be understood that A has given three questions for ET and each of the others has given two questions. Hence, the 2nd question of ET is given by D, the 4th question by C,
6th and 7th questions by A, 9th by E, 10th and 11th by B and 12th by F. Since ET has eight 5
marks questions, three ten marks questions and two 15 marks questions, questions 1 to 8 of
ET carry 8 marks each, 9 to 11 carry ten marks each, 12 and 13 carry 15 marks each. Thus,
we get the following:

In the following the names of the faculties are referred by first letter of their names.
Given that each of MT and ET carries 100 marks. Each of MT and ET has at least four
questions of 5 marks (4 x 5 = 20), at least three questions of 10 marks (3 x 10 =30) and at
least two questions of 15 marks (2 x 15 = 30). These together add up to 80 marks. The
remaining 20 marks can be of the following possible combinations. (Four questions of 5
marks) or (two questions of 10 marks) or (one question of 5 marks and 1question of 15
marks)or (two questions of five marks and one question of ten marks). Hence, the total
number of questions in MT or ET can be 11 or 12 or 13. It is given that ET has more number
of questions than in MT. Hence, MT has 11 or 12 questions.
It is given that the number of questions given by any faculty in both MT and ET together is
the same. If MT has 12 questions and ET has 13 questions, or if MT has 11 questions and ET
has 12 questions, this condition cannot be satisfied. Hence, MT has 11 (Five 5 marks, three
10 marks and three 15 marks) questions and ET has 13 questions (eight 5 marks, three ten
marks and two 15 marks). This implies, each faculty has given four questions in MT and ET
together. Since it is given that faculty A has given only one question in MT and each of the
other faculties has given more than one question, each of the faculties B, C, D, E and F has
given two questions in MT. This implies faculty A has given three questions in ET and all
other faculties have given two questions each in ET. From the given data we get the
following.

Except A, every other faculty gave at least two questions for MT and all the questions of a
faculty appeared consecutively. Hence, 2nd question in MT is given by F, 4th by C, 10th by D. B
also has given two questions and both appeared consecutively. Hence, 6th and 7th questions
are given by B and the 9th question is given by E. In each test first all 5 marks questions
appeared followed by 10 marks questions and then 15 marks questions. It can be
understood that MT has five 5 marks, three 10 marks and three 15 marks. Hence, in MT
questions 1 to 5 carry 5 marks each, 6 to 8 carry 10 marks each and 9 to 11 carry 15 marks
each.
It can be understood that A has given three questions for ET and each of the others has given two questions. Hence, the 2nd question of ET is given by D, the 4th question by C,
6th and 7th questions by A, 9th by E, 10th and 11th by B and 12th by F. Since ET has eight 5
marks questions, three ten marks questions and two 15 marks questions, questions 1 to 8 of
ET carry 8 marks each, 9 to 11 carry ten marks each, 12 and 13 carry 15 marks each. Thus,
we get the following:

In the following the names of the faculties are referred by first letter of their names.
Given that each of MT and ET carries 100 marks. Each of MT and ET has at least four
questions of 5 marks (4 x 5 = 20), at least three questions of 10 marks (3 x 10 =30) and at
least two questions of 15 marks (2 x 15 = 30). These together add up to 80 marks. The
remaining 20 marks can be of the following possible combinations. (Four questions of 5
marks) or (two questions of 10 marks) or (one question of 5 marks and 1question of 15
marks)or (two questions of five marks and one question of ten marks). Hence, the total
number of questions in MT or ET can be 11 or 12 or 13. It is given that ET has more number
of questions than in MT. Hence, MT has 11 or 12 questions.
It is given that the number of questions given by any faculty in both MT and ET together is
the same. If MT has 12 questions and ET has 13 questions, or if MT has 11 questions and ET
has 12 questions, this condition cannot be satisfied. Hence, MT has 11 (Five 5 marks, three
10 marks and three 15 marks) questions and ET has 13 questions (eight 5 marks, three ten
marks and two 15 marks). This implies, each faculty has given four questions in MT and ET
together. Since it is given that faculty A has given only one question in MT and each of the
other faculties has given more than one question, each of the faculties B, C, D, E and F has
given two questions in MT. This implies faculty A has given three questions in ET and all
other faculties have given two questions each in ET. From the given data we get the
following.

Except A, every other faculty gave at least two questions for MT and all the questions of a
faculty appeared consecutively. Hence, 2nd question in MT is given by F, 4th by C, 10th by D. B
also has given two questions and both appeared consecutively. Hence, 6th and 7th questions
are given by B and the 9th question is given by E. In each test first all 5 marks questions
appeared followed by 10 marks questions and then 15 marks questions. It can be
understood that MT has five 5 marks, three 10 marks and three 15 marks. Hence, in MT
questions 1 to 5 carry 5 marks each, 6 to 8 carry 10 marks each and 9 to 11 carry 15 marks
each.
It can be understood that A has given three questions for ET and each of the others has given two questions. Hence, the 2nd question of ET is given by D, the 4th question by C,
6th and 7th questions by A, 9th by E, 10th and 11th by B and 12th by F. Since ET has eight 5
marks questions, three ten marks questions and two 15 marks questions, questions 1 to 8 of
ET carry 8 marks each, 9 to 11 carry ten marks each, 12 and 13 carry 15 marks each. Thus,
we get the following:

In the following the names of the faculties are referred by first letter of their names.
Given that each of MT and ET carries 100 marks. Each of MT and ET has at least four
questions of 5 marks (4 x 5 = 20), at least three questions of 10 marks (3 x 10 =30) and at
least two questions of 15 marks (2 x 15 = 30). These together add up to 80 marks. The
remaining 20 marks can be of the following possible combinations. (Four questions of 5
marks) or (two questions of 10 marks) or (one question of 5 marks and 1question of 15
marks)or (two questions of five marks and one question of ten marks). Hence, the total
number of questions in MT or ET can be 11 or 12 or 13. It is given that ET has more number
of questions than in MT. Hence, MT has 11 or 12 questions.
It is given that the number of questions given by any faculty in both MT and ET together is
the same. If MT has 12 questions and ET has 13 questions, or if MT has 11 questions and ET
has 12 questions, this condition cannot be satisfied. Hence, MT has 11 (Five 5 marks, three
10 marks and three 15 marks) questions and ET has 13 questions (eight 5 marks, three ten
marks and two 15 marks). This implies, each faculty has given four questions in MT and ET
together. Since it is given that faculty A has given only one question in MT and each of the
other faculties has given more than one question, each of the faculties B, C, D, E and F has
given two questions in MT. This implies faculty A has given three questions in ET and all
other faculties have given two questions each in ET. From the given data we get the
following.

Except A, every other faculty gave at least two questions for MT and all the questions of a
faculty appeared consecutively. Hence, 2nd question in MT is given by F, 4th by C, 10th by D. B
also has given two questions and both appeared consecutively. Hence, 6th and 7th questions
are given by B and the 9th question is given by E. In each test first all 5 marks questions
appeared followed by 10 marks questions and then 15 marks questions. It can be
understood that MT has five 5 marks, three 10 marks and three 15 marks. Hence, in MT
questions 1 to 5 carry 5 marks each, 6 to 8 carry 10 marks each and 9 to 11 carry 15 marks
each.
It can be understood that A has given three questions for ET and each of the others has given two questions. Hence, the 2nd question of ET is given by D, the 4th question by C,
6th and 7th questions by A, 9th by E, 10th and 11th by B and 12th by F. Since ET has eight 5
marks questions, three ten marks questions and two 15 marks questions, questions 1 to 8 of
ET carry 8 marks each, 9 to 11 carry ten marks each, 12 and 13 carry 15 marks each. Thus,
we get the following:

The minimum and maximum and possible number of coins (overall) in each slot would be as
follows:

It is given that the average amount of money kept in the nine pouches in any column or any
row is an integer (a multiple of nine).
The total amount of money in the first column must be either 18 or 27 . The minimum value
of the sum of money in the three slots is 8 11 4 23 and the maximum value is
10+13+ 4 = 27.
The number of coins in the first column of the three rows are 10(2 + 4 + 4),13(3+ 5 + 5)
and 4(1+ 2 +1) Similarly in the third row, the sum must be 18 and in the second column, the
sum must be 27.
The number of coins in the second column is 20(6+6 + 8) + 3(1+1+1) and 4(1+1+ 2)
The third column in the first row would be 6(1+ 2 + 3) and the third column in the third row
would be 10(2 + 3 +5)
In the last column, the value in the second row would be 54 -16 = 38(6 +12 +20)
We have the following figure for the number of coins in the pouches in each slot

The minimum and maximum and possible number of coins (overall) in each slot would be as
follows:

It is given that the average amount of money kept in the nine pouches in any column or any
row is an integer (a multiple of nine).
The total amount of money in the first column must be either 18 or 27 . The minimum value
of the sum of money in the three slots is 8 11 4 23 and the maximum value is
10+13+ 4 = 27.
The number of coins in the first column of the three rows are 10(2 + 4 + 4),13(3+ 5 + 5)
and 4(1+ 2 +1) Similarly in the third row, the sum must be 18 and in the second column, the
sum must be 27.
The number of coins in the second column is 20(6+6 + 8) + 3(1+1+1) and 4(1+1+ 2)
The third column in the first row would be 6(1+ 2 + 3) and the third column in the third row
would be 10(2 + 3 +5)
In the last column, the value in the second row would be 54 -16 = 38(6 +12 +20)
We have the following figure for the number of coins in the pouches in each slot

The minimum and maximum and possible number of coins (overall) in each slot would be as
follows:

It is given that the average amount of money kept in the nine pouches in any column or any
row is an integer (a multiple of nine).
The total amount of money in the first column must be either 18 or 27 . The minimum value
of the sum of money in the three slots is 8 11 4 23 and the maximum value is
10+13+ 4 = 27.
The number of coins in the first column of the three rows are 10(2 + 4 + 4),13(3+ 5 + 5)
and 4(1+ 2 +1) Similarly in the third row, the sum must be 18 and in the second column, the
sum must be 27.
The number of coins in the second column is 20(6+6 + 8) + 3(1+1+1) and 4(1+1+ 2)
The third column in the first row would be 6(1+ 2 + 3) and the third column in the third row
would be 10(2 + 3 +5)
In the last column, the value in the second row would be 54 -16 = 38(6 +12 +20)
We have the following figure for the number of coins in the pouches in each slot

In three slots (row 2 , column 1), (row 1 , column 2) and (row 2, column 3), the amount in
the three pouches strictly exceeds 10