CAT 2008 Question Paper

We know the following relation that , revenue = volume * ARDT . To have equal volue the ratio of revenue by ARDT should be constant.
Considering UK we have volume = 12.5/30 and of Spain we have 6.5/15 which is fairly constant as compared to others.
Hence option D. 

We can see that Bhama got calls from all the college hence it is essential for her to clear all 4 sectionals along with overall cutoff. That essentially means that she has to score marks which are greater or equal to the greatest cut-off for that section across the six colleges. from the table we can see that to clear section A the cut offs for colleges 1, 4 and 5 are 42, 43 and 45 respectively. Hence, in order to clear the sectional cut-off of section A for all the colleges, she should have scored at least 45 marks.
Similarly, the minimum marks required sectional cut-off for section B for all colleges = Max{42, 42, 41} = 45.
Similarly, the minimum marks required sectional cut-off for section C for all colleges = Max{42, 45, 46, 43} = 46.
Similarly, the minimum marks required sectional cut-off for section D for all colleges = Max{45, 44} = 45.
So total minimum aggregate marks required is 45+45+46+45 = 181.
Note we have to check for aggregate cut-off as well for all 6 colleges. We can see that 181 is also more than aggregate cut-off marks of all six colleges. Therefore option B is the correct answer. 

We have to minimum marks obtained by him in a section which is possible only when Charlie  must have scored full marks in remaining 3 sections. Along with that we have to ensure that Charlie gets enough marks to clear aggregate cut off as well. 

For college 1, the minimum marks that Charlie needs to get a call = 176. To get the minimum marks in any section he has to score full marks in remaining 3 sections. Since he has to clear the cut off in sections A, B and C we can say that he can score a minimum marks in section D.

Minimum marks in section D despite which he will get call from college 1 = 176 - 3*50 = 26.
Similarly, minimum marks in one section out of  A and D despite which he will get call from college 2 = 175 - 3*50 = 25.
Similarly, minimum marks in one section out of A, B and Ddespite which he will get call from college 3 = 171 - 3*50 = 21.
Similarly, minimum marks in one section out of B and C despite which he will get call from college 4 = 178 - 3*50 = 28.
Similarly, minimum marks in one section out of B and D despite which he will get call from college 5 = 180 - 3*50 = 30.
Similarly, minimum marks in one section out of A and C despite which he will get call from college 6 = 176 - 3*50 = 26.

We can see that Charlie need to score at least 25 marks in one section out of A and D to get call from exactly 2 colleges. Therefore, option C is correct answer. 

Since Aditya didn’t get a call from any of the colleges, so for each college, he either didn’t clear one of the sectional cut-offs or he didn’t clear the aggregate cutoff or both. If he didn’t clear one of the sectional cut-offs, then for that section he scored less marks than the least cut-off among the given cut-offs of all the colleges. 

For example, for section A, it is given that the cut-offs for colleges 1, 4 and 5 are 42, 43 and 45 respectively. The least cut-off among them is 42. So, in order to not clear the sectional cut-off of section A for colleges 1, 4 and 5, he should have scored less than 42.

Similarly for section B, in order to not clear the sectional cut-off for colleges 1, 2 and 6, he should have scored less than 41.
Similarly for section C, in order to not clear the sectional cut-off for colleges 1, 2, 3 and 5, he should have scored less than 42.
Similarly for section D, in order to not clear the sectional cut-off for colleges 4 and 6, he should have scored less than 44. 

We can see that if we consider section C and D then we will cover all six colleges. So the maximum score that one can score in section C and D despite that he misses the sectional cut off are 41 and 43 respectively. Also in order to maximize the total we can assume that the person scored full marks in the both section A and B. 

Maximum marks obtained by Aditya such that he doesn’t get any calls = 41 + 43 + 50 + 50 = 184
Hence, option C is the correct answer. 

There are a total of $$^6 C_2$$ matches => 15 matches. The first 9 matches are held in the first stage and remaining 6 in the second stage.

From the information given, we can conclude that the following matches were held in first stage:

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won)

One team won all matches. As B, C, D E and F have lost at least one match each, A won all three matches. As A, B, D, E have won at least one match, C and F lost both matches.

From the matches already deduced, we can see that A needs to play 2 more matches, B two more matches and C and F one match each. As C and F lose all matches in stage 1, they cannot play against each other. F did not play against the leader i.e. A. Hence, the remaining matches are A-B (A won), A-C (A won), B-F (B won).

Thus, the stage 1 matches are

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won), A-B (A won), A-C (A won), B-F (B won)

Thus Stage 2 matches are D-B, D-E, E-A, F-A, B-C and C-F (all matches - stage 1 matches)

As A lost both matches, F and E must have won the match vs A. As F won against A, F won both its matches and C lost both its matches. One more team lost both its matches. As B, E and F have won at least one match and A and C have been discussed previously, D must have lost both matches. Hence, stage 2 results are:

Stage 2: D-B (B won), D-E (E won), E-A (E won), F-A (F won), B-C (B won) and C-F (F won)

Hence, the two teams that won against stage 1 leader A are E and F.

There are a total of $$^6 C_2$$ matches => 15 matches. The first 9 matches are held in the first stage and remaining 6 in the second stage.

From the information given, we can conclude that the following matches were held in first stage:

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won)

One team won all matches. As B, C, D E and F have lost at least one match each, A won all three matches. As A, B, D, E have won at least one match, C and F lost both matches.

From the matches already deduced, we can see that A needs to play 2 more matches, B two more matches and C and F one match each. As C and F lose all matches in stage 1, they cannot play against each other. F did not play against the leader i.e. A. Hence, the remaining matches are A-B (A won), A-C (A won), B-F (B won).

Thus, the stage 1 matches are

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won), A-B (A won), A-C (A won), B-F (B won)

Thus Stage 2 matches are D-B, D-E, E-A, F-A, B-C and C-F (all matches - stage 1 matches)

As A lost both matches, F and E must have won the match vs A. As F won against A, F won both its matches and C lost both its matches. One more team lost both its matches. As B, E and F have won at least one match and A and C have been discussed previously, D must have lost both matches. Hence, stage 2 results are:

Stage 2: D-B (B won), D-E (E won), E-A (E won), F-A (F won), B-C (B won) and C-F (F won)

Hence, the teams that won both of their stage 2 matches are B, E and F.

There are a total of $$^6 C_2$$ matches => 15 matches. The first 9 matches are held in the first stage and remaining 6 in the second stage.

From the information given, we can conclude that the following matches were held in first stage:

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won)

One team won all matches. As B, C, D E and F have lost at least one match each, A won all three matches. As A, B, D, E have won at least one match, C and F lost both matches.

From the matches already deduced, we can see that A needs to play 2 more matches, B two more matches and C and F one match each. As C and F lose all matches in stage 1, they cannot play against each other. F did not play against the leader i.e. A. Hence, the remaining matches are A-B (A won), A-C (A won), B-F (B won).

Thus, the stage 1 matches are

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won), A-B (A won), A-C (A won), B-F (B won)

Thus Stage 2 matches are D-B, D-E, E-A, F-A, B-C and C-F (all matches - stage 1 matches)

As A lost both matches, F and E must have won the match vs A. As F won against A, F won both its matches and C lost both its matches. One more team lost both its matches. As B, E and F have won at least one match and A and C have been discussed previously, D must have lost both matches. Hence, stage 2 results are:

Stage 2: D-B (B won), D-E (E won), E-A (E won), F-A (F won), B-C (B won) and C-F (F won)

Hence, the wins by each team are A (3), B(4), C(0), D(2), E(4), F(2). Hence, D and F won exactly 2 matches.

There are a total of $$^6 C_2$$ matches => 15 matches. The first 9 matches are held in the first stage and remaining 6 in the second stage.

From the information given, we can conclude that the following matches were held in first stage:

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won)

One team won all matches. As B, C, D E and F have lost at least one match each, A won all three matches. As A, B, D, E have won at least one match, C and F lost both matches.

From the matches already deduced, we can see that A needs to play 2 more matches, B two more matches and C and F one match each. As C and F lose all matches in stage 1, they cannot play against each other. F did not play against the leader i.e. A. Hence, the remaining matches are A-B (A won), A-C (A won), B-F (B won).

Thus, the stage 1 matches are

Stage 1: D-A (A won), D-C (D won), D-F (D won), E-B (B won), E-C (E won), E-F (E won), A-B (A won), A-C (A won), B-F (B won)

Thus Stage 2 matches are D-B, D-E, E-A, F-A, B-C and C-F (all matches - stage 1 matches)

As A lost both matches, F and E must have won the match vs A. As F won against A, F won both its matches and C lost both its matches. One more team lost both its matches. As B, E and F have won at least one match and A and C have been discussed previously, D must have lost both matches. Hence, stage 2 results are:

Stage 2: D-B (B won), D-E (E won), E-A (E won), F-A (F won), B-C (B won) and C-F (F won)

Hence, the wins by each team are A(3), B(4), C(0), D(2), E(4), F(2). Hence, most wins are by B and E.

Percentage increase from 2006 to 2007 is 120/3.8 = 31.5%. approximately

Now if the same % increase is considered over period 2007-2008, we get . 31.5 * 500 = x-500.

X is 657 approximately.

So ,the difference between the estimated subscription in Europe in 2008 and what it would have been if it were computed using the percentage growth rate of 2007 (over 2006), is closest to: 657-600 = 57.  Hence option A.

Suppose there are 100 subscribers from europe in the year 2003.

Out of which 60 are men and 40 are women.

According to given condition in the year 2010 no. of male subscribers would be 60*$$1.05^7$$ = 84 and no. of female subscribers would be 40*$$1.1^7$$ = 78.

So total equal to 162.

Hence % increase is 100*(162-100)/100 = 62.

Hence option A.