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Previous Year Paper

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CAT-2006-Previous Years Paper

Question
19 out of 25
 

Two traders, Chetan and Michael, were involved in the buying and selling of MCS shares over five trading days. At the beginning of the first day, the MCS share was priced at Rs 100, while at the end of the fifth day it was priced at Rs 110. At the end of each day, the MCS share price either went up by Rs 10, or else, it came down by Rs 10. Both Chetan and Michael took buying and selling decisions at the end of each trading day.

The beginning price of MCS share on a given day was the same as the ending price of the previous day.

Chetan and Michael started with the same number of shares and amount of cash, and had enough of both.

Below are some additional facts about how Chetan and Michael traded over the five trading days.

·      Each day if the price went up, Chetan sold 10 shares of MCS at the closing price. On the other hand, each day if the price went down, he bought 10 shares at the closing price.

·      If on any day, the closing price was above Rs 110, then Michael sold 10 shares of MCS, while if it was below Rs 90, he bought 10 shares, all at the closing price.


If Michael ended up with Rs 100 less cash than Chetan at the end of day 5, what was the difference in the number of shares possessed by Michael and Chetan (at the end of day 5)?



A Michael had 10 less shares than Chetan.

B Michael had 10 more shares than Chetan.

C Chetan had 10 more shares than Michael.

D Both had the same number of shares.

On the 1st day, the price was Rs 100 per share. Now we can have the following cases:

Cases

1

2

3

4

5

6

7

8

9

10

Cases

1

2

3

4

5

6

7

8

9

10

At the end of 1st day

90

110

90

110

110

90

110

110

90

110

At the end of 2nd day

100

100

80

120

100

100

120

120

100

120

At the end of 3rd day

90

90

90

110

110

110

130

110

110

110

At the end of 4th day

100

100

100

100

100

100

120

120

120

120

At the end of 5th day

110

110

110

110

110

110

110

110

110

110


Ans. D Obviously from cases 5 and 9, we have:

Chetan

Michael

110

+10

0

120

+10

+10

110

-10

0

100

-10

0

110

+10

0

Chetan

Michael

90

-10

0

100

+10

0

110

+10

0

120

+10

+10

110

-10

0

Using 1st table, Chetan has invested Rs 1300 and Michael has invested Rs 1200.

Michael has Rs 100 less than Chetan, so both have same number of shares.

Hence, option D. is the answer.

CAT-2006-Previous Years Paper Flashcard List

25 flashcards
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In a Class X Board examination, ten papers are distributed over five groups—PCB, Mathematics, Social Science, Vernacular and English. Each of the ten papers is evaluated out of 100. The final score of a student is calculated in the following manner. First, the Group Scores are obtained by averaging marks in the papers within the Group. The final score is the simple average of the Group Scores. The data for the top ten students are presented below. (Dipan’s score in English Paper II has been intentionally removed in the table.) Name of the Student PCB Group Mathemetics Social Science Group Vernacular Group English Group Final Score Phy Chem Bio Hist Geo Paper I Paper II Paper I Paper II Ayesha (G) 98 96 97 98 95 93 94 96 96 98 96.2 Ram B. 97 99 95 97 95 95 96 94 96 98 96.1 Dipan B. 98 98 98 95 96 95 96 94 96 ?? 96.0 Sagnik B. 97 98 99 96 96 98 94 97 92 94 95.9 Sanjiv B. 95 96 97 98 97 96 92 93 95 96 95.7 Shreya (G) 96 89 85 100 97 98 94 95 96 95 95.5 Joseph B. 90 94 98 100 94 97 90 92 94 95 95 Agni B. 96 99 96 99 95 96 82 93 92 93 94.3 Pritam B. 98 98 95 98 83 95 90 93 94 94 93.9 Tima (G) 96 98 97 99 85 94 92 91 87 96 93.7 Note: B or G against the name of a student respectively indicates whether the student is a boy or a girl. Had Joseph, Agni, Pritam and Tirna each obtained Group Score of 100 in the Social Science Group, then their standing in decreasing order of final score would be: A Pritam, Joseph, Tirna, Agni B Joseph, Tirna, Agni, Pritam C Pritam, Agni, Tirna, Joseph D Joseph, Tirna, Pritam, Agni
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In a Class X Board examination, ten papers are distributed over five groups—PCB, Mathematics, Social Science, Vernacular and English. Each of the ten papers is evaluated out of 100. The final score of a student is calculated in the following manner. First, the Group Scores are obtained by averaging marks in the papers within the Group. The final score is the simple average of the Group Scores. The data for the top ten students are presented below. (Dipan’s score in English Paper II has been intentionally removed in the table.) Name of the Student PCB Group Mathemetics Social Science Group Vernacular Group English Group Final Score Phy Chem Bio Hist Geo Paper I Paper II Paper I Paper II Ayesha (G) 98 96 97 98 95 93 94 96 96 98 96.2 Ram B. 97 99 95 97 95 95 96 94 96 98 96.1 Dipan B. 98 98 98 95 96 95 96 94 96 ?? 96.0 Sagnik B. 97 98 99 96 96 98 94 97 92 94 95.9 Sanjiv B. 95 96 97 98 97 96 92 93 95 96 95.7 Shreya (G) 96 89 85 100 97 98 94 95 96 95 95.5 Joseph B. 90 94 98 100 94 97 90 92 94 95 95 Agni B. 96 99 96 99 95 96 82 93 92 93 94.3 Pritam B. 98 98 95 98 83 95 90 93 94 94 93.9 Tima (G) 96 98 97 99 85 94 92 91 87 96 93.7 Note: B or G against the name of a student respectively indicates whether the student is a boy or a girl. Students who obtained Group Scores of at least 95 in every group are eligible to apply for a prize. Among those who are eligible, the student obtaining the highest Group Score in Social Science Group is awarded this prize. The prize was awarded to: A Shreya B Ram C Ayesha D Dipan
10)
In a Class X Board examination, ten papers are distributed over five groups—PCB, Mathematics, Social Science, Vernacular and English. Each of the ten papers is evaluated out of 100. The final score of a student is calculated in the following manner. First, the Group Scores are obtained by averaging marks in the papers within the Group. The final score is the simple average of the Group Scores. The data for the top ten students are presented below. (Dipan’s score in English Paper II has been intentionally removed in the table.) Name of the Student PCB Group Mathemetics Social Science Group Vernacular Group English Group Final Score Phy Chem Bio Hist Geo Paper I Paper II Paper I Paper II Ayesha (G) 98 96 97 98 95 93 94 96 96 98 96.2 Ram B. 97 99 95 97 95 95 96 94 96 98 96.1 Dipan B. 98 98 98 95 96 95 96 94 96 ?? 96.0 Sagnik B. 97 98 99 96 96 98 94 97 92 94 95.9 Sanjiv B. 95 96 97 98 97 96 92 93 95 96 95.7 Shreya (G) 96 89 85 100 97 98 94 95 96 95 95.5 Joseph B. 90 94 98 100 94 97 90 92 94 95 95 Agni B. 96 99 96 99 95 96 82 93 92 93 94.3 Pritam B. 98 98 95 98 83 95 90 93 94 94 93.9 Tima (G) 96 98 97 99 85 94 92 91 87 96 93.7 Note: B or G against the name of a student respectively indicates whether the student is a boy or a girl. Each of the ten students was allowed to improve his/her score in exactly one paper of choice with the objective of maximizing his/her final score. Everyone scored 100 in the paper in which he or she chose to improve. After that, the topper among the ten students was A RamB AgniC PritamD Dipan
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Mathematicians are assigned a number called Erdös number (named after the famous mathematician, Paul Erdös). Only Paul Erdös himself has an Erdös number of zero. Any mathematician who has written a research paper with Erdös has an Erdös number of 1. For other mathematicians, the calculation of his/her Erdös number is illustrated below: Suppose that a mathematician X has co-authored papers with several other mathematicians. From among them, mathematician Y has the smallest Erdös number. Let the Erdös number of Y be y. Then X has an Erdös number of y + 1. Hence, any mathematician with no co-authorship chain connected to Erdös has an Erdös number of infinity. In a seven day long mini-conference organized in memory of Paul Erdös, a close group of eight mathematicians, call them A, B, C, D, E, F, G and H, discussed some research problems. At the beginning of the conference, A was the only participant who had an infinite Erdös number. Nobody had an Erdös number less than that of F. ·      On the third day of the conference F co-authored a paper jointly with A. and C. . This reduced the average Erdös number of the group of eight mathematicians to 3. The Erdös numbers of B. , D. Go through the options. and H remained unchanged with the writing of this paper. Further, no other co-authorship among any three members would have reduced the average Erdös number of the group of eight to as low as 3. ·      At the end of the third day, five members of this group had identical Erdös numbers while the other three had Erdös numbers distinct from each other. ·      On the fifth day, E co-authored a paper with F which reduced the group’s average Erdös number by 0.5. The Erdös numbers of the remaining six were unchanged with the writing of this paper. ·      No other paper was written during the conference.   How many participants in the conference did not change their Erdös number during the conference? A 2 B 3 C 4 D 5
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Mathematicians are assigned a number called Erdös number (named after the famous mathematician, Paul Erdös). Only Paul Erdös himself has an Erdös number of zero. Any mathematician who has written a research paper with Erdös has an Erdös number of 1. For other mathematicians, the calculation of his/her Erdös number is illustrated below: Suppose that a mathematician X has co-authored papers with several other mathematicians. From among them, mathematician Y has the smallest Erdös number. Let the Erdös number of Y be y. Then X has an Erdös number of y + 1. Hence, any mathematician with no co-authorship chain connected to Erdös has an Erdös number of infinity. In a seven day long mini-conference organized in memory of Paul Erdös, a close group of eight mathematicians, call them A, B, C, D, E, F, G and H, discussed some research problems. At the beginning of the conference, A was the only participant who had an infinite Erdös number. Nobody had an Erdös number less than that of F. ·      On the third day of the conference F co-authored a paper jointly with A. and C. . This reduced the average Erdös number of the group of eight mathematicians to 3. The Erdös numbers of B. , D. Go through the options. and H remained unchanged with the writing of this paper. Further, no other co-authorship among any three members would have reduced the average Erdös number of the group of eight to as low as 3. ·      At the end of the third day, five members of this group had identical Erdös numbers while the other three had Erdös numbers distinct from each other. ·      On the fifth day, E co-authored a paper with F which reduced the group’s average Erdös number by 0.5. The Erdös numbers of the remaining six were unchanged with the writing of this paper. ·      No other paper was written during the conference.   The person having the largest Erdös number at the end of the conference must have had Erdös number (at that time) A 5 B 7 C 9 D 14
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Mathematicians are assigned a number called Erdös number (named after the famous mathematician, Paul Erdös). Only Paul Erdös himself has an Erdös number of zero. Any mathematician who has written a research paper with Erdös has an Erdös number of 1. For other mathematicians, the calculation of his/her Erdös number is illustrated below: Suppose that a mathematician X has co-authored papers with several other mathematicians. From among them, mathematician Y has the smallest Erdös number. Let the Erdös number of Y be y. Then X has an Erdös number of y + 1. Hence, any mathematician with no co-authorship chain connected to Erdös has an Erdös number of infinity. In a seven day long mini-conference organized in memory of Paul Erdös, a close group of eight mathematicians, call them A, B, C, D, E, F, G and H, discussed some research problems. At the beginning of the conference, A was the only participant who had an infinite Erdös number. Nobody had an Erdös number less than that of F. ·      On the third day of the conference F co-authored a paper jointly with A. and C. . This reduced the average Erdös number of the group of eight mathematicians to 3. The Erdös numbers of B. , D. Go through the options. and H remained unchanged with the writing of this paper. Further, no other co-authorship among any three members would have reduced the average Erdös number of the group of eight to as low as 3. ·      At the end of the third day, five members of this group had identical Erdös numbers while the other three had Erdös numbers distinct from each other. ·      On the fifth day, E co-authored a paper with F which reduced the group’s average Erdös number by 0.5. The Erdös numbers of the remaining six were unchanged with the writing of this paper. ·      No other paper was written during the conference.   How many participants had the same Erdös number at the beginning of the conference? A 2 B 3 C 4 D 5
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Mathematicians are assigned a number called Erdös number (named after the famous mathematician, Paul Erdös). Only Paul Erdös himself has an Erdös number of zero. Any mathematician who has written a research paper with Erdös has an Erdös number of 1. For other mathematicians, the calculation of his/her Erdös number is illustrated below: Suppose that a mathematician X has co-authored papers with several other mathematicians. From among them, mathematician Y has the smallest Erdös number. Let the Erdös number of Y be y. Then X has an Erdös number of y + 1. Hence, any mathematician with no co-authorship chain connected to Erdös has an Erdös number of infinity. In a seven day long mini-conference organized in memory of Paul Erdös, a close group of eight mathematicians, call them A, B, C, D, E, F, G and H, discussed some research problems. At the beginning of the conference, A was the only participant who had an infinite Erdös number. Nobody had an Erdös number less than that of F. ·      On the third day of the conference F co-authored a paper jointly with A. and C. . This reduced the average Erdös number of the group of eight mathematicians to 3. The Erdös numbers of B. , D. Go through the options. and H remained unchanged with the writing of this paper. Further, no other co-authorship among any three members would have reduced the average Erdös number of the group of eight to as low as 3. ·      At the end of the third day, five members of this group had identical Erdös numbers while the other three had Erdös numbers distinct from each other. ·      On the fifth day, E co-authored a paper with F which reduced the group’s average Erdös number by 0.5. The Erdös numbers of the remaining six were unchanged with the writing of this paper. ·      No other paper was written during the conference.   The Erdös number of C at the end of the conference was A 1 B 2 C 3 D 4
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Mathematicians are assigned a number called Erdös number (named after the famous mathematician, Paul Erdös). Only Paul Erdös himself has an Erdös number of zero. Any mathematician who has written a research paper with Erdös has an Erdös number of 1. For other mathematicians, the calculation of his/her Erdös number is illustrated below: Suppose that a mathematician X has co-authored papers with several other mathematicians. From among them, mathematician Y has the smallest Erdös number. Let the Erdös number of Y be y. Then X has an Erdös number of y + 1. Hence, any mathematician with no co-authorship chain connected to Erdös has an Erdös number of infinity. In a seven day long mini-conference organized in memory of Paul Erdös, a close group of eight mathematicians, call them A, B, C, D, E, F, G and H, discussed some research problems. At the beginning of the conference, A was the only participant who had an infinite Erdös number. Nobody had an Erdös number less than that of F. ·      On the third day of the conference F co-authored a paper jointly with A. and C. . This reduced the average Erdös number of the group of eight mathematicians to 3. The Erdös numbers of B. , D. Go through the options. and H remained unchanged with the writing of this paper. Further, no other co-authorship among any three members would have reduced the average Erdös number of the group of eight to as low as 3. ·      At the end of the third day, five members of this group had identical Erdös numbers while the other three had Erdös numbers distinct from each other. ·      On the fifth day, E co-authored a paper with F which reduced the group’s average Erdös number by 0.5. The Erdös numbers of the remaining six were unchanged with the writing of this paper. ·      No other paper was written during the conference.   The Erdös number of E at the beginning of the conference was A 2 B 5 C 6 D 7
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