Previous CAT Questions - Solved (Logical Reasoning) - Set 9


  • Skadoooosh!!!


    (CAT 2005)
    The year is 2089. Beijing, London, New York, and Paris are in contention to host the 2096 Olympics. The eventual winner is determined through several rounds of voting by members of the IOC with each member representing a different city. All the four cities in contention are also represented in IOC.

    In any round of voting, the city receiving the lowest number of votes in that round gets eliminated. The survivor after the last round of voting gets to host the event.

    A member is allowed to cast votes for at most two different cities in all rounds of voting combined. (Hence, a member becomes ineligible to cast a vote in a given round if both the cities (s)he voted for in earlier rounds are out of contention in that round of voting).

    A member is also ineligible to cast a vote in a round if the city (s)he represents is in contention in that round of voting.

    As long as the member is eligible, (s)he must vote and vote for only one candidate city in any round of voting.

    The following incomplete table shows the information on cities that received the maximum and minimum votes in different rounds, the number of votes cast in their favor, and the total votes that were cast in those rounds.

    0_1503910052970_upload-44836620-1048-41fb-a04d-8d6be53e8e4e

    It is also known that:

    All those who voted for London and Paris in round 1, continued to vote for the same cities in subsequent rounds as long as these cities were in contention. 75% of those who voted for Beijing in round 1, voted for Beijing in round 2 as well.

    Those who voted for New York in round 1, voted either for Beijing or Paris in round 2.

    The difference in votes cast for the two contending cities in the last round was 1.

    50% of those who voted for Beijing in round 1, voted for Paris in round 3.

    Q1) What % of members from among those who voted for New York in round I, voted for Beijing in round 2?
    (1) 33.33
    (2) 50
    (3) 66.67
    (4) 75

    Q2) What is the number of votes cast for Paris in round 1?
    (1) 16
    (2) 18
    (3) 22
    (4) 24

    Q3) What percentage of members from among those who voted for Beijing in round 2 and were eligible to vote in round 3, voted for London?
    (1) 33.33
    (2) 38.10
    (3) 50
    (4) 66.67

    Q4) Which of the following statements must be true?
    a. IOC member from New York must have voted for Paris in round 2.
    b. IOC member from Beijing voted for London in round 3.
    (1) Only a
    (2) Only b
    (3) Both a and b
    (4) Neither a nor b

    Read the question carefully, here a minor detail given in the question is the main clue :)

    Again, this question might eat up some of your time, so be careful, when you start solving such questions. In case, you do not get a clue about the question in 3-4 minutes, leave it, and start the next one, and later come back to the question. The main clue here is a simple harmless sentence which says, that each city is represented by a member.

    Suppose we have “x” members totally, since 4 cities are in contention, total no: of voters in round 1 will be x – 4, as 4 members of the cities in contention are not allowed to vote.

    Similarly, we have x – 3 members voting in the 2nd round.

    In the 3rd round, we have x – 2 – n members voting, where n is the no: of voters who voted for NY and Beijing in Round 1 and 2. As they both are out of the competition, those voters cannot vote again.

    So, we know,
    x – 3 = 83. x = 86. Totally, there were 86 members.
    1st round = 82 voters.
    3rd round = 75 voters. x – 2 – n = 75. So n = 9. That means there were 9 voters who votes for NY in round 1 and Beijing in round 2.

    Q# 1,
    It is given that NY voters in Round 1 voted either for Beijing or for Paris.
    From what we found out, 9 voters who voted for NY in round 1, voted for Beijing in round 2.

    Q# 2,
    It is given that 75% of voters who voted for Beijing in round 1, voted for Beijing in round 2.
    We know 9 NY voters (R1) voted for Beijing in R2.
    Beijing R2 voters = 21, out of which 9 voted for NY in R1, and 12 voters are those who voted or Beijing in R1.
    12 form 75% of total R1 Beijing voters. So R1 Beijing voters = 16.
    So in Round 1,
    Total = 82
    London = 30
    Beijing = 16
    NY = 12
    Hence, Paris = 24.

    Q#3 & Q#4,
    Let us form a table detailing each round now.

    0_1503910139747_upload-c3cbd8f2-48bf-41cd-93bb-7a8b788b6bec

    We know 30 voted for L, 24 for P, 16 for B and 12 for NY in R1.

    Round 2:

    London: Total 83 voters. Paris 32, Beijing 21. Hence L = 30. (only those who voted for L in R1 voted for L in R2 as well, no additional voters.)

    Paris:

    24 who voted for P in R1. (They continue to vote till P gets out) +

    We know that only 12 of 16 Beijing voters in R1 voted for Beijing in R2. So, they would have voted for P in R2 (NY is out and L’s 30 are already the ones from R1).
    So 4 Beijing R1 voters voted for Paris in R2.

    We have already seen that 3 NY voters vote for Paris in R2.

    The remaining 1 vote will be that of the NY representative, who did not take part in R1 voting.

    Beijing:

    12 voters who voted for Beijing in R1.

    9 voters who voted for NY in R1.

    Round 3:

    London:

    We know that London/Paris has 38/37 votes in some order (As vote difference is only 1 in 3rd round)

    30 who voted for London in R1 and R2.

    We know that, out of the 21 who voted for Beijing in R2, only 12 qualify to vote to next round. (9 voters are those who voted for NY in R1 and Beijing in R2, and hence they are disqualified)

    Clue says 50% of Beijing R1 voters, vote for Paris in R3. So 8 votes from Beijing voters. We already know that 4 have voted for Paris in R2. So, additional 4 who voted for Beijing in R1 and R2 votes for Paris in R3.

    So, out of the 12 voters who voted for Beijing in R1 and R2, 4 vote for Paris in R3. Remaining 8 vote for London.

    So, London has 38 votes totally.

    Paris :

    24 who voted in R1 and R2 for Paris.

    4 who voted for Beijing in R1 and Paris in R2.

    4 who voted for Beijing in R1 and R2.

    3 who voted for NY in R1 and Paris in R2.

    1 NY representative.

    1 Beijing representative.

    Totally, 37 voted for Paris.

    All the questions can be answered now.

    (CAT 2005)
    The table below presents the revenue (in million rupees) of four firms in three states. These firms, Honest Ltd., Aggressive Ltd., Truthful Ltd. and Profitable Ltd. are disguised in the table as A, B, C and D, in no particular order.

    0_1503910289849_upload-75a441f1-0fd6-4684-89bf-cf6b85d1c909

    Further, it is known that:
    In the state of MP, Truthful Ltd. has the highest market share.
    Aggressive Ltd.’s aggregate revenue differs from Honest Ltd.’s by Rs. 5 million.

    Q1) What can be said regarding the following two statements?
    Statement 1: Profitable Ltd. has the lowest share in MP market.
    Statement 2: Honest Ltd.’s total revenue is more than Profitable Ltd.
    (1) If Statement 1 is true then Statement 2 is necessarily true.
    (2) If Statement 1 is true then Statement 2 is necessarily false.
    (3) Both Statement 1 and Statement 2 are true.
    (4) Neither Statement 1 nor Statement 2 is true.

    Q2) What can be said regarding the following two statements?
    Statement 1: Aggressive Ltd.’s lowest revenues are from MP.
    Statement 2: Honest Ltd.’s lowest revenues are from Bihar.
    (1) If Statement 2 is true then Statement 1 is necessarily false.
    (2) If Statement 1 is false then Statement 2 is necessarily true.
    (3) If Statement 1 is true then Statement 2 is necessarily true.
    (4) None of the above.

    Q3) What can be said regarding the following two statements?
    Statement 1: Honest Ltd. has the highest share in the UP market.
    Statement 2: Aggressive Ltd. has the highest share in the Bihar market.
    (1) Both statements could be true.
    (2) At least one of the statements must be true.
    (3) At most one of the statements is true.
    (4) None of the above

    Q4) If Profitable Ltd.’s lowest revenue is from UP, then which of the following is true?
    (1) Truthful Ltd.’s lowest revenues are from MP.
    (2) Truthful Ltd.’s lowest revenues are from Bihar.
    (3) Truthful Ltd.’s lowest revenues are from UP.
    (4) No definite conclusion is possible.

    In MP, T has highest share, so T is either firm A/C.
    Ag and H differ by 5 million.
    Ag = 190 million, B = 217 million, C = 222 million, D = 185 million.
    So Ag/H is either firm A/D or firm C/B in any order.

    Q# 1,
    P has lowest share in MP market. So B is P.
    That means Ag/H pair is definitely A/D.
    Hence C is Truthful.
    Whatever order it may be, Ag/H is firm A/D and its income is lesser than P, which is firm B.

    Q# 2,
    Firm B’s lowest revenues are from MP. So firm B is Ag.
    That means firm C is H.
    Firm A is T, and D is P.
    C’s lowest revenue is from Bihar.

    Q#3,
    Question is about highest share in Bihar and UP market. So, it is about Firm B.
    So, both statements together cannot be true.
    Also, if firm B is H, then C will be Agg and A will be T and D will Prof.
    Similarly if B is Ag, then C is H, firm A is T and D is P.
    Only 1 of this can take place maximum. B can either be H or be Ag.

    Q# 4,
    P’s lowest revenue is from UP. So, P can be either firm A or firm D.
    If P is firm A, then T should be firm C. But that makes B/D and Ag/H which is not possible as their revenue difference is not 5 million.
    So P is firm D, A is T, and Ag/H Is firm C/D.
    A’s lowest revenue is from UP.

    (CAT 2005)
    Help Distress (HD) is an NGO involved in providing assistance to people suffering from natural disasters. Currently, it has 37 volunteers. They are involved in three projects: Tsunami Relief (TR) in Tamil Nadu, Flood Relief (FR) in Maharashtra, and Earthquake Relief (ER) in Gujarat. Each volunteer working with Help Distress has to be involved in at least one relief work project.

    A Maximum number of volunteers are involved in the FR project. Among them, the number of volunteers involved in FR project alone is equal to the volunteers having additional involvement in the ER project.

    The number of volunteers involved in the ER project alone is double the number of volunteers involved in all the three projects.

    17 volunteers are involved in the TR project.

    The number of volunteers involved in the TR project alone is one less than the number of volunteers involved in ER Project alone.

    Ten volunteers involved in the TR project are also involved in at least one more project.

    Q1) Based on the information given above, the minimum number of volunteers involved in both FR and TR projects, but not in the ER project is:
    (1) 1
    (2) 3
    (3) 4
    (4) 5

    Q2) Which of the following additional information would enable to find the exact number of volunteers involved in various projects?
    (1) Twenty volunteers are involved in FR.
    (2) Four volunteers are involved in all the three projects.
    (3) Twenty three volunteers are involved in exactly one project.
    (4) No need for any additional information.

    Q3) After some time, the volunteers who were involved in all the three projects were asked to withdraw from one project. As a result, one of the volunteers opted out of the TR project, and one opted out of the ER project, while the remaining ones involved in all the three projects opted out of the FR project.
    Which of the following statements, then, necessarily follows?
    (1) The lowest number of volunteers is now in TR project.
    (2) More volunteers are now in FR project as compared to ER project.
    (3) More volunteers are now in TR project as compared to ER project.
    (4) None of the above

    Q4) After the withdrawal of volunteers, as indicated in Question 3, some new volunteers joined the NGO. Each one of them was allotted only one project in a manner such that, the number of volunteers working in one project alone for each of the three projects became identical. At that point, it was also found that the number of volunteers involved in FR and ER projects was the same as the number of volunteers involved in TR and ER projects. Which of the projects now has the highest number of volunteers?
    (1) ER
    (2) FR
    (3) TR
    (4) Cannot be determined

    Solve such questions using a Venn diagram/pie chart.

    0_1503910514225_upload-b667f026-4c1a-4d95-bb6e-91bf1b617863

    Let common part of FR, TR, ER be x.
    Then according to clues,
    ER alone = 2x.
    TR alone = 2x – 1
    Totally TR = 17. Also 10 involved in TR are involved in other projects.
    So TR alone is 7. ER is 8.
    And all the 3 projects is 4. (FTER)

    We know, FR alone is same as FR with ER (FER). FR = FER.
    Let FR alone be x. FER = x.
    Total number = 37.
    FR (alone) + ER (alone) + TR (alone)+ FTR + FER + TER + FTER = 37.
    In that, TR (alone) + FTR + TER + FTER = 17 (17 involved in TR projects)
    So, 17 + FR (alone) + 8 (ER alone) + FER = 37, which means FR + FER = 8

    We know that FR = FER = x. So, FR = FER = 8/2 = 4.
    So, FR (alone) = 8
    ER (alone) = 8
    TR (alone) = 7
    FTER = 4 (all 3)
    FER = 4 (FR and ER)
    FTR = a (FR and TR)
    ETR = b (FR and ER)
    FR (total) = 8 + 4 + a + 4
    ER (total) = 8 + 4 + b + 4
    TR (total) = 7 + b + a + 4
    TR totally is 17.
    So a + b = 6.
    Also a > b, since FR > ER, as given in question.

    Q # 1,
    We need to find the minimum value of a.
    We know a + b = 6, and a > b. So, minimum value of ‘a’ should be 4.

    Q# 2,
    We already know a or b, or anything involved with that.
    1st option, if we know total FR volunteers, we will be able to calculate ‘a’ and hence ‘b’, and complete the puzzle.
    2nd and 3rd option, we already know.

    Q# 3,
    After opting out, we have
    FR = 14 + a
    ER = 15 + b
    TR = 16.
    We know that minimum value of a should be 4. So, still FR has the highest number of volunteers.

    Q#4,
    After opting out, we have,
    FTR - Instead of a, we have a + 1 (as 1 of them opted out of ER)
    TER - Instead of b, we have b + 2 (as 2 of them opted out of FR)
    And FER becomes 5 instead of 4 (as 1 opted out of TR)
    After adding some more volunteers, each group “alone” has same number and let it be ‘n’.
    Also, FER = TER at this stage. FER = TER = 5. So, b + 2 = 5. And b = 3.
    So, we have a = 3, b = 3.
    Now, FR = n + 4 + 5
    TR = n + 4 + 5
    ER = n + 5 + 5.
    So, TR has the highest volunteers.

    (CAT 2006)
    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, E, G 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.

    Q1) The person having the largest Erdös number at the end of the conference must have had Erdös number (at that time):
    (1) 5
    (2) 7
    (3) 9
    (4) 14
    (5) 15

    Q2) How many participants in the conference did not change their Erdös number during the conference?
    (1) 2
    (2) 3
    (3) 4
    (4) 5
    (5) Cannot be determined

    Q3) The Erdös number of C at the end of the conference was:
    (1) 1
    (2) 2
    (3) 3
    (4) 4
    (5) 5

    Q4) The Erdös number of E at the beginning of the conference was:
    (1) 2
    (2) 5
    (3) 6
    (4) 7
    (5) 8

    Q5) How many participants had the same Erdös number at the beginning of the conference?
    (1) 2
    (2) 3
    (3) 4
    (4) 5
    (5) Cannot be determined

    Read the question thoroughly and carefully, even if it might seem confusing, the question is easy and one can solve this within 2-3 minutes and get all the 5 answers right.

    We know that, F has the smallest Erdos number and A has the largest.

    After 3rd day, Group average is 3. So, total of the group (8 member) is 3*8 = 24.

    Let F have Erdos# of x. After 3rd day meeting, C&A have their Erdos # as x + 1.

    We know that 5 people have the same Erdos# and the other 3 have different Erdos # (different from each other as well) after the 3rd day.

    We know, C&A has the same Erdos # of x+1 and cannot be distinct from each other. So totally, 5 people now have Erdos # of x+1.

    So, now we have,

    F = x ; A = x + 1. ; C = x + 1. ; 3 others = x + 1. (As 5 have the same Erdos # now)

    Other two = Let it be a, b.

    Sum = 24.

    x + 5 * (x + 1) + a + b = 24.

    6x + a + b = 19.

    After 5th day,

    Group average reduces by 0.5. So group average = 2.5 Total (of 8 members) = 20.

    As group average reduces after F’s meeting with E, we know that E does not have a Erdos# of x + 1. Because, if it was x + 1, then Erdos # of E would not have changed after the meeting, and average would have remained the same.

    So we know that E has erdos # other than x + 1.We have already assigned a/b for the other Erdos#. Let E have erdos# of a.

    After meeting, E will have erdos # of x + 1.

    So, now, we have,

    F = x ; A = x + 1 ; C = x + 1 ; 3 others = x + 1 ; E = x + 1 ; Other one = b

    Sum = 20

    x + 6 * (x + 1) + b = 20

    7x + b = 14.

    x cannot be 2, as b will be 0. Only Paul Erdos can have Erods# of 0. So x = 1.

    Hence, b = 7.

    We also know, 6x + a + b = 19. We get a = 6.

    So, at the end of conference, we have,

    A = 2

    C = 2

    E = 2

    F = 1

    3 others = 2

    Remaining 1 = 7.

    Also, we know at the start of meeting C had a different Erdos# than 2, because, it says, after the 3rd meeting Erdos # of B,D,E,G,H did not change. So C’s had changed after the 3rd meeting to 2. Hence, initially he had a different Erdos #, other than 2.

    0_1503911053998_upload-99259f28-3514-4c5b-8204-bb99268b43b2

    We have three 2’s and a 7 assigned to B, D, G, H in some order, which never changes. Also F’s Erdos# never changes.

    All the questions can be answered now.

    (CAT 2006)
    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.

    Q1) If Chetan sold 10 shares of MCS on three consecutive days, while Michael sold 10 shares only once during the five days, what was the price of MCS at the end of day 3?
    (1) Rs 90
    (2) Rs 100
    (3) Rs 110
    (4) Rs 120
    (5) Rs 130

    Q2) 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)?
    (1) Michael had 10 less shares than Chetan.
    (2) Michael had10 more shares than Chetan.
    (3) Chetan had 10 more shares than Michael.
    (4) Chetan had 20 more shares than Michael.
    (5) Both had the same number of shares.

    Q3) If Chetan ended up with Rs 1300 more cash than Michael at the end of day 5, what was the price of MCS share at the end of day 4?
    (1) Rs 90
    (2) Rs 100
    (3) Rs 110
    (4) Rs 120
    (5) Not uniquely determinable

    Q4) What could have been the maximum possible increase in combined cash balance of Chetan and Michael at the end of the fifth day?
    (1) Rs 3700
    (2) Rs 4000
    (3) Rs 4700
    (4) Rs 5000
    (5) Rs 6000

    Q5) If Michael ended up with 20 more shares than Chetan at the end of day 5, what was the price of the share at the end of day 3?
    (1) Rs 90
    (2) Rs 100
    (3) Rs 110
    (4) Rs 120
    (5) Rs 130

    The question can be approached in many ways, however the easiest will be to list all combinations possible (with day 1 starting 100 and day 5 ending 110), and solve the rest of the questions based on that. This will take around 5 minutes initially, but will prove worthy, as all the questions can then be solved immediately. So, let us form a table, in a way by which we can maintain that day 1 start is 100, and day 5 end is 110. I know, it will be difficult, to form a table out of nothing, so let us think of a clue to form the table.

    We know that initially the price was 100, and at the end of 5 days it was 110. So, obviously, 3 days, the price would have gone up, and 2 days the price would have come down. Now, we will form a table in such a way that we will bring down the price in 2 days. These days can be:

    Price falls down on day 1 and day 2.
    Price falls down on day 1 and day 3.
    Price falls down on day 1 and day 4.
    Price falls down on day 1 and day 5.
    Price falls down on day 2 and day 3.
    Price falls down on day 2 and day 4.
    Price falls down on day 2 and day 5.
    Price falls down on day 3 and day 4.
    Price falls down on day 3 and day 5.
    Price falls down on day 4 and day 5.

    For math wizards, this can be found using permutation – combination of 5c3 of 5c2 which is 10 combinations. I have indicated the blue shade in the table on the days the price has fallen down. So, if you notice, it will be just a matter of 2 minutes to form the table.

    0_1503911233799_upload-e8c81dcd-a202-4fd2-b46d-0bc2c8fde31b

    However, the table is not complete. We have listed only the share prices. We need to list Chetan’s and Michael’s cash and shares, as some of the questions are based on it. So, let us expand the table.

    We know Chetan earns cash when the share value goes up, and ‘loses’ when the share value goes down.

    For Michael, he makes if share value goes to 120 (above 110), and ‘loses’ if it goes to 80 (below 90).

    Calculation of Chetan: (Never calculate the full amount, as something or the other cancels off)

    Shares of Chetan after 5th day = -10 always. Because 3 day profit and 2 day loss, as proved earlier. Hence he sells 30 and buys 20.

    Day 1 share value goes down to 90. He buys 10 shares for 90. Loss = 10 * 90
    Day 2 share value goes down to 80. He buys 10 shares for 80. Loss = 10 * 80
    Day 3 share value goes up to 90. He sells 10 shares for 90. Profit = 10 * 90
    Day 4 share value goes up to 100. He sells 10 shares for 100. Profit = 10 * 100
    Day 5 share value goes up to 110. He sells 10 shares for 110. Profit = 10 * 110

    When we calculate, we see 10*90 goes off as he has profit and loss for the same. Rest he has an amount of (10 * 100) + (10 * 110) – (10 * 80) = 1300Rs

    Similarly for combination 1 for Michael,
    Price never goes above 120. So he never sells any share.
    Price comes below 80 once. So he buys 10 shares for 10 * 80 = 800 Rs loss.
    Since he buys 10 shares, and sells none, his number of shares is increased by 10 for the 1st condition

    Similarly, calculate for the other conditions. It will not take more than 2 minutes again here, as many of things cancel off, and others is multiplication by 10. So, we need to calculate only Michael’s cash and shares, and that will change only if the amount goes up to 120 or 80, as stated above.

    0_1503911249825_upload-0347df51-6d83-4ca5-aa30-43e8d4d605aa

    All questions can be answered now. :-)


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