Previous CAT Questions - Solved (Logical Reasoning) - Set 10


  • Skadoooosh!!!


    (CAT 2006)
    A significant amount of traffic flows from point S to point T in the one-way street network shown below. Points A, B, C, and D are junctions in the network, and the arrows mark the direction of traffic flow. The fuel cost in rupees for travelling along a street is indicated by the number adjacent to the arrow representing the street.

    0_1503913511244_upload-155df1f9-bdd1-43db-acf1-6c6380203589

    Motorists travelling from point S to point T would obviously take the route for which the total cost of travelling is the minimum. If two or more routes have the same least travel cost, then motorists are indifferent between them. Hence, the traffic gets evenly distributed among all the least cost routes. The government can control the flow of traffic only by levying appropriate toll at each junction. For example, if a motorist takes the route S-A-T (using junction A alone), then the total cost of travel would be Rs 14 (i.e. Rs 9 + Rs 5) plus the toll charged at junction A.

    Q1) If the government wants to ensure that all motorists travelling from S to T pay the same amount (fuel costs and toll combined) regardless of the route they choose and the street from B to C is under repairs (and hence unusable), then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is:
    (1) 2, 5, 3, 2
    (2) 0, 5, 3, 1
    (3) 1, 5, 3, 2
    (4) 2, 3, 5, 1
    (5) 1, 3, 5, 1

    Q2) If the government wants to ensure that no traffic flows on the street from D to T, while equal amount of traffic flows through junctions A and C, then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is:
    (1) 1, 5, 3, 3
    (2) 1, 4, 4, 3
    (3) 1, 5, 4, 2
    (4) 0, 5, 2, 3
    (5) 0, 5, 2, 2

    Q3) If the government wants to ensure that all routes from S to T get the same amount of traffic, then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is:
    (1) 0, 5, 2, 2
    (2) 0, 5, 4, 1
    (3) 1, 5, 3, 3
    (4) 1, 5, 3, 2
    (5) 1, 5, 4, 2

    Q4) If the government wants to ensure that the traffic at S gets evenly distributed along streets from S to A, from S to B, and from S to D, then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is:
    (1) 0, 5, 4, 1
    (2) 0, 5, 2, 2
    (3) 1, 5, 3, 3
    (4) 1, 5, 3, 2
    (5) 0, 4, 3, 2

    Q5) The government wants to devise a toll policy such that the total cost to the commuters per trip is minimized. The policy should also ensure that not more than 70 per cent of the total traffic passes through junction B. The cost incurred by the commuter travelling from point S to point T under this policy will be:
    (1) Rs 7
    (2) Rs 9
    (3) Rs 10
    (4) Rs 13
    (5) Rs 14

    First, find which all routes are possible.
    SAT, SBCT, SBAT, SDT, SDCT are the routes possible.

    Let a, b, c, d be the tolls levied at A, B, C, and D respectively.
    So the cost for different routes are:
    SAT = 14 + a
    SBCT = 7 + b + c
    SBAT = 9 + b + a
    SDT = 13 + d
    SDCT = 10 + d + c.

    Q# 1,
    SBCT route is not possible as it is under repairs.
    Total cost of all the other routes should be same.
    So, 14 + a = 9 + b + a = 13 + d = 10 + d + c
    14 + a = 9 + b + a. So, b = 5
    13 + d = 10 + d + c. So, c = 3
    14 + a = 13 + d. So, d – a = 1.
    There are 2 options with the same answer. Any one is fine.

    Q#2,
    SDT route is not possible, and total cost of other routes should be same.
    So, 14 + a = 7 + b + c = 9 + b + a = 10 + d + c
    Solving as above, we get, b = 5.
    7 + b + c = 10 + d + c. ----> 7 + b = 10 + d. So, d = 2.
    Also, 14 + a = 12 + c (after d = 2 substitution)
    So, we have b = 5, d = 2, c – a = 2.
    Find the option.

    Q# 3,
    All routes have same cost.
    So, 14 + a = 7 + b + c = 9 + b+ a = 13 + d = 10 + d + c.
    We get b = 5, d = 2 (like above)
    If d = 2, a =1, and finally, we get c = 3.

    Q#4,
    Equal traffic from S to A,B and D. However, through B&D, there is more than 1 way, from which the commuter can choose. We can restrict them to move through 1 way from B&D if we increase the toll in C. So that passengers move through the route without “C” junction involved. So, the routes involved will be SAT, SBAT, SDT, and the tolls of these should be same.
    14 + a = 9 + b + a = 13 + d.
    b = 5, d – a = 1. We can keep a = 0 and d = 1. (As the options have value of a = 0/1 only, and if a = 1, d = 2 which is not there in option)
    So, we have a = 0, b = 5, d = 1. (We can stop the solution here, as we have only 1 option satisfying this).
    If you need to find value of C,
    We know that SBAT < SBCT, so, 14 < 12 + c. c >2.
    Also, SDT < SDCT, so 14 < 11 + c. So, c > 3.
    Hence c > 3.

    Q# 5,
    The least expense route is SBCT, if without toll, cost is 7.
    However, if we do not keep toll, 100% traffic will flow through B, which is not possible.
    The next highest cost route is SDCT which is 10Rs.
    So, if we keep a toll in B for 3rs, then traffic will split 50-50 between SBCT and SDCT, and the total cost of the journey will be 10rs.

    (CAT 2006)
    K, L, M, N, P, Q, R, S, U and W are the only ten members in a department. There is a proposal to form a team from within the members of the department, subject to the following conditions:
    A team must include exactly one among P, R, and S.
    A team must include either M or Q, but not both.
    If a team includes K, then it must also include L, and vice versa.
    If a team includes one among S, U, and W, then it must also include the other two.
    L and N cannot be members of the same team.
    L and U cannot be members of the same team.
    The size of a team is defined as the number of members in the team.

    Q1) What could be the size of a team that includes K?
    (1) 2 or 3
    (2) 2 or 4
    (3) 3 or 4
    (4) Only 2
    (5) Only 4

    Q2) In how many ways a team can be constituted so that the team includes N?
    (1) 2
    (2) 3
    (3) 4
    (4)5
    (5) 6

    Q3) What would be the size of the largest possible team?
    (1) 8
    (2) 7
    (3) 6
    (4) 5
    (5) Cannot be determined

    Q4) Who can be a member of a team of size 5?
    (1) K
    (2) L
    (3) M
    (4) P
    (5) R

    Q5) Who cannot be a member of a team of size 3?
    (1) L
    (2) M
    (3) N
    (4) P
    (5) Q

    Team comprises
    Exactly one among P,R,S
    Exactly one among M,Q
    K < - > L
    S < - > U < - > W
    L ** N
    L ** U

    Q#1,
    One among P,R,S
    One among M,Q
    If K is there, L should be there.
    If L is there, we cannot have N and U.
    If we cannot have U, we cannot have S & W.
    So, only possible team is
    P/R
    M/Q
    K
    L

    Q#2,
    Includes N.
    So no L, hence no K.
    One among P,R,S and one among M,Q
    If S is there, then U and W should be there.
    So, the possible teams are
    P,M,N
    P,Q,N
    R,M,N
    R,Q,N
    S,U,W,M,N
    S,U,W,Q,N

    Q#3,
    Largest possible team – we know 1 among P,R,S and 1 among M,Q.
    We cannot select P,R and we will not be able to select U,W (who comes only with S)
    So we have S,U,W and 1 among M/Q.
    Since U is there, we cannot have L and hence no K.
    We can have N in the team.
    So the team will be S, U, W, M/Q, N. Totally 5 members.

    Q#4,
    As explained in the earlier question, team of 5 includes S,U,Q, M/Q, N

    Q# 5,
    Team of size 3 means S,U,W is not there ( as M/Q should be there in the team).
    We can have P/R as 1 member.
    We can have M/Q as the 2nd member.
    We cannot have K and L, as they need to be together, and it makes the team strength a minimum of 4.
    The 3rd member can be N.

    (CAT 2006)
    In a Class X Board examination, ten papers are distributed over five Groups i.e. 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)

    0_1503914003089_upload-663e099c-e025-4928-b5dd-93bc65c952c2

    Note: B or G against the name of a student respectively indicates whether the student is a boy or a girl.

    Q1) How much did Dipan get in English Paper II?
    (1) 94
    (2) 96.5
    (3) 97
    (4) 98
    (5) 99

    Q2) 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:
    (1) Shreya
    (2) Ram
    (3) Ayesha
    (4) Dipan
    (5) no one from the top ten

    Q3) Among the top ten students, how many boys scored at least 95 in at least one paper from each of the groups?
    (1) 1
    (2) 2
    (3) 3
    (4) 4
    (5) 5

    Q4) 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:
    (1) Ram
    (2) Agni
    (3) Pritam
    (4) Ayesha
    (5) Dipan

    Q5) 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:
    (1) Pritam, Joseph, Tirna, Agni
    (2) Joseph, Tirna, Agni, Pritam
    (3) Pritam, Agni, Tirna, Joseph
    (4) Joseph, Tirna, Pritam, Agni
    (5) Pritam, Tirna, Agni, Joseph

    Q# 1,
    Dipan’s average = 96. So, totally he has scored 96*5 in all the 5 groups together = 480.
    Average mark of Dipan in PCB = 98
    Average mark of Dipan in Maths = 95
    Average mark of Dipan in SS = 95.5
    Average mark of Dipan in Vern. = 95
    Average mark of Dipan in Eng. = x
    x + 95 + 95.5 + 95 + 98 = 480. Hence, x = 96.5
    So, average in English group is 96.5
    He scored 96 in Paper 1. So his score will be 97 in the other paper.

    Q# 2,
    Only Dipan has group average of 95 in all the groups.
    Here, do not calculate each and every group average of all the individuals. It will be really time consuming. Just browse through the scores. For eg:
    Take a quick look in Ayesha’s row. We see that she has 93 and 95 in SS group. So average is 94 < 95.
    Similarly, for Ram, Vernacular group has two 94’s,
    Dipan has 95 in all groups.
    Sagnik – take a quick look – English has 94 and 92.
    Work out like this for others. Never waste time by calculating all the group averages for all the students.

    Q#3,
    Again, just browse through the rows quickly.
    We see that only Dipan has scored 95 in all the groups, atleast 1 paper.

    Q# 4,
    This is calculative intensive. You will have to see, by changing which score to 100, group score gets affected more, and change that, find the new group average and total average.
    For eg: for Ram, it will be Vernacular Paper 1/2, Agni it will be Vernacular paper 1 etc etc.
    I would suggest leaving this question, and coming back and doing if you have time left.

    Q# 5,
    Score of Joseph, Agni, Pritam and Tirna in Social Science Group are 95.5, 95.5, 89 and 89.5.
    Their final scores are 95, 94.3, 93.9, 93.7.
    If Joseph scores 100 in SS group, his SS group score increases by 4.5. This will increase the whole average by 4.5/5 = 0.9% (divided by 5, because of 5 groups)
    For Agni, group score in SS changes from 95.5 to 100. Again change of 4.5 and 0.9% increase.
    For Pritam, change is by 11. So 11/5 = 2.2% increase in final average.
    For Tirna, change is 10.5. So 10.5/5 = 2.1% increase in final average.
    Add the increased amount of final average, and arrange in descending order.

    (CAT 2008)
    There are three houses on each side of the road.
    These six houses are labeled as P, Q, R, S, T and U.
    The houses are of different colours, namely, Red, Blue, Green, Orange, Yellow and White.
    The houses are of different heights.
    T, the tallest house, is exactly opposite to the Red coloured house.
    The shortest house is exactly opposite to the Green coloured house.
    U, the Orange coloured house, is located between P and S.
    R, the Yellow coloured house, is exactly opposite to P.
    Q, the Green coloured house, is exactly opposite to U.
    P, the White coloured house, is taller than R, but shorter than S and Q.

    Q1) What is the colour of the tallest house?
    (1) Red
    (2) Blue
    (3) Green
    (4) Yellow
    (5) none of these

    Q2) What is the colour of the house diagonally opposite to the Yellow coloured house?
    (1) White
    (2) Blue
    (3) Green
    (4) Red
    (5) none of these

    Q3) Which is the second tallest house?
    (1) P
    (2) S
    (3) Q
    (4) R
    (5) cannot be determined

    Read the question fully and thoroughly, work on the “fixed” details first, and then the lesser known details, more importantly, work on 1 set of clues a time. i.e. Do not mix colors with height and assign to houses. 1st try to arrange the houses, then set the color, and then the height.

    We know, U is between P & S. So, P,U,S are in one side.

    RQT on the other side. R is opposite to P, Q is opposite to U. So T is opposite to S.

    Houses can be arranged as (It can be in the reverse order also, but it makes no difference)

    0_1503914137812_upload-711994ff-7e30-4ad4-83d3-a1cff5af06c9

    Now, let us set the colors:

    U = Orange, P = White, Q = Green, R = Yellow.

    T is tallest and opposite to Red colored house. So S is red colored house.

    Hence, T is Blue.

    0_1503914147181_upload-63f95bdb-5520-4f9b-bd9c-b36c12423f4e

    Now the height :

    T is tallest.

    P is taller than R, but shorter than S & Q.

    U is shortest.

    0_1503914156215_upload-f78a7f18-d768-45d4-84e5-57144b447ca6

    Now, all the questions can be answered. :-)

    (CAT 2008)
    In a sports event, six teams (A, B, C, D, E and F) are competing against each other. Matches are scheduled in two stages. Each team plays three matches in Stage-I and two matches in Stage-II. No team plays against the same team more than once in the event. No ties are permitted in any of the matches. The observations after the completion of Stage-I and Stage-II are as given below.

    Stage - I:
    One team won all the three matches.
    Two teams lost all the matches.
    D lost to A but won against C and F.
    E lost to B but won against C and F.
    B lost at least one match.
    F did not play against the top team of Stage- I.

    Stage - II:
    The leader of Stage-I lost the next two matches.
    Of the two teams at the bottom after Stage-I, one team won both matches, while the other lost both matches.
    One more team lost both matches in Stage- II.

    Q1) The team(s) with the most wins in the event is (are):
    (1) A
    (2) A & C
    (3) F
    (4) E
    (5) B & E

    Q2) The two teams that defeated the leader of Stage-I are:
    (1) F & D
    (2) E & F
    (3) B & D
    (4) E & D
    (5) F & D

    Q3) The only team(s) that won both the matches in Stage-II is (are):
    (1) B
    (2) E & F
    (3) A, E & F
    (4) B, E & F
    (5) B & F

    Q4) The teams that won exactly two matches in the event are:
    (1) A, D & F
    (2) D & E
    (3) E & F
    (4) D, E & F
    (5) D & F

    Read the question fully and thoroughly, solve the 1st round first, and then follow to the second round. Take 1 clue at a time and work at it, starting with the fixed “given” details first.

    Let’s work on the first round:

    D – A – Won by A

    D – C – Won by D

    D – F – Won by D

    E – B – Won by B

    E – C – Won by E

    E – F – Won by E

    Now, we know that 1 team won all the 3 matches. It cannot be D, E, C, and F as they already lost. Also, the clue says B lost at least 1 match. So A won all the 3 matches.

    Also, we know that 2 teams have lost all their matches in the 1st round. Since A,D,B,E have already won, C&F are the teams who lost all their matches.

    Till now, we know:

    D & E – played 3 matches

    C & F – played 2 matches

    A & B – played 1 match.

    C cannot play against F, as C & F have to lose all their matches and hence they cannot play against each other. This leaves F playing against A or B, but F does not play against the table topper (A), and so F plays against B and C plays against A. So we have

    B – F – Won by B

    A – C – Won by A

    Now we have only 1 match of A and B pending, and it is A vs B, won by A.

    So the matches in stage 1 are:

    D – A – Won by A

    D – C – Won by D

    D – F – Won by D

    E – B – Won by B

    E – C – Won by E

    E – F – Won by E

    B – F – Won by B

    A – C – Won by A

    A – B – Won by A.

    Now, to stage 2:

    A lost both its matches in stage 2 (leader of stage 1 is A)

    A – F -> Won by F

    A – E -> Won by E.

    Out of 2 teams in bottom after stage 1 (C&F), 1 team won both matches, while the other lost the 2 matches. F has already won 1 match against A. So F wins both matches, and C loses.

    F – C -> Won by F

    C – B -> Won by B

    One more team loses both matches in stage 2. We see that E, F, B has already won at least 1 match. So D has to lose both its matches in the next round.

    D – B -> B

    D – E -> E.

    So, stage 2 has

    A – F -> Won by F

    A – E -> Won by E.

    F – C -> Won by F

    C – B -> Won by B

    D – B -> Won by B

    D – E -> Won by E.

    So, the final table will look like:

    0_1503914336036_upload-66c7a0ac-3d20-4664-a64c-674931212732

    All the questions can be answered now.


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