Previous CAT Questions - Solved (Logical Reasoning) - Set 4

  • (CAT 2001)
    While Balbir had his back turned, a dog ran into his butcher shop, snatched a piece of meat off the counter and ran off. Balbir was mad when he realised what had happened. He asked three other shopkeepers, who had seen the dog, to describe it. The shopkeepers really didn’t want to help Balbir. So each of them made a statement which contained one truth and one lie.
    Shopkeeper #1 said: “The dog had black hair and a long tail.”
    Shopkeeper # 2 said: “The dog had a short tail and wore a collar.”
    Shopkeeper # 3 said: “The dog had white hair and no collar.”
    Based on the above statements, which of the following could be a correct description?
    (a) The dog had white hair, short tail and no collar.
    (b) The dog had white hair, long tail and a collar.
    (c) The dog had black hair, long tail and a collar.
    (d) The dog had black hair, long tail and no collar.

    Such questions can always be approached by “assumption” technique, as it saves time.

    Let us assume the 1st statement of the 1st shopkeeper is true.
    Dog had black hair (true). So Dog had short tail (as long tail is false)
    Dog had short tail (true), and wore no collar (wore collar is false)
    Dog had no collar (true), dog had black hair (as white hair is false).

    This satisfies all the conditions, but such an answer is not present in the solution. Hence let us do a reversal of assumption.

    Dog had long tail (true), so dog had white hair (black hair is false)
    Dog wore collar (true), dog had long tail (short tail is false)
    Dog had white hair (true), and wore collar (no collar is false).

    This is present in the option, so the dog has long tail, wore collar and had white hair.

    (CAT 2001)
    The Bannerjees, the Sharmas and the Pattabhiramans each have a tradition of eating Sunday lunch as a family. Each family serves a special meal at a certain time of day. Each family has a particular set of chinaware used only for this meal. Use the clues below to answer the following question.
    The Sharma family eats at noon.
    The family that serves fried brinjal uses blue chinaware.
    The Bannerjee family eats at 2 o’clock.
    The family that serves sambar does not use red chinaware.
    The family that eats at 1 o’clock serves fried brinjal.
    The Pattabhiraman family does not use white chinaware.
    The family that eats last likes makki-ki-roti.
    Which one of the following statements is true?

    (a) The Bannerjees eat makki-ki-roti at 2 o’ clock, the Sharmas eat fried brinjal at 12 o’ clock and the Pattabhiramans eat sambar from red chinaware.
    (b) The Sharmas eat sambar served in white chinaware, the Pattabhiramans eat fried brinjal at 1 o’ clock and the Bannerjees eat makki-ki-roti in blue chinaware.
    (c) The Sharmas eat sambhar at noon. The Pattabhiramans eat fried brinjal served in blue chinaware and the Bannerjees eat makki-ki-roti served in red chinaware.
    (d) The Bannerjees eat makki-ki-roti served in white chinaware, the Sharmas eat fried brinjal at 12 o’clock and the Pattabhiramans eat sambar from red chinaware.

    Read the question fully and thoroughly and strike off anything that is not important and concentrate on the “fixed” given details, before concentrating on the condition/non fixed details. Let's form a table.


    From the details given:

    Sharma eats at 12, Banerjee at 2, and the other time is 1according to the option.

    Family eating at 1 serves fried brinjal, and the family that eats last eats Makki Ki Roti. The remaining item is Sambar.

    Fried brinjal is served in blue chinaware. Sambar does not use red chinaware, hence Makki Ki Roti is in red chinaware, and the other color is white chinaware.

    From the given options, option c is correct.

    (CAT 2001)
    Mrs Ranga has three children and has difficulty remembering their ages and the months of their birth. The clues below may help her remember.
    The boy, who was born in June, is 7 years old.
    One of the children is 4 years old, but is not Anshuman.
    Vaibhav is older than Supriya.
    One of the children was born in September but it was not Vaibhav.
    Supriya’s birthday is in April.
    The youngest child is only 2 years old.
    Based on the above clues, which one of the following statements is true?

    (a) Vaibhav is the oldest, followed by Anshuman who was born in September, and the youngest is Supriya who was born in April.
    (b) Anshuman is the oldest being born in June, followed by Supriya who is 4-year old, and the youngest is Vaibhav who is 2 years old.
    (c) Vaibhav is the oldest being 7 years old, followed by Supriya who was born in April, and the youngest is Anshuman who was born in September.
    (d) Supriya is the oldest, who was born in April, followed by Vaibhav who was born in June, and Anshuman who was born in September.

    First, work on the fixed details, then assign the random details. Form a table.


    Now consider the rest of the clues. We start with the clue with the “month”, as we have no idea of the age, but we know the month correctly of Supriya (atleast 1 month is fixed)

    One is born in September, and it is not Vaibhav. So, it is Anshuman.

    So, Vaibhav is born in June and he is 7 years old.

    Anshuman is not 4 yrs old, so Supriya is. Hence Anshuman is 2 yrs old.

    That completes the table.


    (CAT 2000)
    There are five machines A, B, C, D, and E situated on a straight line at distances of 10 metres, 20 metres, 30 metres, 40 metres and 50 meters respectively from the origin of the line. A robot is stationed at the origin of the line. The robot serves the machines with raw material whenever a machine becomes idle. All the raw material is located at the origin. The robot is in an idle state at the origin at the beginning of a day. As soon as one or more machines become idle, they send messages to the robot- station and the robot starts and serves all the machines from which it received messages. If a message is received at the station while the robot is away from it, the robot takes notice of the message only when it returns to the station. While moving, it serves the machines in the sequence in which they are encountered, and then returns to the origin. If any messages are pending at the station when it returns, it repeats the process again. Otherwise, it remains idle at the origin till the next message (s) is received.

    Q1) Suppose on a certain day, machines A and D have sent the first two messages to the origin at the beginning of the first second, and C has sent a message at the beginning of the 5th second and B at the beginning of the 6th second, and E at the beginning of the 10th second. How much distance in metres has the robot travelled since the beginning of the day, when it notices the message of E? Assume that the speed of movement of the robot is 10 metres per second.
    a. 140
    b. 80
    c. 340
    d. 360

    Q2) Suppose there is a second station with raw material for the robot at the other extreme of the line which is 60 metres from the origin, that. is, 10 meters from E. After finishing the services in a trip, the robot returns to the nearest station. If both stations are equidistant, it chooses the origin as the station to return to. Assuming that both stations receive the messages sent by the machines and that all the other data remains the same, what would be the answer to the above question?
    a. 120
    b. 140
    c. 340
    d. 70

    For Q # 1,
    R --10m--A--10m ---B---10m ----C---10m ----D---10m -----E
    Initially the robot would have received messages from A and D.
    He goes to A 1st, and then to D and returns to the station at R. He would have travelled 80m and hence has taken 8 seconds.
    When he is back at R, he would have received messages from B and C, and would travel to B, then C and come back, thus travelling additional 60m. This will be a total of 140m and 14 seconds, when he reaches back his station to receive the message from E.

    For Q#2, R --10m--A--10m ---B---10m ----C---10m ----D---10m -----E ---10m -----S
    Initially the robot would have received messages from A and D.
    He would travel to A, then to D, and from there to S travelling a total of 60m. (As S is nearer)
    He would have received messages from B and C then. He would travel to C, then to B, and from there to R, as R is nearby, again travelling 60m. (As R is nearer to B now)
    So the total distance will be 120m, 12 seconds, when he sees the message of E at R.

    (CAT 2000)
    There are three bottles of water, A, B, C, whose capacities are 5 litres, 3 litres, and 2 litres respectively. For transferring water from one bottle to another and to drain out the bottles, there exists a piping system. The flow through these pipes is computer controlled. The computer that controls the flow through these pipes can be fed with three types of instructions, as explained below:
    FILL (X, Y) - Fill bottle labeled X from the water in bottle labeled Y, where the remaining capacity of X is less than or equal to the amount of water in Y.
    EMPTY (X, Y) - Empty out the water in bottle labeled X into bottle labeled Y, where the amount of water in X is less than or equal to remaining capacity of Y.
    DRAIN (X) - Drain out all the water contained in bottle labeled X.Initially, A is full with water, and B and C are empty.

    Q1) After executing a sequence of three instructions, bottle A contains one litre of water. The first and the third of these instructions are shown below:
    First instruction: FILL (C, A)
    Third instruction FILL (C, A)
    Then which of the following statements about the instruction is true?
    a. The second instruction is FILL (B, A)
    b. The second instruction is EMPTY (C, B )
    c. The second instruction transfers water from B to C
    d. The second instruction involves using the water in bottle A.

    Q2) Consider the same sequence of three instructions ‘and the same initial state mentioned above. Three more instructions are added at the end of the above sequence to have A contain 4 litres of water. In this total sequence of six instructions, the fourth one is DRAIN (A). This is the only DRAIN instruction in the entire sequence. At the end of the execution of the above sequence, how much water (in litres) is contained in C?
    a. One
    b. Two
    c. Zero
    d. None of these

    For Q # 1,
    1st instruction is Fill C from A, which can fill C upto 2 litres. After this A will have 3 litres, and C will have 2 litres. (Initially, A has 5 litres, as A is full)
    Third instruction is again Fill C from A, which means C has to be empty before the 3rd instruction is executed.. That means either the 2nd instruction deals with draining from C or transferring from C to B, and the correct option is b.

    For Q# 2,
    There are a total of 6 instructions with the same 1st 3 instructions of the above question and 4th instruction as the “only” drain instruction in the 6 instructions. So the 2nd instruction will be to empty C to B.
    So we have the 1st 4 instructions as:
    A – C, after this A = 3L, B = 0L, C = 2L (Fill C, A)
    C – B, after this A = 3L, B = 2L, C = 0L (Empty C, B )
    A - C, after this A = 1L, B = 2L, C = 2L
    Drain A, after this A = 0L, B = 2L, C = 2L
    Now, it is given after 6th instruction, A will have 4L. This is possible, only if we empty out contents of B and C to A. Thus C will have no water left after the 6th step.

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