Logical Reasoning Capsules - Shashank Prabhu - CAT 100 Percentiler - Set 1

  • CAT 100%iler, 5 times AIR 1, Director - learningroots.in, Ex ITC, Pagalguy, TAS

    Question 1

    Mrs. Sharma has a house which she wants to convert to a hostel and rent it out to students of a nearby women's college. The house is a two story building and each floor has eight rooms. When one looks from the outside, three rooms are found facing North, three found facing East, three found facing West and three found facing South. Expecting a certain number of students, Mrs. Sharma wanted to follow certain rules while giving the sixteen rooms on rent:

    • All sixteen rooms must be occupied.
    • No room can be occupied by more than three students.

    Six rooms facing north is called north wing. Similarly six rooms facing east, west and south are called as east wing, west wing and south wing. Each corner room would be in more than one wing. Each of the wings must have exactly 11 students. The first floor must have twice as many students as the ground

    However Mrs. Sharma found that three fewer students have come to rent the rooms. Still. Mrs. Sharma could manage to allocate the rooms according to the rules.

    Q.1. How many students turned up for renting the rooms?
    A. 24
    B. 27
    C. 30
    D. 33
    E. None of these

    Q.2. If Mrs. Sharma allocates the north-west corner room on the ground floor to 2 students, then the number of students in the corresponding room on the first floor, and the number of students in the middle room in the first floor of the east wing are:
    A. 2 and 1 respectively
    B. 3 and 1 respectively
    C. 3 and 2 respectively
    D. Both should have 3 students
    E. Such an arrangement is not possible.

    Q.3. If all the student that Mrs. Sharma expected initially had come to rent the rooms, and if Mrs. Sharma had allocated the north west corner room in the ground floor to 1 student, then the number of students in the corresponding room on the first floor, and the number of students in the middle room in the first of the east wing would have been:
    A. 1 and 2 respectively
    B. 2 and 3 respectively
    C. 3 and 1 respectively
    D. Both should have 2 students.
    E. Such an arrangement is not possible.


    The arrangement of the rooms will be as shown in the figure. Lets try and find the minimum and maximum number of students possible. Minimum will be 8 students on ground floor + 16 on first floor = 24. But with this, number of students in a wing will be 3+6 = 9 which is less than 11. Hence, minimum > 24.

    Lets look at the maximum now. If each wing has 11 students, the total number of students will be 11*4 = 44 students. However, we are counting the corner rooms twice. So taking at least 1 student in the 8 corner rooms, the number will be 44-8 = 36. With this, we have 12 students on the ground floor and 24 on the first. But if there are 24 on the first, that means each room will have 3 students. On the ground floor, there are 12 students meaning each room has to have at least 1 student. So in a wing there will be at least 9+3=12 students thus negating the condition of 11 students. Hence, even 36 is not possible. So maximum < 36.

    With the condition of first floor having twice as many as ground, the number of students will be a multiple of 3. Hence, possible values of number of students are 27,30 and 33.

    Lets take 33 as number of students. Hence, on ground floor no = 11 and first floor = 22. If the number of students on the ground floor is 11, every room will have at least 1 student. So for a wing, number of students will be at least 3 on the ground. That means on the first floor, number of students has to be less than or equal to 11-3 = 8. For this, keep 2 students in diagonally opposite corner rooms and 3 students elsewhere. This will satisfy the condition of 22 on the first floor. But with this arrangement, all rooms on ground floor have to have only 8 students (1 in each room ) to satisfy wing condition. But this is not possible as total number of students on the ground floor is 11. Hence, 33 is not possible.

    That means 30 was the original number of students expected and 27 was the number which came. With this, the questions can be solved.


    Question 2

    Five men are sitting around a circular table in such a way that all of them can see each other. Each of these five men is wearing a hat the colour of which is not known to him. However, all of them are aware that the hats have to be either black or white in colour and there are at least two hats of each colour. How many of them can deduce the colour of their hats if they are not allowed to communicate with each other?
    (a) 1
    (b) 2
    (c) 3
    (d) 4


    Consider, 5 members including you, now there would be 3 cases which you can observe
    2w 2b hats
    3w 1b hats
    3b 1w hats
    Consider first case
    You can't deduce, which hat do you have,
    Consider second case, here you know black should be at least 2, so have black hat.
    So sequence would be like
    1.w 2.w 3.w 4.b 5.(you) b
    Consider 1 person, he can see 2 w 2b, so can't deduce
    Consider 2. Person, he will see 2b 2w, so can't deduce
    Same case with 3. Person
    Consider 4. Person
    He sees 3w 1 b, so he is sure that he will be black.
    Hence 2 persons in total.

    Question 3

    There are four persons Himesh, Honey, Badshah and Taher out of whom two always lie and the other two always speak the truth. Each of the four persons makes a statement which is given below.
    Himesh: Honey lies.
    Honey: Badshah lies.
    Badshah: Himesh speaks the truth.
    Taher: Exactly two out of Himesh, Honey and Badshah lie.
    Who can be the liars?
    (a) Himesh and Honey
    (b) Honey and Taher
    (c) Himesh and Badshah
    (d) Either (b) or (c)


    Let Himesh be true. Honey is liar, Badshah speaks the truth and so, Taher is a liar.
    Second case, let Himesh be a liar. Honey speaks the truth, Badshah lies and Taher speaks the truth.
    So, b and c are both possible.

    Question 4

    Two numbers among A, B, C, D, E and F are positive odd integers. Two of these numbers are positive even integers. The remaining two are both equal to zero. The numbers satisfy the following conditions:
    I. 5(A + B + C + D + F) is odd.
    II. A + C is not even.
    III. EC = 2AD, but AD is greater than BE.

    (1) The two odd numbers are
    a. A and E
    b. B and E
    c. D and F
    d. A and C

    (2) The two even numbers are
    a. F and D
    b. C and E
    c. C and D
    d. B and F

    (3) The numbers which are both zero, are
    a. B and E
    b. B and F
    c. D and F
    d. A and C


    From the first statement, a+b+c+d+f is odd. So, we cannot have both the odd numbers in the formation. So, e is odd. From statement 3, ec is even and so, c has to be even. From statement 2, a+c is odd and so, a has to be odd. From statement 3, we can see that b and f have to be zero.
    Zero: b, f
    Odd: a, e
    Even: c, d

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