LR Capsules - Shashank Prabhu - CAT 100 Percentiler

  • Set 1

    Alok, Bharti, Chaman, Dinu, Ekant and Faisal are the only people available for selection in a team. The team should have at least two people subject to the following conditions.
    • If Bharti is selected then Dinu should also be selected while Ekant should be rejected.
    • If Alok is selected then exactly one from Bharti and Chaman should also be selected.
    • If Chaman is selected then Ekant should also be selected while Faisal should be rejected.
    • If the size of the team is less than 4 then Dinu and Faisal cannot be selected together.

    (1) If Bharti and Faisal are not selected in the team then who should definitely be selected?
    (a) Chaman
    (b) Ekant
    (c) Dinu
    (d) Alok

    (2) Which of the following pairs of two people cannot be selected along with any one else out of the remaining 4 people?
    (a) Bharti and Dinu
    (b) Chaman and Ekant
    (c) Dinu and Ekant
    (d) Ekant and Faisal

    (3) If the team selected is of the maximum possible size, then who is/are definitely selected in the team?
    (a) Alok
    (b) Dinu
    (c) Bharti
    (d) Both Alok and Dinu


    Teams with 2 members:
    A cannot be in any such team as otherwise one of B and C must be selected and then subsequently one of D and E must be selected as well (hence exceeding 2 member size).
    Possible teams are:
    B, D
    C, E
    D, E
    E, F

    Teams with 3 members:
    A, C, E
    A, B, D
    C, E, D

    Teams with 4 members:
    A, B, D, F
    A, C, E, D

    Note: No other team size is possible.

    (1) b Possible teams without B and F are:
    C, E
    D, E
    A, C, E
    C, E, D
    A, C, E, D

    We can see that Ekant (E) should definitely be selected.

    (2) d Ekant (E) and Faisal (F) (team number 4) appear together only once and their pair is not selected with any one else out of the remaining 4 people.

    (3) d Team selected must be either
    A, B, D, F
    A, C, E, D
    We can see that Alok (A) and Dinu (D) both are definitely selected in the team.

    Set 2

    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.


    Set 3

    Jack, Davy and Hector stole a few pieces of Aztec Gold from the flying Dutchman. As it was late in evening and as they were fairly intoxicated, they decided to divide the loot equally among themselves in the morning and went to sleep. During the night however, Hector tip toed to the chest, counted the pieces and took one. From the remaining pieces, he took a precise third and went back to sleep. After sometime Davy woke up. He counted the pieces, took one and then took exactly third of the remaining pieces and went to sleep. A little before sunrise Jack woke up, took one piece and like the other two pirates, took a precise third of the remainder. In the morning, all three sat together to divide the loot among themselves. Among them they found one piece, which was not real Aztec gold and so, they threw it away. From the remaining pieces they made an exact division. What could be the minimum number of pieces of Aztec gold (real or otherwise) that were initially stolen?


    Let total be 3x+1, after first person takes it, 2x left, after second person takes it 2(2x-1)/3 left, after third person takes it, 2/3[2(2x-1)/3-1] left ie (8x-10)/9 left. After one is removed, (8x-10)/9-1 left. This is in the form of 3k and so, 8x-19=27k. x=(27k+19)/8. As 19 gives remainder 3, 27k should give remainder 5 when divided by 8. The sequence of remainders are 3, 6, 1, 4, 7, 2, 5 (easier to do if you can notice the difference of 3 and treat it as a cycle from 1 to 8 and back to 1). So, k=7, x=26 and initial amount is 79 at least.

    Set 4

    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.

    Set 5

    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.

    Set 6

    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

    Set 7

    Sixteen candies are to be distributed among four boys Raja, Ram, Mohan and Roy such that each boy receives at least one candy and no two boys receive the same number of candies. Roy should receive 4 more candies than Ram. The number of candies received by Ram should be less than that received by Raja but more than that received by Mohan. What is the difference between the maximum and the minimum number of candies that Raja can receive?
    (a) 1
    (b) 2
    (c) 3
    (d) 4


    Let Ram get x candies. Raja gets x+k candies, Mohan gets x-m candies and Roy gets x+4 candies. So, we can see that Mohan gets the least number of candies, followed by Ram and the top two candy holders are Raja and Roy in no particular order. The possible cases are: (7,2,1,6)(5,3,1,7)(4,3,2,7). So, the difference between max and min of Raja is 3.

    Set 8

    There were 200 students who gave a competitive exam. The exam had three sections English, Maths and General Knowledge (GK). Ten students could not pass in any of the subjects. Number of students passed in English was 125, Maths was 95 and General Knowledge was 95. 55 students passed in English and Mathematics. 45 students passed in GK and Maths and 60 students passed in GK and English. 20 students passed in English and Maths but failed in GK.

    1. How many students passed in GK and Maths, but failed in English?
    2. How many students passed in English and GK but failed in Maths?
    3. How many students passed only in Maths?
    4. How many passed in all three Subjects?


    Let a, b and c be the number of students who study exactly 1 subject, 2 subjects and 3 subjects respectively.
    a + b + c = 190
    a + 2b + 3c = 125 + 95 + 95 = 315
    b + 3c = 160
    Using these, we get c = 35, b = 55, a = 100
    Plugging these in the given data, you will get all the answers. 10, 25, 30, 35 are the right answers.

  • Can somebody explain me solution / better solution for question no. 3?

  • @Rohit-Rathore
    Q3) Let say we had x coins in the beginning
    After 1st loot : x - 1 - (x-1)/3 = 2/3(x-1) = (2x-2)/3
    After 2nd loot : (2x - 2)/3 - 1 - [(2x - 2)/3 - 1]/3 = 2/3[(2x-2)/3 - 1] = (4x - 10)/9
    After 3rd loot : (4x - 10)/9 - 1 - [((4x - 10)/9 - 1]/3 = 2/3 [(4x - 10)/9 - 1] = (8x - 38)/27
    Now the final value (after throwing away the fake coin) = (8x - 38)/27 - 1 = (8x - 65)/27 is a multiple of 3.
    So we can write, (8x - 65)/27 = 3k
    x = (81k + 65)/8
    x is an integer, 81K + 65 should be a multiple of 8, means (81K + 65) Mod 8 = 0
    We know 81 Mod 8 = 1 and 65 mod 8 = 1. So to get remainder 0, we need to adjust K in a way that 81K mod 8 = 7.
    Least value would be K = 7
    For K = 7, x = (81 * 7 + 65)/8 = 79, which should (could) be our answer.
    Let me know in case of any Logical/Calculation errors.

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