Quant Boosters - Hemant Yadav - Set 1

  • Being MBAtious!

    Number of Questions - 30
    Topic - Quant Mixed Bag
    Solved ? : Yes
    Source : Collection of solved questions from Hemant Yadav, Math faculty - FIITJEE

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    Q1) There are 5 boxes and 5 balls, in how many ways can all the balls be put in boxes if,
    a) boxes and balls are all distinct and only 1 ball in one box
    b) boxes are same and balls are distinct and only 1 ball per box
    c) balls are same and boxes are different and only 1 ball per box
    d) balls and boxes are same and only 1 ball per box
    e) repeat all above questions with any number of balls in any box

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    Q2) The expression (x + y + z)^2006 + (x - y - z)^2006 is simplified by expanding it & combining like terms. How many terms are in the simplified expression ?
    (a) 6018
    (b) 671676
    (c) 1007514
    (d) 1008016
    (e) 2015028

    [OA : 1008016]

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    Q3) Find all positive integers n such that the set {n, n + 1, n + 2, n + 3, n + 4, n + 5} can be partitioned into two subsets so that the product of the numbers in each subset is equal.

    [OA : 0]

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    Every perfect square is of the form: 7n or (7n + 1) or (7n + 2) or (7n + 4).
    That means if a number leaves 3, 5 or 6 as remainder with 7, it can never be a perfect square.

    Here, all the 6 six numbers when divided by 7 will leave a remainder of 1, 2, 3, 4, 5, 6 in some order. So product of the 6 numbers = 7k + 720 = 7n - 1
    But no perfect sq = -1(mod 7)
    Hence, no solution

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    Q4) For each natural number N, define S(N) be the sum of digits of N.
    For eg: If N = 9801, S(N) = 9 + 8 + 0 + 1 = 18
    N = 25, S(N) = 2 + 5 = 7
    How many natural numbers, N, exist such that S(N) + S(N^2) = 2011

    [OA: 0]

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    2011 is in 3k + 1 form.
    N can be of the form 3k, 3k + 1 or 3k + 2
    for 3k the expression is clearly divisible by 3 but 2011 is 3k + 1.
    For 3k + 1 square is also in the form 3k + 1. So sum is 3k + 2.
    For N of the form 3k + 2 the square is 3k + 4 or 3k + 1.Hence sum is 3n
    So no solutions for S(N) + S(N^2) = 2011

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    Q5) Suppose Pk denotes kth prime number, e.g. P1 = 2, P2 = 3, P3 = 5, P4 = 7, P5 = 11 etc. Find all ordered pairs (x, n) of positive integers satisfying following equation: x^2 = 1 + (P1)(P2) ... (Pn)

    [OA : 0]

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    x^2 = 1 + P(1)P(2)P(3)…P(n)
    So x^2 – 1 = (x - 1)(x + 1) = P(1)P(2)P(3)…P(n) = 2k
    But (x – 1) and (x + 1) are either both odd or both even. So RHS must be odd or multiple of 4. As RHS is neither odd nor multiple of 4, hence no positive integral solution.

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    Q6) A and B play a simple number game. Alternately, A chooses a digit and B substitutes it for one of the place in the difference "---- – ----" A is trying to maximise the final difference, B to minimise it. What difference will be arrived at with best play (all the digits should be distinct)?

    [OA : 2258]

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    Let mod [X-Y]=[x4x3x2x1 - y4y3y2y1]

    There are total 10 distinct no’s, out of which 2 will not be used. Alice chooses the no’s and Bob places them

    Now, if Alice chooses any no. that is not 0 or 9 on her first call, Bob will manage to keep the difference within 2000. (suppose Alice chooses 3, then bob replace one of the stars in the 4th place with 3 like this ---- - 3---. Now whenever Alice chooses 2 or 4, Bob will replace the corresponding place with it like this 2--- - 3--- or 4--- - 3---.

    So she will choose neither, then she must choose any of 5 or 6 and bob will pick it because he knows Alice will not choose 2 or 4 as long as the star in the corresponding place is not replaced)

    Now, if Alice chooses any of 0 and 9, say 0, and bob places it in the 4th place, she will not choose 1 and 2 and will be able to make the difference cross 2000. so bob will never put 0 in the 4th place. He will place it in x3 in order to make the difference between X and Y minimum. now if the next choice is neither 1 nor 9 and greater than or equal to 5, he will place it in x4 and otherwise in y4 and thus will be able to keep the difference down below 2000. so the next choice must be 1 or 9. Bob will place it in y3 to keep the difference between x3 and y3 minimum, proceeding like this, the largest difference that can be obtained is 6012-3754=2258

    If the first chosen no. is 9, then the largest difference is 3987-6245=-2258
    So their best play would give 2258 as the difference.

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    Q7) Six persons A,B,C,D,E & F went to cinema. There are six consecutive seats. A sits in one of the seats followed by B, followed by C and so on. If A takes one of the six seats , then B should sit adjacent to A. C should sit adjacent to A or B. D should sit adjacent to A, B,or C and soon. How many possibilities are there?

    [OA : 32]

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    If f(n) is number of arrangements possible for n persons, then f(n + 1) = 2f(n), as the next person can sit at either end as a continuous chain will form and there is no possibility of a middle chair remaining vacant.
    Also, f(1) = 1, f(2) = 2
    So, f(n) = 2^(n – 1)
    Hence, f(6) = 2^5 = 32

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    Q8) There are N secret agents each know a different piece of secret information. They can telephone each other and exchange all the information they know. After the telephone call, they both know anything that either of them knew before the call. What are the minimum number of telephone calls needed so that all of the them know everything?

    [OA : 2N - 4]

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    (2N - 3) telephone calls, for N = 2,3
    (2N - 4) telephone calls, for N > 3

    Divide the N secret agents into two groups. If N is odd, one group will contain one extra agent.

    Consider first group: agent 1 will call up agent 2, agent 2 will call up agent 3 and so on. Similarly in second group, agent 1 will call up agent 2, agent 2 will call up agent 3 and so on. After (N - 2) calls, two agents in each the group will know anything that anyone knew in his group, say they are Y1 & Y2 from group 1 and Z1 & Z2 from group 2.

    Now, Y1 will call up Z1 and Y2 will call up Z2. Hence, in next two calls total of 4 agents will know everything.

    Now (N - 4) telephone calls are reqiured for remaining (N - 4) secret agents.

    Total telephone calls require are
    = (N - 2) + 2 + (N - 4)
    = 2N - 4

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    Q9) A point P is chosen at random on segment AB of a square ABCD. Point Q is chosen at random on CD. Segment PQ divides ABCD into two regions. What is the probability that one of the two regions is more than twice the area of the other?

    [OA : 4/9]

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    Plotting it on a graph would be eaisier I guess.

    After joining PQ we will get a trapezium. So to fulfil the required condition sum of the parallel sides of the trapezium should be less than 2a/3 or greater than 4a/3, where a is the side of the square.

    For AP = 0 to a/3, probability of choosing Q will constantly decrease from 2/3 to 1/3
    For AP = a/3 to 2a/3, it will be 1/3
    For AP = 2a/3 to a, it will constantly increase from 1/3 to 2/3

    Now, we can get the area as 4/9

    Another way can be let AB = 1.
    The requirement is the event (AP + DQ) > 4/3 or (AP + DQ) < 2/3. That is the mean of the two sides is either > 2/3 or < 1/3.

    Let X be the random variable for the continuous uniform distribution of AP. The distribution of DQ is independent but also belongs to X.

    Let Y = X + X (the sum of the two opposite sides).

    P(Y > 4/3) is the area of the triangle from Y = 4/3 to Y = 2.
    P(Y > 4/3) = (1/2) * (2/3) * (2/3)
    = 2/9

    Similarly P(Y < 2/3) is the area from Y = 0 to Y = 2/3 and
    P(Y < 2/3) = 2/9

    The events Y > 4/3 and Y < 2/3 are mutually exclusive so the probability of the union is the sum of the probabilities of both events.

    P(Y > 4/3 or Y < 2/3) = 2/9 + 2/9 = 4/9

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    Q10) Let k ≥ 1 be a positive integer. How many natural numbers 'n' exists with the following property:
    i) n has exactly k digits (in decimal representation)
    ii) all the digits of n are odd.
    iii) n is divisible by 5
    iv) the number m = n/5 has k odd digits.

    [OA : 3^k - 1]

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    From the 4th point, m = 2n/10

    As 'n' has 'k' digits, 2n will have either 'k' digits or 'k + 1' digits. Now 'm' has got to have 'k' odd digits, hence 2n must have 'k + 1' digits with the last digit obviously 0 (as 'n' is divisible by 5) and all the remaining digits odd. So, 'm' is also a 'k' digit number and combining that with the 4th point we can say that 'm' is a 'k' digit number with all of its digits odd.

    Hence, 'n' must have its digits such that each digit of 'n' when multiplied by 2 gives a carry of 1, if this doesn't hold then we will get an even digit in 'm', which is not acceptable.

    As each digit of 'n' when multiplied by 2, gives a carry forward of 1, each digit of 'n' should be greater than equal of 5 with the last digit always 5 (as 'n' is a multiple of 5)

    For k=1, possible cases =1

    For k=2, last digit can be selected in 1 way and 2nd last digit can be selected in 3 ways ( 5 or 7 or 9)

    For k=3, last digit in 1 way, other 2 digits in 3*3 ways

    Generalizing, 'n' can be formed in 3^(k - 1) ways

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    Q11) For how many natural numbers 'p', the three numbers (p - 1)/4, (p + 1)/2 and p are always prime?

    [OA : 1]

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