Quant Boosters - Anubhav Sehgal, NMIMS Mumbai - Set 3


  • NMIMS, Mumbai (Marketing)


    Q26) Two different two-digit natural numbers are written beside each other such that the larger number is written on the left. When the absolute difference of the two numbers is subtracted from the four-digit number so formed, the number obtained is 5481. What is the sum of the two two-digit numbers?


  • NMIMS, Mumbai (Marketing)


    abcd - (ab -cd ) = 5481
    1000a +100b + 10c + d - 10a + 10c-b+d = 5481
    990a + 99b + 20c + 2d = 5481
    Clearly a=5 as b, c and d all are digits and a cd).
    And sum = 55 + 18 = 73


  • NMIMS, Mumbai (Marketing)


    Q27) A natural number x leaves a remainder 1 when divided by p. The resultant quotient, when divided by q, leaves a remainder 2. The resultant quotient, when divided by r, leaves a remainder 3 and the quotient, thus obtained, is exactly divisible by 5. If p, q and r are all natural numbers, which of the following is the least possible value of x?


  • NMIMS, Mumbai (Marketing)


    It is a case of successive division. In such a case the number is written as:
    x= (p(q(r(5k) + 3) + 2) + 1.
    As the number needs to be smallest, we must take as small values as possible.
    Now, remainder with p is 1, so p >1. So, p = 2.
    Remainder with q = 2, so, q > 2, so q = 3.
    Remainder with r = 3, so, r > 3, so, r = 4.
    And, k = 0. x = 2(3(4(0) + 4) + 2) + 1 = 2(3(4) + 2) + 1 = 2(14) + 1 = 29.


  • NMIMS, Mumbai (Marketing)


    Q28) One of the smaller sides of a right angled triangle is (2^2) * (3^3) * (4^4) * (5^5) * (6^6) * (7^7). It is known that other two sides are integers. How many triangles of this type are possible?


  • NMIMS, Mumbai (Marketing)


    b^2 = (2^4) * (3^6) * (4^8 ) * (5^10) * (6^12) * (7^14)
    = (2^32) * (3^18 ) * (5^10) * (7^14)
    => (a + c)(a - c) = (2^32) * (3^18 ) * (5^10) * (7^14)
    Since both (a + c) and (a - c) are even, say 2k and 2n, then
    k * n = (2^30) * (3^18 ) * (5^10) * (7^14)

    Now, k is greater than n, so we just have to write (2^30) * (3^18 ) * (5^10) * (7^14) as product of two numbers where k is the greater one and n is smaller one, then we can get values for a and c
    So, answer will be (31 * 19 * 11 * 15 - 1)/2 = 48592


  • NMIMS, Mumbai (Marketing)


    Q29) In a test consisting of 15 questions, 3 marks are awarded for a correct answer, 1 mark is deducted for an incorrect answer and no mark is awarded for an unattempted question. If a student attempts at least one question in the paper, what is the number of distinct scores that he can get?


  • NMIMS, Mumbai (Marketing)


    Basically if no of marks are +n for correct and -1 for incorrect, then
    Number of score which are not possible is given by (n - 1) + (n - 2) + ... + 1 = n * (n - 1)/2 = C (n, 2)
    Assume a +6 and -1 marking scheme.
    When one gets 30 correct; he gets 180.
    And with 29 correct, he gets 174.
    So, for a score of more than174, one needs 30 correct.
    And with 30 correct i.e. all correct, you can have 0 wrong.
    So, 30 correct, 1 wrong or 2 wrong or 3 wrong or 4 wrong or 5 wrong is not possible.
    With 29 correct, 1 wrong is possible. But 2, 3 , 4 or 5 is not.
    With 28 correct, 3, 4, 5 wrong is not possible.
    So, basically, when we have +6 and -1: Number of invalid scores are: 1 + 2 + 3 + 4 + 5 = 15

    When we have +5 and -1:
    Number of invalid scores are: 1 + 2 + 3 + 4 = 10.

    When we have +4 and -1:
    Number of invalid scores are: 1+2+3 = 6.

    Here, 15 questions are there with +3 and -1 marking scheme.
    Highest marks possible = 15 * 3 = 45
    Lowest possible marks = 15 * -1 = -15
    Total marks = 45 – (-15) + 1 = 61 possible scores.
    Out of which, C (3, 2) = 3 are not possible.
    So, number of valid scores = 61 – C (3, 2) = 58.

    How many integers ‘n’ are there between 0 and 10^99, such that the unit digit of n^3 is 1?

    Each number from 1 to 10 results in distinct unit digit for cube.
    So, each digit in cube appears 1/10 times.
    So, from 1 to 10^99 it will appear 10^98 times


  • NMIMS, Mumbai (Marketing)


    Q30) How many subsets of the set {1, 2, 3, 4 ... 30} have the property that the sum of the elements of a subset is less than or equal to 232?


  • NMIMS, Mumbai (Marketing)


    Sum of all elements of set S = 465.
    For any subset having sum of elements x, there will be one complimentary set which is having sum of elements equal to (465 – x).

    Example let there be a set {1, 2, 3, 4}.
    Then subset which is having sum =3 will be {1, 2} {3}: (2)
    and subset having sum (1+2+3+4) -3 =7 will also be 2: {3,4} {1,2,4}

    So, number of subsets having sum of elements x = Number of subsets having sum of elements (465 - x)
    => No of subsets having sum 1 = No of subsets having sum 464
    Also, No of subsets having sum 2 = No of subsets having sum 463
    ….
    No of subsets having sum 232 = No of subsets having sum 233
    Adding them we will get,
    No of subsets having sum less than or equal to 232
    = No of subsets having sum more than 233
    = 2^30/2
    = 2^29


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