Quant Boosters - Soumil Jain, CAT Quant 100 percentiler, IIM Calcutta - Set 1


  • IIM Calcutta | IIT Kanpur | CAT Quant 100 percentiler


    Three positive integers a, b, and c are such that their average is 20 and a ≤ b ≤ c. If the median is (a + 11), what is the least possible value of c?

    Theoretically, the least value of c is when c = b.
    Therefore, a + (a + 11) + (a + 11) = 60 (b and c are equal and b, the median, is a + 11)
    Or 3a = 38 or a = 12.66
    So, b = c = 12.66 + 11 = 23.66

    However, we know that these numbers are all integers.
    Therefore, a, b, and c cannot take these values.
    So, the least value for c with this constraint is NOT likely to be when c = b

    Let us increment c by 1. Let c = (b + 1)
    In this scenario, a + (a + 11) + (a + 12) = 60
    Or 3a = 37. The value of the numbers is not an integer in this scenario as well.

    Let us increment c again by 1. i.e., c = b + 2
    Now, a + (a + 11) + (a + 13) = 60
    Or 3a = 36 or a = 12.
    If a = 12, b = 23 and c = 25.
    The least value for c that satisfies all these conditions is 25

    A sequence of numbers is defined as 2 = a_n – a_(n-1) . S_n is sum upto n terms in this sequence and a_3 = 5. How many values m, n exist such than S_m – S_n = 65?

    a3 = 5, a4 = 7, a5 = 9, a2 = 3, a1 = 1
    So, the sequence is nothing but 1, 3, 5, 7, 9....
    S1 = 1
    S2 = 4
    S3 = 9
    S4 = 16
    Sn = n2
    Sm – Sn
    => m2 – n2 = 65
    => (m + n) ( m – n) = 65, m, n are natural numbers.
    => This could be 65 × 1 or 13 × 5. Two possibilities. Note that this cannot be written as 1 x 65 or 5 x 13 as m + n > m - n {as m, n are natural numbers}

    4x^3 + ax^2 – bx + 3 divided by x – 2 leaves remainder 2, divided by x + 3 leaves remainder 3. Find remainder when it is divided by x + 2

    Let f(x) = 4x3 + ax2 – bx + 3.
    Remainder on division of f(x) by x - 2 = f(2)
    Or, f(2) = 32 + 4a – 2b + 3 = 2
    4a – 2b = -33 ----Eqn (1)
    Remainder on division of f(x) by x + 3 = f(-3)
    Or, f(-3) = –108 + 9a + 3b + 3 = 3
    9a + 3b = 108
    Dividing by 3, the above equation becomes
    3a + b = 36. Multiplying this equation by 2, we get
    6a + 2b = 72 ----Eqn (2)
    Adding equations (1) and (2), we have 10a = 39
    a = 3.9
    4(3.9) – 2b = –33
    b = 24.3
    f(x) = 4x3 + 3.9x2 – 24.3x + 3
    Remainder on division of f(x) by x + 2 = f(-2)
    f(–2) = 4(–8) + 3.9(4) – 24.3(–2) + 3 = 35.2

    If x is a positive integer such that (x-1)(x-3)(x-5)....(x-93) < 0, how many values can x take?

    The expression takes negative values when x = 4, 8, 12, 16 ..... 92 (multiples of 4).
    92, the last value that x can take is the 23rd multiple of 4. Hence, number of such values of x = 23

    A number when divided by 18 leaves a remainder 7. The same number when divided by 12 leaves a remainder n. How many values can n take?

    Number can be 7, 25, 43, 61, 79.
    Remainders when divided by 12 are 7 and 1.
    n can take exactly 2 values

    If a three digit number ‘abc’ has 3 factors, how many factors does the 6-digit number ‘abcabc’ have?

    abc’ has exactly 3 factors, so ‘abc’ should be square of a prime number. (This is an important inference, please remember this).

    Any number of the form p^a q^b r^c will have (a + 1) (b + 1) (c + 1) factors, where p, q, r are prime. So, if a number has 3 factors, its prime factorization has to be p2.

    ‘abcabc’ = ‘abc’ * 1001 or abc * 7 * 11 * 13 (again, this is a critical idea to remember)

    Now, ‘abc’ has to be square of a prime number. It can be either 121 or 169 (square of either 11 or 13) or it can be the square of some other prime number.

    When abc = 121 or 169, then ‘abcabc’ is of the form p^3q^1r^1 1, which should have 4 * 2 * 2 = 16 factors.

    When ‘abc’ = square of any other prime number (say 172 which is 289) , then ‘abcabc’ is of the form p^1q^1r^1s^2 , which should have 2 * 2 * 2 * 3 = 24 factors

    So, ‘abcabc’ will have either 16 factors or 24 factors

    Working alone, A can complete a task in ‘a’ days and B in ‘b’ days. They take turns in doing the task with each working 2 days at a time. If A starts they finish the task in exactly 10 days. If B starts, they take half a day more. How long does it take to complete the task if they both work together?

    A starts and works for 2 days. So, A will work on day 1 and day 2.
    Then B will work for the next 2 days. B will work on day 3 and day 4.
    A will continue for the next 2 days. i.e., on day 5 and day 6.
    B will work on day 7 and day 8.
    A will for the last 2 days i.e., day 9 and day 10.

    Therefore, A will work on day 1, day 2, day 5, day 6, day 9, and day 10. i.e., for 6 days.
    And B will work on day 3, day 4, day 7, and day 8. i.e., for 4 days.

    In 6 days A will complete 6/a of the task.
    In 4 days B will complete 4/b of the task.

    With A working 6 days and B working 4 days, the task is completed.
    i.e., 6/a + 4/b = 1 .... eqn (1)

    Therefore, B will work on day 1, day 2, day 5, day 6, day 9, and day 10. i.e., for 6 days.
    And A will work on day 3, day 4, day 7, day 8 and half a day on day 11. i.e., for 4.5 days.

    In 6 days B will complete 6/b of the task.
    In 4.5 days A will complete 4.5/b of the task.

    With B working 6 days and A working 4.5 days, the task is completed.
    i.e., 4.5/a + 6/b = 1 .... eqn (2)

    Solving the two equations we get a = 9 days and b = 12 days.

    The question: How long does it take to complete the task if they both work together?

    Working together A and B will complete 7/36th of the task in a day.

    Hence, they will complete the task in 36/7 days

    What is the area enclosed in the region defined by y = |x – 1| + 2, line x = 1, X–axis and Y–axis?

    The point of intersection of y = |x – 1| + 2 where the y–axis can be found by substituting x = 0 in the equation. Thus we get y = 3.

    Required area = Area of the trapezium formed by the points (0, 0), (1, 0), (1, 2) and (0, 3).

    Area of a trapezium = 12 × height × sum of the parallel sides = 12 × 1 × (3 + 2) = 52 sq units.

    If x and y are integers and |x - y| = 12, what is the minimum possible value of xy?

    Square both sides and solve

    Squaring both sides, we get (x - y)^2 = 144
    x^2 + y^2 - 2xy = 144

    Add, 4xy to both sides of the equation.
    x^2 + y^2 - 2xy + 4xy = 144 + 4xy
    x^2 + y^2 + 2xy = 144 + 4xy
    Or (x + y)^2 = 144 + 4xy

    (x + y)^2 will NOT be negative for real values of x and y.
    i.e., (x + y)^2 ≥ 0

    ∴ 144 + 4xy ≥ 0

    Or 4xy ≥ -144
    So, xy ≥ -36

    The least value that xy can take is -36

    If | r-6| = 11 and | 2q-12| = 8, then what is the minimum possible value of q/r ?

    There are clearly only 4 combinations, r = 17, q = 10 # r = -5 , q= 10 # r = 17, q= 2 and r = -5, q = 2. Now the minimum value therefore would be 10/(-5) = -2



  • In this question :

    "If x is a positive integer such that (x-1)(x-3)(x-5)....(x-93) < 0, how many values can x take? "

    How did you get that for all multiples of 4 the expression will take negative value ?


  • Being MBAtious!


    @Piyus-Jain
    (x-1)(x-3)(x-5)....(x-93) --> we have 47 terms
    When x is a multiple of 4, we will have an odd number of negative values which yields a negative result.
    For example, if x = 4, except (x-1) and (x-3) all the other 45 values starting from (x-5) would be negative and the overall product is a negative value < 0
    Also, If x = 8, except (x-1), (x-3), (x-5) and (x-7) all the other 43 values starting from (x-9) would be negative resulting in a negative value.. again < 0.
    Similarly we can extend the logic to all the multiple of 4 till 92 for the given expression.
    Hope it is clear.



  • @zabeer

    Thanks for the explanation, what I wanted to know was how did we decide on the number 4, thought process. TIA :)


  • Being MBAtious!


    @Piyus-Jain
    (x-1)(x-3)(x-5)....(x-93) has to be negative.

    First hint - x has to be less than 93 otherwise the value will not be negative
    Second hint - x cannot be an odd number less than or equal to 93 - as the value will be 0 in this case.
    Conclusion - 1 - x has to be a subset of even numbers less than 93
    Third hint - There are 47 terms in the expression and we need odd numbers of negative values to get a final negative value as two negatives cancels out to yield a positive value
    If x = 2, 46 terms are negative (overall value - positive)
    if x = 4, 45 terms are negative (overall value - negative)
    if x = 6, 44 terms are negative (overall value - positive)
    if x = 8, 43 terms are negative (overall value - negative)
    and so on.
    Conclusion 2 - x can take values - 4, 8, 12 ... 88, 92 - 23 values
    Clear ? :thumbsup_tone2:



  • Awesome, Thanks :)


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