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Algebra and Quadratic Equations Actual Question MBA CET 2024  Slot 1

Quant in MBA CET 2024 Slot-1 Actual Questions with Solutions

  1. Question

If a, b and c are positive integers such that ab = 432 bc = 96 and c < 9, then the smallest possible value of a + b + cis:


a)59
b)38
c)49
d)56
e)46

Explanation

SOLUTION[E]-

bc = 96

c < 9

Possible factors can be 48*2,32*3,24*4, 16 * 6 ,12*8

ab = 432

Possible factors can be (closest observation 9 * 48 ,18*24

From the above

a = 18

b = 24

c = 4

So, a + b + c = 46

.. The smallest value can be 46


Question-2

    A, B and C have a few coins with them. 7 times the number of coins that A has is equal to 5 times the number of coins B has while 6 times the number of coins B has is equal to 11 times the number of coins C has.

    What is the minimum number of coins that A, B and C put together have with them?


    a)112
    b)174
    c)154
    d)164
    e)110

    Explanation

    SOLUTION[B]

    The minimum number of coins with A, B, and C put together would be 174.

    Step-by-step explanation:

    Let the number of coins A, B, and C have

      respectively.

    • Then, Seven times a = Five times b

    7a = 5b

    • Now, Six times b = Eleven times c

    6b = 11c

    • Now, the total number of coins =  a+b+c

    Substituting the values of a and c from above, we obtain

    • The total number of coins
    • Thus, if b has 77 coins, then the total number of coins would become 174


    Question-3

      Determine, as to how many distinct positive integer-valued solutions exist to the equation:

      (x ^ 2 – 7x + 11)(x ^ 2 – 13x + 42) = 1


      a)0
      b)6
      c)4
      d)2
      e)8

      Explanation

      SOLUTION-[B]-6


      Question-4

        Mathematician, Dr. Arnav has devised a magic 3 * 3 square box, shown below, where in the sum of each row, column and diagonal is the same number, W. The entries in each of the 9 cells are given by x, y, z. Find the value of W.

        a)18
        b)48
        c)12
        d)24
        e)36

        Explanation

        SOLUTION[E]


        Question-5

          If given that:

          i. – y ^ 2 + x ^ 2 = 20

          ii. – y ^ 3 – 2x ^ 2 – 4z >= – 12/8

          iii. x, y, and z are all positive integers

          Find the value of z ?
          a)6
          b)1
          c)24
          d)3
          e)12

          Explanation

          SOLUTION[B]-1

          Since x ^ 2 – y ^ 2 = 20 and x, y, z are positive integers,

          (x+y)^ * (x – y) = 20 , Hence x – y, x + y are factors of 20.

          Since x, y are positive integers, x+y is always positive, and for the product of (x+y)^ * (x – y) to be positiv x-y must be positive.

          x, y are positive integers and x-y is positive x must be greater than y.

          The possible cases are: (x + y = 10, x – y = 2) (x + y = 5, x – y = 4)

          The second case fails because we get x = 9/2 y = 1/2 but x, y are integral values

          For case one x = 6 y = 4 y ^ 3 – 2x ^ 2 – 4z >= – 12 Substituting the values of x and y, we have: 64 – 72 -4^ * z>-12 -8-4^ * z >= – 12

          z < 1

          Since x, y, z are positive integers, the only possible value for z is 1.

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