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A two digit positive number is such that the product of its digits is 6. If 9 is added to the number, the digits interchange their places. Find the number.

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This question has been taken from Book:- ML aggarwal, Avichal publication, class10th, quadratic equation in one variable, chapter 5, exercise 5.5
This is an important ques and asked in 2014

A two digit positive number is such that the product of its digits is 6.

If 9 is added to the number, the digits interchange their places.

Find the number.

Question no.12 , ML Aggarwal, chapter 5, exercise 5.5, quadratic equation in one variable, ICSE board,

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1 Answer

  1. Solution:

    Let us consider 2-digit number be ‘xy’ = 10x + y

    Reversed digits = yx = 10y + x

    So according to the question,

    10x + y + 9 = 10y + x

    It is given that,

    xy = 6

    y = 6/x

    so, by substituting the value in above equation, we get

    10x + 6/x + 9 = 10(6/x) + x

    By taking LCM,

    10x2 + 6 + 9x = 60 + x2

    10x2 + 6 + 9x – 60 – x2 = 0

    9x2 + 9x – 54 = 0

    Divide by 9, we get

    x2 + x – 6 = 0

    let us factorize,

    x2 + 3x – 2x – 6 = 0

    x(x + 3) – 2 (x + 3) = 0

    (x + 3) (x – 2) = 0

    So,

    (x + 3) = 0 or (x – 2) = 0

    x = -3 or x = 2

    Value of x = 2 [since, -3 is a negative value]

    Now, substitute the value of x in y = 6/x, we get

    y = 6/2 = 3

    ∴ 2-digit number = 10x + y = 10(2) + 3 = 23

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