Math people, I need some help!

Discussion in 'Random Ramblings' started by Ec_Prokta, Jun 8, 2010.

  1. Ec_Prokta

    Ec_Prokta Continuum Shift Anomaly

    Jan 14, 2009
    A math problem Mom and I found while we were doing out math was impossible to figure out how to do. Could you help, please?

    A teacher wrote 2 numbers on the blackboard for his pupils to calculate the product of these two numbers. The last digit, 8 (Aka, the digit in the ones place.) was not written clearly so John mistook it for 6. As a result, the answer he got was 4740, which was wrong. Mary mistook the 8 for 3 and, after doing the problem, got 4695, which was also wrong. What is the correct answer?
    Last edited by a moderator: Jun 8, 2010
  2. chicknerd

    chicknerd Songster

    Apr 28, 2010
    You have two equations with two unknowns.

    For John, let x + 6 represent the first number, whose last digit is unknown but is known as 6 for now, let y represent the second number

    Then John's equation is

    (x+6)*y = 4740

    Mary's equation is then

    (x+3)*y = 4695

    Solve for x and y

    Your actual real first number will be (x+8)

    So the real answer is


    Hope this helps and it better not be an exam question!
  3. Ec_Prokta

    Ec_Prokta Continuum Shift Anomaly

    Jan 14, 2009
    Thanks ChickNerd! And it's not an exam question. [​IMG]
  4. Maidservant

    Maidservant Songster

    Feb 20, 2008
    Norwich, Norfolk, UK
    There is actually not enough information to complete the problem. We need to know at least how many digits were in each of those two numbers. Sounds like one of the SAT questions I tutor students with.

    Emily in NC
  5. HeatherLynn

    HeatherLynn Songster

    May 11, 2009
    Kentucky, Cecilia
    I was busy with my pencil and paper and linked this problem to the hubby, he found the answer I had labored over and had started trying to type out on yahoo so I just pasted theirs instead of continuing to try to type. You almost made it where I could not go to bed.

    the formula X*(Y - 2) = 4740 comes about because he multiplied a number that was 2 shy of being the actual number. they gave us the hint that it was in the ones place. If it had been in the 10
    s it would have been 20 off ... anywho so we know what x and the number less 2 results in and x and the number less 5 results in. We take that and solve for Y. Then plug the number back into the formula to find what x is.

    Let Ms. White's first number be "X", and her second number be "Y".

    John's equation:

    X*(Y - 2) = 4740

    Mary's equation:

    X*(Y - 5) = 4695

    Take the ratio of these two equations:

    X*(Y - 2) 4740
    ----------- = --------
    X*(Y - 5) 4695

    The "X"s cancel:

    (Y - 2) 4740
    -------- = ------
    (Y - 5) 4695

    Cross multiply:

    (4695)*(Y - 2) = 4740*(Y - 5)

    4695Y - 2*4695 = 4740Y - 5*4740

    Collect terms:

    45Y = 5*4740 - 2*4695 = 14310

    Y = 14310/45 = 318

    X(318 - 5) = 4695

    X = 4695/313 = 15

    The correct answer is X*Y = 15*318 = 4770

    Copied from yahoo
  6. chicknerd

    chicknerd Songster

    Apr 28, 2010
    Quote:Two equations and two unknowns. It is solvable. You do not need to know the number of digits either number is. X is a variable which represents any number, which means any number of digits.

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