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Richard Wiseman’s Friday puzzle 04-11-2014 solution.

12 Apr

Richard Wiseman's Friday puzzle 04-11-2014 solution.

Richard Wiseman’s Friday puzzle 04-11-2014 solution.

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6 Responses to “Richard Wiseman’s Friday puzzle 04-11-2014 solution.”

  1. pineapple01 April 13, 2014 at 12:48 pm #

    m_w_1=m_1*r_1 (1)

  2. pineapple01 April 13, 2014 at 12:54 pm #

    It’s easy to say it in math.

    Given
    m_1=100lb\\ r_1=99\%\\ r_2=98\%\\
    Find out
    m_{e-w}=?\\

    System of equations is translated from given conditions.
    \begin{cases}    m_{w1} = m_1 * r_1 (1)\\    m_{w2} = m_2 * r_2 (2)\\    m_2 = m_1 - m_{e-w} (3)\\    m_{e-w} = m_{w1} - m_{w2} (4)\\   \end{cases}
    Solution for system of equations is:
    m_{w2} = (m_1 - m_{e-w}) * r_2 (5) (from 2, 3) => \\ m_{e-w}= m_1*r_1 - (m_1 - m_{e-w})*r_2 (from 5, 1, 4) => \\ m_{e-w} = m_1*r_1 - m_1*r_2 + m_{e-w}*r_2 => \\ m_{e-w} * (1-r_2)= m_1 * (r_1 - r_2) \\

    Final result:
    m_{e-w} = m_1 * \frac{r_1 - r_2}{1 - r_2}
    Add numbers:
    m_{e-w} = 100 * \frac{0.99 - 0.98}{1 - 0.98} = 100 * \frac{0.01}{0.02} = 100 * \frac{1}{2} = 50 lb

    Any questions?

  3. pineapple01 June 1, 2014 at 12:28 pm #

    $e=m*c^2$

  4. pineapple01 June 9, 2014 at 10:43 pm #

    Given,
    c < 100  (1)\\ c = (10*a + b)  (2)
    ,where
    \begin{cases}     0 <= a < 10 \\     0 <= b < 10  \\ \end{cases}
    based on equation (1), (2) and, a, b are integers.

    c_{reverse} = (10*b + a)  (3) \\ c_{reverse} = (1\frac{1}{5}) * c (4)
    is given as well.

    Let's find a solution from what is given by combining (2), (3) and (4)

    (10*b + a) = (1\frac{1}{5}) * (10*a + b) \\ or\\ (10*b + a) = (\frac{6}{5}) * (10*a + b)\\ or\\ (10*b + a) = \frac{60*a + 6*b}{5}\\ or\\ 50*b + 5*a = 60*a + 6*b\\ or\\ 50*b - 6*b  = 60*a - 5*a\\ or\\ 44*b = 55*a\\ or\\ 4*b = 5*a\\  or\\ \begin{cases}     a=4 \\     b=5  \\ \end{cases}

    Final answer is 45 based on (1).

  5. pineapple01 July 1, 2014 at 7:51 am #

    18 * \pi

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