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Thread: math: alternative to profit/loss calculations ..

  1. #1
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    math: alternative to profit/loss calculations ..

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    I have stumbled upon an easier alternative to the successive gain/loss types of sums. Instead of doing calculations like 1.15 x 1.15 for a sum what involves successive profits of 15% you can try the following :

    Question: Product P is marked up for a profit/loss of 15% and another profit/loss of 15% is made during a season sale etc..What
    is the total profit on the product - instead of multiplying numbers like 1.15 x 1.15 (which can be painful if you are in a hurry,
    you can follow this general observation that I have come up with while working on such calcuations one day!!

    a. profit 15% + profit of 15% = add (15 +15) + (1.5 x 1.5) = 32.25 (total profile)
    b. 15 profit + 15% loss = subtract loss from profit (15 -15) - (1.5 x 1.5) = -2.25 ( loss of 2.25)
    c. 15% loss + 15 % profit = same as b.
    d 15% loss + 15 % loss = 15 + 15 - (1.5x 1.5) = 27.75 ( total loss )


    For the first part: Add/subtract the whole numbers beased on whether they are gain/loss.
    a. gain, gain = gain + gain
    b. gain, loss (and vise-versa) = gain - loss
    c. loss, loss = loss + loss
    Whenever there is a LOSS invloved always subtract the second part, (1.5 x 1.5), from the first part.
    Also, the second part is derived by moving the decimal on the original numbers one place to the left.

    I have found it to work with all numbers. It even can be used for compunt interest problems.

    Disclaimer: I have not tested this against any mathematical principle. It works for me so thought I would share you you folks

    Let me know your thoughs!

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    Grrr... arjmen's Avatar
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    Re: math: alternative to profit/loss calculations ..

    Thanks for the post mba_06!!

    General Case:

    1) x% increase followed by a y% increase on a base amount of 'P'
    Final amount = P(1+x)(1+y) ---> 'x' and 'y' in their decimal form
    Initial amount = P

    => Change in Amount = P [1+x+y+xy] - P = P[(x+y)+(xy)]
    => Total percentage change = (x+y)+(xy)

    2) x% increase followed by a y% decrease on a base amount of 'P'
    Final amount = P(1+x)(1-y) ---> 'x' and 'y' in their decimal form
    Initial amount = P

    => Change in Amount = P[1+x-y-xy] - P = P[(x-y)-(xy)]
    => Total percentage change = (x-y)-(xy)

    3) x% decrease followed by y% decrease on a base amount of 'P'
    Final amount = P(1-x)(1-y) ---> 'x' and 'y' in their decimal form
    Initial amount = P
    => Change in Amount = P [1-x-y+xy] - P = P[-(x+y)+(xy)]
    => Total percentage change = -(x+y)+(xy)
    Last edited by arjmen; 06-15-2005 at 05:50 PM.

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    rum
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    HI,

    Could you please tell me how to solve this problem in the alternate methods mentioned above.

    2. Mr. Adams sold 2 properties, X and Y, for 30 000 each. She sold property X for 20% more than she paid for it and sold property Y for 20% less than she paid for it. If expenses are disregarded, what was her total net gain or loss, if any, on the 2 properties?
    Loss of 1250
    Loss of 2500
    Gain of 1250
    Gain of 2500
    No gain, no loss

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    Quote Originally Posted by rum
    HI,

    Could you please tell me how to solve this problem in the alternate methods mentioned above.

    2. Mr. Adams sold 2 properties, X and Y, for 30 000 each. She sold property X for 20% more than she paid for it and sold property Y for 20% less than she paid for it. If expenses are disregarded, what was her total net gain or loss, if any, on the 2 properties?
    Loss of 1250
    Loss of 2500
    Gain of 1250
    Gain of 2500
    No gain, no loss
    Rum, I do not think the alternate method mentioned above applies to this case. In this case there are two different transactions taking place at two different costs/sale prices. The alternate method I had mentioned really applies to transaction done multiple times starting with one cost/sale price with successive profits/losses. That method basically substitutes difficult multiplications with easier additions/substractions.

    hope that help!

    good luck

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    Quote Originally Posted by mba_06
    I have stumbled upon an easier alternative to the successive gain/loss types of sums. Instead of doing calculations like 1.15 x 1.15 for a sum what involves successive profits of 15% you can try the following :

    Question: Product P is marked up for a profit/loss of 15% and another profit/loss of 15% is made during a season sale etc..What
    is the total profit on the product - instead of multiplying numbers like 1.15 x 1.15 (which can be painful if you are in a hurry,
    you can follow this general observation that I have come up with while working on such calcuations one day!!

    a. profit 15% + profit of 15% = add (15 +15) + (1.5 x 1.5) = 32.25 (total profile)
    b. 15 profit + 15% loss = subtract loss from profit (15 -15) - (1.5 x 1.5) = -2.25 ( loss of 2.25)
    c. 15% loss + 15 % profit = same as b.
    d 15% loss + 15 % loss = 15 + 15 - (1.5x 1.5) = 27.75 ( total loss )


    For the first part: Add/subtract the whole numbers beased on whether they are gain/loss.
    a. gain, gain = gain + gain
    b. gain, loss (and vise-versa) = gain - loss
    c. loss, loss = loss + loss
    Whenever there is a LOSS invloved always subtract the second part, (1.5 x 1.5), from the first part.
    Also, the second part is derived by moving the decimal on the original numbers one place to the left.

    I have found it to work with all numbers. It even can be used for compunt interest problems.

    Disclaimer: I have not tested this against any mathematical principle. It works for me so thought I would share you you folks

    Let me know your thoughs!
    I agree with this trick. Also, i wanted to point out based on your numbers that if you have to sqaure a number ending with 5, then this is the trick

    lets say the number is: a5. Then (a5)^2 = a(a+1)25. Just place 25 next to a(a+1). ex: 45^2 = 4(4+1)25 ==> 2025

    --C

  6. #6
    An Urch Guru Pundit Swami Sage SangFroid's Avatar
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    Hey Mba, Arjmen and rayara,
    why didn't u tell me this trick earlier ??
    Thanks a lot for this.... really a genious beeline
    by the way thanksa lot to Rum too...who sifted this from the old ones...
    if he had not, then it would have remained hidden
    This post really deserves a "Thumbs up"

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