Calculating discount not working Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Find the sale price given the 40% markup over wholesale and later 20% discountCalculating percentage to compensate for percent discount.Profit-Loss : DiscountHow to work out this discount?Calculating a 20% discount by multiplying by 0.8333?Finding discount % given original price and discount value?GRE percentages / discount question$(1+x)(1-x)=1-x^2$Calculate remain discount for an amount .Find Original Value using result of discounted discount

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Calculating discount not working



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Find the sale price given the 40% markup over wholesale and later 20% discountCalculating percentage to compensate for percent discount.Profit-Loss : DiscountHow to work out this discount?Calculating a 20% discount by multiplying by 0.8333?Finding discount % given original price and discount value?GRE percentages / discount question$(1+x)(1-x)=1-x^2$Calculate remain discount for an amount .Find Original Value using result of discounted discount










7












$begingroup$


To increase a number by a percentage I was taught the formula no x 1.percentage, so for example:



100 increased by 15%:
100 x 1.15 = 115


But today, I learned the hard way that doing the inverse of the above (i.e. no / 1.percentage), isn't the correct way of decreasing a number by a percentage (aka "applying a discount"). E.g:



100 decreased by 15%:
100 / 1.15 = 86.95652 <- WRONG
100 - (100 x 0.15) = 85 <- RIGHT


My Question: Why doesn't the formula no / 1.percentage work, when decreasing a number by a percentage?



Side Question: Is there a simpler formula to decrease a number by a percentage than no - (no x 0.percent)?



Note: My math knowledge is super basic, so maybe pretend you're explaining this to a twelve year old when answering. But if you want to give sophisticated answers for future readers, that's fine too, I might just not understand it though.










share|cite|improve this question









$endgroup$







  • 3




    $begingroup$
    $86.95652$ is the answer to the question : If a product is worth $100 after a 15% increase, how much was it worth before the increase?
    $endgroup$
    – Eric Duminil
    Apr 1 at 18:35










  • $begingroup$
    Please don't use codeblocks for anything that is not actual code. There are tutorials available on using MathJax typesetting, such as that in the answers below.
    $endgroup$
    – Nij
    Apr 2 at 7:35















7












$begingroup$


To increase a number by a percentage I was taught the formula no x 1.percentage, so for example:



100 increased by 15%:
100 x 1.15 = 115


But today, I learned the hard way that doing the inverse of the above (i.e. no / 1.percentage), isn't the correct way of decreasing a number by a percentage (aka "applying a discount"). E.g:



100 decreased by 15%:
100 / 1.15 = 86.95652 <- WRONG
100 - (100 x 0.15) = 85 <- RIGHT


My Question: Why doesn't the formula no / 1.percentage work, when decreasing a number by a percentage?



Side Question: Is there a simpler formula to decrease a number by a percentage than no - (no x 0.percent)?



Note: My math knowledge is super basic, so maybe pretend you're explaining this to a twelve year old when answering. But if you want to give sophisticated answers for future readers, that's fine too, I might just not understand it though.










share|cite|improve this question









$endgroup$







  • 3




    $begingroup$
    $86.95652$ is the answer to the question : If a product is worth $100 after a 15% increase, how much was it worth before the increase?
    $endgroup$
    – Eric Duminil
    Apr 1 at 18:35










  • $begingroup$
    Please don't use codeblocks for anything that is not actual code. There are tutorials available on using MathJax typesetting, such as that in the answers below.
    $endgroup$
    – Nij
    Apr 2 at 7:35













7












7








7


0



$begingroup$


To increase a number by a percentage I was taught the formula no x 1.percentage, so for example:



100 increased by 15%:
100 x 1.15 = 115


But today, I learned the hard way that doing the inverse of the above (i.e. no / 1.percentage), isn't the correct way of decreasing a number by a percentage (aka "applying a discount"). E.g:



100 decreased by 15%:
100 / 1.15 = 86.95652 <- WRONG
100 - (100 x 0.15) = 85 <- RIGHT


My Question: Why doesn't the formula no / 1.percentage work, when decreasing a number by a percentage?



Side Question: Is there a simpler formula to decrease a number by a percentage than no - (no x 0.percent)?



Note: My math knowledge is super basic, so maybe pretend you're explaining this to a twelve year old when answering. But if you want to give sophisticated answers for future readers, that's fine too, I might just not understand it though.










share|cite|improve this question









$endgroup$




To increase a number by a percentage I was taught the formula no x 1.percentage, so for example:



100 increased by 15%:
100 x 1.15 = 115


But today, I learned the hard way that doing the inverse of the above (i.e. no / 1.percentage), isn't the correct way of decreasing a number by a percentage (aka "applying a discount"). E.g:



100 decreased by 15%:
100 / 1.15 = 86.95652 <- WRONG
100 - (100 x 0.15) = 85 <- RIGHT


My Question: Why doesn't the formula no / 1.percentage work, when decreasing a number by a percentage?



Side Question: Is there a simpler formula to decrease a number by a percentage than no - (no x 0.percent)?



Note: My math knowledge is super basic, so maybe pretend you're explaining this to a twelve year old when answering. But if you want to give sophisticated answers for future readers, that's fine too, I might just not understand it though.







arithmetic percentages






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 1 at 14:02









PrintlnParamsPrintlnParams

655




655







  • 3




    $begingroup$
    $86.95652$ is the answer to the question : If a product is worth $100 after a 15% increase, how much was it worth before the increase?
    $endgroup$
    – Eric Duminil
    Apr 1 at 18:35










  • $begingroup$
    Please don't use codeblocks for anything that is not actual code. There are tutorials available on using MathJax typesetting, such as that in the answers below.
    $endgroup$
    – Nij
    Apr 2 at 7:35












  • 3




    $begingroup$
    $86.95652$ is the answer to the question : If a product is worth $100 after a 15% increase, how much was it worth before the increase?
    $endgroup$
    – Eric Duminil
    Apr 1 at 18:35










  • $begingroup$
    Please don't use codeblocks for anything that is not actual code. There are tutorials available on using MathJax typesetting, such as that in the answers below.
    $endgroup$
    – Nij
    Apr 2 at 7:35







3




3




$begingroup$
$86.95652$ is the answer to the question : If a product is worth $100 after a 15% increase, how much was it worth before the increase?
$endgroup$
– Eric Duminil
Apr 1 at 18:35




$begingroup$
$86.95652$ is the answer to the question : If a product is worth $100 after a 15% increase, how much was it worth before the increase?
$endgroup$
– Eric Duminil
Apr 1 at 18:35












$begingroup$
Please don't use codeblocks for anything that is not actual code. There are tutorials available on using MathJax typesetting, such as that in the answers below.
$endgroup$
– Nij
Apr 2 at 7:35




$begingroup$
Please don't use codeblocks for anything that is not actual code. There are tutorials available on using MathJax typesetting, such as that in the answers below.
$endgroup$
– Nij
Apr 2 at 7:35










7 Answers
7






active

oldest

votes


















14












$begingroup$

The method of increasing a value $V$ by $p%$ is actually adding $Vtimesfrac p100$:
$$V_textnew = V + Vtimesfrac p100 = Vtimesleft(1 + frac p100right)$$
For $p=15$ you have a nice multiplier
$$left(1 + frac p100right) = 1 + 0.15 = 1.15$$



Remember, however, that '1.p' is a shortcut or mnemonic, not a method. It wouldn't even work for $p$ greater than 99, say for $+120%$. And it certainly won't work for $p <0$.



For decreasing you need to apply the method, which is:
$$V_textnew=V - Vtimesfrac p100 = Vtimesleft(1 - frac p100right)$$
so for 15-percent decrement you get:
$$V_textnew=V - Vtimesfrac 15100 = Vtimes(1 - 0.15) = Vtimes 0.85$$



EDIT



To answer your main question directly, forget percentages. Suppose you need to rise a value from 100 to 125. The new value is $$125 = 100timesfrac 54=100times( 1+frac 14) $$ hence an increase by 25%.
Now, if you want to reduce it back to 100, you get $$100=125times frac 45=120times(1-frac 15)$$ which is 20% decrease.
Why was it 1/4 before, and 1/5 now? Because the same difference $pm 25$ was taken relative to 100 in the former case and to 125 in the latter one. When we added a fourth part, we got five fourths of the initial value, so we needed to take away one of those five, i.e. a fifth part to get back.



And here you have the difference in percentages: 1/4=25%, while 1/5=20%.



So, a division like $value/1.p$ reduces the value by p% of the resulting value, not of the original one.






share|cite|improve this answer











$endgroup$




















    11












    $begingroup$

    The reason that your suggested formula doesn't work is that "increase by $k$%" and "decrease by $k$%" do not undo one another like $+5$ and $-5$ or $times 2$ and $div 2$ do (we would say that they are not inverses).



    For example $100$ increased by $50%$ is $150$. However, $150$ decreased by $50%$ is 75.



    Therefore, if we want to decrease a number by a percentage, we can't just attempt to do the opposite operation of increasing by that percentage.






    share|cite|improve this answer









    $endgroup$








    • 5




      $begingroup$
      It's also important to understand this applies to any ratio and can be used to play psychological tricks. For example, if A is 50% faster than B, then B is only about 30% slower than A. You could use either depending on whether you want to emphasize or downplay this gap.
      $endgroup$
      – MooseBoys
      Apr 1 at 20:04


















    5












    $begingroup$

    What you did is the right answer to a different problem. The general method, "multiply by $1+$ change" is a good one when used properly.



    To increase a number by $15%$ you multiply by
    $$
    1 + 0.15 = 1.15.
    $$

    To decrease a number by $15%$ you multiply by
    $$
    1 - 0.15 = 0.85.
    $$

    But to undo a $15%$ increase you have to "unmultiply" by $1.15$. So you divide by $1.15$. Since
    $$
    frac11.15 = 0.86956521739 approx 0.87 = 1 - 0.13,
    $$

    to undo a $15%$ increase you make a $13%$ decrease.






    share|cite|improve this answer









    $endgroup$




















      2












      $begingroup$

      To find $15 %$ of some number $N$, we solve an equation (proportion) $fracN100=fracx15$ or $x=0.15N$. That's why a $15%$ increase can be found as $N+0.15N=1.15N$ and decrease is found as $N-0.15N=0.85N$






      share|cite|improve this answer









      $endgroup$




















        2












        $begingroup$

        It works the same way...



        Increase : to increase of $15 %$ means to multiply by $(1+0,15)=1,15$.



        Decrease : to decrease of $15 %$ means to multiply by $(1-0,15)=0,85$.




        Unfortunately, the $15 %$ factors get similar results; thus it can be misleading.



        Consider instead an increase of $20 %$. Why the "inverse" operation (i.e. to divide by $1,2$ to decrease does not produce the correct result ?



        With a $20 %$ increase the updated price will be $120$.



        Now, what is the amount of a decrease of the updated price by $20 %$ ? It is not $20$ but $24$.



        And we have that $120 times (1-0,2)=96$ while $dfrac 1201,2=100$.






        share|cite|improve this answer











        $endgroup$




















          2












          $begingroup$

          Consider the literal meaning of "per cent"



          Increase $X$ by 15%:
          $$
          X + X times frac15100 = frac100X100 + frac15X100 = frac(100+15)X100 = frac115X100 = 1.15X
          $$



          Decrease X by 15%:
          $$
          X - X times frac15100 = frac100X100 - frac15X100 = frac(100-15)X100 = frac85X100 = 0.85X
          $$






          share|cite|improve this answer











          $endgroup$




















            1












            $begingroup$

            If we want to increase $n$ by $p$%, we work out what $p$% of $n$ is, and then add it to $n$ itself: $$0.ptimes n+n=(0.p+1)times n=1.ptimes n$$This is how you arrived to your formula.



            If we want to decrease $n$ by $p$%, we work out what $p$% of $n$ is, and then subtract it from $n$ itself:$$n-0.ptimes n=(1-0.p)times n$$This is how to derive the formula for a decrease.






            share|cite|improve this answer









            $endgroup$













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              7 Answers
              7






              active

              oldest

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              7 Answers
              7






              active

              oldest

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              active

              oldest

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              active

              oldest

              votes









              14












              $begingroup$

              The method of increasing a value $V$ by $p%$ is actually adding $Vtimesfrac p100$:
              $$V_textnew = V + Vtimesfrac p100 = Vtimesleft(1 + frac p100right)$$
              For $p=15$ you have a nice multiplier
              $$left(1 + frac p100right) = 1 + 0.15 = 1.15$$



              Remember, however, that '1.p' is a shortcut or mnemonic, not a method. It wouldn't even work for $p$ greater than 99, say for $+120%$. And it certainly won't work for $p <0$.



              For decreasing you need to apply the method, which is:
              $$V_textnew=V - Vtimesfrac p100 = Vtimesleft(1 - frac p100right)$$
              so for 15-percent decrement you get:
              $$V_textnew=V - Vtimesfrac 15100 = Vtimes(1 - 0.15) = Vtimes 0.85$$



              EDIT



              To answer your main question directly, forget percentages. Suppose you need to rise a value from 100 to 125. The new value is $$125 = 100timesfrac 54=100times( 1+frac 14) $$ hence an increase by 25%.
              Now, if you want to reduce it back to 100, you get $$100=125times frac 45=120times(1-frac 15)$$ which is 20% decrease.
              Why was it 1/4 before, and 1/5 now? Because the same difference $pm 25$ was taken relative to 100 in the former case and to 125 in the latter one. When we added a fourth part, we got five fourths of the initial value, so we needed to take away one of those five, i.e. a fifth part to get back.



              And here you have the difference in percentages: 1/4=25%, while 1/5=20%.



              So, a division like $value/1.p$ reduces the value by p% of the resulting value, not of the original one.






              share|cite|improve this answer











              $endgroup$

















                14












                $begingroup$

                The method of increasing a value $V$ by $p%$ is actually adding $Vtimesfrac p100$:
                $$V_textnew = V + Vtimesfrac p100 = Vtimesleft(1 + frac p100right)$$
                For $p=15$ you have a nice multiplier
                $$left(1 + frac p100right) = 1 + 0.15 = 1.15$$



                Remember, however, that '1.p' is a shortcut or mnemonic, not a method. It wouldn't even work for $p$ greater than 99, say for $+120%$. And it certainly won't work for $p <0$.



                For decreasing you need to apply the method, which is:
                $$V_textnew=V - Vtimesfrac p100 = Vtimesleft(1 - frac p100right)$$
                so for 15-percent decrement you get:
                $$V_textnew=V - Vtimesfrac 15100 = Vtimes(1 - 0.15) = Vtimes 0.85$$



                EDIT



                To answer your main question directly, forget percentages. Suppose you need to rise a value from 100 to 125. The new value is $$125 = 100timesfrac 54=100times( 1+frac 14) $$ hence an increase by 25%.
                Now, if you want to reduce it back to 100, you get $$100=125times frac 45=120times(1-frac 15)$$ which is 20% decrease.
                Why was it 1/4 before, and 1/5 now? Because the same difference $pm 25$ was taken relative to 100 in the former case and to 125 in the latter one. When we added a fourth part, we got five fourths of the initial value, so we needed to take away one of those five, i.e. a fifth part to get back.



                And here you have the difference in percentages: 1/4=25%, while 1/5=20%.



                So, a division like $value/1.p$ reduces the value by p% of the resulting value, not of the original one.






                share|cite|improve this answer











                $endgroup$















                  14












                  14








                  14





                  $begingroup$

                  The method of increasing a value $V$ by $p%$ is actually adding $Vtimesfrac p100$:
                  $$V_textnew = V + Vtimesfrac p100 = Vtimesleft(1 + frac p100right)$$
                  For $p=15$ you have a nice multiplier
                  $$left(1 + frac p100right) = 1 + 0.15 = 1.15$$



                  Remember, however, that '1.p' is a shortcut or mnemonic, not a method. It wouldn't even work for $p$ greater than 99, say for $+120%$. And it certainly won't work for $p <0$.



                  For decreasing you need to apply the method, which is:
                  $$V_textnew=V - Vtimesfrac p100 = Vtimesleft(1 - frac p100right)$$
                  so for 15-percent decrement you get:
                  $$V_textnew=V - Vtimesfrac 15100 = Vtimes(1 - 0.15) = Vtimes 0.85$$



                  EDIT



                  To answer your main question directly, forget percentages. Suppose you need to rise a value from 100 to 125. The new value is $$125 = 100timesfrac 54=100times( 1+frac 14) $$ hence an increase by 25%.
                  Now, if you want to reduce it back to 100, you get $$100=125times frac 45=120times(1-frac 15)$$ which is 20% decrease.
                  Why was it 1/4 before, and 1/5 now? Because the same difference $pm 25$ was taken relative to 100 in the former case and to 125 in the latter one. When we added a fourth part, we got five fourths of the initial value, so we needed to take away one of those five, i.e. a fifth part to get back.



                  And here you have the difference in percentages: 1/4=25%, while 1/5=20%.



                  So, a division like $value/1.p$ reduces the value by p% of the resulting value, not of the original one.






                  share|cite|improve this answer











                  $endgroup$



                  The method of increasing a value $V$ by $p%$ is actually adding $Vtimesfrac p100$:
                  $$V_textnew = V + Vtimesfrac p100 = Vtimesleft(1 + frac p100right)$$
                  For $p=15$ you have a nice multiplier
                  $$left(1 + frac p100right) = 1 + 0.15 = 1.15$$



                  Remember, however, that '1.p' is a shortcut or mnemonic, not a method. It wouldn't even work for $p$ greater than 99, say for $+120%$. And it certainly won't work for $p <0$.



                  For decreasing you need to apply the method, which is:
                  $$V_textnew=V - Vtimesfrac p100 = Vtimesleft(1 - frac p100right)$$
                  so for 15-percent decrement you get:
                  $$V_textnew=V - Vtimesfrac 15100 = Vtimes(1 - 0.15) = Vtimes 0.85$$



                  EDIT



                  To answer your main question directly, forget percentages. Suppose you need to rise a value from 100 to 125. The new value is $$125 = 100timesfrac 54=100times( 1+frac 14) $$ hence an increase by 25%.
                  Now, if you want to reduce it back to 100, you get $$100=125times frac 45=120times(1-frac 15)$$ which is 20% decrease.
                  Why was it 1/4 before, and 1/5 now? Because the same difference $pm 25$ was taken relative to 100 in the former case and to 125 in the latter one. When we added a fourth part, we got five fourths of the initial value, so we needed to take away one of those five, i.e. a fifth part to get back.



                  And here you have the difference in percentages: 1/4=25%, while 1/5=20%.



                  So, a division like $value/1.p$ reduces the value by p% of the resulting value, not of the original one.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Apr 12 at 21:39

























                  answered Apr 1 at 14:09









                  CiaPanCiaPan

                  10.4k11248




                  10.4k11248





















                      11












                      $begingroup$

                      The reason that your suggested formula doesn't work is that "increase by $k$%" and "decrease by $k$%" do not undo one another like $+5$ and $-5$ or $times 2$ and $div 2$ do (we would say that they are not inverses).



                      For example $100$ increased by $50%$ is $150$. However, $150$ decreased by $50%$ is 75.



                      Therefore, if we want to decrease a number by a percentage, we can't just attempt to do the opposite operation of increasing by that percentage.






                      share|cite|improve this answer









                      $endgroup$








                      • 5




                        $begingroup$
                        It's also important to understand this applies to any ratio and can be used to play psychological tricks. For example, if A is 50% faster than B, then B is only about 30% slower than A. You could use either depending on whether you want to emphasize or downplay this gap.
                        $endgroup$
                        – MooseBoys
                        Apr 1 at 20:04















                      11












                      $begingroup$

                      The reason that your suggested formula doesn't work is that "increase by $k$%" and "decrease by $k$%" do not undo one another like $+5$ and $-5$ or $times 2$ and $div 2$ do (we would say that they are not inverses).



                      For example $100$ increased by $50%$ is $150$. However, $150$ decreased by $50%$ is 75.



                      Therefore, if we want to decrease a number by a percentage, we can't just attempt to do the opposite operation of increasing by that percentage.






                      share|cite|improve this answer









                      $endgroup$








                      • 5




                        $begingroup$
                        It's also important to understand this applies to any ratio and can be used to play psychological tricks. For example, if A is 50% faster than B, then B is only about 30% slower than A. You could use either depending on whether you want to emphasize or downplay this gap.
                        $endgroup$
                        – MooseBoys
                        Apr 1 at 20:04













                      11












                      11








                      11





                      $begingroup$

                      The reason that your suggested formula doesn't work is that "increase by $k$%" and "decrease by $k$%" do not undo one another like $+5$ and $-5$ or $times 2$ and $div 2$ do (we would say that they are not inverses).



                      For example $100$ increased by $50%$ is $150$. However, $150$ decreased by $50%$ is 75.



                      Therefore, if we want to decrease a number by a percentage, we can't just attempt to do the opposite operation of increasing by that percentage.






                      share|cite|improve this answer









                      $endgroup$



                      The reason that your suggested formula doesn't work is that "increase by $k$%" and "decrease by $k$%" do not undo one another like $+5$ and $-5$ or $times 2$ and $div 2$ do (we would say that they are not inverses).



                      For example $100$ increased by $50%$ is $150$. However, $150$ decreased by $50%$ is 75.



                      Therefore, if we want to decrease a number by a percentage, we can't just attempt to do the opposite operation of increasing by that percentage.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Apr 1 at 14:09









                      JamesJames

                      4,4651822




                      4,4651822







                      • 5




                        $begingroup$
                        It's also important to understand this applies to any ratio and can be used to play psychological tricks. For example, if A is 50% faster than B, then B is only about 30% slower than A. You could use either depending on whether you want to emphasize or downplay this gap.
                        $endgroup$
                        – MooseBoys
                        Apr 1 at 20:04












                      • 5




                        $begingroup$
                        It's also important to understand this applies to any ratio and can be used to play psychological tricks. For example, if A is 50% faster than B, then B is only about 30% slower than A. You could use either depending on whether you want to emphasize or downplay this gap.
                        $endgroup$
                        – MooseBoys
                        Apr 1 at 20:04







                      5




                      5




                      $begingroup$
                      It's also important to understand this applies to any ratio and can be used to play psychological tricks. For example, if A is 50% faster than B, then B is only about 30% slower than A. You could use either depending on whether you want to emphasize or downplay this gap.
                      $endgroup$
                      – MooseBoys
                      Apr 1 at 20:04




                      $begingroup$
                      It's also important to understand this applies to any ratio and can be used to play psychological tricks. For example, if A is 50% faster than B, then B is only about 30% slower than A. You could use either depending on whether you want to emphasize or downplay this gap.
                      $endgroup$
                      – MooseBoys
                      Apr 1 at 20:04











                      5












                      $begingroup$

                      What you did is the right answer to a different problem. The general method, "multiply by $1+$ change" is a good one when used properly.



                      To increase a number by $15%$ you multiply by
                      $$
                      1 + 0.15 = 1.15.
                      $$

                      To decrease a number by $15%$ you multiply by
                      $$
                      1 - 0.15 = 0.85.
                      $$

                      But to undo a $15%$ increase you have to "unmultiply" by $1.15$. So you divide by $1.15$. Since
                      $$
                      frac11.15 = 0.86956521739 approx 0.87 = 1 - 0.13,
                      $$

                      to undo a $15%$ increase you make a $13%$ decrease.






                      share|cite|improve this answer









                      $endgroup$

















                        5












                        $begingroup$

                        What you did is the right answer to a different problem. The general method, "multiply by $1+$ change" is a good one when used properly.



                        To increase a number by $15%$ you multiply by
                        $$
                        1 + 0.15 = 1.15.
                        $$

                        To decrease a number by $15%$ you multiply by
                        $$
                        1 - 0.15 = 0.85.
                        $$

                        But to undo a $15%$ increase you have to "unmultiply" by $1.15$. So you divide by $1.15$. Since
                        $$
                        frac11.15 = 0.86956521739 approx 0.87 = 1 - 0.13,
                        $$

                        to undo a $15%$ increase you make a $13%$ decrease.






                        share|cite|improve this answer









                        $endgroup$















                          5












                          5








                          5





                          $begingroup$

                          What you did is the right answer to a different problem. The general method, "multiply by $1+$ change" is a good one when used properly.



                          To increase a number by $15%$ you multiply by
                          $$
                          1 + 0.15 = 1.15.
                          $$

                          To decrease a number by $15%$ you multiply by
                          $$
                          1 - 0.15 = 0.85.
                          $$

                          But to undo a $15%$ increase you have to "unmultiply" by $1.15$. So you divide by $1.15$. Since
                          $$
                          frac11.15 = 0.86956521739 approx 0.87 = 1 - 0.13,
                          $$

                          to undo a $15%$ increase you make a $13%$ decrease.






                          share|cite|improve this answer









                          $endgroup$



                          What you did is the right answer to a different problem. The general method, "multiply by $1+$ change" is a good one when used properly.



                          To increase a number by $15%$ you multiply by
                          $$
                          1 + 0.15 = 1.15.
                          $$

                          To decrease a number by $15%$ you multiply by
                          $$
                          1 - 0.15 = 0.85.
                          $$

                          But to undo a $15%$ increase you have to "unmultiply" by $1.15$. So you divide by $1.15$. Since
                          $$
                          frac11.15 = 0.86956521739 approx 0.87 = 1 - 0.13,
                          $$

                          to undo a $15%$ increase you make a $13%$ decrease.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Apr 1 at 15:16









                          Ethan BolkerEthan Bolker

                          46.2k555121




                          46.2k555121





















                              2












                              $begingroup$

                              To find $15 %$ of some number $N$, we solve an equation (proportion) $fracN100=fracx15$ or $x=0.15N$. That's why a $15%$ increase can be found as $N+0.15N=1.15N$ and decrease is found as $N-0.15N=0.85N$






                              share|cite|improve this answer









                              $endgroup$

















                                2












                                $begingroup$

                                To find $15 %$ of some number $N$, we solve an equation (proportion) $fracN100=fracx15$ or $x=0.15N$. That's why a $15%$ increase can be found as $N+0.15N=1.15N$ and decrease is found as $N-0.15N=0.85N$






                                share|cite|improve this answer









                                $endgroup$















                                  2












                                  2








                                  2





                                  $begingroup$

                                  To find $15 %$ of some number $N$, we solve an equation (proportion) $fracN100=fracx15$ or $x=0.15N$. That's why a $15%$ increase can be found as $N+0.15N=1.15N$ and decrease is found as $N-0.15N=0.85N$






                                  share|cite|improve this answer









                                  $endgroup$



                                  To find $15 %$ of some number $N$, we solve an equation (proportion) $fracN100=fracx15$ or $x=0.15N$. That's why a $15%$ increase can be found as $N+0.15N=1.15N$ and decrease is found as $N-0.15N=0.85N$







                                  share|cite|improve this answer












                                  share|cite|improve this answer



                                  share|cite|improve this answer










                                  answered Apr 1 at 14:12









                                  VasyaVasya

                                  4,4771618




                                  4,4771618





















                                      2












                                      $begingroup$

                                      It works the same way...



                                      Increase : to increase of $15 %$ means to multiply by $(1+0,15)=1,15$.



                                      Decrease : to decrease of $15 %$ means to multiply by $(1-0,15)=0,85$.




                                      Unfortunately, the $15 %$ factors get similar results; thus it can be misleading.



                                      Consider instead an increase of $20 %$. Why the "inverse" operation (i.e. to divide by $1,2$ to decrease does not produce the correct result ?



                                      With a $20 %$ increase the updated price will be $120$.



                                      Now, what is the amount of a decrease of the updated price by $20 %$ ? It is not $20$ but $24$.



                                      And we have that $120 times (1-0,2)=96$ while $dfrac 1201,2=100$.






                                      share|cite|improve this answer











                                      $endgroup$

















                                        2












                                        $begingroup$

                                        It works the same way...



                                        Increase : to increase of $15 %$ means to multiply by $(1+0,15)=1,15$.



                                        Decrease : to decrease of $15 %$ means to multiply by $(1-0,15)=0,85$.




                                        Unfortunately, the $15 %$ factors get similar results; thus it can be misleading.



                                        Consider instead an increase of $20 %$. Why the "inverse" operation (i.e. to divide by $1,2$ to decrease does not produce the correct result ?



                                        With a $20 %$ increase the updated price will be $120$.



                                        Now, what is the amount of a decrease of the updated price by $20 %$ ? It is not $20$ but $24$.



                                        And we have that $120 times (1-0,2)=96$ while $dfrac 1201,2=100$.






                                        share|cite|improve this answer











                                        $endgroup$















                                          2












                                          2








                                          2





                                          $begingroup$

                                          It works the same way...



                                          Increase : to increase of $15 %$ means to multiply by $(1+0,15)=1,15$.



                                          Decrease : to decrease of $15 %$ means to multiply by $(1-0,15)=0,85$.




                                          Unfortunately, the $15 %$ factors get similar results; thus it can be misleading.



                                          Consider instead an increase of $20 %$. Why the "inverse" operation (i.e. to divide by $1,2$ to decrease does not produce the correct result ?



                                          With a $20 %$ increase the updated price will be $120$.



                                          Now, what is the amount of a decrease of the updated price by $20 %$ ? It is not $20$ but $24$.



                                          And we have that $120 times (1-0,2)=96$ while $dfrac 1201,2=100$.






                                          share|cite|improve this answer











                                          $endgroup$



                                          It works the same way...



                                          Increase : to increase of $15 %$ means to multiply by $(1+0,15)=1,15$.



                                          Decrease : to decrease of $15 %$ means to multiply by $(1-0,15)=0,85$.




                                          Unfortunately, the $15 %$ factors get similar results; thus it can be misleading.



                                          Consider instead an increase of $20 %$. Why the "inverse" operation (i.e. to divide by $1,2$ to decrease does not produce the correct result ?



                                          With a $20 %$ increase the updated price will be $120$.



                                          Now, what is the amount of a decrease of the updated price by $20 %$ ? It is not $20$ but $24$.



                                          And we have that $120 times (1-0,2)=96$ while $dfrac 1201,2=100$.







                                          share|cite|improve this answer














                                          share|cite|improve this answer



                                          share|cite|improve this answer








                                          edited Apr 1 at 15:03









                                          John Doe

                                          12.1k11339




                                          12.1k11339










                                          answered Apr 1 at 14:07









                                          Mauro ALLEGRANZAMauro ALLEGRANZA

                                          68.1k449117




                                          68.1k449117





















                                              2












                                              $begingroup$

                                              Consider the literal meaning of "per cent"



                                              Increase $X$ by 15%:
                                              $$
                                              X + X times frac15100 = frac100X100 + frac15X100 = frac(100+15)X100 = frac115X100 = 1.15X
                                              $$



                                              Decrease X by 15%:
                                              $$
                                              X - X times frac15100 = frac100X100 - frac15X100 = frac(100-15)X100 = frac85X100 = 0.85X
                                              $$






                                              share|cite|improve this answer











                                              $endgroup$

















                                                2












                                                $begingroup$

                                                Consider the literal meaning of "per cent"



                                                Increase $X$ by 15%:
                                                $$
                                                X + X times frac15100 = frac100X100 + frac15X100 = frac(100+15)X100 = frac115X100 = 1.15X
                                                $$



                                                Decrease X by 15%:
                                                $$
                                                X - X times frac15100 = frac100X100 - frac15X100 = frac(100-15)X100 = frac85X100 = 0.85X
                                                $$






                                                share|cite|improve this answer











                                                $endgroup$















                                                  2












                                                  2








                                                  2





                                                  $begingroup$

                                                  Consider the literal meaning of "per cent"



                                                  Increase $X$ by 15%:
                                                  $$
                                                  X + X times frac15100 = frac100X100 + frac15X100 = frac(100+15)X100 = frac115X100 = 1.15X
                                                  $$



                                                  Decrease X by 15%:
                                                  $$
                                                  X - X times frac15100 = frac100X100 - frac15X100 = frac(100-15)X100 = frac85X100 = 0.85X
                                                  $$






                                                  share|cite|improve this answer











                                                  $endgroup$



                                                  Consider the literal meaning of "per cent"



                                                  Increase $X$ by 15%:
                                                  $$
                                                  X + X times frac15100 = frac100X100 + frac15X100 = frac(100+15)X100 = frac115X100 = 1.15X
                                                  $$



                                                  Decrease X by 15%:
                                                  $$
                                                  X - X times frac15100 = frac100X100 - frac15X100 = frac(100-15)X100 = frac85X100 = 0.85X
                                                  $$







                                                  share|cite|improve this answer














                                                  share|cite|improve this answer



                                                  share|cite|improve this answer








                                                  edited Apr 1 at 15:36

























                                                  answered Apr 1 at 15:23









                                                  MagooMagoo

                                                  1213




                                                  1213





















                                                      1












                                                      $begingroup$

                                                      If we want to increase $n$ by $p$%, we work out what $p$% of $n$ is, and then add it to $n$ itself: $$0.ptimes n+n=(0.p+1)times n=1.ptimes n$$This is how you arrived to your formula.



                                                      If we want to decrease $n$ by $p$%, we work out what $p$% of $n$ is, and then subtract it from $n$ itself:$$n-0.ptimes n=(1-0.p)times n$$This is how to derive the formula for a decrease.






                                                      share|cite|improve this answer









                                                      $endgroup$

















                                                        1












                                                        $begingroup$

                                                        If we want to increase $n$ by $p$%, we work out what $p$% of $n$ is, and then add it to $n$ itself: $$0.ptimes n+n=(0.p+1)times n=1.ptimes n$$This is how you arrived to your formula.



                                                        If we want to decrease $n$ by $p$%, we work out what $p$% of $n$ is, and then subtract it from $n$ itself:$$n-0.ptimes n=(1-0.p)times n$$This is how to derive the formula for a decrease.






                                                        share|cite|improve this answer









                                                        $endgroup$















                                                          1












                                                          1








                                                          1





                                                          $begingroup$

                                                          If we want to increase $n$ by $p$%, we work out what $p$% of $n$ is, and then add it to $n$ itself: $$0.ptimes n+n=(0.p+1)times n=1.ptimes n$$This is how you arrived to your formula.



                                                          If we want to decrease $n$ by $p$%, we work out what $p$% of $n$ is, and then subtract it from $n$ itself:$$n-0.ptimes n=(1-0.p)times n$$This is how to derive the formula for a decrease.






                                                          share|cite|improve this answer









                                                          $endgroup$



                                                          If we want to increase $n$ by $p$%, we work out what $p$% of $n$ is, and then add it to $n$ itself: $$0.ptimes n+n=(0.p+1)times n=1.ptimes n$$This is how you arrived to your formula.



                                                          If we want to decrease $n$ by $p$%, we work out what $p$% of $n$ is, and then subtract it from $n$ itself:$$n-0.ptimes n=(1-0.p)times n$$This is how to derive the formula for a decrease.







                                                          share|cite|improve this answer












                                                          share|cite|improve this answer



                                                          share|cite|improve this answer










                                                          answered Apr 1 at 14:12









                                                          John DoeJohn Doe

                                                          12.1k11339




                                                          12.1k11339



























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