Calculating discount not workingFind 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

Why is consensus so controversial in Britain?

What is the word for reserving something for yourself before others do?

Why does Kotter return in Welcome Back Kotter

Watching something be written to a file live with tail

Brothers & sisters

I'm flying to France today and my passport expires in less than 2 months

UK: Is there precedent for the governments e-petition site changing the direction of a government decision?

Will google still index a page if I use a $_SESSION variable?

How can saying a song's name be a copyright violation?

Twin primes whose sum is a cube

Can I use a neutral wire from another outlet to repair a broken neutral?

Has there ever been an airliner design involving reducing generator load by installing solar panels?

Stopping power of mountain vs road bike

What to put in ESTA if staying in US for a few days before going on to Canada

What's the point of deactivating Num Lock on login screens?

How to show the equivalence between the regularized regression and their constraint formulas using KKT

How much of data wrangling is a data scientist's job?

prove that the matrix A is diagonalizable

Is there a hemisphere-neutral way of specifying a season?

Did converts (ger tzedek) in ancient Israel own land?

What exploit are these user agents trying to use?

Combinations of multiple lists

Can I make "comment-region" comment empty lines?

intersection of two sorted vectors in C++



Calculating discount not working


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
    2 days ago















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
    2 days ago













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
    2 days ago












  • 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
    2 days ago







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
2 days ago




$begingroup$
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
2 days ago










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 wouln't even work for $p$ greater than 99, say for $+120%$. And it won't work for $p <0$, either.



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, $value/1.p$ division 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










          New contributor




          Magoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$




















            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









            $endgroup$













              Your Answer





              StackExchange.ifUsing("editor", function ()
              return StackExchange.using("mathjaxEditing", function ()
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              );
              );
              , "mathjax-editing");

              StackExchange.ready(function()
              var channelOptions =
              tags: "".split(" "),
              id: "69"
              ;
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function()
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled)
              StackExchange.using("snippets", function()
              createEditor();
              );

              else
              createEditor();

              );

              function createEditor()
              StackExchange.prepareEditor(
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader:
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              ,
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              );



              );













              draft saved

              draft discarded


















              StackExchange.ready(
              function ()
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3170647%2fcalculating-discount-not-working%23new-answer', 'question_page');

              );

              Post as a guest















              Required, but never shown

























              7 Answers
              7






              active

              oldest

              votes








              7 Answers
              7






              active

              oldest

              votes









              active

              oldest

              votes






              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 wouln't even work for $p$ greater than 99, say for $+120%$. And it won't work for $p <0$, either.



              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, $value/1.p$ division 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 wouln't even work for $p$ greater than 99, say for $+120%$. And it won't work for $p <0$, either.



                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, $value/1.p$ division 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 wouln't even work for $p$ greater than 99, say for $+120%$. And it won't work for $p <0$, either.



                  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, $value/1.p$ division 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 wouln't even work for $p$ greater than 99, say for $+120%$. And it won't work for $p <0$, either.



                  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, $value/1.p$ division 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 1 at 22:20

























                  answered Apr 1 at 14:09









                  CiaPanCiaPan

                  10.3k11248




                  10.3k11248





















                      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

                      4,4301822




                      4,4301822







                      • 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

                          45.6k553120




                          45.6k553120





















                              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

                                  4,2771618




                                  4,2771618





















                                      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

                                          11.7k11239




                                          11.7k11239










                                          answered Apr 1 at 14:07









                                          Mauro ALLEGRANZAMauro ALLEGRANZA

                                          67.7k449117




                                          67.7k449117





















                                              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










                                              New contributor




                                              Magoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                              Check out our Code of Conduct.






                                              $endgroup$

















                                                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










                                                New contributor




                                                Magoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.






                                                $endgroup$















                                                  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










                                                  New contributor




                                                  Magoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                  Check out our Code of Conduct.






                                                  $endgroup$



                                                  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










                                                  New contributor




                                                  Magoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                  Check out our Code of Conduct.









                                                  share|cite|improve this answer



                                                  share|cite|improve this answer








                                                  edited Apr 1 at 15:36





















                                                  New contributor




                                                  Magoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                  Check out our Code of Conduct.









                                                  answered Apr 1 at 15:23









                                                  MagooMagoo

                                                  1213




                                                  1213




                                                  New contributor




                                                  Magoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                  Check out our Code of Conduct.





                                                  New contributor





                                                  Magoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                  Check out our Code of Conduct.






                                                  Magoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                  Check out our Code of Conduct.





















                                                      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

                                                          11.7k11239




                                                          11.7k11239



























                                                              draft saved

                                                              draft discarded
















































                                                              Thanks for contributing an answer to Mathematics Stack Exchange!


                                                              • Please be sure to answer the question. Provide details and share your research!

                                                              But avoid


                                                              • Asking for help, clarification, or responding to other answers.

                                                              • Making statements based on opinion; back them up with references or personal experience.

                                                              Use MathJax to format equations. MathJax reference.


                                                              To learn more, see our tips on writing great answers.




                                                              draft saved


                                                              draft discarded














                                                              StackExchange.ready(
                                                              function ()
                                                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3170647%2fcalculating-discount-not-working%23new-answer', 'question_page');

                                                              );

                                                              Post as a guest















                                                              Required, but never shown





















































                                                              Required, but never shown














                                                              Required, but never shown












                                                              Required, but never shown







                                                              Required, but never shown

































                                                              Required, but never shown














                                                              Required, but never shown












                                                              Required, but never shown







                                                              Required, but never shown







                                                              Popular posts from this blog

                                                              រឿង រ៉ូមេអូ និង ហ្ស៊ុយលីយេ សង្ខេបរឿង តួអង្គ បញ្ជីណែនាំ

                                                              Crop image to path created in TikZ? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Crop an inserted image?TikZ pictures does not appear in posterImage behind and beyond crop marks?Tikz picture as large as possible on A4 PageTransparency vs image compression dilemmaHow to crop background from image automatically?Image does not cropTikzexternal capturing crop marks when externalizing pgfplots?How to include image path that contains a dollar signCrop image with left size given

                                                              Romeo and Juliet ContentsCharactersSynopsisSourcesDate and textThemes and motifsCriticism and interpretationLegacyScene by sceneSee alsoNotes and referencesSourcesExternal linksNavigation menu"Consumer Price Index (estimate) 1800–"10.2307/28710160037-3222287101610.1093/res/II.5.31910.2307/45967845967810.2307/2869925286992510.1525/jams.1982.35.3.03a00050"Dada Masilo: South African dancer who breaks the rules"10.1093/res/os-XV.57.1610.2307/28680942868094"Sweet Sorrow: Mann-Korman's Romeo and Juliet Closes Sept. 5 at MN's Ordway"the original10.2307/45957745957710.1017/CCOL0521570476.009"Ram Leela box office collections hit massive Rs 100 crore, pulverises prediction"Archived"Broadway Revival of Romeo and Juliet, Starring Orlando Bloom and Condola Rashad, Will Close Dec. 8"Archived10.1075/jhp.7.1.04hon"Wherefore art thou, Romeo? To make us laugh at Navy Pier"the original10.1093/gmo/9781561592630.article.O006772"Ram-leela Review Roundup: Critics Hail Film as Best Adaptation of Romeo and Juliet"Archived10.2307/31946310047-77293194631"Romeo and Juliet get Twitter treatment""Juliet's Nurse by Lois Leveen""Romeo and Juliet: Orlando Bloom's Broadway Debut Released in Theaters for Valentine's Day"Archived"Romeo and Juliet Has No Balcony"10.1093/gmo/9781561592630.article.O00778110.2307/2867423286742310.1076/enst.82.2.115.959510.1080/00138380601042675"A plague o' both your houses: error in GCSE exam paper forces apology""Juliet of the Five O'Clock Shadow, and Other Wonders"10.2307/33912430027-4321339124310.2307/28487440038-7134284874410.2307/29123140149-661129123144728341M"Weekender Guide: Shakespeare on The Drive""balcony"UK public library membership"romeo"UK public library membership10.1017/CCOL9780521844291"Post-Zionist Critique on Israel and the Palestinians Part III: Popular Culture"10.2307/25379071533-86140377-919X2537907"Capulets and Montagues: UK exam board admit mixing names up in Romeo and Juliet paper"Istoria Novellamente Ritrovata di Due Nobili Amanti2027/mdp.390150822329610820-750X"GCSE exam error: Board accidentally rewrites Shakespeare"10.2307/29176390149-66112917639"Exam board apologises after error in English GCSE paper which confused characters in Shakespeare's Romeo and Juliet""From Mariotto and Ganozza to Romeo and Guilietta: Metamorphoses of a Renaissance Tale"10.2307/37323537323510.2307/2867455286745510.2307/28678912867891"10 Questions for Taylor Swift"10.2307/28680922868092"Haymarket Theatre""The Zeffirelli Way: Revealing Talk by Florentine Director""Michael Smuin: 1938-2007 / Prolific dance director had showy career"The Life and Art of Edwin BoothRomeo and JulietRomeo and JulietRomeo and JulietRomeo and JulietEasy Read Romeo and JulietRomeo and Julieteeecb12003684p(data)4099369-3n8211610759dbe00d-a9e2-41a3-b2c1-977dd692899302814385X313670221313670221