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
Tht Aain’t Right... #2
Generate an RGB colour grid
What would be the ideal power source for a cybernetic eye?
Why did the US and UK choose different solutions to the problem of an undemocratic upper house?
Why are there no cargo aircraft with "flying wing" design?
Seeking colloquialism for “just because”
Sci-Fi book where patients in a coma ward all live in a subconscious world linked together
Gordon Ramsay Pudding Recipe
How to answer "Have you ever been terminated?"
Can we see the USA flag on the Moon from Earth?
Why are both D and D# fitting into my E minor key?
Denied boarding although I have proper visa and documentation. To whom should I make a complaint?
Delete nth line from bottom
How widely used is the term Treppenwitz? Is it something that most Germans know?
How much time will it take to get my passport back if I am applying for multiple Schengen visa countries?
Dating a Former Employee
51k Euros annually for a family of 4 in Berlin: Is it enough?
What's the purpose of writing one's academic biography in the third person?
In predicate logic, does existential quantification (∃) include universal quantification (∀), i.e. can 'some' imply 'all'?
Why didn't this character "real die" when they blew their stack out in Altered Carbon?
2001: A Space Odyssey's use of the song "Daisy Bell" (Bicycle Built for Two); life imitates art or vice-versa?
When were vectors invented?
How come Sam didn't become Lord of Horn Hill?
tcolorbox: Potential bug with duplicate label for hyperref link
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
$begingroup$
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.
arithmetic percentages
$endgroup$
add a comment |
$begingroup$
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.
arithmetic percentages
$endgroup$
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
add a comment |
$begingroup$
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.
arithmetic percentages
$endgroup$
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
arithmetic percentages
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
add a comment |
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
add a comment |
7 Answers
7
active
oldest
votes
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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 $N-0.15N=0.85N$
$endgroup$
add a comment |
$begingroup$
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 $dfrac 1201,2=100$.
$endgroup$
add a comment |
$begingroup$
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%:
$$
X - X times frac15100 = frac100X100 - frac15X100 = frac(100-15)X100 = frac85X100 = 0.85X
$$
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
Your Answer
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
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited Apr 12 at 21:39
answered Apr 1 at 14:09
CiaPanCiaPan
10.4k11248
10.4k11248
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Apr 1 at 15:16
Ethan BolkerEthan Bolker
46.2k555121
46.2k555121
add a comment |
add a comment |
$begingroup$
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 $N-0.15N=0.85N$
$endgroup$
add a comment |
$begingroup$
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 $N-0.15N=0.85N$
$endgroup$
add a comment |
$begingroup$
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 $N-0.15N=0.85N$
$endgroup$
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 $N-0.15N=0.85N$
answered Apr 1 at 14:12
VasyaVasya
4,4771618
4,4771618
add a comment |
add a comment |
$begingroup$
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 $dfrac 1201,2=100$.
$endgroup$
add a comment |
$begingroup$
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 $dfrac 1201,2=100$.
$endgroup$
add a comment |
$begingroup$
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 $dfrac 1201,2=100$.
$endgroup$
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 $dfrac 1201,2=100$.
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
add a comment |
add a comment |
$begingroup$
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%:
$$
X - X times frac15100 = frac100X100 - frac15X100 = frac(100-15)X100 = frac85X100 = 0.85X
$$
$endgroup$
add a comment |
$begingroup$
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%:
$$
X - X times frac15100 = frac100X100 - frac15X100 = frac(100-15)X100 = frac85X100 = 0.85X
$$
$endgroup$
add a comment |
$begingroup$
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%:
$$
X - X times frac15100 = frac100X100 - frac15X100 = frac(100-15)X100 = frac85X100 = 0.85X
$$
$endgroup$
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%:
$$
X - X times frac15100 = frac100X100 - frac15X100 = frac(100-15)X100 = frac85X100 = 0.85X
$$
edited Apr 1 at 15:36
answered Apr 1 at 15:23
MagooMagoo
1213
1213
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Apr 1 at 14:12
John DoeJohn Doe
12.1k11339
12.1k11339
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
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