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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
$begingroup$
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
2 days ago
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
2 days ago
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
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
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 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.
$endgroup$
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%:
$$
X - X times frac15100 = frac100X100 - frac15X100 = frac(100-15)X100 = frac85X100 = 0.85X
$$
New contributor
$endgroup$
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 |
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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 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.
$endgroup$
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 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.
$endgroup$
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 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.
$endgroup$
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.
edited Apr 1 at 22:20
answered Apr 1 at 14:09
CiaPanCiaPan
10.3k11248
10.3k11248
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
4,4301822
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
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
45.6k553120
45.6k553120
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
4,2771618
4,2771618
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
11.7k11239
11.7k11239
answered Apr 1 at 14:07
Mauro ALLEGRANZAMauro ALLEGRANZA
67.7k449117
67.7k449117
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%:
$$
X - X times frac15100 = frac100X100 - frac15X100 = frac(100-15)X100 = frac85X100 = 0.85X
$$
New contributor
$endgroup$
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%:
$$
X - X times frac15100 = frac100X100 - frac15X100 = frac(100-15)X100 = frac85X100 = 0.85X
$$
New contributor
$endgroup$
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%:
$$
X - X times frac15100 = frac100X100 - frac15X100 = frac(100-15)X100 = frac85X100 = 0.85X
$$
New contributor
$endgroup$
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%:
$$
X - X times frac15100 = frac100X100 - frac15X100 = frac(100-15)X100 = frac85X100 = 0.85X
$$
New contributor
edited Apr 1 at 15:36
New contributor
answered Apr 1 at 15:23
MagooMagoo
1213
1213
New contributor
New contributor
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
11.7k11239
11.7k11239
add a comment |
add a comment |
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$begingroup$
$86.95652$ 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