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Case 1:

Subtracting a percentage $p$ (represented as a number between 0 and 1) from a value $X$ results in the original value becoming $X(1-p)$.

Adding $ p\% $ back to it gives $\{X(1-p)\}*(1+p)$.

Case 2:

Now, adding $p\%$ to $X$ first, we get $X(1+p)$ and subtracting $p\%$ from this result gives $\{X(1+p)\}(1-p)$.

Of course, both of these are equal and the reason they are lower than X is because the result is $X(1-p^2)$.

This makes total sense to me mathematically, but how do you explain it to a child in an intuitive way?

rahs
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2 Answers2

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In whichever order you go (increase followed by decrease; or vice versa) you are always subtracting a percentage from a larger amount than you are adding a percentage.

If you increase first, then the amount you are decreasing from is greater than the amount you increased from.

If you decrease first, then the amount you are increasing from is less than the amount you decreased from.

So in either case the amount you are decreasing from is greater than the amount you are increasing from; and since its the same percentage both times, you end up losing more than you gain.

paw88789
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0

You can play a candy game with a child.

Let us use $G(m,n) =m \cdot n$ to represent $m$ groups of $n$ candies.You have 100 candies, you group them into 10 groups with 10 each, G(10,10)=100 is what you start with.

Round 1: You take away 10%, which is one group of 10. You are left with 9 groups of 10 G(9,10)=90. Then you take 1 candy from every group to form a new group of 9. You have 10 groups each with 9 candies now G(10,9)=90. You add 10%, which is one group of 9. You have G(11,9)=99.

$G(10,10)\to G(9,10)\to G(10,9)\to G(11,9)$

So, you first take away a group of 10, then add back a group of 9, you lose 1.

Round 2: Going the other direction, you add 10%, which is again a group of 10, you have G(11,10)=110. You break down 1 group and distribute its 10 candies 1 each to the other 10 groups, you have now G(10,11)=110. You finally take away 10%, which is one group, you are left with G(9,11)=99.

$G(10,10)\to G(11,10)\to G(10,11)\to G(9,11)$

Here, you first add a group of 10, then take away a group of 11, again you lose 1.

At the end of the games, hope the child can catch the essence, the group size of candies is different when you add and take away in both cases.

Star Bright
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