A percentage change problem for debate

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vincent1

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I came across this question:

If A increases by 35%, then B increases by 25%. Given that A increases 1%, find the percentage increase of B.

Here is 1 possible solution

Assumed (B1/B0) = (A1/A0)^k , ( this is sort of like compound interest, with 1.35 maps to 1.25, 1.35^2 maps to 1.25^2. So k is (log1.25/log1.35).

sub in (A1/A0)=1.01 to obtain (B1/B0).

Here is another suggested solution,

Assume (deltaB/B0) = k (deltaA/A0), ( this is sort of like return on investing a certain portion of your money), 35% maps to 25%, and 7% maps to 5%, so k=5/7.

There was a suggestion on letting B=f(A), then 1.25f(A)=f(1.35A) , 1.25^n f(A) =f(1.35^n A), for integral values n.


Can you analysis which method is appropriate and effective, and what implicit assumptions are made in each method.



Thanks in advance.
 
You first need to clarify what the question itself even means. If A increases by 35%, how can it have increased by only 1%? And if the latter refers to a subsequent increase, how can you possibly know what has happened subsequently to B?

You seem to be making some assumptions not stated in the question. Please explain.
 
That's how the question reads. We are just discussion what us the natural/correct way to interpret the statement in the question.
 
What does "A increases by 35%" mean?

1) A + 35%
2) A * (1 + 35%)
 
You are asking to find a relationship between a and b and the only thing you know is that

[MATH]a_1 = 1.35a_2 \implies b_1 = 1.25b_2[/MATH].

There are an infinite number of possible relationships.
 
That's how the question reads. We are just discussion what us the natural/correct way to interpret the statement in the question.
When a question is unclear, the correct way to interpret it is to ask the author, not to speculate.
 
Intuitive, increase as a percentage of original. Let's interpret it this way to avoid further diversion in interpretation.
Not intuitive. Interpret the question however you like, just state your interpretation clearly.
 
Intuitive, increase as a percentage of original. Let's interpret it this way to avoid further diversion in interpretation.

The real issue is not what increase by 35% means; it's what assumptions you make about the relationship between the quantities. In a mathematical question, you don't make unjustified assumptions!

You suggested two different proportional relationships, which result in different answers; neither can be justified by what we were told. This shows that there is not enough information. If this were an engineering problem, you would not just make an assumption and move on; you would find out what is true, from your boss or from an experiment or whatever!
 
Get your point.

Just out of curiosity, would the suggestion of ' B=f(A), then 1.25f(A)=f(1.35A) , 1.25^n f(A) =f(1.35^n A), for integral values n ' get us closer to an explanation to an answer for what the question asks?
 
Get your point.

Just out of curiosity, would the suggestion of ' B=f(A), then 1.25f(A)=f(1.35A) , 1.25^n f(A) =f(1.35^n A), for integral values n ' get us closer to an explanation to an answer for what the question asks?
What you're implicitly assuming here, and may be what is intended, is that the question means,

Whenever A increases by 35%, then B increases by 25%. Given that A increases 1%, find the percentage increase of B.​

I think your argument is reasonable, and we can extend it to non-integers if we make some further assumptions.

When [MATH]A[/MATH] is multiplied by [MATH]1.35[/MATH], [MATH]\log(A)[/MATH] is increased by [MATH]\log(1.35)[/MATH]. So we can suppose that [MATH]\log(B)[/MATH] is a linear function of [MATH]\log(A)[/MATH], with slope [MATH]\frac{\log(1.25)}{\log(1.35)}[/MATH]: [MATH]\log(B) = \log(B_0) + \frac{\log(1.25)}{\log(1.35)}\log(A)[/MATH].

This turns into [MATH]B = B_0A^\frac{\log(1.25)}{\log(1.35)}[/MATH].

So the first solution you suggested would fit. I just wish the problem had been a little clearer.
 
I would also want to know even if the word 'whenever' is assumed in the statement, is it sufficient to derive that Log(A) and Log(B) are linear for a continous domain.
 
I would also want to know even if the word 'whenever' is assumed in the statement, is it sufficient to derive that Log(A) and Log(B) are linear for a continous domain.

Mathematically, I would say no. Practically, perhaps. Is this from a mathematical context (e.g. proofs expected), or something else (e.g. financial)?
 
No, someone just posted it online. I know you would not like me to use the word guess?, but I am guessing that this is from a middle school maths exercise on the topic of proportions. Although the question didnt say, I think it implicitly assumes Change in A is proportional to Change in B.

It's interesting to see how people with difference experience in maths see the problem.

I know where you are getting at when you ask me what context is this question from. There are different models of thinking between relationship of 2 number in different fields.

One other thing I want to confirm with you, is this statement 1.25 f(A) =f(1.35 A), for B=f(A), equivalent to the general equation of B=kA^(log1.25/log1.35) ?
 
Frankly, I keep going back and forth in my thinking about what it should mean. I'm not at all sure about the answer.

But I have to say that one thing that really bothers me is when proportions are taught without teaching students how to determine whether two quantities really are proportional, so they learn to see proportions everywhere, uncritically.

I think your functional equation probably has the exponential function as its unique solution under some conditions, which would include continuity, and perhaps something more. I'm not prepared to say it definitely is, without having a proof, which I don't feel like taking the time for.
 
Two things here I want to check with you.

1. Is the following sufficient to show equivalence the statement 1.25 f(A) =f(1.35 A), for B=f(A) and the general equation of B=kA^(log1.25/log1.35)?

Let A0 and B0 be an arbitrary pair of values such that B0=f(A0),

we have the parametric form
A=1.35^n A0
B=1.25^n B0

then
(logA - logA0)/ (log1.35)= n
(logB - logB0)/ (log1.25)= n

then,
log1.25logA - log1.35logB = log1.25logA0 - log1.35logB0

so,
logB=(log1.25/log1.35)logA + (a constant)

then
B=10^(a constant) (A^(log1.25/log(1.35))

so
B=k A^(log1.25/log1.35)




2. Even with the interpretion with implicit assumption 'whenever A increase 35%, B increases 25%' , it does not help generate a definite solution of the increase of B when increase of A is 1%.

The required solution of f(1.01A) depends on the value of f(1.35^(-n) 1.01A) in terms of f(A) for integral value n. Which is not given in the question.

So I think the interpretion with the word 'whenever' added will not be fruitful to generate a definition solution to the question.

I think with the aim at generating a definite solution to the question, the added implicit interpretion needs to connect the pairing of cases of 'increase of A by 35%' and 'increase of A by 1%',

Is that the key to a meaningful implicit assumption to solve the problem?

eg
A increase 35%, then B increase by 25%, find the increase in B when increase of A is 1% , with increase of A is proportional to increase of B for the domain under consideration.

Which is probably the intention of the writer if this question is taken from a middle school maths book.
 
Two things here I want to check with you.

1. Is the following sufficient to show equivalence the statement 1.25 f(A) =f(1.35 A), for B=f(A) and the general equation of B=kA^(log1.25/log1.35)?

Let A0 and B0 be an arbitrary pair of values such that B0=f(A0),

we have the parametric form
A=1.35^n A0
B=1.25^n B0

then
(logA - logA0)/ (log1.35)= n
(logB - logB0)/ (log1.25)= n

then,
log1.25logA - log1.35logB = log1.25logA0 - log1.35logB0

so,
logB=(log1.25/log1.35)logA + (a constant)

then
B=10^(a constant) (A^(log1.25/log(1.35))

so
B=k A^(log1.25/log1.35)

What's missing here is a full justification of your parametric form for all n, not just for integers. But given that, your work is good. (And your arbitrary k is better than my (quickly written in #11) [MATH]B_0[/MATH], whose name is not quite appropriate. Note that k is the value of B when A = 1.

2. Even with the interpretion with implicit assumption 'whenever A increase 35%, B increases 25%' , it does not help generate a definite solution of the increase of B when increase of A is 1%.

The required solution of f(1.01A) depends on the value of f(1.35^(-n) 1.01A) in terms of f(A) for integral value n. Which is not given in the question.

So I think the interpretion with the word 'whenever' added will not be fruitful to generate a definition solution to the question.

No, this does yield a definite answer (assuming you are okay with the formula you got above for all n).

We've found that [MATH]B(A) = k A^\frac{\log1.25}{\log1.35}[/MATH]; so the ratio [MATH]\frac{B(1.01A)}{B(A)} = \frac{k (1.01A)^\frac{\log1.25}{\log1.35}}{k A^\frac{\log1.25}{\log1.35}} = (1.01)^\frac{\log1.25}{\log1.35} = 1.007426[/MATH]. So B increases by 0.7426%.

I think with the aim at generating a definite solution to the question, the added implicit interpretion needs to connect the pairing of cases of 'increase of A by 35%' and 'increase of A by 1%',

Is that the key to a meaningful implicit assumption to solve the problem?

eg
A increase 35%, then B increase by 25%, find the increase in B when increase of A is 1% , with increase of A is proportional to increase of B for the domain under consideration.

Which is probably the intention of the writer if this question is taken from a middle school maths book.

The trouble is that a percent increase is different from an increase. It is not the [absolute] increase of B that is proportional to the [absolute] increase of A; rather, the relative increase of B is proportional to the relative increase of A, as I read it. Well, no, not even that. The one is just a function of the other, not proportional. Nothing in the problem says "proportional".

Let's take an example, to make sure I'm right. Suppose that A and B both start at 100, to keep things simple. If A increases by 35% and B increases by 25%, then A = 135 and B = 125. Now if they both increase again by 35% and 25% respectively, we end up with A = 182.25 and B = 156.25. The new increases are respectively 47.25 and 31.25. The ratio of these is not 35:25 = 1.4, but 1.512. The increases are not proportional. And for that matter, then total percent increases are 82.25% and 56.25%, and even those are not proportional (the ratio is 1.462).
 
1.
For the case with the "Whenever" interpretation, the relation of relative increases only applies for 35% of A. Is it sufficient to imply that a fractional increase that is less than 35% will also follow the equation, i.e., can n be a non integer?



2.
You are right. We are trying to pack the question, and I was still adding words to it that's not concise. The precise wording needs to include that word relative increase to avoid ambiguity.

For this part, if the word whenever is removed, would it be natural to interpret it as a one-off increase for the statement 'When A has a relative increase of 35%, then B has a relative increase of 25%, find the relative increase of B, when the relative increase of A is 1%". The reason I think that is if successive relative increases are considered, then the new relative increase of A would involve numbers like 82.25% on order to use the condition.

The intention of the question is really unknown. It's like take other people's word out of context. It doesn't really do anything.

I will probably end here if there is nothing more to gain from discussing it.
 
To begin with, the question itself is a little bit ambiguous on its assumption. However, if we intend to treat this problem as a mathematical problem, then the statements of JeffM and Dr.Peterson are equivalent ("if A, then B" and "whenever A occurs, then B occurs" are logically equivalent). Thus it is a reasonable interpretation for this problem.

You are asking to find a relationship between a and b and the only thing you know is that

[MATH]a_1 = 1.35a_2 \implies b_1 = 1.25b_2[/MATH].

What you're implicitly assuming here, and may be what is intended, is that the question means,

Whenever A increases by 35%, then B increases by 25%. Given that A increases 1%, find the percentage increase of B.​

Now, allow me to elaborate further on the assumptions, which are not specifically given by the problem itself.

Since we are only given a relationship between A and B, it makes sense to describe this relation as a function, namely f(A) = B. Moreover, the given relation implies that f(1.35 A) = 1.25 f(A). Since the question talks about increment, it is reasonable to assume both A and B are real numbers, not just integers or positive numbers. In other words, we treat f as a function whose input and output values are real numbers. To sum up, f is a real-valued function with a single real variable, i.e. the following statement (a):

f: R -> R, subject to
condition (b): f(1.35 x) = 1.25 f(x), for every x in R.

---

Under this assumption, we can make precise what JeffM's conclusion means as follows:

A moment of thought will lead one to see that any real valued function g defined on [1, 1.35] union [-1.35, -1], subject to 1.25 g(1) = g(1.35) and 1.25 g(-1) = g(-1.35), can be extended to a function G : R -> R with G(0) = 0, which satisfies statement (a). Because there are infinitely many such functions g, there are infinitely many solutions G that answer G(1.01 x) / G(x).

As a matter of fact, both methods suggested in vincent1's post (#1) are just two of infinitely many such functions g, yielding two possible answers to the question. The first method, however, leads to some interesting discussion.

---

Not to offend anyone, but I have a few words about the method of using Log and the model assuming "(B1/B0 = A1/A0)^k" (I suppose that 1's should be k though).

- To follow deductive reasoning, I would avoid using Log because the input and output values could be zero and negative under general assumption. For the same reason, I will not assume a particular model or relation between A and B.

- I think by "continuous domain", vincent1 referred to "f is a continuous function" since the domain is already assumed to be the real number line or the positive reals. In the case when continuity is assumed, in addition to statement (a), then as Dr.Peterson would have almost had it, the solution will be of polynomial form in x. Here is a sketch of proof without "cheating" with Log and assuming additional relations between A and B.

1. By induction on condition (b) we have f(1.35^k x) = 1.25^k f(x), for every natural number k and every x in R. By setting x' = 1.35^(-1) x, via condition (b) we have 1.25^(-1) f(x) = f(1.35^(-1) x). By induction again, we conclude f(1.35^k x) = 1.25^k f(x), for every non-zero integer k and every x in R.

2. By making use of the real domain, we further deduce that f(1.35^t x) = 1.25^t f(x), for every non-zero rational number t and every x in R because every such t = k/m for non-zero integers k and m.

3. Now, by continuity of f and the fact that the set of rational numbers is dense in reals, we deduce that f(1.35^t x) = 1.25^t f(x) for all t in reals (including t = 0), and all x in reals.

4. To this end, any arbitrary non-zero x in R can be uniquely determined either by x = 1.35^t or x = -1.35^t. Therefore, simple algebra gives us

f(x) = x^(Log 1.25 / Log 1.35) f(1), when x > 0,
f(x) = x^(Log 1.25 / Log 1.35) f(-1), when x <0, and
f(0) = 0 by continuity.

In conclusion, under the additional assumption of continuity (i.e. relation between A and B is continuous), then f is necessarily of piecewise polynomial form that is uniquely determined by its values at x = 1 and x = -1 (I guess some people would like to call these initial values?). This answers the original question regarding f(1.01 x) / f(x) = 1.01^(Log 1.25 / Log 1.35) as a consistent value for all non-zero x, which also matches our immediate intuition toward this problem in the first sight.

Nevertheless, we cannot take continuity for granted if it is not given. Hence, I think JeffM made the most general conclusion to this problem; whereas the "natural" method would have been true otherwise. If this problem is meant to be an elementary or secondary or high school math problem, I would have taken the "natural" method without any justification, just assume everything and use the formula, easy-to-go, no-brainer. If it is a problem in a proof-based math course or analysis course, I would say the "natural" method must be justified very carefully in a deductive reasoning manner. Finally, I found this problem and its discussions demonstrated very well how our natural belief in everyday math (i.e. things are continuous, be it function or domain) might have failed easily in rigorous mathematics and lead to some very counter-intuitive and interesting observations.
 
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I came across this question:

If A increases by 35%, then B increases by 25%. Given that A increases 1%, find the percentage increase of B.

Here is 1 possible solution

Assumed (B1/B0) = (A1/A0)^k , ( this is sort of like compound interest, with 1.35 maps to 1.25, 1.35^2 maps to 1.25^2. So k is (log1.25/log1.35).

sub in (A1/A0)=1.01 to obtain (B1/B0).

Here is another suggested solution,

Assume (deltaB/B0) = k (deltaA/A0), ( this is sort of like return on investing a certain portion of your money), 35% maps to 25%, and 7% maps to 5%, so k=5/7.

There was a suggestion on letting B=f(A), then 1.25f(A)=f(1.35A) , 1.25^n f(A) =f(1.35^n A), for integral values n.


Can you analysis which method is appropriate and effective, and what implicit assumptions are made in each method.



Thanks in advance.

I kept thinking about the methods and these [MATH]A_k[/MATH] and [MATH]B_k[/MATH] here. I hope vincent1 will clarify my understanding on his methods before I can further determine which one is more "appropriate" and "effective".

Method 1.

Question 1. What are these [MATH]A_0, A_1, B_0, B_1[/MATH], etc. supposed to mean? Did you assume there are some fixed values of [MATH]A[/MATH] and [MATH]B[/MATH] to begin with? But I do not see this information given in the question itself.

Question 2. How did you make additional assumption [MATH]\frac{B_1}{B_0} = \left( \frac{A_1}{A_0} \right)^k[/MATH]? What if [MATH]A_1 \neq 1.35 A_0[/MATH] and [MATH]B_1 \neq 1.25 B_0[/MATH]? You mentioned that it is like compound interest, but the question does not guarantee it.

Correct me if this is not your intention. I suppose you meant to pick an "initial value" of [MATH]A[/MATH], say [MATH]A_0[/MATH], and its corresponding value [MATH]B_0[/MATH]. By the relation, when [MATH]A_1 = 1.35A_0[/MATH], we get [MATH]B_1 = 1.25B_0[/MATH], and so on to generate a sequence of [MATH](A_k, B_k)[/MATH], where [MATH]A_k = 1.35^k A_0, B_k = 1.25^k B_0[/MATH]. But then, it goes back to what I concerned, the way of construction is not relevant to getting [MATH]1.01A[/MATH], because no k will give [MATH]\frac{A_k}{A_0} = 1.01[/MATH] by this process.

On the other hand, if these values are allowed to be arbitrary as long as they satisfy the relation (in your words, 1.35 maps to 1.25, 1.35k maps to 1.25k, etc.), but isn't it really just

[MATH] (\dagger) \qquad \qquad 1.25 f(A) = f(1.35 A), \text{where } B = f(A), \forall A \in \mathbb{R}? [/MATH]
In this case, your method 1 is no different than determining [MATH]\frac{f(1.01A)}{f(A)}[/MATH], but it turns out that one must first determine what f is. Thus it reduces to the case of JeffM's conclusion.

Method 2.

Question 3. I do not understand why the assumption [MATH] \frac{\Delta B}{B_0} = k \frac{\Delta A}{A_0}[/MATH] was made. Here you mentioned return on investment, but again the question does not guarantee it.

Firstly, in this method you mentioned about "35% maps to 25%", etc. In fact, from the above formula you set up, you were talking about a function [MATH]\Delta A \mapsto \Delta B[/MATH]. However, in order for this function to make sense, one must fix a value [MATH]A_0[/MATH] (resp. [MATH]B_0[/MATH]) to define precisely [MATH]\Delta A[/MATH] (resp. [MATH]\Delta B)[/MATH] because the concept of change is relative. Your function might not be well-defined if there were no fixed values [MATH]A_0[/MATH] and [MATH]B_0[/MATH] to begin with.

Secondly, suppose these fixed values [MATH]A_0[/MATH] and [MATH]B_0[/MATH] were given (which is not the case from the question).

Question 4. How do you justify this above assumption from the question? If you could make this assumption, can somebody else make another one, such as
[MATH] (\dagger \dagger) \qquad \qquad \frac{\Delta B}{B_0} = k^3 \frac{\Delta A}{A_0} ? [/MATH]Then one can just calculate [MATH]k = \left(\frac{5}{7}\right)^{1/3}[/MATH] and claim that it is another solution to the modified problem.

With these questions uncleared, I cannot determine which one of the two methods is more "appropriate" and "effective" than another, because none of them really justify the usage of [MATH]A_0, B_0[/MATH], etc. without making additional assumption about initial values. (Once again, if the initial value [MATH]A_0[/MATH] is made arbitrary, then it is no different than [MATH](\dagger)[/MATH]).

Even with initial values given, none of these two methods justify the usage of the model you set up (be it called compound interest or return on investment). If your method 2 "works", then [MATH](\dagger \dagger)[/MATH] is another one and there are infinitely many more.

Therefore, I do not think any of these two methods can be called "appropriate" without additional information or assumption. Hence, along with my conclusion in the previous comment, the conclusion from JeffM that there are infinitely many solutions is the most "appropriate" with the given information. As for "effectiveness", I am not sure what type of measure we are talking about here. If one is to "solve" this problem by running an algorithm on a computer, any polynomial relation one assumes between [MATH]A[/MATH] and [MATH]B[/MATH] (i.e. [MATH]f(A)[/MATH] is a polynomial) would be an "effective" candidate in terms of computational complexity. However, I am not sure if this is at all relevant to your original intention.
 
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