(A new question of the week)
Some problems can be done either by algebra or by basic arithmetic methods and some creativity; and although algebra generally makes work easier by making it routine, sometimes special-purpose thinking (once you have thought it!) can be quicker. Here we have a problem where a creative method didn’t quite make sense. Can we make sense of it? Can we fix it?
How old is John?
Here is Kalyan’s question, from early August:
Hello Doctor:
The question:
10 years ago from the present age, ratio of John’s age to his father was 1:5.
After 6 years from the present age, ratio becomes 3:7.
Find the present age of son and father.
Everyone knows to solve this problem using x and y and using simultaneous equations to find the x and y. But, how to solve the same problems using ratios, specially for Olympiads or any competitive exam to save time.
I saw a video on YouTube (it was not in English, so cannot post it here). But, I can tell you the trick.
First he took
John : Father 10 years ago 1 : 5 6 years hence 3 : 7Then, he said the difference between 1 and 3 in the ratio is 2 units, and the difference between 10 ago years and 6 years hence is 6-(-10) = 16 years.
So, 2 units = 16 years
Therefore, 1 unit = 8 years.
So, therefore, John’s age 10 years ago, 8 * 1 = 8 years.
So, after 10 years John’s age is 8 + 10 = 18 years.
Correspondingly, I can find father’s age.
My question:
Why is 3-1 unit difference has been taken, because 3-1 is not the actual age difference? How this process was derived from the traditional x and y method, could you tell me something regarding this?
He’s right to question that! As we’ll see, taking the difference between terms in two different ratios is not really meaningful!
We’ll be looking at an algebraic solution; but an intuitive approach may or may not parallel that.
I answered:
Hi, Kalyan.
I’ll first show you how I worked it out myself without looking at what you showed; then I’ll see if there’s anything to say about the latter; and then I’ll try to relate it to the algebra, if I can.
Here’s the problem (slightly reworded):
10 years before the present time, the ratio of John’s age to his father’s was 1:5.
6 years after the present time, the ratio becomes 3:7.
Find the present age of son and father.
A correct intuitive approach
I first tried my own usual approach to such problems, knowing from experience a useful fact about ages of people at different times.
My method:
The key will be that the difference in their ages is a constant.
10 years ago, the ratio 1:5 tells us that that difference was 4 times John’s age, since it’s 5-1 = 4 parts to John’s 1.
In 6 years, 16 years later, the ratio 3:7 tells us that the difference will be 7-3 = 4 parts to John’s 3 parts, making it 4/3 of John’s age then.
Inverting both facts: 10 years ago, John’s age was 1/4 of the (constant) difference; 16 years later it will be 3/4 of that same difference.
So adding 16 to his earlier age multiplies his age by 3 (from 1/4 to 3/4); that is, the increase in age is 2 times the original age. So his age then must be 1/2 of 16, namely 8.
Their ages then were 8 and 40 (1:5 ratio, the difference being 32). Now, 10 years later, they are 18 and 50; and in 6 more years they will be 24 and 56 (ratio 3:7).
By finding a constant to compare both ages to, we have a way to turn ratios into absolute numbers.
The provided (non-)intuitive approach
Did the unknown YouTuber use a similar approach? Hopefully it will be easier to analyze now that we have a simple solution to compare it to. We know that his answer, at least, is correct (18 years old).
Now I’ll read the method you’re asking about, and paraphrase it here:
The data are:
John : Father 10 years ago 1 : 5 6 years hence 3 : 7The difference between 1 and 3 in the ratio is 2 units.
The difference between 10 ago years ago and 6 years from now is 6-(-10) = 16 years.
So, 2 units = 16 years.
Therefore, 1 unit = 8 years.
John’s age 10 years ago, 8 * 1 = 8 years.
After 10 years John’s age is 8 + 10 = 18 years.
I see that I did the same 3 – 1 = 2, but I did it near the end, and that came from 3/4 – 1/4, which would not have been justified if the two denominators had not both been 4 (5-1 and 7-3).
He makes no mention (at least in your retelling) of the important fact that 5-1 and 7-3 are both 4 (or, equivalently, that the difference 7-5 is also 2). I don’t think this would work without that. For example, if the first ratio had been written as 2:10, then subtracting 3-2 = 1 and 7-10 = -3 would not give the same answer!
In general, the “units” or “parts” used in a ratio are arbitrary, and can’t be compared between different ratios. In my work, I used such “parts”, but in one ratio at a time.
Did he mention any of these other ideas, but you omitted them?
It is important to realize that the concept of “parts”, as in “1 part to 5 parts”, is local, referring only to one ratio; the “parts” will in general be different in other ratios. So it appears that the YouTuber either was lucky that the units were the same in both ratios, in some sense, or recognized that and failed to communicate this essential part of his thinking.
Later we’ll do a different example and see that the method as stated really does fail when the conditions of the problem change.
An algebraic approach
Kalyan had asked whether a simple method (such as the YouTuber’s) could be derived from the algebraic work, which he implied he found easy. That’s possible, so it was worth trying. First, I had to lay out a version of the algebraic solution for discussion:
Now let’s solve it by algebra, and see if any of what he or I did is reflected in that.
Let John’s age now be J, and his father’s F. Then the two statements become
(J-10):(F-10) = 1:5 [Ratio 10 years ago]
(J+6):(F+6) = 3:7 [Ratio 6 years from now]
Cross-multiplying,
5(J – 10) = F – 10
7(J + 6) = 3(F + 6)
Simplifying,
5J – F = 40
7J – 3F = -24
Multiplying the first equation by -3 and adding,
-8J = -144
J = 18
Solving for F,
5(18 – 10) = F – 10, F = 50
I see no 2 or 4 anywhere in here. Possibly we could see more similarity in a different approach.
That gave the correct answer, which is encouraging; but we never subtracted \(3-1\) or \(6-(-10)\). It is likely that those are hidden inside the work.
Making a general formula
One way to find such a hidden operation is to do the same work with variables, so that every operation is visible.
So now I want to make a general formula, and see if that has anything in common with the other methods. Let’s change the problem to this:
m years before the present time, the ratio of John’s age to his father’s was a:b.
n years after the present time, the ratio becomes c:d.
Find the present age of son and father in terms of a, b, c, d.
My algebraic solution:
Let John’s age now be J, and his father’s F. Then the two statements become
(J-m):(F-m) = a:b
(J+n):(F+n) = c:d
Cross-multiplying,
b(J – m) = a(F – m)
d(J + n) = c(F + n)
Simplifying,
bJ – aF = m(b – a)
dJ – cF = n(c – d)
Multiplying the first equation by c and the second by -a,
bcJ – acF = cm(b – a)
-adJ + acF = an(d – c)
and adding,
(bc – ad)J = cm(b – a) + an(d – c)
J = [cm(b – a) + an(d – c)] / [bc – ad] = [bcm – acm + adn – acn] / [bc – ad]
Applying this with m=10, n=6, a=1, b=5, c=3, d=7, we get
J = [3*10(5 – 1) + 6*1(7 – 3)] / [5*3 – 1*7] = (120 + 24)/(15 – 7) = 144/8 = 18
Notice the presence of both b-a and d-c, which both happen to be the same for our problem.
That shows that \(5-1=4\) and \(7-3=4\) are both used; but the \(3-1=2\) on which the solution in question relied is \(c-a\). That still isn’t visible in the algebra.
Our formulas for both unknowns are $$J=\frac{cm(b-a)+an(d-c)}{bc-ad}\\F=\frac{dm(b-a)+bn(d-c)}{bc-ad}$$
What if it were given that \(c-a=d-b\), as in our specific problem? Then also \(d-c=b-a\), and we could factor that out from our formula: $$J=\frac{(cm+an)(b-a)}{bc-ad}.$$ We can also replace \(d=b+c-a\) and get a formula without \(d\): $$J=\frac{(cm+an)(b-a)}{bc-ab-ac+a^2}=\frac{(cm+an)(b-a)}{(b-a)(c-a)}=\frac{cm+an}{c-a}.$$
We’ll be seeing that again …
Solving the general problem intuitively
What I did without algebra, replacing numbers with variables, looks like this:
m years ago, the ratio a:b tells us that the difference was (b-a)/a times John’s age, since it’s b-a parts to John’s a.
In n years, m+n years later, the ratio c:d tells us that the difference will be d-c parts to John’s c parts, making it (d-c)/c times John’s age then.
Reversing both facts, m years ago, John’s age was a/(b-a) of the (constant) difference; m+n years later it will be c/(d-c) times that same difference.
So adding m+n to his earlier age multiplies his age by [c/(d-c)]/[a/(b-a)] = [c(b-a)]/[a(d-c)]; that is, the m+n years is [c(b-a)]/[a(d-c)]-1 = [c(b-a)-a(d-c)]/[a(d-c)]= [bc-ad]/[a(d-c)] times his age then. So his age then must be [a(d-c)]/[bc-ad] times m+n, namely [a(d-c)(m+n)]/[bc-ad].
Now, m years later, it is [a(d-c)(m+n)]/[bc-ad] + m = [a(d-c)(m+n)+m(bc-ad)]/[bc-ad] = [adm – acm + adn – acn + bcm – adm] / [bc – ad] = [bcm – acm + adn – acn] / [bc – ad].
And that’s the same formula.
With variables, the algebra was a lot easier to follow than the “intuitive method”, wasn’t it? But at least we see that my intuitive method agrees fully with the algebra. Does the other?
Where the provided approach fails
What he did amounts to
c – a units = m+n years.
So 1 unit = (m+n)/(c – a) years.
So John’s age m years ago was a(m+n)/(c – a), and
J = a(m+n)/(c – a) + m = [an + cm]/[c – a].
Applying this with a=1, b=5, c=3, d=7, we get
J = [1*6 + 3*10]/[3 – 1] = 36/2 = 18
That is not the same formula, since b and d are absent; he is wrong in general, but just happens to get the correct answer here. I hope in a different problem he would do something different to account for b and d!
I don’t mind looking through a non-English video to see what I can extract from it; please provide the link.
The formula I got here agrees with what I found above given that \(c-a=d-b\).
Might the YouTuber have actually made an easily overlooked comment explaining that he was using the fact that \(c-a=d-b\)? We never heard more on that.
A visual approach
There is another way to think about problems like this that involve ratios, and would commonly be solved by algebra. It uses what are variously called “tape diagrams” or “bar models”, and is commonly associated with what is called “Singapore math”. I had images similar to these in my mind when I devised my intuitive method, and on second thought, this seemed like a better way to explain it. So I wrote again:
Thinking about this again, it occurred to me that since I like thinking of ratio problems in visual terms, I should try that here.
Here is the ratio 10 years ago (1:5, reading bottom to top):
+---+---+---+---+---+ | | Father +---+---+---+---+---+ | | John +---+Here it is 6 years from now (3:7):
+---+---+---+---+---+---+---+ | | Father +---+---+---+---+---+---+---+ | | John +---+---+---+Since the latter is obtained by adding 16 years to both ages, I can line them up:
........+---+---+---+---+---+ ........| | Father........ ...16...+---+---+---+---+---+ ........| | John ........+---+ +---+---+---+---+---+---+---+ | | Father +---+---+---+---+---+---+---+ | | John +---+---+---+Clearly 16 years equals 2 parts, so each part is 8 years, and the initial ages are 8 and 40.
The idea here is to make bars representing the two quantities, showing the “parts” used in each ratio without knowing their actual size, and relate them to one another. For this problem, little extra thinking was needed because the “parts” in both ratios represented the same number of years (the difference in the ages being 4 parts in both cases), so that we could just add the 16 years and see that both ages increased by two “parts” from one time to the other. But that won’t always happen:
And this is exactly what your video maker did …
But I could do that only because the difference of terms in each ratio was 4 units.
Again, if he said something about the differences both being 4, then his work is valid.
One might well solve the problem this way and never make note of the “happy coincidence” that made it easy. With less helpful numbers, we’d be forced to pay attention:
Modifying the problem so his method fails
Now, what if we had a problem where the differences in the ratios are not the same? Here is one:
Today, the ratio of John’s age to his father’s is 1:6. In 6 years, it will be 1:3. How old are they?
I can make the same sort of bars, but they don’t work:
........+---+---+---+---+---+---+ ........| | Father ...6....+---+---+---+---+---+---+ ........| | John ........+---+ +---+---+---+ | | Father +---+---+---+ | | John +---+
The bars don’t have the same scale in each ratio, as we see from the fact that the difference of ages is not the same; so we can’t just add years. We need to rescale the diagrams so that the difference is the same size in each; for that purpose, we can do the same sort of thing we do to add, or compare, fractions using a common denominator (though here it is not a denominator).
To make them line up, I need to make equivalent ratios such that the differences, rather than being 5 and 2, will both be 10 (the LCM). So I multiply by 2 and 5, respectively, rewriting 1:6 as 2:12, and 1:3 as 5:15. The new bars do line up correctly:
............+---+---+---+---+---+---+---+---+---+---+---+---+ ............| | Father ......6.....+---+---+---+---+---+---+---+---+---+---+---+---+ ............| | John ............+---+---+ +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ | | Father +---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+ | | John +---+---+---+---+---+Now 6 years corresponds to 3 parts; each part is 2 years, so John was 4 and his father was 24. (In 6 years, they will be 10 and 30, with the required ratio.)
This visual method helps to make sense of everything, without needing the structure that algebra provides.
We can, of course, solve this problem with algebra, too. Here’s the work equivalent to what I did before:
Let John’s age now be J, and his father’s F. Then the two statements become
\(J:F = 1:6\) [Ratio now]
\((J+6):(F+6) = 1:3\) [Ratio 6 years from now]
Cross-multiplying,
\(6J=F\)
\(3(J+6)=F+6\)
Simplifying the second,
\(3J+18=F+6\)
Substituting from the first equation into the second,
\(3J+18=6J+6\)
\(12 = 3J\)
\(J = 4\)
Solving for F, \(F=6(4) = 24\).
This time it was probably less work with algebra, in part because I gave data for “now” as one of the equations. But each method can be good, as long as we pay attention to the reason we can take each step.