0
$\begingroup$

The question is:

Lou has three daughters: Wen, Mildred and Tyla. Three year ago, when Lou was twice as old as Tyle, he was thirty years older than Mildred. Now, he is forty-seven years old than Wen. In four years, Wen will be half as old as Tyla. What is Lou's, Wen's, Mildred's and Tyla's combined age

I started with letting x = Tyla's age.

Three year ago, when Lou was twice as old as Tyle

Ok. T-3 -> x-3 and L - 3 -> 2(x-3)

In four years, Wen will be half as old as Tyla

Sure. W + 4 -> (x+4)/2. Here is the interesting part.

Now, he is forty-seven years old than Wen

We can move Lou to present time

So, L -> 2(x-3) + 3

In that case, move Wen to present

W -> (x+4)/2 - 4

So, 2(x-3)+3 = (x+4)/2 - 4 + 47

x=32, matches the answer

But here is the weired part,

If Now, Lou is forty-seven years old than Wen, in 4 years, he is 47+3 = 51 years old than Wen

We can move Lou to present time + 4

So, L -> 2(x-3) + 3 + 4

In that case, Wen value remains same.

W -> (x+4)/2

So, 2(x-3)+3 +4 = (x+4)/2 + 51

which does not yield correct answer. What gives ?

$\endgroup$
8
  • $\begingroup$ For consistency sake, the first thing I would do is to use $T$ for Tyla's age, rather than $x$ $\endgroup$
    – Bram28
    Oct 24, 2016 at 18:35
  • $\begingroup$ You made a mistake with the 'In four years, Wen will be half as old as Tyla'. That should be $W + 4 = (T + 4)/2$ $\endgroup$
    – Bram28
    Oct 24, 2016 at 18:37
  • $\begingroup$ made a typo. Sorry. edited @Bram28 $\endgroup$
    – cowboysaif
    Oct 24, 2016 at 18:40
  • $\begingroup$ You can solve for W once you have T: W = (T + 4)/2 - 4 = (32+4)/2-4 = 18-4=14. And then L = W+47=14+47=61. $\endgroup$
    – Bram28
    Oct 24, 2016 at 18:45
  • 1
    $\begingroup$ You say that if now, Lou is 47 years older than Wen, then in 4 years he is 51 years older ...?! No, he will always be exactly 47 years older. I think maybe there is your mistake $\endgroup$
    – Bram28
    Oct 24, 2016 at 18:47

2 Answers 2

2
$\begingroup$

Yes, that was your mistake: Lou remains exactly 47 years older than Wen. So, the last equation should be $2(x-3)+3+4=(x+4)/2+47$, which indeed works out for $x = 32$.

$\endgroup$
1
$\begingroup$

We can set up a system of four linear equations with four variables and solve that system of linear equations for the four ages. This is simpler than "moving" characters to present time; the "moving" you do is really arithmetic manipulation, but the wording adds some confusion.

Let $L$ denote Lou's age now. Let $M$ denote Mildred's age now. Let $W$ denote Wen's age now. Let $T$ denote Tyla's age now.

We begin by constructing a linear equation from each part of the problem.

Three years ago [...] Lou was twice as old as Tyla

So we have $L - 3 = 2(T - 3).$

$\therefore L-2T=-3$.

Three years ago [Lou] was thirty years older than Mildred

So we have $L - 3 = (M - 3) + 30$.

$\therefore L - M = 30$.

Now, [Lou] is forty-seven years older than Wen

So we have $L = W + 47$.

$\therefore L - W = 47$.

In four years, Wen will be half as old as Tyla.

So we have $W + 4 = .5(T + 4)$.

$\therefore .5T - W = 2$.

We can manipulate each of these equations so that they have a term for each of our four unknowns:

$L-2T = 1L + 0M + 0W -2T = -3$.

$L - M = 1L - 1M + 0W + 0T = 30$.

$L - W = 1L + 0M - 1W + 0T = 47$.

$.5T - W = 0L + 0M - 1W + .5T = 2$.

We can convert that system of linear equations to a matrix of the coefficients of our four unknowns, and row-reduce to solve for the four unknowns:

We get the matrix $\begin{bmatrix}1&0&0&-2&-3\\1&-1&0&0&30\\1&0&-1&0&47\\0&0&-1&.5&2\end{bmatrix}$, which row reduces to $\begin{bmatrix}1&0&0&0&61\\0&1&0&0&31\\0&0&1&0&14\\0&0&0&1&32\end{bmatrix}$.

So we conclude $L = 61$, $M = 31$, $W = 14$, and $T = 32$, which can be verified against the original system of equations given by the word problem.

Thus, the desired sum is $L + M + W + T = 138$.

$\endgroup$
3
  • $\begingroup$ this literally blew my mind $\endgroup$
    – cowboysaif
    Oct 24, 2016 at 19:05
  • 1
    $\begingroup$ Glad to be of help :) $\endgroup$
    – lukass
    Oct 24, 2016 at 19:10
  • $\begingroup$ 'literally'?! I hope not! :) But yeah, this is pretty cool! $\endgroup$
    – Bram28
    Oct 24, 2016 at 19:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .