# Petra is 4 times as old as Philippa. In 8 years time she will be 5 times as old as Philippa. How old are Philippa and Petra?

Oct 17, 2015

The question as given has no positive solutions.

If the $4$ and $5$ were the other way round then Petra's current age is $120$ and Philippa's $24$.

#### Explanation:

The question as given has no positive solutions.

Let Petra's current age be $x$ and Philippa's $y$.

We are given:

$x = 4 y$

$x + 8 = 5 \left(y + 8\right) = 5 y + 40$

Subtract the first of these equations from the second to get:

$8 = x + 8 - x = 5 y + 40 - 4 y = y + 40$

Subtract $40$ from both sides to get:

$y = - 32$

Then

$x = 4 y = 4 \left(- 32\right) = - 128$

So Petra is $- 128$ and Philippa is $- 32$

Alternative problem

Suppose the $4$ times and $5$ times were supposed to be the other way round.

Then we are given:

$x = 5 y$

$x + 8 = 4 \left(y + 8\right) = 4 y + 32$

Subtract the second equation from the first to get:

$- 8 = x - \left(x + 8\right) = 5 y - \left(4 y + 32\right) = y - 32$

Add $32$ to both ends to get:

$y = - 8 + 32 = 24$

Then

$x = 5 y = 5 \cdot 24 = 120$

So Petra is $120$ and Philippa is $24$

Oct 17, 2015

Petra = -32, Philippa = 94
Given that initial condition year is ${t}_{0}$, Petra had not been born. She would be in 32 years that is, ${t}_{32}$. Consequently Petra's age is negative.

#### Explanation:

I gave the wrong answer originally. Submission deleted!
Hopefully I will not get interrupted this time!!! Wrong, 4th attempt!

Let current time be ${t}_{0}$
Let time in 8 years be ${t}_{8}$
Let Petra be $\text{Pe}$
Let Philippa be $\text{Ph}$

At ${t}_{o} \to \text{Pe" = 4"Ph}$
Then $\text{Pe - 4"Ph} = 0$.................. ( 1 )

At t_8 -> "Pe" + 8 = 5("Ph" + 8)
Then $\text{Pe" -5"Ph} = 40 - 8 = 32$....... ( 2 )

Subtract ( 1 ) from ( 2 ) giving:

$- \text{Ph} = 32$
So $\text{Ph} = - 32$ .................... ( 3 )

( at first this looks as though it is not possible. However if you view negative age as to be born in the future then it works. So -32 means that $\text{Ph is to born at } {t}_{32}$
If at ${t}_{0} \text{ Pe}$ is four times older then, stepwise we have from $\text{Ph's}$ age:

(-32), 32, 32, 32

So $\text{Pe's}$ age at ${t}_{0}$ is $32 + 32 + 32 = 94$ because $P h ' s$ age is 32 years the other side of 0 giving a total of $4 \times 32 = 128$ years difference. ( 94 - (-32)) = 128