Find the distance?

Now this is an interesting and challenging question.

Let the speed of Ford be $p$ and the speed of Audi be $q$ .

The ratio of their speeds is given as $4:3$ . Therefore:

$\dfrac{p}{q} = \dfrac{4}{3}\implies q=\dfrac{3}{4}p$

Now, let the distance of AB be $d$ , and let point M be the point that Ford car and Audi car meets. Let the distance traveled by Ford car from A to M be $x$ . Therefore, the distance from B to M would be $d-x$ . The diagram below illustrates the map:

A--------------------------------------------M----------------------------------B

<-------------------- $(x)$ -------------------><------------ $(d-x)$ ----------->

<-------------------------------------- $(d)$ --------------------------------------->

Let the time taken for Ford to drive from A to M, and the time taken for Audi to drive from B to M be $t_1$ . Assuming that the speed here is constant, we have the equation:

Distance = Speed $\times$ Time

From here, we can form the equations:

$pt_1=x$ ----------> (Eq. 1)

$qt_1=d-x \implies \dfrac{3}{4}pt_1=d-x$ ------------> (Eq. 2)

After they passed each other, Ford reduces its speed by 25%, which gives:

$p_{new}=p\times \dfrac{75}{100} = \dfrac{3}{4}p$

On the other hand, Audi increases its speed by 25%, which gives:

$q_{new}= q \times \dfrac{125}{100} = \dfrac{3}{4}p \times \dfrac{5}{4} = \dfrac{15}{16}p$

Now, at the new speeds, let the time taken for Ford to drive from M to B be $t_2$ . In this time period, Audi drove to a point $20$ km away from A, which means that Audi drove $x-20$ km within time period $t_2$ . This gives us the next two equations:

$p_{new}t_2 = d-x \implies \dfrac{3}{4}pt_2 = d-x$ -------------> (Eq. 3)

$q_{new}t_2 = x-20 \implies \dfrac{15}{16}pt_2 = x-20$ ------------------------> (Eq. 4)

Now, the hardest part of this question is forming the sets of equations above. Once you have these, finding the distance between A and B (which is $d$ ) is not much of a problem anymore.

Eq. 2 $\div$ Eq. 1: $\dfrac{3}{4} = \dfrac{d-x}{x}$

Eq. 4 $\div$ Eq. 3: $\dfrac{5}{4} = \dfrac{x-20}{d-x}$

We are now left with $2$ linear equations with $2$ variables. I've done most of the work already, so the rest is up to you. Solve this and you should get $d=\boxed{560}$ .

Let the time when they meet t. The speed of ford =4x, audi 3x, So the distance AB = 4xt + 3xt = 7xt. After they meet the speed of ford 0.75(4x) = 3x, Audi 1.25(3x) = 3.75 x. 4xt = 3.75 xt + 20, xt = 80. So AB= 7xt = 560.