Algebraic Speed And Time Question a) On Monday Albert runs 3 km. One week later he runs the same distance, but his average
speed is 20% faster. If his total running time is reduced by 2 minutes 24 seconds, what
was his original running time (in minutes and seconds)?
$speed = \frac{distance}{time}$
Week 1, $s = \frac{3}{t}$ 
Week 2, $s\frac{120}{100} = \frac{3}{t-144}$
$=> \frac{3}{t}\frac{120}{100} = \frac{3}{t-144}\\
=>t=864 s$ 
Therefore original running time was 14 mins 24 secs
b) Cath runs the same distance on Thursday, Friday and Saturday. On Saturday she runs
15% faster than on Friday and her total running time is  minutes less. On Thursday she
runs 15% slower than on Friday and her total running time is  minutes more. Show, with
clear justification, that 23 =  for an integer  which you should calculate.
Thursday, $s\frac{85}{100} = \frac{d}{t+z}$ 
Friday, $s = \frac{d}{t}$ 
Saturday, $s\frac{115}{100} = \frac{d}{t-y}$
Substituting $s = \frac{d}{t}$ 
Based on @David K insight:
=>
$\frac{d}{t}\frac{85}{100} = \frac{d}{t+z}
=>17z=3t$
=>
$\frac{d}{t}\frac{115}{100} = \frac{d}{t-y}
=>23y=3t$
$=>23y=17z$ (QED)
So $n=17$
Doesn't feel intuitive and insightful somehow? What does it say about y and z?
 A: You already have the correct answers in the body of the question itself (after some edits), so I'll just discuss what kind of intuition one might develop around problems like these.
It usually is desirable to express percentages as the true fractions that they represent (e.g. $0.2$ or $1/5$ instead of $20\%$).
In problems like these, dealing with multiple rates,
it's often even better to convert the fractions from fractions you have to multiply and add into ratios you just multiply;
for example, "$20\%$ more" doesn't just translate to "$1/5$ more"; it translates to "$1.2$ times as much."
It helps for much the same reason that it helps to use the ratios $1.5$ and $0.5$ to explain why you don't arrive at the original price after the price goes up $50\%$ one week and down $50\%$ the next week.

For the first problem, if Albert runs $20\%$ faster in the second week than in the first week, he runs $1.2$ times as fast in the second week.
Let's call that $6/5$ times as fast, since it's a nice fraction that expresses the exact same ratio, and because we can easily take the reciprocal of that fraction to find that Albert took $5/6$ as much time to complete the course in the first week as in the second week.
At this point we find it's helpful to be able to switch back and forth between the "$5/6$ times as long" way of thinking and the "$1/6$ shorter" way of thinking.
The reduction in running time, $2$ minutes $24$ seconds, represented $1/6$ of the first week's running time. Therefore the first week's running time was $6$ times $2$ minutes $24$ seconds. Answer:
$$ 6 \times 144 = 864. $$

For the second problem, to see that it is solvable at all,
we know Cath's speed on Saturday was $1.15$ times her speed on Friday,
which means she finished the course on Saturday in $k_1$ times Friday's time,
where $k_1$ is a constant we can compute from the given $15\%$
(in fact $k_1 = 1/1.15,$ but I just want to remember that we can find the number if we want, not to worry yet about what the number is).
For similar reasons, Cath finished the course on Thursday in $k_2$ times Friday's time, where $k_2$ is some constant we can compute from the given $15\%.$
So $z$ is $k_2 - 1$ times Friday's time, and $y$ is $1 - k_1$ times Friday's time, so we know $z$ and $y$ are proportional.
In fact,
$$ \frac{z}{k_2 - 1} = (\text{Friday's time}) = \frac{y}{1 - k_1}, $$
which is enough to solve the problem by plugging terms into formulas and simplifying.
However, for the actual calculation I still find it easier to just
take the standard formulas and play with them until they come out.
I would have used the formula $d = vt$ to set up the equations since I prefer to postpone dealing with division, but $v = \frac dt$ also works, as you showed.
Remembering (from the intuition above) that $z$ and $y$ are both somehow going to be proportional to $t,$ you just need to work on those proportions (one from each of the two equations), make sure the parts that involve $t$ are equal, and simplify ... as you correctly did.
