Here is a problem that I found. It states the following:
Susan drives from city A to city B. After two hours of driving she noticed that she covered $80$ km and calculated that, if she continued driving at the same speed, she would end up been $15$ minutes late. So she increased her speed by $10$ km/hr and she arrived at city B $36$ minutes earlier than she planned.
Find the distance between cities A and B.
My attempt:
Let $x$ be the original time. So $x-2-\frac{36}{60}$ will be the time that the person traveled with the speed of $50$ km/h, because she has already driven for $2$ hours and arrived $36$ minutes early. If we add $80$ miles to the total distance (the time by which Susan traveled), we will get the entire distance. So the left-side equation would be $50\left(x-2-\frac{36}{60}\right)+80$. The second part of the question states that if she continues to drive in such speed, she would be $15$ minutes late. Since $x$ is the entire time, we can write $40x$ is the entire distance. Now we can set them equal to each other and solve. $50\left(x-2-\frac{36}{60}\right)+80=40x$
Solving the equation, I get $x=5$
Since $x$ is the original time that it will take Susan to drive the entire distance, we can figure out the distance by multiply $40$ by $5$. Doing so we will get $200$. However, the answer is $250$. I do not know why my answer is wrong. Here is the link to the question:
https://www.math10.com/en/algebra/word-problems.html
Best Answer
Let's assume we start at $t=0$ and she must arrive at $B$ at $t=T$ where $t$ is measured in hours. We know that she is traveling at $40$ kilometers/hour because she travels $80$ kilometers in two hours.
Using the distance formula we get that $$\frac{d}{T+\frac{1}{4}}=40\implies d=40T+10$$ where $d$ is the distance between $A$ and $B$.
After changing speed to $50$ kilometers/hour (original speed plus $10$) we get $$\frac{d-80}{50}=T-2-\frac{3}{5}\implies d=50T-50$$
Combining this we get that $T=6$ and $d=250$