Is my answer to the word problem correct?

Question

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

Answer 1

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 $d=250$

Answer 2

There's a desired arrival time in the mix, let's let $T$ be the desired drive time (the time between departure and desired arrival). And let $L$ denote the remaining distance.

We are told that $$2+\frac L{40}=T+\frac {15}{60}$$

and that $$2+\frac L{50}=T-\frac {36}{60}$$

This is easily solved to yield $L=170$ (we don't care about $T$) and now we just add $80$.

Answer 3

Your equation is not correct because RHS must be $40x+10$.

Notice, if you assume $x$ to be the planned time for the entire distance between cities A and B, the remaining distance can be covered by following two cases as per question

Case-1: $$\text{Remaining distance }= \text{Speed }\times \text{Time taken} $$ $$= 40\underbrace{\left(x-2+\dfrac{15}{60}\right)}_{\text{time taken is 15 minutes more}}\tag 1$$ Case-2: $$\text{Remaining distance }= \text{Speed }\times \text{Time taken} $$ $$= 50 \underbrace{\left(x-2-\dfrac{36}{60}\right)}_{\text{time taken is 36 minutes less}}\tag 2$$ Now, equating the remaining distance from (1) and (2), one should obtain $$40\left(x-2+\dfrac{15}{60}\right)=50\left(x-2-\dfrac{36}{60}\right)$$or $$50\left(x-2-\dfrac{36}{60}\right)+80=40x+10$$ which gives $x=6$ hrs which is the total time planned for the journey between cities A and B.

Now, the remaining distance can be computed by substituting $x=6$ in any of (1) or (2) which gives $170\ \mathrm{km}$

As Susan has already covered $80$ km distance, therefore the total distance between the cities will be $170+80=250\ \mathrm{km}$

Answer 4

You have already got $3$ answers, here is a method I offer in case you are appearing for competitive exams where problems are solvable, but time per question is severely limited, as in MBA entrance tests.

If you draw a V-T diagram, it will be evident that

$t$ hrs @ extra speed of $10$mph
= difference in arrival times of $0.85$h @ $40$ mph

ie $10t = 40*0.85 \Longrightarrow t = 3.4$

and distance $= 80+50*3.4 = 250$ mi

Answer 5

(Since the source of your difficulty with the solution has been traced, I'll add another approach.)

The problem statement indicates that Susan had thus far been driving at $ \ 40 \ \frac{\text{km.}}{\text{hr.}} \ $ for those first two hours. Had she been driving the $ \ 80 \ \text{km.} \ $ at $ \ 10 \ \frac{\text{km.}}{\text{hr.}} \ $ faster, or $ \ 50 \ \frac{\text{km.}}{\text{hr.}} \ , \ $ she would have covered that distance in $ \ 1.6 \ $ hours, or $ \ 96 \ $ minutes. At the faster speed, then, she gains $ \ 120 - 96 \ = \ 24 \ $ minutes for every $ \ 80 \ \text{km.} \ $ traveled, or $ \ 3 \ $ minutes for every $ \ 10 \ \text{km.} \ \ $ (This serves to "calibrate" the effect of the higher speed of travel.)

Over the remaining portion of the trip, she gains $ \ +15 - (-36) \ = \ 51 \ $ minutes, so that distance is $ \ \frac{51 \ \text{min.}}{3 \ \text{min.} / 10 \ \text{km.}} \ = \ 170 \ \text{km.} \ \ $ The total distance between the two cities is then $ \ 80 + 170 \ = \ 250 \ \text{km.} $