Rate and Work Problems: Free Response

1. Two pumps, one pool . Foundational, 9 points. Question 1 of 5.

A garden pool is empty. Pump A can fill it alone in $8$ hours; pump B can fill it alone in $12$ hours. The gardener runs both pumps at once, starting from empty.

Part A.
Find pump A's rate and pump B's rate, each in pools per hour, then add them to find the combined rate.

Solve and show your work Write each step out, and end with the value and its units. 3 points
Part B.
Using the combined rate, find how long the two pumps take to fill the pool together, to the nearest tenth of an hour.

Carry your own answer forward Continue from the combined rate you found in part A, whatever fraction that came out to be.

Solve and show your work Write each step out, and end with the value and its units. 3 points
Part C.
Pump A, at $8$ hours, is the faster of the two. Check whether the team's fill time from part B is at least HALF of that $8$ hours. Then argue that this floor can never be broken by any pair of pumps: whichever of the two is the faster, the team's time is always at least half of THAT pump's solo time. Finish by saying what would have to be true of the two pumps for the team to reach the floor exactly.

Carry your own answer forward Compare against the fill time you found in part B, whatever value it came out to be.

Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
Turn each solo time into a RATE before you combine anything. A rate is one job over the time it takes, and when two workers act at once it is the rates, never the times, that add.
Hint 2 of 3 · Part B
Once you have a combined rate, getting back to a TIME is the same flip you used for each pump alone: take the reciprocal.
Hint 3 of 3 · Part C
The second pump is the slower one, so ask what the combined rate would be at its very largest, if that partner were as strong as pump A but no stronger. A ceiling on the rate becomes a floor on the reciprocal time.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

Each pump's rate is the reciprocal of its solo time: pump A does $\frac{1}{8}$ of the pool each hour, and pump B does $\frac{1}{12}$ each hour. Add the rates over the common denominator $24$ :

$\frac{1}{8} + \frac{1}{12} = \frac{3}{24} + \frac{2}{24} = \frac{5}{24} \text{ pool per hour}.$

Part B

The time together is the reciprocal of the combined rate:

$T = \frac{1}{\,5/24\,} = \frac{24}{5} = 4.8 \text{ hours}.$

Since $0.8 \times 60 = 48$ , that is $4$ hours and $48$ minutes.

Part C

Call the two rates $r_A$ and $r_B$ , both positive. Pump A has the SMALLER solo time, so it has the LARGER rate: $r_B \le r_A$ . Replacing $r_B$ by the largest thing it could be gives a ceiling on the combined rate:

$r_A + r_B \le r_A + r_A = 2 r_A.$

Time is the reciprocal of a positive rate, and a reciprocal gets BIGGER as the number it inverts gets SMALLER. A combined rate capped at $2 r_A$ therefore has a reciprocal that cannot fall below the reciprocal of $2 r_A$ :

$T = \frac{1}{r_A + r_B} \ge \frac{1}{2 r_A} = \frac{T_A}{2}.$

With $T_A = 8$ hours, that floor is $4$ hours: no partner that is not itself faster than pump A can bring the fill time under $4$ hours, and part B's time does sit above the floor. (A partner faster than pump A would simply be the faster pump, and the same bound would then be measured against ITS solo time.) The whole argument turns on the single inequality $r_B \le r_A$ , which is strict unless the two pumps are equally fast. So the floor is reached exactly when the two pumps have the SAME solo time, each doing half the pool; any genuinely slower partner leaves the team's time strictly above half.

In one line

Pump A's rate is $\frac{1}{8}$ and pump B's rate is $\frac{1}{12}$ pool per hour, combining to $\frac{5}{24}$ pool per hour; the team therefore fills the pool in $\frac{24}{5} = 4.8$ hours. That time cannot fall below half of pump A's $8$ hours: the slower pump's rate is at most the faster pump's, so the combined rate is at most $2 \times \frac{1}{8}$ , and a rate that is at most doubled gives a time that is at least halved. The floor of $4$ hours is reached only by two pumps with the same solo time.

Another way: A direct shortcut for exactly two workers

For exactly two contributors, the two steps of adding rates and then flipping the sum can be collapsed into a single formula, obtained by carrying the same algebra through in letters:

$T = \frac{T_1 T_2}{T_1 + T_2}.$

Here that gives $T = \dfrac{8 \times 12}{8 + 12} = \dfrac{96}{20} = 4.8$ hours, the same answer reached the long way.

When it is worth it A fast check for exactly two workers. It does not generalize the way adding rates does once a third worker, or a drain, joins in.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Converts each pump's solo time into a rate before combining anything, rather than adding the two times directly. . Worth 2 points.

Adds the two rates over a correct common denominator and reports the combined rate with its units (pool per hour). . Worth 1 point.

Part B 3 points

Takes the reciprocal of the combined rate to get a time, rather than the reciprocal of either pump's individual rate. . Worth 2 points.

Reports the time with correct units and converts the fractional hour into minutes correctly. . Worth 1 point.

Part C 3 points

Compares the two pumps' rates from their solo times and uses that comparison to put a ceiling on the combined rate, stated as a general consequence of one pump being the faster rather than as a check on this one pair. . Worth 2 points. needs an explanation, not just an answer

Turns the ceiling on the rate into a floor on the time by using that a reciprocal reverses order on positive numbers, and names the condition on the two solo times under which that floor is reached exactly. . Worth 1 point.

Try a similar problem (Optional)

Same idea, different numbers. Work it on paper, then check yourself the same way.

Pump C can fill the same pool alone in $6$ hours, and pump D alone in $9$ hours. Running together, how long do pumps C and D take to fill the pool?

2. The addition law, extended . Reasoning, 12 points. Question 2 of 5.

The proof that two workers' rates add rests on one fact: in a single unit of time, whatever the first worker finishes is work the second one never has to redo, so the two fractions simply add. Nothing in that reasoning is really about the number two.

Part A.
Let three workers, working alone, take $T_1$ , $T_2$ , and $T_3$ units of time to finish the same job. Reasoning exactly as in the two-worker case, state what fraction of the job each of the three completes in one unit of time, argue why the three fractions can simply be added, and conclude that the time $T$ for all three working together satisfies $\frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2} + \frac{1}{T_3}$ .

Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 4 points
Part B.
Suppose now that, instead of a third worker, the job is a tank with two pipes filling it, with individual times $T_1$ and $T_2$ , and one drain that, left alone, would empty a full tank in $T_3$ units of time. All three run at once. Using the same one-unit-of-time reasoning as part A, explain why the drain's contribution must enter the combined rate as $-\frac{1}{T_3}$ rather than $+\frac{1}{T_3}$ , and write the resulting net-rate equation.

Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 4 points
Part C.
Using the addition law from parts A and B, prove that adding any further POSITIVE-rate worker to an existing team can never increase the time the team takes, no matter what that new worker's own solo time is. State clearly where in your argument you use the fact that the new worker's rate is strictly positive.

Justify your claim State the claim, then give the reason it has to be true. 4 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
Everything here comes from one idea: in a single unit of time, add up what each contributor does, or undoes, to the whole job. That idea never mentions how many contributors there are.
Hint 2 of 3 · Part B
Ask what the drain accomplishes in one unit of time: does it move the tank toward full, or away from it? Whichever direction that is settles the sign its term carries.
Hint 3 of 3 · Part C
Compare the OLD combined rate to the NEW one after a positive rate joins it. A bigger positive number always has a smaller reciprocal, so once you know the rate only grew, you already know something about the time.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

In one unit of time, working alone, the first worker completes $\frac{1}{T_1}$ of the job, the second $\frac{1}{T_2}$ , and the third $\frac{1}{T_3}$ , exactly as in the two-worker case. Whatever any one of them finishes is simply work that the other two no longer need to do; none of the three progress figures depends on what the others are doing, so the three fractions of the job completed in that unit of time can be added with no adjustment:

$\frac{1}{T_1} + \frac{1}{T_2} + \frac{1}{T_3}.$

But the fraction of the job a team finishes in one unit of time is, by definition, the team's combined rate. So this sum IS the combined rate, and its reciprocal is the time $T$ the three take together:

$\frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2} + \frac{1}{T_3}.$

Nothing in this argument singled out having exactly two or exactly three helpers; the same one-unit-of-time reasoning would add a fourth, fifth, or any further worker's rate the same way.

Part B

Apply the same one-unit-of-time reasoning as part A, but look carefully at what each contributor DOES to the tank in that unit of time. Each pipe adds $\frac{1}{T_1}$ and $\frac{1}{T_2}$ of a full tank, exactly as an ordinary worker would. The drain, however, is undoing work rather than doing it: left alone for one unit of time, it removes $\frac{1}{T_3}$ of a full tank, so its contribution to how much fuller the tank gets in that unit of time is $-\frac{1}{T_3}$ , not $+\frac{1}{T_3}$ . Adding the three contributions, exactly as in part A, gives the net rate:

$\text{net rate} = \frac{1}{T_1} + \frac{1}{T_2} - \frac{1}{T_3}.$

The fill time is the reciprocal of this net rate, on the understanding that the net rate is positive; if the drain were strong enough to make it negative, the tank would never fill at all.

Part C

Let the existing team have combined rate $R > 0$ and time $\frac{1}{R}$ . A new worker with solo time $T_n$ contributes a rate $\frac{1}{T_n}$ , and the addition law gives the new combined rate as

$R' = R + \frac{1}{T_n}.$

Because $T_n$ is a finite solo time, $\frac{1}{T_n}$ is strictly positive, so $R' = R + \frac{1}{T_n} > R$ : this is exactly where the positivity of the new worker's rate is used, since adding zero or a negative amount would not force the inequality. Time is the reciprocal of a positive rate, and the reciprocal function reverses order on positive numbers (a bigger input gives a smaller output), so

$\frac{1}{R'} < \frac{1}{R}.$

The left side is the new team's time and the right side is the old team's time, so joining a new, genuinely positive-rate worker to any existing team can only shrink the time, never grow it, whatever that worker's own solo time happens to be.

In one line

For three workers, $\frac{1}{T} = \frac{1}{T_1} + \frac{1}{T_2} + \frac{1}{T_3}$ , by the same one-unit-of-time reasoning as the two-worker case; replacing one worker by a drain flips its term's sign, giving $\frac{1}{T_1} + \frac{1}{T_2} - \frac{1}{T_3}$ . In general, adding any further worker with a genuinely positive rate strictly increases the combined rate and therefore strictly decreases the time, because the reciprocal of a positive number shrinks as that number grows.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Argues, from what each worker independently completes in one unit of time, why three fractions of work can be added with no adjustment, rather than merely asserting that the two-worker rule extends. . Worth 3 points. needs an explanation, not just an answer

States the concluded equation for the combined rate and identifies $T$ as the time for all three working together, not as any one worker's own time. . Worth 1 point.

Part B 4 points

Explains that the drain REMOVES a fraction of the tank in one unit of time rather than adding one, so its contribution to the sum is the negative of a positive rate. . Worth 2 points. needs an explanation, not just an answer

Writes the complete net-rate equation with the drain's term subtracted, and notes that the fill time is its reciprocal only when that net rate is positive. . Worth 2 points.

Part C 4 points

Argues from the addition law that appending a positive rate strictly increases the combined rate, for ANY finite solo time the new worker has, rather than for chosen examples. . Worth 3 points. needs an explanation, not just an answer

Identifies exactly where the strict positivity of the new rate is used, and connects the increased rate to a decreased time via the order-reversing property of the reciprocal. . Worth 1 point.

Try a similar problem (Optional)

Same idea, different numbers. Work it on paper, then check yourself the same way.

Prove the corresponding law for a job with TWO pipes, each with its own solo time, and TWO drains, each with its own solo time, all four running at once: write the equation for the net rate, and say which of the four times you would need to make smaller to slow the whole system down.

3. Closing the gap, two ways . Application, 10 points. Question 3 of 5.

Two cyclists start at the same moment from towns that are $63$ miles apart on a straight road, riding toward each other. The first rides at $12$ mph and the second at $9$ mph.

Part A.
Find how long it takes the two cyclists to meet.

Solve and show your work Write each step out, and end with the value and its units. 3 points
Part B.
Using the meeting time from part A, find how far from the first cyclist's starting town the two riders meet.

Carry your own answer forward Continue using the meeting time you found in part A, whatever value it came out to be.

Solve and show your work Write each step out, and end with the value and its units. 4 points
Part C.
Suppose instead the two riders start $63$ miles apart on the same road and travel in the SAME direction: the front rider goes $9$ mph, and a second rider $12$ mph chases from behind, starting from the back position. Explain how the equation for the catch-up time differs from the meeting-time equation in part A, and use it to find how long the catch-up takes.

Compare the two methods Say what each one costs you, and when you would reach for it. 3 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
When two riders move toward each other, the gap between them shrinks at the SUM of their speeds; multiplied by the time elapsed, that sum accounts for the whole original distance.
Hint 2 of 3 · Part B
Once you know how long the riders travelled before meeting, applying $d = rt$ to just ONE of them tells you how far from THAT rider's own starting point the meeting happens.
Hint 3 of 3 · Part C
Ask how fast the gap ITSELF shrinks in each situation. Riders closing on each other shrink it using both speeds together; a rider chasing another only gains ground by however much faster it is.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Models the closing gap as the SUM of the two riders' own distances, each written as $d = rt$ over a shared unknown time. . Worth 2 points.

Solves the resulting linear equation correctly and reports the time with correct units. . Worth 1 point.

Part B 4 points

Uses the meeting time to find ONE rider's own distance via $d = rt$ , rather than dividing the total distance in some other way. . Worth 2 points.

Reports the distance with correct units and checks that the two riders' distances together add to the full gap given in the stem. . Worth 2 points.

Part C 3 points

Identifies that the chase scenario uses the DIFFERENCE of the two speeds (a relative speed) rather than their sum, and explains why only that leftover speed closes the gap. . Worth 2 points. needs an explanation, not just an answer

Writes the catch-up equation with the correct combination of speeds and solves it correctly for the time. . Worth 1 point.

Try a similar problem (Optional)

Same idea, different numbers. Work it on paper, then check yourself the same way.

Two hikers start $100$ miles apart on a straight trail and walk toward each other, one at $15$ mph and the other at $10$ mph. Find how long until they meet and how far each has walked. Then, if instead they walked the SAME direction with the faster hiker at $15$ mph chasing the slower $10$ mph hiker starting $100$ miles behind, find how long the catch-up takes.

4. The line that breaks the chain . Application, 10 points. Question 4 of 5.

A pipe fills an empty tank in $5$ hours. A drain, left open the whole time, would empty a full tank in $20$ hours. Both run at once. Here is a five-line solution for how long the tank takes to fill. Read it line by line: exactly one line is the first to go wrong, and every line after it follows correctly from that wrong line.

Line 1: The pipe fills $\frac{1}{5}$ of the tank each hour, and the drain would empty $\frac{1}{20}$ of a full tank each hour.

Line 2: Since both act on the tank at once, the combined rate is $\frac{1}{5} + \frac{1}{20} = \frac{4}{20} + \frac{1}{20} = \frac{5}{20} = \frac{1}{4}$ tank per hour.

Line 3: So the tank fills at a rate of $\frac{1}{4}$ tank per hour.

Line 4: The time to fill is the reciprocal of the rate, $T = \dfrac{1}{1/4} = 4$ hours.

Line 5: So the tank fills in $4$ hours.

Part A.
Identify the first line that is not justified, say exactly what is wrong with it, and rewrite that line correctly.

Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 4 points
Part B.
Using the corrected rate from part A, find how long the tank actually takes to fill.

Carry your own answer forward Continue from the corrected net rate you found in part A, whatever fraction it came out to be.

Solve and show your work Write each step out, and end with the value and its units. 3 points
Part C.
The original Line 5 concluded a fill time of $4$ hours. Compare that to the correct time you found in part B, and explain what it is about an ADDED, rather than subtracted, drain rate that put the flawed time on the wrong side of the truth, too short or too long.

Carry your own answer forward Compare against the corrected fill time you found in part B, whatever value it came out to be, rather than recomputing it here.

Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
Check each line as its own claim: does it follow from ordinary algebra applied to the line right before it? The line you want is the first one where that test fails, even if every later line is arithmetically flawless applied to what came before.
Hint 2 of 3 · Part A
A drain works against the pipe rather than alongside it. Ask which side of a plus-or-minus sign its rate belongs on, compared to the pipe's rate.
Hint 3 of 3 · Part C
Think about what an artificially bigger rate does to a reciprocal time, in general, before you look at these particular numbers again.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

Test the lines in order rather than judging by the final answer.

Line 1 correctly reads each rate off the given solo times, so it stands. The trouble is in Line 2: it treats the drain exactly like a second pipe, adding its rate to the pipe's rate. But a drain undoes work rather than doing it, so in one hour it should be SUBTRACTED from what the pipe contributes, not added to it. The corrected line reads

$\frac{1}{5} - \frac{1}{20} = \frac{4}{20} - \frac{1}{20} = \frac{3}{20} \text{ tank per hour}.$

Lines 3, 4, and 5 make no NEW mistake: each one correctly restates, inverts, or reports the rate handed to it, so they are simply built on Line 2's wrong number rather than introducing an error of their own.

Part B

The time is the reciprocal of the corrected net rate:

$T = \frac{1}{\,3/20\,} = \frac{20}{3} \text{ hours}.$

Since $\frac{20}{3} = 6 + \frac{2}{3}$ and $\frac{2}{3} \times 60 = 40$ , this is $6$ hours and $40$ minutes.

Part C

The flawed Line 2 used $\frac{1}{5} + \frac{1}{20} = \frac{1}{4}$ , while the corrected rate is the smaller $\frac{1}{5} - \frac{1}{20} = \frac{3}{20}$ :

$\frac{1}{5} + \frac{1}{20} = \frac{1}{4} \quad \text{(flawed)} \qquad\qquad \frac{1}{5} - \frac{1}{20} = \frac{3}{20} \quad \text{(correct)}.$

Adding the drain's rate instead of subtracting it can only make the resulting number BIGGER than it should be, since it added a positive amount that should have been removed. Time is the reciprocal of the rate, and a reciprocal shrinks as the number it inverts grows, so a rate that came out too big produces a time that comes out too SMALL. That is exactly the direction of the error here: the flawed $4$ hours is smaller than the true time, because the sign mistake inflated the rate rather than deflating it.

In one line

Line 2 is the first that is wrong: it adds the drain's rate instead of subtracting it. The corrected net rate is $\frac{1}{5} - \frac{1}{20} = \frac{3}{20}$ tank per hour, so the tank actually fills in $\frac{20}{3} \approx 6.67$ hours, not the flawed $4$ hours. Because the sign error inflated the rate, and a bigger rate always gives a smaller reciprocal time, the flawed answer was bound to come out too short.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Names one specific line as the first that is unjustified, and confirms that the lines before it are genuinely correct. . Worth 2 points.

Attaches a reason to the diagnosis, naming what the drain actually does to the rate in one hour, and rewrites the flawed step correctly. . Worth 2 points. needs an explanation, not just an answer

Part B 3 points

Takes the reciprocal of the CORRECTED net rate found in part A, not the flawed rate from the original Line 2. . Worth 2 points.

Reports the time with correct units and converts the fractional hour into minutes correctly. . Worth 1 point.

Part C 3 points

Connects an artificially large net rate to a reciprocal time that is too small, using the general fact that a bigger positive rate always gives a smaller time, rather than only re-describing the sign error. . Worth 2 points.

States, and correctly signs, which direction the flawed answer missed the truth (too short rather than too long), tying that direction back to the size of the mistaken rate. . Worth 1 point.

Try a similar problem (Optional)

Same idea, different numbers. Work it on paper, then check yourself the same way.

A pipe fills a tank in $6$ hours; a drain, left open, would empty a full tank in $24$ hours; both run at once. Here are three lines of work.

Line 1: The pipe fills $\frac{1}{6}$ of the tank each hour, and the drain would empty $\frac{1}{24}$ of a full tank each hour.

Line 2: The combined rate is $\frac{1}{6} + \frac{1}{24} = \frac{4}{24} + \frac{1}{24} = \frac{5}{24}$ tank per hour.

Line 3: So the tank fills in $\frac{24}{5} = 4.8$ hours.

Find the first line that is wrong, correct it, and give the true fill time.

5. Same two speeds, two different splits . Reasoning, 11 points. Question 5 of 5.

A trip is covered using only two speeds, $20$ mph and $30$ mph, but split two different ways: one split gives the trip two legs of EQUAL TIME at those speeds, and the other gives it two legs of EQUAL DISTANCE at those same two speeds.

Part A.
For the equal-time split, a vehicle travels $2$ hours at $20$ mph, then $2$ hours at $30$ mph. Find its average speed for the whole trip, and compare it to the plain mean of $20$ and $30$ .

Solve and show your work Write each step out, and end with the value and its units. 3 points
Part B.
For the equal-distance split, the same vehicle travels $60$ miles at $20$ mph, then $60$ miles at $30$ mph. Find its average speed for the whole trip, and compare it to the plain mean of the two speeds.

Solve and show your work Write each step out, and end with the value and its units. 3 points
Part C.
Compare your answers to parts A and B with the plain mean of the two speeds, $\dfrac{20 + 30}{2}$ . Say which of the two splits gives an average speed equal to that mean and which does not, then explain, in terms of how total time is built up in each case, exactly what property of the split, equal time or equal distance, decides the outcome. Finally, say which of the two speeds your part B answer sits closer to, and why.

Carry your own answer forward Use your own average speeds from parts A and B, whatever they came out to be, to make the comparison.

Explain why it works A sentence or two. Reasons, not steps. 5 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
Average speed is always total distance divided by total time; the real question in each split is how that total TIME gets divided between the two speeds.
Hint 2 of 3 · Part B
Because $20$ and $30$ are different speeds, covering the SAME distance at each one cannot possibly take the same amount of time at each. Work out which leg eats up the bigger share.
Hint 3 of 3 · Part C
Ask which quantity is literally split evenly between the two legs in each scenario, time or distance. Whichever one is NOT split evenly is the one that ends up weighted unevenly in the final average.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

The two legs cover $20 \times 2 = 40$ miles and $30 \times 2 = 60$ miles, for a total distance of $100$ miles over a total time of $2 + 2 = 4$ hours:

$\text{average speed} = \frac{100}{4} = 25 \text{ mph}.$

The plain mean of the two speeds is $\frac{20 + 30}{2} = 25$ mph as well, so here the average speed and the mean agree exactly.

Part B

Each leg's own time comes from $t = \dfrac{d}{r}$ : the first leg takes $\dfrac{60}{20} = 3$ hours, and the second takes $\dfrac{60}{30} = 2$ hours, for a total time of $5$ hours over a total distance of $60 + 60 = 120$ miles:

$\text{average speed} = \frac{120}{5} = 24 \text{ mph}.$

That is not the plain mean of $20$ and $30$ , which is $25$ mph.

Part C

Total time is what the average is really divided by, so ask how that total time is built up in each split.

In the equal-time split, each speed gets the SAME share of the total time by construction, so the average speed is literally a weighted blend of the two speeds with equal weights, which is exactly the plain mean, $\frac{20+30}{2}$ . That is why part A's answer matched it.

In the equal-distance split, the shared quantity is distance, not time, and because time is distance over speed, the SLOWER leg automatically eats up more time than the faster leg to cover the same distance. So the two legs do not get equal shares of the total time; the slower speed is overrepresented in the time-weighted total, and the average is pulled toward it:

$|24 - 20| = 4 \qquad \text{versus} \qquad |24 - 30| = 6.$

That is why part B's average sits closer to the slower $20$ mph than to the faster $30$ mph.

In one line

The equal-time split gives an average of $\frac{100}{4} = 25$ mph, exactly the plain mean of $20$ and $30$ . The equal-distance split instead gives $\frac{120}{5} = 24$ mph, below the mean, because the slower leg eats up more of the total time when the distances, not the times, are held equal; that is also why $24$ sits closer to $20$ than to $30$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Builds total distance from each leg's own $d = rt$ , and total time as the sum of the two EQUAL leg times, rather than assuming the average in advance. . Worth 2 points.

Reports the average speed with correct units, and states explicitly how it compares to the numerical mean of the two speeds. . Worth 1 point.

Part B 3 points

Finds each leg's OWN time from $t = d/r$ , recognizing that the two times are unequal even though the two distances are equal. . Worth 2 points.

Reports the average speed with correct units, and states explicitly how it compares to the numerical mean of the two speeds. . Worth 1 point.

Part C 5 points

Identifies which quantity is genuinely held equal across the two legs of each split, time or distance, and ties that to how total time gets built up, rather than only reporting the two numeric averages again. . Worth 3 points. needs an explanation, not just an answer

Explains, in terms of time spent at each speed, why the equal-distance average sits closer to one of the two speeds than to the other, and correctly identifies which one. . Worth 2 points.

Try a similar problem (Optional)

Same idea, different numbers. Work it on paper, then check yourself the same way.

A trip uses only the speeds $40$ mph and $60$ mph. First, for a trip split into two legs of $1$ hour each at those speeds, find the average speed and compare it to the plain mean. Then, for a trip split into two legs of $120$ miles each at those speeds, find the average speed, and say which of the two speeds it sits closer to.

1. Two pumps, one pool . Foundational, 9 points. Question 1 of 5.

2. The addition law, extended . Reasoning, 12 points. Question 2 of 5.

3. Closing the gap, two ways . Application, 10 points. Question 3 of 5.

4. The line that breaks the chain . Application, 10 points. Question 4 of 5.

5. Same two speeds, two different splits . Reasoning, 11 points. Question 5 of 5.

Marking rubrics

1. Two pumps, one pool 9 points

2. The addition law, extended 12 points

3. Closing the gap, two ways 10 points

4. The line that breaks the chain 10 points

5. Same two speeds, two different splits 11 points