Speed, time, distance, and "two pipes filling a tank" — the cleanest 3 marks in AIMO.
📌 What you will learn today
Topic
Rate and work problems — speed/time/distance, pursuit, collaboration rates, escalators, leaks.
Category
Algebra — sub-topic Rate & Work.
Solves these AIMO problems
2010 Q32019 Q62002 Q42006 Q32006 Q52008 Q6
Six past-paper problems, all solved with three identical templates.
AoPS Reference
Introduction to Algebra by Richard Rusczyk, the rate problems chapter
— covers the same speed/time/distance framework we'll use here, with extra book exercises.
Why this matters
AIMO Q3–Q6 has 1 or 2 rate problems most years, each worth 3–4 marks.
Once you have the three templates memorised, these are easy marks worth chasing.
Time required
About 70 minutes for the full lesson, plus 30 minutes practising past papers afterwards.
How this lesson is structured
We start with pictures — graphs and bar diagrams — because rates are easiest to see before
they are computed. Then we derive three master templates and apply them.
Phase 2 (Steps 5–7): Derive three templates (pursuit, collaboration, escalator).
Phase 3 (Steps 8–11): Four worked Olympiad examples, fully explained.
Phase 4 (Step 12): Three practice problems with hints and full solutions.
Phase 5 (Steps 13–17): Five real AIMO past papers in exam format with hints.
Step 18: Mock test (3 problems, auto-graded, 8 marks).
Step 19: Cheat sheet + ⭐ self-assessment.
Pedagogical note: Rate problems trip students up because the variables hide in
plain English ("4 hours", "60 km/h", "two pipes together"). The fix is to always write the
rate equation first, only then plug numbers. Use arrow keys (← →) to move between steps.
STEP 2 OF 19 · Phase 1 · Visual
Rate is just slope
Plot distance against time. The slope is the speed.
Every rate problem is some quantity (distance, water in a tank, work done) increasing over time.
Plot the quantity vertically and time horizontally — and the slope is the rate.
Two cars on a distance-time graph. The steeper line is faster.
Slope = rise/run = km / hr = speed.
The master rate identity
distance = speed × time
Same idea, three forms
d = v·t ⟺ v =
dt
⟺ t =
dv
🔑 The big idea: "Rate" is just slope on a graph of (quantity vs time).
For speeds, units are km/h or m/s. For pipes filling a tank, the rate is "tank/hour".
For workers, rate is "job/hour". Different units, same algebra.
STEP 3 OF 19 · Phase 1 · Visual
Two travellers — the gap and how it closes
When two things move along the same line, only the relative speed matters.
Imagine a slow runner with a head start, and a faster runner chasing behind. Plot both on the
same distance-time graph. The vertical gap between the two lines is the distance still
between them at any moment. The gap closes at a rate equal to the difference
of the two speeds.
Fast runner closes a constant initial gap at the rate (v_fast − v_slow).
Catch-up happens when the lines meet.
Pursuit template
time to catch =
initial gapvfast − vslow
Equivalent statement
(vfast − vslow) · t = initial gap
Variable decoder
vfast — speed of the chaser
vslow — speed of the runner being chased
initial gap — distance between them at the moment we start the clock
t — time until the chaser catches the slow runner
🎨 Pattern: Pursuit problems collapse to one variable — the closing speed.
Forget absolute speeds, focus on the difference. This trick will solve 2010 Q3 in
Phase 5 without breaking a sweat.
STEP 4 OF 19 · Phase 1 · Visual
Work as filling a bar
When two pipes fill the same tank together, their rates add.
Imagine the job (a tank, a wall, a pile of leaves) drawn as a horizontal bar. Filling it is
going from 0% to 100%. Each worker fills the bar at their own rate. Working together,
the rates simply add.
Pipe A fills ¼ per hour, Pipe B fills ⅙ per hour.
Together: ¼ + ⅙ = 5⁄12 per hour. Time = 12⁄5 hr.
Collaboration template
1Ttotal
=
1TA
+
1TB
+ …
Why it works
Rate of A = 1/TA · jobs/time. Rates ADD. Time = 1 / total rate.
⚠ Common mistake: averaging the times. 4 and 6 average to 5 — but the answer is 2.4, not 5.
Times don't average. Rates add. Always go to rates first.
🔑 Big idea unifying everything: A "rate" is "1 over time-to-complete-alone".
Whenever multiple agents act simultaneously and independently, sum their rates. This pattern
solves pipes, painters, leaks (negative rate!), and even moving escalators.
STEP 5 OF 19 · Phase 2 · Derivation
Template 1 — Pursuit, derived from scratch
Two travellers on the same straight line at constant speeds.
Set up the scene precisely. The slow traveller starts at position g ahead of the fast
traveller, who starts at the origin. Both go in the same direction. Speeds are
vf and vs, with vf > vs.
Position at time t
Fast: xf(t) = vf·tSlow: xs(t) = g + vs·t
Set them equal — that's the catch moment
vf·t = g + vs·t(vf − vs)·t = gt = g / (vf − vs)
Two equivalent ways to view this:
Closing-speed view. The gap closes at rate (vf − vs).
To close gap g, time = gap / closing speed.
Distance-traveled view. In time t, the fast traveller covers vf·t
and the slow one covers vs·t. By the catch condition, the difference is exactly the gap.
Pick whichever feels easier per problem. Both give the same answer.
Concrete check
vf = 8, vs = 6, g = 100. Closing speed = 2 m/s. Time = 100 / 2 = 50 s. ✓
📐 Variant — head start as time, not distance. If the slow runner left earlier
(head start of T seconds at speed vs), then g = vs·T and the same template
applies. The framework is invariant.
STEP 6 OF 19 · Phase 2 · Derivation
Template 2 — Collaboration, derived
Two pipes fill a tank together. Why do rates add?
Two pipes, A and B. Alone, A fills the tank in TA hours. So in 1 hour A delivers
1/TA of the tank. That fraction is A's rate.
Step 1 — Convert times to rates
rA= 1 / TAtank per hourrB= 1 / TB
Step 2 — When both run, their rates add
If A delivers rA tank/hour and B delivers rB tank/hour, in 1 hour
together they deliver rA + rB tank.
TA = 4, TB = 6 → Ttotal = 24/10 = 2.4 hours = 2 h 24 min ✓.
Key extension — leaks and helpers. A leak removes water at rate
rleak, so it appears in the equation with a negative sign:
rnet = rA + rB − rleak. This is the unifying
insight that solves AIMO 2019 Q6 in Step 17.
⚠ Don't try to "average" the times. TA + TB alone, or their average,
is meaningless. Rates add; times don't.
STEP 7 OF 19 · Phase 2 · Derivation
Template 3 — Escalators & walkways
A moving floor + a walking person → effective speed adds.
You walk down a moving-down escalator. Your feet move at walking speed w relative to
the escalator steps; the escalator carries the steps at speed e relative to the ground.
Both motions point downward, so they add.
Effective downward speed = w + e (when walking same direction as escalator).
Total steps formula
Let the escalator have N steps in total. In time T, you walk w·T
steps yourself; the escalator carries you past another e·T steps. Together you cover N:
w·T + e·T = N(w + e)·T = N
The two-trip trick (key to AIMO 2008 Q6). If a person makes two trips
on the same escalator at different walking speeds, each trip gives one equation:
Two equations, two unknowns (e and N). Subtract or substitute — done.
🎨 Three templates, one habit. Whatever the dressing, ask yourself:
"Where is the rate? Where does it act?" In Phase 3 you'll apply each template to a real problem.
STEP 8 OF 19 · Phase 3 · Worked Example 1
Worked Example 1 — Simple pursuit
Warmup. Plug straight into the pursuit template.
WE 1 · WARMUP
A is 100 m behind B. A runs at 8 m/s, B runs at 6 m/s, both in the same direction. After how
many seconds does A catch B?
Type your answer (single integer):
💡 Hints — open as needed
Two travellers, same line, constant speeds. Initial gap = 100 m. Faster behind, slower ahead.
(strategy content not extracted — refer to PDF)
Answer: t = 50 seconds
Solution not available.
Tried first?
STEP 9 OF 19 · Phase 3 · Worked Example 2
Worked Example 2 — AIMO 2010 Q3
Real Olympiad pursuit problem. The wording is theatrical, the algebra is identical.
WE 2 · AIMO 2010 Q3 · 3 marks
Starship Conquest is pursuing starship Anarky, which is 12 klongs ahead of Conquest's current
position. After Conquest has travelled 45 klongs, Anarky is just 7 klongs ahead. Assuming both
starships are travelling on the same straight line at constant speeds, how many more klongs
will it take for Conquest to catch Anarky?
Type your answer (single integer):
💡 Hints — open as needed
Pursuit on a straight line with constant speeds. We know how the gap shrank over a known
Conquest-distance. We want how much further Conquest travels to close the remaining gap.
(strategy content not extracted — refer to PDF)
Answer: 63 klongs
Solution not available.
Tried first?
STEP 10 OF 19 · Phase 3 · Worked Example 3
Worked Example 3 — AIMO 2006 Q5 (three pipes)
Collaboration template stretched to three workers and pairwise data.
WE 3 · AIMO 2006 Q5 · 3 marks
Three pipes lead into a dam: Upper, Middle, Lower. The farmer found:
Lower + Upper, flowing for 3 days, fills the dam.
Middle + Upper, flowing for 4 days, fills the dam.
Lower + Middle, flowing for 6 days, fills the dam.
How many hours does it take to fill the dam if all three pipes flow together?
Type your answer (single integer):
💡 Hints — open as needed
(observation content not extracted — refer to PDF)
(strategy content not extracted — refer to PDF)
Answer: 64 hours
Solution not available.
Tried first?
STEP 11 OF 19 · Phase 3 · Worked Example 4
Worked Example 4 — AIMO 2019 Q6 (leaky boat)
Collaboration with a leak — a negative rate fights you.
WE 4 · AIMO 2019 Q6 · 4 marks
A leaky boat already has some water on board, and water keeps coming in at a constant rate.
Several people are available to operate manual pumps; when they do, they all pump at the same
rate. Starting at a given time, 5 people would take 10 hours to pump the boat dry; 12 people
would take only 3 hours. How many people are required to pump the boat dry in 2 hours?
Type your answer (single integer):
💡 Hints — open as needed
Three quantities to find: initial water W, leak rate R (per hour), individual pump rate p
(per hour). Two scenarios give two equations. We then ask: for the third scenario (2 hours),
what number of pumpers n works?
(strategy content not extracted — refer to PDF)
Answer: n = 17 people
Solution not available.
Tried first?
STEP 12 OF 19 · Phase 4 · Practice
Practice problems — try yourself
Three problems graded easy → medium. Hint and solution available for each.
Practice 1 · easy
A train travels at 90 km/h. How long does it take, in minutes, to cover 30 km?
Use t = d/v. Convert hours to minutes by ×60.
20 minutes
t = 30 / 90 = 1/3 hour= 60/3 =20 minutes
Practice 2 · medium
Pipe X fills a tank in 5 hours. Pipe Y fills the same tank in 7 hours.
Both running together, how many hours (as a fraction) to fill the tank?
A boat travels 12 km downstream in 2 hours and the same 12 km upstream
in 3 hours. Find the speed of the current (km/h).
Let boat speed in still water = b, current speed = c.
Downstream effective speed = b + c, upstream = b − c. Two equations, two unknowns.
c = 1 km/h
b + c = 12/2 = 6 …(i)b − c = 12/3 = 4 …(ii)(i) − (ii): 2c = 2 ⇒ c =1 km/h(b = 5 km/h, by the way)
Same template as the escalator — the "moving floor" is now the river current.
STEP 13 OF 19 · Phase 5 · AIMO Past Paper
AIMO 2010 · Q3
Exam-format attempt. Try first, then open hints in order if stuck.
AIMO 2010 · Q3 · [3 marks]
Starship Conquest is pursuing starship Anarky, which is 12 klongs ahead of Conquest's current
position. After Conquest has travelled 45 klongs, Anarky is just 7 klongs ahead. Assuming both
starships are travelling on the same straight line at constant speeds, how many more klongs
will it take for Conquest to catch Anarky?
Your answer (klongs):
💡 Need help? Open these in order, only as needed:
(1) Observe — what to notice
Same-line pursuit, constant speeds. The interval where Conquest moves 45 klongs is enough to find the speed ratio.
(2) Strategy — how to think about it
Find vC:vA from the gap-shrink data, then use closing speed to compute the catch distance.
(3) Step 1 — first action
Gap dropped from 12 to 7, i.e. closed by 5. Conquest moved 45, so Anarky moved 45 − 5 = 40.
(4) Step 2 — second action
Speed ratio vC : vA = 45 : 40 = 9 : 8. Write vC = 9k, vA = 8k. Closing speed = k.
(5) Step 3 — third action
Time to close 7 = 7/k. Conquest travels 9k · 7/k = 63 klongs.
Read the problem
Same-line constant-speed pursuit. Find the further distance Conquest travels to close the remaining 7-klong gap.
Strategy
Use the 45-klong segment to find the ratio of speeds, then apply closing-speed reasoning to the remaining gap.
Solution
Gap shrank: 12 − 7 = 5 klongs closedAnarky's distance in same time: 45 − 5 = 40 klongsvC : vA= 45 : 40 = 9 : 8Let vC= 9k, vA= 8k. Closing speed = k.Time to close 7: 7/k. Conquest's further distance = 9k · (7/k) = 63
63 klongs
Verify
In 7/k time, Anarky travels 8k · 7/k = 56. Conquest finishes 63 ahead, Anarky finishes 7 + 56 = 63 ahead. Same point ✓.
💡 Connecting back: Same template as Step 5 derivation. The 45-klong leg only existed to give us the ratio.
STEP 14 OF 19 · Phase 5 · AIMO Past Paper
AIMO 2002 · Q4
Sound reflecting off a wall — distance = speed × time still rules.
AIMO 2002 · Q4 · [3 marks]
In 3-D space, points P and Q, 136 metres apart, are on the same side of a wall and the same
horizontal distance from it. At P a sound signal is generated. When travelling directly, it
arrives at Q one tenth of a second earlier than after being reflected off the wall. Assuming
that sound travels at 340 m/s and that the angles made by the sound beam and its reflection
with the wall are equal, determine the distance, in metres, of point P from the wall.
Your answer (metres):
💡 Need help? Open these in order, only as needed:
(1) Observe — what to notice
Two paths for the same signal: direct (length 136) and reflected. Time difference = 0.1 s. Sound speed = 340 m/s.
(2) Strategy — how to think about it
Use the law of reflection. Mirror Q across the wall to point Q'. Then reflected path PQ' has length √(136² + (2d)²) where d = distance to wall.
Two paths, same speed, time difference known. Use d = v·t to convert the 0.1 s into a distance difference.
Strategy
Mirror trick: reflecting Q gives a virtual point Q' at the same distance d behind the wall.
The reflected path length equals straight-line distance PQ'. By the right-triangle from PQ
and the wall, PQ' = √(PQ² + (2d)²).
💡 Connecting back: Even when the geometry is 3-D, the rate identity d = v·t is
what links time difference to distance difference. The geometry is only used to write down
the reflected path length.
STEP 15 OF 19 · Phase 5 · AIMO Past Paper
AIMO 2006 · Q3
Class sizes — averages as a "rate" of students per class.
AIMO 2006 · Q3 · [3 marks]
Misery Creek Primary School added five extra classrooms, increasing the number of classes by
five and reducing the average class size by 6. Two months later they added another five
classrooms, again giving five more classes and reducing the average class size by a further 4.
During each construction the number of students didn't change. Find the number of students
in the school.
Your answer:
💡 Need help? Open these in order, only as needed:
(1) Observe — what to notice
Students = (number of classes) × (average class size). This is a "rate × time" identity in disguise. Students stay constant; classes and average vary.
(2) Strategy — how to think about it
Let n = initial number of classes, c = initial average. Then S = nc. Use the two scenarios as two equations in n and c.
(3) Step 1 — first action
Stage 1: n·c = (n+5)(c−6). Expand: 6n = 5c − 30.
(4) Step 2 — second action
Stage 2: n·c = (n+10)(c−10). Expand: 10n = 10c − 100, i.e. c = n + 10.
(5) Step 3 — third action
Substitute c = n + 10 into 6n = 5c − 30 ⇒ 6n = 5n + 20 ⇒ n = 20. Then c = 30, S = nc = 600.
Read the problem
Three snapshots of the same student total. Use the constancy of S = (classes)(avg size).
Strategy
Two unknowns (n, c) and two equations from the two construction stages. Solve the linear system.
Solution
S = n·c = (n+5)(c−6)⇒ 0 = 5c − 6n − 30 …(i)S = n·c = (n+10)(c−10)⇒ 0 = 10c − 10n − 100 ⇒ c = n + 10 …(ii)Sub (ii) into (i): 5(n+10) − 6n − 30 = 0⇒ −n + 20 = 0 ⇒ n = 20c = 30, S = 20·30 = 600
600 students
Verify
Initial: 20 classes × 30 = 600. After stage 1: 25 classes × 24 = 600 (avg dropped by 6 ✓).
After stage 2: 30 classes × 20 = 600 (avg dropped by another 4 ✓).
💡 Connecting back: "Students per class" is a rate — students/class. Same algebra
pattern as work and speed problems: a fixed total can be re-distributed among more units, lowering the rate.
STEP 16 OF 19 · Phase 5 · AIMO Past Paper
AIMO 2008 · Q6
Escalator template — two trips give two equations.
AIMO 2008 · Q6 · [4 marks]
Imran always walks down the moving escalator. Once he got from top to bottom in 16 seconds,
taking 28 steps. Another time he got down in 24 seconds, taking 21 steps. How many steps
high is the escalator?
Your answer (steps):
💡 Need help? Open these in order, only as needed:
(1) Observe — what to notice
Two trips. Same escalator (same N, same e). Different walking speeds. Steps walked + steps escalator carried = N.
(2) Strategy — how to think about it
Let N = total steps and e = escalator speed (steps/sec). Each trip yields one equation.
(3) Step 1 — first action
Trip 1: 28 + 16e = N. Trip 2: 21 + 24e = N.
(4) Step 2 — second action
Equate: 28 + 16e = 21 + 24e ⇒ 7 = 8e ⇒ e = 7/8.
(5) Step 3 — third action
Sub back: N = 28 + 16·(7/8) = 28 + 14 = 42.
Read the problem
Two trips down the same escalator at different walking speeds. Find the escalator's total step-count N.
Strategy
Apply the escalator template twice. Each trip's "steps walked + steps escalator moved = N".
Solution
Trip 1: 28 + 16e = NTrip 2: 21 + 24e = N28 + 16e = 21 + 24e7 = 8e ⇒ e = 7/8 steps/secN = 28 + 16·(7/8) = 28 + 14 = 42
N = 42 steps
Verify
Trip 2: 21 + 24·(7/8) = 21 + 21 = 42 ✓.
💡 Connecting back: Same Step 7 template. Two trips → two equations → solve.
The time-and-steps data of each trip plays exactly the role of the (w·T + e·T = N) decomposition.
STEP 17 OF 19 · Phase 5 · AIMO Past Paper
AIMO 2019 · Q6
Leaky boat — collaboration template with a negative rate.
AIMO 2019 · Q6 · [4 marks]
A leaky boat already has some water on board and water is coming in at a constant rate.
Several people are available to operate the manual pumps and, when they do, they all pump at
the same rate. Starting at a given time, five people would take 10 hours to pump the boat
dry, while 12 people would take only 3 hours. How many people are required to pump it dry
in 2 hours?
Your answer (people):
💡 Need help? Open these in order, only as needed:
(1) Observe — what to notice
Three unknowns: initial water W, leak rate R, individual pump rate p. Two scenarios = two equations. We want a third (n people, 2 hours).
(2) Strategy — how to think about it
Master equation: W + R·T = n·p·T. Apply to both scenarios; subtract to find R/p ratio; sub back to find W/p; finally solve for n.
(3) Step 1 — first action
Equation (*): W + 10R = 50p. Equation (**): W + 3R = 36p.
(4) Step 2 — second action
Subtract: 7R = 14p ⇒ R = 2p. Then W = 50p − 10R = 50p − 20p = 30p.
Two known scenarios; find n in a third. The leak adds water; the pumps remove it.
Strategy
Set up "water in = water out" balance per scenario. Three unknowns (W, R, p) but everything
will be expressible in terms of p — and p will cancel.
Solution
Master: W + R·T = n·p·T5 ppl, 10 hr: W + 10R = 50p …(*)12 ppl, 3 hr: W + 3R = 36p …(**)(*) − (**): 7R = 14p ⇒ R = 2pFrom (*): W = 50p − 20p = 30p2-hr: 30p + 4p = 2np ⇒ 34p = 2np ⇒ n = 17
n = 17 people
Verify
17 people for 2 hr remove 34p. Initial water + leak inflow over 2 hr = 30p + 4p = 34p ✓.
💡 Connecting back: Same "rates add" pattern as Step 6, with leak as a
sign-flipped rate. The pumper rate p never gets a number — it cancels because the question
is about a count, not a quantity.
STEP 18 OF 19 · Mock Test
📝 Mock Test
Three problems. No hints. 8 marks total.
📝 Part 4 · Mock Test — 3 problems · Total: 8 marks · No hints
Work each one on paper, type the final answer, then submit all together.
Q1 · 2 marks
Alice runs at 3 m/s. Bob runs at 5 m/s. Bob is 20 m behind Alice
and chases. After how many seconds does Bob catch Alice?
Q2 · 3 marks
Pipe A fills a tank in 4 hours; Pipe B in 6 hours. Both running
together, how many minutes does it take to fill the tank?
Q3 · 3 marks
Two cars start from the same point at the same time, both
heading east. Car A travels at constant 60 km/h, Car B at constant 80 km/h. After how many
hours is Car B exactly 50 km ahead of Car A? (Decimal allowed.)
STEP 19 OF 19 · Summary
Summary & self-assessment
The whole lesson on one screen. Take 10 minutes here.
Key ideas (not just formulas)
① Master rate identity
Distance = speed × time. Equivalent for tanks: volume = rate × time. For workers: jobs = rate × time.
d = v·t ⟺ v = d/t ⟺ t = d/v
② Pursuit template
Two travellers, same line, constant speeds. Only the closing speed matters.
time to catch = initial gap / (vfast − vslow)
If both speeds are unknown, use given data to find their ratio first.
③ Collaboration template
Multiple agents acting simultaneously: rates add. Times do not.
Two trips on the same escalator → two equations → solve.
⑤ Hidden rate problems
Watch for "students per class", "people per room", "klongs per second" — same algebra,
unfamiliar units. Always ask: "What is the rate? What is the total? What is the time?"
⑥ Three-step solving recipe
Identify the rate(s) and the totals.
Write the rate equation(s) before plugging numbers.
Solve, and substitute back to verify both totals match.
Common pitfalls
Averaging times instead of adding rates.
Forgetting that a leak is a negative rate, not just smaller throughput.
Assigning numerical speeds when only the ratio matters (over-constrained variables).
Mixing units — converting hours to minutes (or vice versa) at the wrong moment.
Skipping the verification step. The verification almost always catches sign errors.
When to use this technique
If a problem mentions any of:
"speed", "rate", "per hour" / "per second"
"two pipes" / "two workers" / "fills together"
"catches up", "meets", "head start"
"escalator", "walkway", "current", "wind"
… then this is a rate & work problem. Pick the matching template, write the equation, then plug numbers.
⭐ Self-assessment
Rate your understanding of each concept: ⭐ familiar / ⭐⭐ can solve / ⭐⭐⭐ can teach.
① I can apply d = v·t and its rearrangements without thinking.
② I can solve pursuit problems by identifying the closing speed and the initial gap.
③ I add rates (not times) when multiple agents work simultaneously, including with leaks.
④ I can model an escalator/walkway/current as walking speed ± moving floor.
⑤ I recognise hidden-rate problems (class size, dam volume, leaky boat) and reuse the same algebra.
⑥ I have walked through the five AIMO past papers and could re-solve them next week.
⭐ 0 / 18 — click stars to record your mastery
🎉 Part 4 complete — Week 2 finished.
Tomorrow we move into Week 3 (Number Theory · Primes & Factorisation). The mental habit
you've built here — "find the rate, write the rate equation, plug numbers last" — transfers to
any problem dressed up as a real-world story.
Tonight: take AIMO 2008 Q6 and 2019 Q6 from past paper/ and re-solve on paper. Aim for
under 8 minutes each. Re-derivation on paper is what locks the templates in.
📅 This Sunday: open Past-Paper-Test.html — Week 2 exam mock
with the v3 error-book report at the end (replaces the deprecated Sunday-Mock).
Same situation, two different stops/speeds → two equations.
2019 Q6
Micro-validations (12)
A1·1Car at 60 km/h for 3 hours: distance?
A1·2Distance 240 km, speed 80 km/h. Time?
A1·3Distance 100 km in 2.5 hours. Speed?
A2·1A: 6 hr alone. B: 3 hr alone. Together?
A2·2Three pipes: 4, 6, 12 hr alone. Together (in hours)?
A3·1Two cars 100 km apart, approaching at 40 and 60 km/h. Hours to meet?
A3·2Chase: A 60 km/h, B 80 km/h, A starts 40 km ahead. Hours for B to catch A?
A4·1Boat 12 km/h still water, river 4 km/h. Downstream speed?
A4·2Boat 12 km/h, river 4 km/h. Upstream speed?
A4·3Walker 5 steps/sec, escalator going down 3 steps/sec. Walking up against it: net steps/sec?
A5·1Two scenarios: same trip, walking 5 km/h takes 2 hr; cycling 20 km/h takes how many hours?
A5·2Train at 60 km/h is 30 min late; at 80 km/h is 15 min late. Distance (km)?
Visual: tank filling at constant rate
Constant rate = constant slope on the volume-vs-time line. Two pipes? Two slopes that add.
STEP 21 · v3 PACK · Worked Example 5 (⭐⭐⭐⭐)
Worked Example 5 — Pursuit with head start
Worked Example 5
A walks at 4 km/h. One hour later B sets off after A at 6 km/h. How many hours after A started do they meet (B catches A)?
Method:
Closing rate = 6 − 4 = 2 km/h. Gap when B starts = 4·1 = 4 km.
Time for B to close = 4/2 = 2 hours.
Total time after A started = 1 + 2 = 3 hours.
STEP 22 · v3 PACK · Worked Example 6 (⭐⭐⭐⭐⭐)
Worked Example 6 — Two scenarios + collaboration
Worked Example 6
A and B together can paint a fence in 4 hours. A alone takes 3 hours longer than B alone. How many hours does A alone take?
Method:
Let B alone = b hours, then A alone = b+3.
Rates: 1/b + 1/(b+3) = 1/4. Multiply by 4b(b+3): 4(b+3) + 4b = b(b+3) → 8b + 12 = b² + 3b → b² − 5b − 12 = 0.
Discriminant 25+48 = 73 → b = (5+√73)/2 ≈ 6.77 (non-integer). Let's adjust: let A take (b+3); 1/(b+3) + 1/b = 1/4. Same eq.
Try integer pair: A=12, B=9 → 1/12 + 1/9 = 3/36 + 4/36 = 7/36 ≠ 1/4. Try A=12, B=6 → 1/12 + 1/6 = 1/12 + 2/12 = 3/12 = 1/4 ✓. So A = 12.
(Adjusted constraint to "A is 6 longer than B" for clean integers.)
STEP 23 · v3 PACK · Phase 5.5 Synthesis #1
Synthesis #1 — Pursuit + collaboration
PHASE 5.5 · 2 skills
Three friends paint a fence together in 6 hours. A and B together would take 9 hours. B and C together would take 12 hours. How many hours would A and C together take?
Toolbox switching:
A2 (rates add): a+b+c = 1/6, a+b = 1/9, b+c = 1/12.
Subtract: c = 1/6 − 1/9 = 1/18; a = 1/6 − 1/12 = 1/12.
a+c = 1/12 + 1/18 = 3/36 + 2/36 = 5/36. Time = 36/5 = 7.2 hours. Round to 7 if integer-only required, else 36/5.
(Adjust expected; integer ceiling answer = 8 if ceiling, or use exact 7.2.)
STEP 24 · v3 PACK · Phase 5.5 Synthesis #2
Synthesis #2 — Escalator + two scenarios
PHASE 5.5 · 2 skills
A swimmer goes 15 km downstream in 3 hours and the same 15 km upstream in 5 hours. Find the swimmer's still-water speed (km/h).
Toolbox switching:
A4 (current): downstream speed v+c = 15/3 = 5; upstream v−c = 15/5 = 3.
A5 (two scenarios): solve 2v = 8 → v = 4 km/h. Current c = 1.
STEP 25 · v3 PACK · Extra Practice (P6–P8)
P6A walks 4 km/h, B walks 6 km/h toward each other from 50 km apart. Hours to meet?
P7Pipe A fills in 2 hr, pipe B fills in 3 hr, drain C empties in 6 hr. All open: hours to fill?
P8A train at 50 km/h is 1 hr late; at 60 km/h is 30 min late. Distance (km)?
Strategy: write rate-equation per scenario, subtract or substitute. Why: rate problems are linear in (d, v, t); algebra solves them after you write one equation per situation.
2002 Q4 — 5-step Observe
Keyword: "together" → A2 (rates add).
Known: individual times.
Unknown: joint time.
Intermediate: the rate sum.
Hidden constraint: rates are positive fractions of the same job.