AMC 10 · 2011 · #12

Grade 8 rate-ratio

Archetype Meeting Time with Piecewise Rates └ Sub-archetype Constant-Rate Closing Speed

ratelinear-equations-two-varratio-proportion convert-to-algebradimensional-analysis ↑ Prerequisites: rate

📏 Long solution 💡 2 insights

Problem

A drifting object and a powered one leave together, and the powered one turns and catches it later. Find the outbound time.

Pick an answer.

(A)

(B)

3.5

(C)

(D)

4.5

(E)

How to solve

Strategy Convert to Algebra

The whole difficulty is that two of the three quantities in the story — the current speed and the boat's speed — are never given. So the real claim to establish is not just "the time is some number" but "the time is the same number no matter what those two speeds are." Tool #13 (Convert to Algebra) is the anchor because writing the meeting condition as one equation in the letters p, r, t settles that claim outright: both p and r cancel, leaving a single value of t. Tool #4 (Introduce a Variable) names the unknown speeds so they can cancel, and Tool #8 (Analyze the Units) keeps the two legs honest by tracking which speed is measured against the bank and which against the water — the distinction the problem is built on. Tool #16 (Change Focus) then supplies a second, structurally different route: stop watching from the bank and watch from the water, where the raft is motionless and the boat's out-and-back becomes plainly symmetric. Tool #3 (Eliminate Possibilities) closes by checking the value is consistent with the story and matching it to a choice.

1STEP 1

Name the speeds nobody gave us

Two unnamed speeds describe everything.

r=current speed=raft speed, p=boat speed relative to the water, t=hours from A to B

2STEP 2

Speed against water, speed against bank

Going with and against the flow differ by that current.

v_down=p+r, v_up=r-p

3STEP 3

Locate both craft at hour 9

At the meeting time both positions agree.

raft(9)=9r, boat(9)=(p+r)t+(r-p)(9-t)

4STEP 4

Both unknown speeds cancel

Both unknown speeds cancel completely.

(p+r)t+(r-p)(9-t)=9r ; pt+rt+9r-9p-rt+pt=9r ; 2pt-9p=0 ⟹ p(2t-9)=0 ⟹ t=9/2

5STEP 5

Same answer, seen from the water

From the drifting frame the trip is plainly symmetric.

x(h)=x(h)-rh ; raft: x(h)=0 for all h; boat: v=+p then -p ; → out time=back time → 2t=9 → t=9/2

6STEP 6

Check it fits the story

Either way the outbound leg is 4.5 hours, choice (C).

t=4.5: 0 ＜ 4.5 ＜ 9; gap at the turn=(p+r)t-rt=pt ＞ 0 ; (r,p)=(1,3): 18-2(4.5)=9=9r ✓ ; (r,p)=(4,1): 22.5+3(4.5)=36=9r ✓

Answer

4.5

The answer is exactly half of 9, and the water's-eye view says why it must be: relative to the water, the boat spends the trip going away at speed p and the trip returning at the same speed p over the same gap, so the 9 hours split evenly. Two concrete rivers, (r,p)=(1,3) and (r,p)=(4,1), both check out at t=4.5, which is the independence the algebra proved. It is also worth seeing where the wrong choices come from. The phrase "constant speed with respect to the river" is load-bearing: if instead the boat held a constant speed over the ground, the meeting condition would read pt-p(9-t)=9r, giving t=4.5+9r/2p — a value that is always larger than 4.5 and that changes with the river, so the problem would have no single answer at all. Every choice above 4.5 is that misreading pushed to some particular river. Note also that the boat is never required to be faster than the current: even when p ＜ r it drifts backwards relative to the water and still closes the gap, so no hidden condition p ＞ r is being used.

💡Key takeaway

When the water carries everything, measure from the water: the raft stops moving, the boat's trip out and trip back become mirror images, and the 9 hours split exactly in half.

Name the speeds nobody gave us
Speed against water, speed against bank
Locate both craft at hour 9
Both unknown speeds cancel
Same answer, seen from the water
Check it fits the story