Two Roads, Three Ratios, and a Surprising Time Relationship

On the whiteboard, a simple problem was framed:

Road A is 20% longer than Road B. It is in good condition throughout, with an average speed of 60 km/h.

Road B is shorter, but 80% of it is in good condition and 20% is in poor condition. The speed on the good portion is 60 km/h, while the speed on the bad portion drops to 20 km/h.

Which road gets you to the destination faster?

"Road B, obviously," said Rahul immediately. "You're only on the bad patch for one-fifth of the trip. The rest of the time you're matching Road A's speed, but on a shorter total distance."

"Not so fast," counter-argued Priya. "That 20 km/h speed drop is severe. You lose way more time on the bad patch than you gain from the shorter distance."

Instead of settling the argument directly, the teacher turned to the board. "Let's test both theories with actual numbers."

Episode 1: The Concrete Numbers

"Suppose Road B is 100 km long," the teacher said. "What happens?"

Priya took the marker and wrote down the calculations:

Road B = 100 km → Road A = 120 km.
Road A Time: 120 / 60 = 2 hours (120 minutes).
Road B Time:
- Good portion (80 km): 80 / 60 = 1 hour 20 minutes (80 minutes).
- Bad portion (20 km): 20 / 20 = 1 hour (60 minutes).
- Total time: 80 + 60 = 2 hours 20 minutes (140 minutes).

"Road A wins by 20 minutes," Priya announced. "The slow patch pulls Road B down."

"Wait," Rahul said, leaning forward. "That’s because 100 km is a long distance. What if the roads are much shorter? Say Road B is only 40 km long?"

The class ran the numbers for 40 km:

Road B = 40 km → Road A = 48 km.
Road A Time: 48 / 60 = 48 minutes.
Road B Time:
- Good portion (32 km): 32 / 60 = 32 minutes.
- Bad portion (8 km): 8 / 20 = 24 minutes.
- Total time: 32 + 24 = 56 minutes.

"Road A is still faster," Priya pointed out. "By 8 minutes."

Episode 2: The Ratio Pattern

Rahul looked at the two results side by side:

Case 1 (100 km): Road A takes 120 min, Road B takes 140 min.
Case 2 (40 km): Road A takes 48 min, Road B takes 56 min.

"Look at the ratio of times," Rahul noticed. "In Case 1, 120 : 140 simplifies to 6 : 7. In Case 2, 48 : 56 also simplifies to 6 : 7!"

"Is that just a coincidence because 40 and 100 are nice numbers?" someone asked from the back row.

"It is not a coincidence," the teacher said. "Whether Road B is 10 km, 40 km, 100 km, or 500 km long, the ratio of the journey times between Road A and Road B will always be 6 : 7."

The class went quiet for a moment. Distance didn't change who won, nor did it change by what proportion they won.

"So, the actual distance is completely irrelevant?" asked Priya. "Then what does the time ratio depend on?"

Episode 3: Stripping Away the Units

"It depends on three fundamental ratios hidden inside the problem," the teacher explained, writing three symbols on the board:

p (Distance ratio between roads):

p = (Distance of Road B) / (Distance of Road A) = 100 / 120 = 5/6
q (Bad portion of Road B as a fraction of its total length):

q = (Bad distance of Road B) / (Total distance of Road B) = 20 / 100 = 1/5
r (Speed ratio on Road B):

r = (Speed on bad section) / (Speed on good section) = 20 / 60 = 1/3

The teacher said. "No kilometers, no hours—just pure proportions."

"Can we find the time ratio using only p, q, and r?" Rahul asked.

"Let's build it," the teacher said. "Let the distance of Road A be D, and the speed on a good road be v."

Distance of Road A = D → T_A = D / v
Total distance of Road B = pD.
Since q is the fraction of bad road, Road B splits into:
- Bad portion = q × (pD)
- Good portion = (1 - q) × (pD)
Since the speed on the bad portion is r × v, the total time for Road B is:
T_B = [(1 - q) × pD / v] + [q × pD / (r × v)] = (pD / v) × [1 - q + (q / r)]

When you divide T_A by T_B, the distance D and speed v vanish completely:

T_A / T_B = 1 / [p × (1 - q + q / r)] = r / [p × (r - rq + q)]

Priya plugged in the numbers on her notepad:

T_A / T_B = (1/3) / [(5/6) × (1/3 - 1/15 + 1/5)]

T_A / T_B = (1/3) / [(5/6) × (7/15)] = (1/3) / (7/18) = 6/7

"It worked!" she said. "No actual speeds or distances involved!"

Episode 4: The Break-Even Challenge

"Now for the real question," said the teacher. "We know Road A wins when the bad section speed is 20 km/h (r = 1/3). How fast would you have to drive on the bad section of Road B for both roads to take the exact same time?"

"That means T_A / T_B must equal 1," Rahul realized.

He took the formula and set it equal to 1:

r / [p × (r - rq + q)] = 1

Rearranging the algebra to solve for r:

r = p × (r - rq + q)

r = pr - prq + pq

r - pr + prq = pq

r × (1 - p + pq) = pq

r = pq / [1 - p × (1 - q)]

"That's the break-even speed ratio!" Priya said.

The class substituted p = 5/6 and q = 1/5:

r = [(5/6) × (1/5)] / [1 - (5/6) × (4/5)]

r = (1/6) / [1 - 4/6] = (1/6) / (2/6) = 1/2

"To tie with Road A, the speed on the bad section of Road B must be at least half the good-road speed," Priya concluded. "Since the good speed is 60 km/h, the break-even speed is 30 km/h."

"And in our original problem," Rahul said with a smile, "the speed on the bad patch was only 20 km/h—which corresponds to r = 1/3, less than 1/2. That's why Road B lost!"

Episode 5: What Happens When the Bad Portion Grows?

Priya stared at the break-even formula on the board:

r = pq / [1 - p × (1 - q)]

"In our problem, Road B was 20% bad (q = 1/5) and 20% shorter (p = 5/6), which gave us r = 1/2. What if the bad patch gets longer or shorter? How does r change with q?"

"Let's test extreme cases first," Rahul suggested. "That usually reveals the boundary behavior."

Case 1: q → 0 (Zero Bad Road)

If q = 0, there is no bad road at all:

r = p × (0) / [1 - p × (1)] = 0

"If there’s zero bad road, Road B is fully good. Since it's shorter (p = 5/6 < 1), Road B wins no matter how slow the non-existent bad section would have been!"

Case 2: q = 1/10 (10% Bad Road)

Keeping p = 5/6:

r = [(5/6) × (1/10)] / [1 - (5/6) × (9/10)] = (1/12) / (1 - 3/4) = 1/3

"If only 10% of Road B is bad, you only need r = 1/3 (a speed of 20 km/h) to tie with Road A!" Priya exclaimed. "Our original 20 km/h speed would have been enough to break even!"

Case 3: q = 1/3 (33.3% Bad Road)

r = [(5/6) × (1/3)] / [1 - (5/6) × (2/3)] = (5/18) / (1 - 5/9) = (5/18) / (4/9) = 5/8 = 0.625

"Now you need r = 5/8, or 62.5% of the good speed (37.5 km/h on a 60 km/h road) just to tie!"

The Sensitivity Table

The teacher summarized their findings in a simple table on the board, keeping p = 5/6 (Road B is 20% shorter than Road A):

Fraction of Bad Road (q)	Break-Even Speed Ratio (r)	Required Speed on Bad Road (v_bad at v = 60 km/h)
0% (0)	0	0 km/h (Road B always wins)
10% (1/10)	1/3 ≈ 0.333	20 km/h
20% (1/5)	1/2 = 0.500	30 km/h (Original Problem)
33.3% (1/3)	5/8 = 0.625	37.5 km/h
50% (1/2)	5/7 ≈ 0.714	42.86 km/h

"Look at that trend," Rahul noticed. "As q increases, r increases too—but not linearly."

"Why does r grow so fast as q grows?" asked Priya.

"Because driving slowly hurts you twice over," the teacher explained. "Not only are you spending more time on the bad patch as q increases, but the lower speed r magnifies that time penalty non-linearly since Time = Distance / Speed. As q gets larger, Road B loses its distance advantage very quickly unless r stays nearly equal to 1."

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