In short
You step out of your house and have a choice. The local autorickshaw charges ₹30 to start the meter plus ₹15 per kilometre. The Ola Mini asks ₹50 just to begin the trip but only ₹12 per kilometre afterwards. For a short ride the auto is cheaper (lower base). For a long ride Ola is cheaper (lower per-km rate). The break-even distance is where the two fares are equal — the kilometre at which it stops mattering which one you take. Set the fares equal: 30 + 15d = 50 + 12d, so 3d = 20, giving d = 6.67 km. For shorter rides, take the auto. For longer ones, book the Ola.
This is the most concrete real-world application of a system of linear equations you will ever meet — and it shows up in CBSE Class 10 board exams, Olympiad shortlists, and the back of a Kota coaching book under "comparison problems". Two services, two pricing models, one decision: which one is cheaper for your trip?
Setting up the two equations
Let d be the distance of your trip in kilometres and F be the fare in rupees.
Autorickshaw: the meter drops at ₹30 the moment you sit. Every kilometre after that adds ₹15.
Ola Mini: the app charges ₹50 as a base — convenience fee, app fee, the company's cut. Every kilometre adds ₹12.
These are two linear equations of the form F = \text{(base)} + \text{(rate)} \cdot d. Plot each on a graph with d on the x-axis and F on the y-axis, and you get two straight lines. The auto line starts lower (intercept 30) but climbs faster (slope 15). The Ola line starts higher (intercept 50) but climbs more gently (slope 12). Two lines with different slopes must cross somewhere. Why: that crossing point is the unique distance at which both services charge you the same amount — the boundary between "auto is cheaper" territory and "Ola is cheaper" territory.
Solving the system
You want to find d such that both fares are equal:
Subtract 12d from both sides:
Subtract 30:
The fare at break-even is F = 30 + 15(6.67) = 30 + 100 = ₹130. Or check via Ola: F = 50 + 12(6.67) = 50 + 80 = ₹130. Both give ₹130, which confirms the algebra. Why: at d = 6.67 km, both services bill you ₹130 — the rupee value where the rivalry is a perfect tie. Any further, and the gentler Ola slope wins. Any shorter, and the lower auto base wins.
Two lines crossing — the picture
Try it yourself — drag the fares
The widget below lets you change the base fare and per-km rate of either service and watch the break-even distance update in real time. Try lowering the Ola base to ₹35 — the lines barely cross, and the break-even drops below 2 km. Push the auto's per-km rate down to ₹13 — Ola starts losing across most of the range.
Three worked examples
Example 1: The canonical auto-vs-Ola problem
Set the fares equal:
Subtract 12d from both sides:
Subtract 30:
Why: collecting all d-terms on one side and all constants on the other reduces a two-variable system (because F was the same on both sides, the system collapsed to one equation in d).
The fare at break-even: F = 30 + 15 \cdot \frac{20}{3} = 30 + 100 = ₹130.
Interpretation. For any trip shorter than 6.67 km — your local market, the school, the chai stall two stops away — the auto charges less. For anything longer — the airport, your cousin's house in the next neighbourhood, a Sunday outing — Ola is the better deal. The same trip in two different ways: the algebra has just told you exactly where to switch.
Example 2: When one service is *always* cheaper
Suppose your area also has Uber Premier (the comfy option) at ₹100 base + ₹18/km, and you compare it with the same Ola Mini at ₹50 base + ₹12/km. Set fares equal:
Why: a negative break-even distance has no physical meaning — you cannot drive backwards through the origin. The graph would show the two lines crossing on the wrong side of the y-axis. In the real world (d \geq 0), they never meet.
Interpretation. Ola Mini has both the lower base (50 < 100) AND the lower per-km rate (12 < 18). It is cheaper at d = 0 and gets relatively cheaper as d grows. There is no distance at which Uber Premier wins on price. If you only care about cost, the answer is "always Ola, never Uber" — and that is exactly what the negative d is whispering. Why: in a real comparison problem, an impossible solution is often the most useful answer. It tells you the decision has already been made.
This is the same algebraic situation as a system with no positive-region intersection — geometrically meaningful, decision-wise even more so.
Example 3: Two services with very similar pricing
Suppose a new app, "Namma Auto", advertises ₹35 base + ₹14/km, and the local autorickshaw still charges ₹30 base + ₹15/km. Set the fares equal:
Subtract 14d:
Subtract 30:
The fare at break-even: F = 30 + 15(5) = 30 + 75 = ₹105. Check via Namma Auto: F = 35 + 14(5) = 35 + 70 = ₹105. Both agree.
Why: when the slopes are close (15 vs 14), the lines are nearly parallel, and the difference at any single distance is small. But "small" is not "zero" — even a one-rupee per-km gap eventually accumulates.
Interpretation. Namma Auto charges ₹5 more upfront but saves ₹1 per km after that. By 5 km, the saving cancels the extra base. After 5 km, every additional kilometre saves you another ₹1 — modest, but real. For a 10 km ride, Namma Auto saves you ₹5; for a 20 km ride, ₹15. When two services have similar-looking pricing, the break-even distance is the diagnostic that tells you whether the difference is worth caring about for your typical trip length.
Why this matters
This is the canonical real-world application of a system of linear equations — and it goes far beyond cab fares.
Comparison shopping. Two mobile plans: ₹199/month with 1.5 GB/day vs ₹299/month with 3 GB/day. At what monthly data usage does the costlier plan win? Same algebra.
Rent vs buy. Renting an apartment for ₹15{,}000/month forever vs buying with a ₹2{,}50{,}000 down payment plus ₹12{,}000/month EMI. After how many months does buying save you money?
Subscriptions vs pay-as-you-go. Netflix ₹199/month flat vs renting individual movies at ₹49 each. How many movies per month justify the subscription?
Manufacturing. Two machines: one with high setup cost and low per-unit cost, one with low setup and high per-unit. The break-even production volume decides which to install.
In every case the structure is identical: \text{cost}_1 = a_1 + b_1 x and \text{cost}_2 = a_2 + b_2 x. Setting them equal and solving for x gives the crossover. The graph always tells you which option owns each region. Why: linear comparison problems are everywhere because most pricing in the real world is "fixed fee plus variable rate" — a structure that maps directly onto y = mx + c.
CBSE Class 10 connection. Comparison-of-pricing problems are the most popular real-world application of pair of linear equations in two variables in NCERT and the CBSE board exam. They appear under names like "auto-fare vs taxi-fare", "two railway tickets", "shopkeeper's discount schemes". They are not random — they are the most concrete way to make the abstract idea "two equations meet at a point" feel like something you would actually do at 11pm before booking a ride home.
A quick sanity check
When you solve a real-world comparison problem and get d = -8.33 or d = 1000 km, do not panic. The algebra is fine. The negative or absurdly large answer is a finding: it means the two services do not realistically compete in the consumer range. One simply dominates the other.
Likewise, if you get d = 0, the two services charge the same base fare and the same per-km rate — they are mathematically identical pricing models. The "break-even" is everywhere because every d satisfies both equations. (This is the infinitely many solutions case from Systems of Linear Equations.)
And if the per-km rates are equal but the bases are different (15d + 30 vs 15d + 35), you get a contradiction like 30 = 35 when you try to set them equal. No solution — the lines are parallel and never cross. The cheaper-base service wins for every distance, no contest.
These three cases — unique crossover, no positive solution, infinitely many — are exactly the three cases of a 2 \times 2 system, dressed in real-world clothing.
References
- NCERT Mathematics Textbook for Class 10, Chapter 3: Pair of Linear Equations in Two Variables — ncert.nic.in
- Karnataka State Transport Authority — Autorickshaw fare structure notification — transport.karnataka.gov.in
- Ola Cabs — Fare structure overview — olacabs.com
- Khan Academy: Systems of equations word problems — khanacademy.org
- Cut the Knot: Linear equations in real life — cut-the-knot.org
- Wikipedia: Break-even (economics) — en.wikipedia.org/wiki/Break-even_(economics)