Unitary method — direct and inverse proportion: Question bank

This page assumes only the two laws from the parent note:

Direct — the ratio $y / x$ stays fixed (both quantities move the same way).
Inverse — the product $x y$ stays fixed (they move opposite ways).

True or false — justify

Every item is a claim. Decide true/false, then give the one-line reason before revealing.

More workers always finish a job in fewer days.

Usually true for a fixed job (inverse proportion), but false if the job itself grows with the workers, or if extra workers can't fit / interfere — the "total work constant" assumption must hold.

y

is directly proportional to

x

, then doubling

x

doubles

y

True — direct means

y / x = k

is fixed, so

y = k x

; replacing

x

2 x

gives

2 k x = 2 y

exactly.

y

is inversely proportional to

x

, then doubling

x

doubles

y

False — inverse means

x y = k

; doubling

x

halves

y

so the product stays

k

. The two quantities move opposite ways.

In a direct proportion, if you add 3 to

x

you must add 3 to

y

False — direct proportion preserves ratios (multiply), not differences (add). Adding a fixed amount breaks

y / x = k

unless

k = 1

"A costs ₹8, so 5 of A cost ₹40" is the unitary method in action.

True — we knew the value of ONE unit (₹8) and scaled up by multiplying, which is exactly the "one-then-many" step.

Speed and time for a fixed route are directly proportional.

False — the distance

= speed \times time

is the fixed product, so speed and time are inversely proportional; go faster and the time drops.

Doubling the length of a wall doubles the days one worker needs.

True — with worker count fixed, more work means proportionally more time, so days are directly proportional to the wall's size.

If 3 machines make 300 parts, then 1 machine makes 100 parts (same time).

True if machines are identical and independent — this is the unitary "go to one" step, dividing the output equally.

x y = k

, the constant

k

has no units.

False —

k

carries the product of the units, e.g. worker-days or km. Naming its units ("worker-days") is exactly what stops you from dropping it.

Direct proportion graphs as a straight line through the origin.

True —

y = k x

passes through

(0, 0)

with slope

k

; at zero units you have zero value, which is why the line starts at the origin.

Spot the error

Each item is a flawed line of reasoning. Name what went wrong and why.

"15 workers → 8 days, so 20 workers →

\frac{15}{8} = \frac{20}{y}

, giving

y \approx 10.7

The direct cross-multiply form was used on an inverse problem. More workers → fewer days, so use

x_{1} y_{1} = x_{2} y_{2}

15 (8) = 20 y \Rightarrow y = 6

"7 pens cost ₹56, so 1 pen costs

56 \times 7 = ₹392

Going from many to one you must divide, not multiply. One pen costs

56 \div 7 = ₹8

; multiplying inflated the price instead of splitting it.

"More taps fill the tank faster, and cost is proportional to taps, so cost is inversely proportional to time."

Two separate relations got mashed together. Cost∝taps is direct; taps∝(1/time) is inverse. You can't declare cost∝(1/time) without a genuine chain — always test each pair on its own.

"5 pumps empty a pool in 6 h, so 10 pumps do it in 12 h."

The product was ignored; the answer moved the wrong way. More pumps → less time:

5 (6) = 30

, so

10 t = 30 \Rightarrow t = 3

h, not 12.

"Cost per person for a shared taxi is directly proportional to the number of passengers."

The opposite — a fixed fare split among more people gives each person less, so per-person cost is inversely proportional to passenger count.

"Since

y / x

is constant in direct proportion,

y - x

is also constant."

A constant ratio does not force a constant difference. For

y = 3 x

: differences

2, 4, 6

grow while the ratio stays

3

"20 men dig a trench in 5 days, so 20 men dig two such trenches in 5 days."

The job doubled, so time should double to 10 days (direct in job size with workers fixed). The worker count being unchanged doesn't keep the days unchanged when the work grows.

Why questions

Answer the reason, not just the fact.

Why do we divide (not multiply) when going from many units to one?

Because the total value is spread equally across all the units, so splitting it means dividing the total by the number of units.

Why must you run the up/down test before writing any equation?

Because it decides which law applies — ratio-equal (direct) or product-equal (inverse). Pick wrong and cross-multiplication silently gives a wrong-direction answer.

Why is "worker-days" a product and not a sum?

Because the whole job equals workers × days working together; each of the

x

workers contributes for

y

days, and it's the product that measures total labour done.

Why does the direct-proportion line pass through the origin but a "flat fee + per-item" cost does not?

True proportion means zero units → zero value, hitting

(0, 0)

. A fixed fee adds a constant

c

, giving

y = k x + c

, which starts at

c > 0

— that's no longer pure proportion.

Why can't you use

\frac{y _{1}}{x _{1}} = \frac{y _{2}}{x _{2}}

for workers-and-days?

That form assumes a constant ratio, but for a fixed job the product

x y

is constant instead. The correct relation is

x_{1} y_{1} = x_{2} y_{2}

Why does naming the constant's units help avoid mistakes?

If you can say "120 worker-days" out loud, you're forced to keep the product intact; people who track only "days" quietly lose the 120 and mis-scale.

Why is the unitary method called "unitary"?

Because everything routes through the value of a single unit (one item, one worker, one hour) — you find that first, then scale to any quantity.

Edge cases

The boundaries where the rules bend or break.

What is the "cost of 0 pens" in a direct proportion, and why?

Exactly ₹0 — the line

y = k x

gives

y = 0

x = 0

, matching the intuition that buying nothing costs nothing.

Can

x = 0

appear in an inverse proportion

x y = k

with

k > 0

No — it would require

0 \cdot y = k > 0

, impossible. Zero workers can never complete a positive job, so

y

would blow up to infinity (never finishes).

If 1 worker takes 60 days, how long do 0 workers take?

Infinite / never — with no one working, the fixed job is never done; the inverse relation

y = k / x

shoots to infinity as

x \to 0

Two painters each finish a room in 6 h alone; do two together finish in 3 h?

Only if their work adds cleanly with no interference — then the combined "rate" doubles and time halves to 3 h. Real overlap (one wall, shared space) can break this idealisation.

Is a proportion still "direct" if

k = 0

Technically

y = 0

for all

x

(a flat line on the axis) — the ratio is constant at 0, but it's a degenerate case where the quantity is always zero regardless of

x

If doubling

x

leaves

y

unchanged, what kind of proportion is it?

Neither direct nor inverse —

y

is independent of

x

(constant). Direct would double

y

; inverse would halve it; here nothing moves.

Recall

Recall One-line reflexes

Same direction → direct → ratio $y / x$ constant.
Opposite direction → inverse → product $x y$ constant.
Many → one → divide; one → many → multiply.
Speed & time (fixed distance) → inverse, product = distance.
Direct graph → straight line through the origin.

Connections

Ratio and Proportion — direct proportion is a constant ratio, literally.
Speed Distance Time — the classic "fixed distance ⇒ speed & time inverse" trap.
Time and Work — worker-days is the product to never drop.
Percentages — "per 1" scaled to 100, a cousin of the unitary step.
Linear Equations — $y = k x$ vs $y = k x + c$ explains the origin edge case.