Unitary method — direct and inverse proportion: Visual walkthrough

We use only two ideas the whole way: sharing equally and rebuilding. Everything grows from a picture.

Step 1 — What "value of ONE" even means

WHAT. We start with a pile: many things, and one total number attached to them. Say 7 pens together cost ₹56. We want to break that total into equal one-pen pieces.

WHY. If every pen is identical, then the total ₹56 must be shared equally — each pen "carries" the same slice of the money. Finding that one slice is the whole trick; once we know one pen, we can rebuild any number of pens.

PICTURE. Look at the row of 7 pens below. The single tall bar of ₹56 gets sliced into 7 equal chalk blocks. Each block is one pen's share.

Figure — Unitary method — direct and inverse proportion

$whole total 56 \div how many pieces 7 = value of ONE piece 8$

$56$ — the total we started with (all pens together).
$7$ — the number of equal pieces we cut it into.
$8$ — the size of a single piece: ₹8 per pen. This number has a name coming up: $k$ .

Step 2 — Naming the one-piece value: the constant $k$

WHAT. Give that one-pen value a permanent letter. Call it $k$ .

WHY. Because it never changes as long as the pens are the same price. Whether you buy 3 pens or 300, each still carries ₹8. A number that stays fixed while everything around it scales deserves its own symbol — that fixed number is what makes the problem solvable.

PICTURE. Below, three different pile-sizes (3, 7, 12 pens) all use the same yellow unit block of height $k = 8$ . The piles differ; the block does not.

Step 3 — Direct proportion: rebuild by multiplying

WHAT. Now go the other way. We know one pen = $k = 8$ . We want 12 pens. Stack 12 copies of the unit block.

WHY. Rebuilding is the reverse of splitting. Splitting used division ( $56 \div 7$ ); rebuilding uses multiplication. Twelve identical ₹8 blocks stacked together is just $8 \times 12$ .

PICTURE. Watch the yellow block copy itself 12 times into a taller bar. Height climbs in step with the count — both rise together. That "both rise together" is exactly direct proportion.

$y = value of one k \times how many x \Rightarrow y = 8 \times 12 = 96$

$x$ — the number of units you want (here $12$ pens).
$k$ — the fixed per-unit value ( $8$ ).
$y$ — the rebuilt total (₹ $96$ ).

Step 4 — Why "same ratio" is the same picture as "same block"

WHAT. Show that $\frac{y _{1}}{x _{1}} = \frac{y _{2}}{x _{2}}$ is not a new fact — it is Step 2 twice.

WHY. Each side of the equation is a height-to-count ratio, and that ratio is the block height $k$ . If both piles are built from the same block, both ratios equal $k$ , so they equal each other. Seeing this stops you from memorising a separate formula.

PICTURE. Two staircases, one with 7 steps, one with 12. Both steps have identical rise ( $k$ ). The slope (rise ÷ run) is the same line — that shared slope is $k$ .

$= k (pile 1’s step) \frac{y _{1}}{x _{1}} = = k (pile 2’s step) \frac{y _{2}}{x _{2}}$

Both fractions are just "one step's height", so of course they're equal.

Step 5 — Inverse proportion: a DIFFERENT thing being shared

WHAT. Switch problems. 15 workers finish a wall in 8 days. Now more workers means fewer days. What's constant here?

WHY. Not the ratio this time — the whole job. The job is a fixed heap of effort: $15 \times 8 = 120$ worker-days. Whether few workers grind slowly or many workers finish fast, the heap is always 120. The thing that stays put is a product, not a ratio.

PICTURE. A rectangle of area 120. Its width = workers, its height = days. Squash it wider (more workers) and it must get shorter (fewer days) to keep the same area.

$workers (width) x \times days (height) y = total job (area) k = 120$

Step 6 — Solving the inverse case, and reading the curve

WHAT. Use the fixed area 120 to find the days for 20 workers.

WHY. If area = width × height and we know the area and the new width, height is just area ÷ width — division again, but now dividing the product, not a simple total.

PICTURE. On the same area-120 curve, slide from the point $(15, 8)$ to the point $(20, ?)$ . As width grows to 20, the point drops down the curve.

$20 \times y = 120 \Rightarrow y = \frac{120}{20} = 6 days$

$120$ — fixed area (total worker-days).
$20$ — new width (workers).
$6$ — new height (days) read off the curve.

Compare the shapes: direct is a straight line through the origin (Step 4); inverse is a bending curve (a hyperbola) that never touches the axes.

Step 7 — Edge and degenerate cases (never get ambushed)

WHAT. Check the weird inputs so no problem surprises you.

WHY. A rule you trust must survive its extremes. Here are the corners:

Zero units (direct). $x = 0 \Rightarrow y = k \cdot 0 = 0$ . Zero pens cost ₹0. The line passes exactly through the origin — buy nothing, pay nothing.
One unit (direct). $x = 1 \Rightarrow y = k$ . The total is the unit value. Sanity anchor.
Zero workers (inverse). $x = 0 \Rightarrow y = k /0$ , which is undefined / infinite. Picture the area-120 rectangle with width 0: to keep area 120 the height must shoot to infinity. Zero workers means the wall is never finished — the curve flies up the vertical axis but never lands on it.
Fractional / huge inputs. Both laws still hold: $2.5$ pens cost $2.5 k$ ; $600$ workers finish in $120/600 = 0.2$ days. The line and curve extend smoothly in both directions.

PICTURE. Direct line hitting the origin (buy 0 → pay 0) beside the inverse curve blowing up as workers → 0.

The one-picture summary

Everything on this page is one diagram: on the left, splitting a total into equal blocks then restacking (ratio stays fixed → straight line → direct); on the right, a fixed-area rectangle squashing wider and shorter (product stays fixed → curve → inverse).

Recall Feynman retelling (cover and explain it yourself)

I have 7 pens costing ₹56 total. I cut that money into 7 equal blocks — each block is ₹8, and that block never changes. To find any number of pens I just stack that many blocks: 12 pens = 12 blocks = ₹96. Because every pile is built from the same block, the height-to-count is always the same (₹8 per pen) — that's why the "cost vs pens" picture is a straight line rising from zero. That's direct.

Workers and days are different: the job is a fixed rectangle of area 120 worker-days. Width is workers, height is days. Add workers → the rectangle gets wider, so to keep the same area it must get shorter — fewer days. 20 workers → width 20 → height 120/20 = 6 days. The picture is a bending curve, and it never touches the axes: with 0 workers the height would have to be infinite (never done). That's inverse. One splits a total (line), the other shares a fixed area (curve) — that's the whole chapter.

Connections

Ratio and Proportion — the constant slope $k$ in Steps 3–4 is a proportion.
Linear Equations — $y = k x$ is the straight line through the origin drawn in Step 4.
Speed Distance Time — Step 5's fixed-area idea is distance = speed × time.
Time and Work — the worker-days rectangle is this topic's home turf.
Percentages — "per cent" is just a unit block scaled to 100.