Exercises — Unitary method — direct and inverse proportion

Up/down test: more chocolates → more cost. Both rise together → direct. Value of one unit: divide the total by the number. $\frac{₹80}{4}=₹20\text{ per chocolate.}$ Answer: direct; ₹20 each.

Up/down test: more pipes → less time to fill. Opposite directions → inverse. For inverse, the product is the constant. Here it is best named pipe-hours: $6\times 12=72\text{ pipe-hours.}$ Answer: inverse; the constant is 72 pipe-hours.

Direction: more books → more cost → direct. Value of one (many → one, so divide): $\frac{210}{7}=₹30\text{ per book.}$ Scale to 11 (one → many, so multiply): $30\times 11=₹330.$ Answer: ₹330.

Direction: fewer workers → more days → inverse. Constant total work (multiply): $18\times 10=180\text{ worker-days.}$ Solve for new days $y$ using $x_{2}y=180$: $15\times y=180\Rightarrow y=\frac{180}{15}=12\text{ days.}$ Answer: 12 days. (Sanity check: fewer workers, more days — matches.)

The distance is fixed — that's our constant, and speed–time is a classic speed–distance–time inverse pair. $\text{Distance}=72\times 5=360\text{ km.}$ More time available → we can go slower (inverse). New speed: $\text{speed}=\frac{360}{6}=60\text{ km/h.}$ Answer: 60 km/h.

The total amount of food is fixed — measure it in soldier-days: $40\times 30=1200\text{ soldier-days.}$ More mouths → food lasts fewer days → inverse. New number of soldiers $=40+10=50$. $50\times y=1200\Rightarrow y=\frac{1200}{50}=24\text{ days.}$ Answer: 24 days. (More soldiers, fewer days — correct direction.)

Careful: here more machines → more bottles, so this is direct, not inverse! (The "machines" word tempts you toward inverse, but the output rises with machines.) Bottles per machine (many → one, divide): $\frac{900}{12}=75\text{ bottles per machine.}$ Scale to 20 machines (multiply): $75\times 20=1500\text{ bottles.}$ Answer: 1500 bottles.

One turn of the wheel covers its circumference. A bigger wheel covers more per turn, so it needs fewer turns for the same distance → inverse between "wheel size" and "number of revolutions". The fixed quantity is the total distance. Doubling the radius doubles the distance-per-revolution, so the count is halved: $\frac{240}{2}=120\text{ revolutions.}$ Answer: 120 revolutions.

Two influences change at once, so we go through one worker, one wall, one day — the pure unitary idea, stretched. Measure the total effort in worker-days per wall.

Step 1 — total worker-days for 3 walls: $6\times 8=48$ worker-days. Step 2 — worker-days for ONE wall: $\frac{48}{3}=16$ worker-days per wall. Step 3 — worker-days for 5 walls: $16\times 5=80$ worker-days. Step 4 — spread 80 worker-days over 4 workers → days: $4\times y=80\Rightarrow y=\frac{80}{4}=20\text{ days.}$ Answer: 20 days.

Direction check: fewer workers (6→4) pushes days up; more walls (3→5) pushes days up — both effects raise the day-count, and indeed $20>8$. ✓ (This "worker-days" bookkeeping is the heart of Time and Work.)

Step 1 — old price per kg (direct, many → one): $\frac{300}{5}=₹60\text{ per kg.}$ Step 2 — apply the 20% rise (a percentage is just "per 100" scaled): $60\times \left(1+\frac{20}{100}\right)=60\times 1.2=₹72\text{ per kg.}$ Step 3 — scale to 8 kg (direct, one → many): $72\times 8=₹576.$ Answer: ₹576.

You cannot add or average the times (a classic trap). Times are inverse to rate, so we convert to rate = tank per hour, add rates, then invert back.

Think of the tank as 1 whole job.

• A's rate: $\frac{1}{6}$ tank per hour.
• B's rate: $\frac{1}{3}$ tank per hour.

Add rates (rates do add, since both pour into the same tank at once): $\frac{1}{6}+\frac{1}{3}=\frac{1}{6}+\frac{2}{6}=\frac{3}{6}=\frac{1}{2}\text{ tank per hour.}$ Invert to get time (one → whole, so time $=\frac{1\text{ job}}{\text{rate}}$): $y=\frac{1}{\frac{1}{2}}=2\text{ hours.}$ Answer: 2 hours. Sanity check: together must be faster than the fastest tap alone (3 h). $2<3$ ✓.

Let the original plan be for $D$ days. Total food $=24D$ labourer-days — this constant never changes. After some leave, food lasts $1.5D$ days for $n$ labourers, with the same total: $n\times (1.5D)=24\times D.$ The $D$ cancels (it appears on both sides — the actual number of days doesn't matter): $n=\frac{24}{1.5}=16\text{ labourers remain.}$ So the number who left $=24-16=8$. Answer: 8 labourers left.

The equation is $y=\frac{60}{x}$ — this connects to equations but is a curve, not a straight line. See the two-panel figure below.

• As $x\to$ very large: $y=60/x\to 0$. The curve hugs the horizontal axis but never touches it. (More and more workers → time keeps shrinking toward zero, never negative.)
• As $x\to 0^{+}$ (approaching zero from the positive side): $y=60/x\to \infty$. Almost no workers → the job takes almost forever. The curve shoots up along the vertical axis.
• At $x=60$: $y=60/60=1$.
• At $x=120$: $y=60/120=0.5$. Answer: $y\to 0$ for huge $x$; $y\to \infty$ for tiny $x$; $y=1$ at $x=60$; $y=0.5$ at $x=120$.

Notice on the left panel of the figure: doubling $x$ from 60 to 120 halves $y$ from 1 to 0.5 — the signature of inverse proportion. Compare the right panel (direct, a straight line through the origin) where doubling $x$ doubles $y$.

• L1: name the type (up/down test) + find one unit.
• L2: run a full direct or inverse calculation.
• L3: don't be fooled by keywords — the relationship sets the type.
• L4: chain two changes through a single-unit checkpoint.
• L5: invert questions, add rates not times, and read limiting behaviour off the graph.