Solving Variation Problems

A recipe for $4$ people needs $600$ g of flour. Work out the flour for $6$ people. Now do it a second way, without ever calculating the amount per person. Which of your two routes was quicker, and would that still be true if the numbers were $4$ and $7$?

Four steps solve every variation problem: decide the type, write the equation, find $k$ from the known pair, then substitute to answer. When the question compares two situations, a shortcut skips the constant entirely.

1. Decide the type. Read the wording, or test the data. "Directly proportional" and "varies directly" mean divide; "inversely proportional" and "varies inversely" mean multiply. If given a table, divide to test for direct and multiply to test for inverse.

2. Write the equation with an unknown constant: $y = kx$ or $y = \dfrac{k}{x}$.

3. Find $k$ by substituting the known pair.

4. Substitute again to answer the question, then check the direction.

Worked on the hook: six machines take ten hours, so more machines means less time and the variation is inverse. Write $t = \dfrac{k}{m}$. Substituting $m = 6$, $t = 10$ gives $k = 60$. Then $m = 9$ gives $t = \dfrac{60}{9} \approx 6.67$ hours.

Direction check: more machines, less time. $6.67 < 10$, so the answer points the right way. A result larger than $10$ would have signalled an error immediately, whatever the arithmetic looked like.

Many questions give one situation and ask about a second, without ever needing the constant itself. In that case steps 2 to 4 collapse.

Direct variation. The ratio is the same in both situations:

$$\frac{y_1}{x_1} = \frac{y_2}{x_2}$$

because both equal $k$. So from $600$ g of flour for $4$ people, the flour $F$ for $6$ people satisfies $\dfrac{600}{4} = \dfrac{F}{6}$, giving $F = 900$ g.

Inverse variation. The product is the same in both situations:

$$x_1y_1 = x_2y_2$$

So from six machines taking ten hours, nine machines take $t$ where $6 \times 10 = 9 \times t$, giving $t = \dfrac{60}{9} \approx 6.67$ hours, the same answer as before in one line.

Both shortcuts are the four-step method with $k$ cancelled out rather than computed. Use the shortcut when comparing exactly two situations, and the full method when the question asks for the equation, asks several separate parts, or gives a graph.

A third route is often the fastest of all, and it is worth recognising even if you do not use it.

In direct variation, multiplying the input by a factor multiplies the output by the same factor. In inverse variation, multiplying the input by a factor divides the output by that factor.

From $4$ people to $6$ people the scale factor is $\dfrac{6}{4} = 1.5$. Direct variation, so the flour is also multiplied by $1.5$: $600 \times 1.5 = 900$ g.

From $6$ machines to $9$ machines the scale factor is $\dfrac{9}{6} = 1.5$. Inverse variation, so the time is divided by $1.5$: $\dfrac{10}{1.5} \approx 6.67$ hours.

This is quickest when the factor is simple, such as doubling, halving or tripling. It becomes awkward when the factor is not a neat number, as it would be moving from $4$ people to $7$: the factor $\dfrac{7}{4}$ is perfectly usable but the comparison method is tidier.

All three routes are the same mathematics. Choose by which numbers you have been given, not by habit.

The constant carries the units it was calculated in, so mismatched units produce silently wrong answers.

If a cyclist covers $15$ km in $50$ minutes and speed is required in km/h, the $50$ minutes must become $\dfrac{5}{6}$ of an hour before the constant is used, not after. Substituting minutes and then labelling the answer km/h is the standard error.

The reliable habit: convert first, then calculate. Decide what units the answer must be in, convert every given value to match, and only then substitute.

Multi-step problems chain two relationships. If $y \propto x$ and $z \propto y$, then $z \propto x$, as shown in Lesson 1's stretch task, so the two can be combined into a single relationship with a single constant. In practice it is usually clearer to handle them one at a time:

find the first constant, use it to get the middle quantity, then find the second constant and use it to get the answer.

Doing it in one combined step is faster but harder to check, and it hides which of the two relationships an error came from.

Five variation problems. Work each one, then reveal the answer.

1. 1 $y \propto x$, and $y = 20$ when $x = 8$. Find $y$ when $x = 14$.

   Show answer
   Compare: $\tfrac{20}{8} = \tfrac{y}{14}$.$y = 35$

2. 2 $y$ varies inversely as $x$, and $y = 9$ when $x = 4$. Find $y$ when $x = 12$.

   Show answer
   Products match: $4(9) = 12y$.$y = 3$

3. 3 $5$ workers take $12$ days. How long for $10$ workers?

   Show answer
   Twice the workers, so half the time.$6$ days

4. 4 $3$ kg costs $\$21$. Find the cost of $8$ kg.

   Show answer
   Direct: $\tfrac{21}{3} = 7$ per kg.$\$56$

5. 5 $y \propto x$. If $x$ is halved, what happens to $y$?

   Show answer
   Both scale by the same factor.It is halved