10 workers take 9 hours. How long do 15 workers take?

6 hours, not 13.5. This is inverse proportion: more workers means less time. The product is fixed at 10 times 9 = 90 worker-hours. For 15 workers: 90 divided by 15 = 6 hours. The trap of 13.5 hours comes from treating it as direct proportion and multiplying the time by 15 divided by 10, but that makes more workers take longer, which is physically impossible.

How do you find k in y = k/x?

Multiply x by y. If the curve passes through (6, 21), then k = 6 times 21 = 126. So y = 126/x. At x = 12, y = 126 divided by 12 = 10.5. The trap is using k = y divided by x = 3.5, which is the direct-proportion formula. You can check any pair: x times y should always equal k.

How can I tell if a proportion relationship is direct or inverse?

Ask: does more of x mean more or less of y? If more x means more y, it is direct and y = kx (the ratio y/x is fixed). If more x means less y, it is inverse and y = k/x (the product xy is fixed). For workers and time: more workers means less time, so it is inverse. For pay and hours worked: more hours means more pay, so it is direct.

GCSE Maths Higher · AQA · Proportion

Inverse proportion treated as direct: fix the product, not the ratio

The AQA Higher proportion question that catches most students: 10 workers take 9 hours, how long for 15 workers? The trap is to scale the time up: $9 \times \tfrac{15}{10} = 13.5$ hours. But more workers means less time, not more. This is inverse proportion, where the product is fixed, not the ratio.

The fix: multiply the two given values to get $10 \times 9 = 90$ worker-hours, then divide by the new quantity: $90 \div 15 = 6$ hours. The same habit works for algebraic inverse proportion: from a point on $y = k/x$ , compute $k = x \times y$ , then evaluate at the new $x$ .

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How to spot it in your own work

You multiplied the time by the ratio of workers: $9 \times \tfrac{15}{10} = 13.5$ hours (direct-proportion scaling on an inverse relationship).
You computed $k = y/x$ (direct formula) for a curve given as $y = k/x$ , getting $k = 3.5$ instead of $k = 6 \times 21 = 126$ .
You filled an inverse-proportion table by scaling the output upward as the input grows, rather than fixing the product and dividing.
You did not check whether more of one quantity should mean more or less of the other before deciding which proportion formula to use.

An exam question that triggers it

Here is NOV24 Paper 1 Question 8a, reproduced in structure:

It takes 10 workers 9 hours to complete a job.

How long would it take 15 workers to complete the same job?

The misconception produces 13.5 hours. The correct answer is 6 hours. More workers means the job gets done faster: the total work is fixed at $10 \times 9 = 90$ worker-hours, so $15 \times t = 90$ gives $t = 6$ hours.

Why students fall for this

The default mental model for proportion is direct: “more of one means more of the other.” That model is correct for many everyday relationships (more hours worked, more pay; more fuel used, more miles driven) so it fires automatically. When the relationship is inverse, the same reflex produces the wrong direction of change.

The algebraic form reinforces the confusion. Students who have just practised $y = kx$ (direct) encounter $y = k/x$ (inverse) and reach for the same formula to find $k$ : they compute $k = y/x$ instead of $k = xy$ , because that is what they did for the direct case.

The tell is simple: check whether bigger $x$ gives bigger or smaller $y$ . If bigger $x$ gives smaller $y$ , the product $xy$ is fixed, not the ratio $y/x$ .

The fix: Fix the product, divide to find the missing value

Step 1: decide the direction. Ask whether more of the first quantity means more or less of the second. More workers, less time: inverse. More time working, more pay: direct.

Step 2: fix the product. Multiply the two known values: $10 \times 9 = 90$ worker-hours. This is $k$ , the constant in $y = k/x$ .

Step 3: divide. Divide $k$ by the new value of $x$ : $90 \div 15 = 6$ hours.

Step 4: sense-check. More workers (10 to 15) should give less time (9 to 6). If your answer is bigger than the starting time, you scaled the wrong way.

Worked example

NOV24·1·8a structure: 10 workers take 9 hours. How long for 15 workers? The direct-proportion trap is $9 \times 1.5 = 13.5$ hours.

Direction: more workers, less time. Inverse proportion.
Fix the product.
$k = 10 \times 9 = 90 \text{ worker-hours}$
Divide by the new number of workers.
$t = \frac{90}{15} = 6 \text{ hours}$
Sense-check: 15 workers, 6 hours $(15 > 10,\; 6 < 9)$ . Inverse confirmed.

JUN24·2·24a structure: $y = k/x$ , point $(6,\; 21)$ . Find $k$ and $y$ at $x = 12$ .

Fix k. $k = x \times y = 6 \times 21 = 126$ .
Evaluate at x = 12. $y = 126 \div 12 = 10.5$ .
Product check: $12 \times 10.5 = 126$ . Correct.
Trap: $k = 21/6 = 3.5$ (direct formula) gives $y = 3.5 \times 12 = 42$ . Product $12 \times 42 = 504 \neq 126$ . Wrong.

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Common questions

10 workers take 9 hours. How long do 15 workers take?: 6 hours, not 13.5. This is inverse proportion: more workers means less time. The product is fixed at $10 \times 9 = 90$ worker-hours. For 15 workers: $90 \div 15 = 6$ hours. The trap of 13.5 hours multiplies the time by $15/10$ , which is the direct-proportion rule and makes more workers take longer, which is physically impossible.
How do you find $k$ in $y = k/x$ ?: Multiply $x$ by $y$ . If the curve passes through $(6,\; 21)$ , then $k = 6 \times 21 = 126$ . So $y = 126/x$ . At $x = 12$ , $y = 126 \div 12 = 10.5$ . The trap is $k = y/x = 3.5$ , which is the direct-proportion formula. Check any pair: $x \times y$ should always equal $k$ .
How can I tell whether a proportion relationship is direct or inverse?: Ask: does more $x$ mean more or less $y$ ? If more means more, it is direct and $y = kx$ (ratio $y/x$ is fixed). If more means less, it is inverse and $y = k/x$ (product $xy$ is fixed). Workers and time: more workers, less time — inverse. Pay and hours: more hours, more pay — direct.

Related misconceptions

Proportion to a power scales linearlyFor y proportional to x squared, doubling x quadruples y, not doubles it. The scale factor carries the power.
Constant of proportionality and graph shapeFinding k = y/x for an inverse law, or expecting a reciprocal curve to be a straight line.

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