10 people build a wall in 9 hours. How long do 15 people take?

6 hours. The job is a fixed amount of work: 10 × 9 = 90 person-hours. Share that total among 15 people: 90 ÷ 15 = 6 hours. More people means LESS time, so the answer must be smaller than 9 — and 6 is. The common wrong answer 13.5 scales the time up with the people (9 × 15 ÷ 10), which would mean more workers take longer — impossible.

y is inversely proportional to x. When x = 4, y = 15. What is y when x = 12?

5. For inverse proportion the product x × y is the fixed constant k, so k = 4 × 15 = 60. When x = 12, y = 60 ÷ 12 = 5 — y goes down as x goes up. The trap answer 45 treats it as direct (y = kx) and scales y up: 15 × 12 ÷ 4 = 45.

Does a slower speed over the same distance take more or less time?

More time. Over a fixed distance, speed and time are inversely proportional. 60 miles at 30 mph takes 60 ÷ 30 = 2 hours; the same 60 miles at 20 mph takes 60 ÷ 20 = 3 hours — more, not less. Saying a slower speed gives less time is the direct-thinking trap.

GCSE Maths Foundation · AQA · Proportion, rates & compound measures

Direct vs inverse proportion: why more workers means less time

Proportion questions trip students who treat every relationship as direct — one quantity up, the other up in the same ratio. Told 10 people build a wall in 9 hours, they answer 13.5 hours for 15 people — scaling the time up, $9 \times (15 \div 10)$ — instead of seeing that more workers must finish sooner. The right answer is $90 \div 15 = 6$ hours.

The thirty-second fix: for a fixed job the work is conserved — the product is fixed. Multiply to find the total (10 × 9 = 90 person-hours), then divide by the new amount. For inverse proportion, x × y = k is constant, so more of the resource always means less time.

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How to spot it in your own work

You scaled the time UP with the workers — writing $9 \times (15 \div 10) = 13.5$ instead of $90 \div 15 = 6$ .
Your “more workers” answer came out bigger, when more workers on the same job should finish sooner.
You filled an inverse table as if it were direct ( $y = kx$ ), scaling y up with x rather than dividing into the fixed product.
You read the constant k as a multiplier, not as the fixed total — the product $x \times y$ (person-hours, pump-minutes).
You said a slower speed over the same distance takes less time, when it takes more.

An exam question that triggers it

Here is the canonical AQA Foundation trigger (Nov24 P1 Q27a shape, calculator):

10 people build a wall in 9 hours. Working at the same rate, how long would 15 people take?

The misconception answer is $9 \times (15 \div 10) = 13.5$ — scaling the time up with the people. But more workers on the same job must take less time, so an answer above 9 hours cannot be right.

The work is fixed: $10 \times 9 = 90$ person-hours, so $90 \div 15 = 6$ hours.

Why students fall for this

Direct proportion is the first model students meet — double the recipe, double the flour — so they apply it everywhere. But a work-rate problem is different: the job is a fixed amount of work, and adding workers does not change the work, only how fast it is shared out. The conserved quantity is the product (person-hours), not either number on its own.

The same belief shows up on inverse tables. Students fill $D = k \div b$ as if it were $D = kb$ , scaling D up as b grows, because they never read k as the fixed product $D \times b$ . And it shows up in speed-time reasoning: over a fixed distance, halving the speed doubles the time, but a student thinking “direct” expects a slower speed to give a smaller time.

AQA Foundation calculator papers exploit every face of this: a work-rate change (Nov24 P1 Q27a), recognising which relationship is inverse (Nov24 P2 Q18), interpreting and using the constant k in $D = k \div b$ (Jun24 P2 Q21a/b), and the slower-speed-more-time slip (Jun24 P3 Q24b).

The fix: The product is fixed: find k by multiplying, then divide by the new amount

For a fixed job, the work is conserved. 10 people × 9 hours = $90$ person-hours. With 15 people: $90 \div 15 = 6$ hours. The answer is smaller than 9, because more workers finish sooner. Scaling up (13.5) is bigger — the tell that it is wrong.

Inverse proportion: x × y = k. The product is the fixed constant. If $x = 4, y = 15$ then $k = 60$ , so when $x = 12$ , $y = 60 \div 12 = 5$ . Find k by multiplying, then divide.

Read k as the fixed total. In $D = 240 \div b$ the 240 is the product $D \times b$ : b = 4 gives 60, b = 6 gives 40, b = 8 gives 30 — every column multiplies back to 240. And over a fixed distance, a slower speed always means more time.

Worked example

6 pumps empty a tank in 80 minutes. Working at the same rate, how long do 8 pumps take — and how long do 4 pumps take?

Find the fixed total (the constant k). The work is the product:
$k = 6 \times 80 = 480 \text{ pump-minutes}$
Divide by the new number of pumps.
$8 \text{ pumps: } 480 \div 8 = 60 \text{ minutes}$
$4 \text{ pumps: } 480 \div 4 = 120 \text{ minutes}$
Sense-check with the direction. More pumps (8) means less time (60 < 80); fewer pumps (4) means more time (120 > 80).

Check every column multiplies back to k: $8 \times 60 = 480$ and $4 \times 120 = 480$ — a constant product is what makes it inverse. The trap is to scale the time up: $80 \times (8 \div 6) = 106.67$ , bigger than 80, which would mean more pumps take longer. Find k, then divide.

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Common questions

Why isn’t this just direct proportion like everything else?: Because the job is a fixed amount of work, and adding workers does not change the work — only how fast it is done. The conserved quantity is the product (person-hours). For a fixed job, more workers means less time: $90 \div 15 = 6$ , not $9 \times 15 \div 10 = 13.5$ .
How do I tell direct and inverse proportion apart?: Look at the direction. Direct: when one goes up, the other goes up in the same ratio ( $y = kx$ ). Inverse: when one goes up, the other goes down, and the product stays fixed ( $x \times y = k$ ). Workers double and time halves is inverse; both doubling is direct.
What does the constant k mean for inverse proportion?: It is the fixed product $x \times y$ — the total amount of work, or the constant in $D = k \div b$ . Find it by multiplying a known pair ( $4 \times 15 = 60$ ), then divide it by any new x to get the new y.
Does driving slower over the same distance take less time?: No — it takes more. Over a fixed distance, speed and time are inversely proportional. 60 miles at 30 mph is $60 \div 30 = 2$ hours; at 20 mph it is $60 \div 20 = 3$ hours. Slower means more time.

Related misconceptions

Speed, distance, time & rates: convert the time before you divideThe neighbouring rates misconception — getting the units right (minutes versus hours) before dividing distance by time.
Comparing rates & density: convert to a common unit, and density multipliesThe compound-measure companion — putting rates on a common basis to compare, and remembering mass = density × volume.

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