The capacity of an elevator is either 15 children or 11 adults? If 9 children are currently on the elevator how many adults can still get in?

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If 9 children are already in the elevator, I would wait for the next one. – Théophile Apr 23 '12 at 14:27
I think the assumption that all children weigh one amount and all adults weigh some other amount (made by both responders) is a reasonable one, given the problem statement. Given that, it is soluble. – Ross Millikan Apr 23 '12 at 14:29
@Neil Downvoting for "missing info" isn't fair to the OP if - as likely - the problem was posed this way. It is not uncommon that elementary word problems are posed poorly. – Bill Dubuque Apr 23 '12 at 15:04
In addition to editing the problem might benefit from a modicum of effort by the OP? – Jyrki Lahtonen Apr 23 '12 at 16:45
At first, I downvoted because this question sucks! But after a minute or two, I un-downvoted because the OP doesn't have any control over that. But now I'm thinking about downvoting again, as he hasn't shown any effort. – The Chaz 2.0 Apr 23 '12 at 19:04

Assume that each child weighs $1$ unit, whence the capacity of the elevator is $15$ units. This makes the weight of each adult $\frac{15}{11}\approx 1.36$ units.

If the elevator is occupied by $9$ children, i.e. $9$ units, you're left with $6$ units. How many adults would fit into those units?

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I assume that children have equal weights , and adults have equal weights also . Let's denote weight of children as $x$ , weight of adults as : $y$ , and capacity of the elevator as $C$ , then:

$C=15x=11y \Rightarrow x=\frac{11}{15}y$

$C \geq9x+ny \Rightarrow 6x \geq ny$

$6\cdot \frac{11}{15} y \geq ny \Rightarrow n=\left \lfloor \frac{66}{15} \right \rfloor =4$

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Very helpful post. – El Cholo Grande Apr 23 '12 at 14:27

Hint wlog, by scaling, max weight = $15\!\cdot\! 11,\:$ with weights: $\:\!$ child $= 11,\:$ adult $= 15.$

With $9\:\!$ children, remaining weight $= 15\!\cdot\! 11-9\!\cdot\! 11 = 6\!\cdot\! 11,\:$ enough for $66/15 = 22/5$ adults.

Note how the arithmetic simplifies by exploiting an innate scaling symmetry. This allows us to choose any convenient unit of weight measurement. Thus we can assume that the unit of weight is chosen so that the elevator weight capacity is divisible by both $15$ and $11,\:$ for example $15\!\cdot\!11.\:$ Doing so helps, for as long as possible, to keep the arithmetic computations simple (involving only integers vs. fractions). For some less trivial examples of solving word problems by way of exploiting innate symmetries see V. Arnold's famous sunrise problem.

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