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?
Word problem about elevator capacity of children vs adults.
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1At 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. – 2012-04-23
3 Answers
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?
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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0Very helpful post. – 2012-04-23
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.