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A child has n toys, when we remove the $4$ heaviest toys, the total weight of the toys decreases by $32\%$. Of the rest of the toys, take the $4$ lightest ones, and then the total weight decreases another $\frac{6}{17}$. Therefore, determine the sum of the possible values ​​of n. (Answer:$29$)

What would be wrong with this resolution? I try:

Let T = total weight P = heavies weight L = lightest weight

$$P=\dfrac{32T}{100}=\dfrac{8T}{25}\qquad\qquad\qquad\quad\space$$

$$L =\dfrac{6}{17}⋅(1−\dfrac{8}{25})T⇒L=\dfrac{6}{25}T$$

$$\dfrac{T}{n}≤\frac{P}{4}\qquad\qquad\qquad\qquad\qquad\quad\enspace \tag{I}$$

$$\dfrac{T}{n}≥\frac{L}{4}\qquad\qquad\qquad\qquad\qquad\quad\enspace\tag{II}$$

From (I) and (II): $\enspace\dfrac{25}{2}≤n≤\dfrac{50}{3} \implies n = 13,14,15,16$

${\qquad\qquad\qquad\large{\therefore}}\enspace\enspace 13+14+15+16 = 58??$

YNK
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peta arantes
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1 Answers1

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The inequalities are not tight enough. They can be tightened by saying I=in between weights: $$ P+L = \frac{14}{25}T\implies I=\frac{11}{25}T $$ and so: $$ \frac L4\leq\frac I{n-8}\leq\frac P4 $$ which implies: $$ \frac{100}{6}\geq\frac{25n-200}{11}\geq\frac{100}{8} $$ which reduces to $13.5\leq n\leq 15.\overline 3$ and so $n=14$ and $n=15$ are the only two solutions.

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