Finding the total number of objects(stones).

Question

I have $N$ stones. Then the stones are arranged in ascending order of weights. If I remove three stones that are heaviest, then the total weight of the stones decreases by $35$%. Now if I remove the three lightest stones, the total weight of the stones further decreases by a factor of $\frac{5}{13}$.

Find the value of $N$.

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What is tried:Let the total weight be $S$.

After removing three stones that are heaviest the total weight is $0.65 S$ and after further removing three stones that are lightest the total weight is $0.40 S$.

Now using average weights we have:

$$\frac{0.65}{N-3} < \frac1N \text{ and } \frac{0.65}{N-3} < \frac{0.4}{N-6}$$

from which we can conculde that $N>8$ and $N< 11$. So we are having two possible values of $N$ which are $9$ or $10$.

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But how to narrow it down between $9$ and $10$ which one is correct?

Source: Homework Question.

Answer 1

The total weight of the heaviest ones is $0.35 S$, the total weight of lightest ones is $0.25 S$ and the total weight of middle average ones are $0.40 S$.

If $N = 9$, there are $3$ stones remaining whose average weight is $0.1333 S$

but the average weight of the heaviest ones is $0.1166 S$ which is not possible.

So $N$ must be $10$.

Answer 2

If I follow the steps you tried, I get $N>8.57$ and $N<10.8$. My answer would be N can be both 9 and 10.

Answer 3

For another approach: The three heaviest stones weigh $\frac{7}{20} S$. The three lightest stones weigh $\frac{5}{20} S$. The $N-6$ other stones weigh $\frac{8}{20} S$. The averages of these three groups must be in order:

$$ \frac{5 S}{3 \cdot 20} < \frac{8 S}{(N-6)20} < \frac{7 S}{3 \cdot 20} $$ $$ \frac{24}{5} > N-6 > \frac{24}{7} $$

There's only one integer in that range. $N=10$.