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I found the following problem (#2 of SIMOC 2015 Sample Papers – Grade 10(Secondary 4)).

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The answer key is $V \gt 18 \pi$.

In order to get the said answer, I can use brute force by setting up $$\dfrac {V}{(4/6)\pi 10^3} \gt (\dfrac {3}{10})^3$$

1) Can anyone tell me why the above (using ratio of volumes of “similar” objects) work?

2) Would it be a bit unfair to the Secondary 4 students because the relation could have been coming from either $V \lt \pi (10^2 – 7^2) \times 3$ or $\pi (10^2 – 7^2) \times 7 \lt (4/6)\pi 10^3 – V$ or some other comparisons with the respective cylindrical objects. Maybe eventually the students have found that none of the comparisons meet the given options, but time has been wasted. Afterall, this is just a MC question.

Mick
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  • Did you mistype your equation? $3^3(6/4)\sim 40$ While this is greater than $18$, it is substantially so and I'm not sure where the $6$ is coming from. – Stella Biderman Nov 09 '17 at 18:22
  • @StellaBiderman 6 is from $(\dfrac 12)(\dfrac 43)$. The whole denominator is half the volume of a hemi-sphere. Sorry for the late reply. – Mick Jan 08 '18 at 03:06

2 Answers2

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Consider hemisphere of radius $3$. The volume of such a hemisphere is $\frac{1}{2}\frac{4}{3}(3)^3\pi = 18\pi$. This hemisphere has volume smaller than the volume of the water, and so $V>18\pi$.

It doesn't sound outlandish to me to pose this question to fourth graders, especially fourth graders doing competition mathematics. The problem is computationally straightforward, requires almost no reasoning, and the only requires knowing the volume of a sphere. I fully expect clever 10-year olds to be able to solve this, especially with practice.

I think your equation is supposed to read: $$\frac{\text{volume of smaller}}{\text{volume of larger}}=\dfrac {V}{(4/3)\pi 10^3} \gt (\dfrac {3}{10})^3 = \text{(scale factor)}^3$$

Notice that the water is not similar to the whole shape, because the water is not hemispherical. But, as my explanation shows, $V$ is greater than the volume of the hemisphere of radius $3$ that would be similar to the hemisphere. When converting between the volumes of similar shapes, you need to multiply by the scale factor cubed, so if $V$ was the volume of the similar hemisphere then the above inequality would be an equality. Since $V$ is in fact larger, it is in inequality with the left hand side being larger. That's why the inequality holds. Solving for $V$ gives $V>18\pi$ as desired. I’m not sure why you describe this as “brute force.”

Stella Biderman
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  • In order get the suggested answer, I made up that inequality without any supporting reason. That is why I called my approach as brute force. 2. I totally agree with your solution given in 1st paragraph. The question seems (according to the options) to ask for inequalities and there are many ways to generate them … your solution is one and my proposed methods (comparing the volumes with corresponding cylindrical objects) are also feasible approaches. This (MC) question is then a bit unfair to a S4 student because he/she has to spend time before getting the right choice.
    • – Mick Nov 10 '17 at 03:14
  • @Mick Does my explanation of how to use a similar inequality to answer the question make sense? I don’t know what people who are 10 or 11 years old are taught in your country, but I think it’s very reasonable to give people multiple choice questions where the answer requires a little thought, especially when thinking about it cleverly can give you the answer in seconds (it took me 30s to solve the problem). But I’ve heard that how little kids are taught wildly varies from country to country. – Stella Biderman Nov 10 '17 at 04:00