• M0

    In the answer to Day 6, challenge explanation 3 of 4, it says

    If the big ball had diameter 12, then its volume would 12/9 all squared times the volume of the smaller ball,"

    Is that a mistake and should it be cubed?

    Also, is it possible to show how you could work this problem out not by using the answers in the multiple choice to work backwards so to speak, but to take the problem and work out the answer directly (ie work out the diameter of the bigger ball directly rather than working backwards and taking the multiple choice answer and see if it works)? Hope the question makes sense but finding this module a bit difficult so I want to put more effort into trying to understand it.
    Thank you

  • MOD

    Hi @ingeniousosprey!

    Wow, thank you! You are right: there should be a cubed instead of squared. You are very attentive, well done! Keep it up. 🙂

    It is possible to solve this problem without trying the multiple-choice answers, but you will need to use a cube root and volume formula to do it.
    Here: $$3=\frac{V_2}{V_1}=\frac{(^4/_3)\times\pi\times (r_2)^3}{(^4/_3)\times\pi\times (r_1)^3}=\frac{(r_2)^3}{(4.5)^3}$$ $$(r_2)^3 =3\times (4.5)^3$$ $$r_2 = \sqrt[\bf 3]{3\times (4.5)^3} = 4.5\times\sqrt[\bf 3]{3}$$ $$d_2 = 2\times r_2=9\times\sqrt[\bf 3]{3}={\color{blue}12.980...}\approx \boxed{\color{blue}13}.$$