# Chapter 36 Assorted problems

Solutions to problems which have yet to be categorised.

## 36.0.1 Cones problem

Problem: Two similar cones, $X$ and $Y$, have surface areas $270\text{cm}^{2}$ and $120\text{cm}^{2}$ respectively.

The volume of cone $X$ is $1215\text{cm}^{3}$.

Show that the volume of cone Y is $360\text{cm}^{3}$.

Solution:

We have two cones, let’s let $X$ be this one, with radius $r$ and height $h$

Then we can draw $Y$ (which is smaller than $X$, but not drawn to scale here). As they are similar cones, the radius of $Y$ is equal to $ar$ (where $a$ is the scale factor between the two cones) and the height of $Y$ is equal to $ah$

First, the volume of a cone is $\frac{\pi r^{2}h}{3}$ and the surface area of a cone is $\pi r(r+\sqrt{h^{2}+r^{2}})$. These can be derived using integration (but not in Single Maths A Level at least).

Using this, we can write this equation relating the radius and height of $X$ to the value given in the question

We can also do the same for $Y$

This can be simplified a bit by factoring the $a^{2}$ from the square rooted part of the equation, giving

and thus that

We can solve this for $a$ by dividing the two equations, giving that

and thus, after cancelling, that

which means that

Given this, we can then turn to the volume formula.

We know that the volume of $X$ is

and the formula for the volume of $Y$ is equal to

Thus, by multiplying Equation 36.8 through by $a^{3}$, we have

The left-hand side is just the volume of $Y$, and the right-hand side is just

$\displaystyle a^{3}1215$ | $\displaystyle=\left(\frac{2}{3}\right)^{3}\cdot 1215$ | (36.11) | ||

$\displaystyle=360$ | (36.12) |

Which is what we needed to show.

## 36.0.2 Divisibility of $abc_{10}$

###### Example 36.0.1

\sayWhen we represent 789 in decimal, we mean 7 hundreds plus 8 tens plus 9 units. If $abc_{10}$ is taken to represent a decimal number (not $a$ times $b$ times $c$ as in algebra), we mean the value $100a+10b+c$. Show that if $a+b+c$ is divisible by 9, then $abc_{10}$ is divisible by 9.

Solution: Anything that is divisible by 9 can be written as $9\cdot\text{something}$. As this is algebra, we can use a letter (e.g. $m$) to represent \saysomething. As we are assuming that $a+b+c$ is divisible by 9, we can write that

Where $m$ is some number (which depends on the value of $a$, $b$ and $c$). Then we can consider $abc_{10}$. First, we can write this (as given in the question) as

To show that this is divisible by $9$ we need to write it as $9\cdot\text{something}$. We need to apply the fact that $a+b+c=9m$ here (because we are trying to show that, given this assumption, $abc_{10}$ is divisible by $9$). We want to somehow extract an $a+b+c$, here. Currently, we have a $c$, which is great (because it’s what we’re after), and all we need is a $b$ and a $c$, which we can get by breaking $100a$ into $99a+a$ and $10b$ into $9b+b$. When we apply this to the expression as a whole, our desired result pops out fairly quickly.

$\displaystyle abc_{10}$ | $\displaystyle=100a+10b+c$ | (36.15) | ||

$\displaystyle=99a+a+9b+b+c$ | (36.16) | |||

$\displaystyle=9(11a+b)+a+b+c$ | (36.17) | |||

$\displaystyle=9(11a+b)+9m$ | (36.18) | |||

$\displaystyle=9(11a+b+m)$ | (36.19) |

Therefore, if $a+b+c=9m$ (i.e. $a+b+c$ is divisible by $9$) $abc_{10}$ is also divisible by $9$.

## 36.0.3 Rationalising

###### Example 36.0.2

Show that $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$, and hence rationalise the denominator of

Solution

The first part is just algebraic manipulation, i.e.

$\displaystyle(a-b)(a^{2}+ab+b^{2})$ | $\displaystyle=a^{2}+a^{2}b+ab^{2}-ba^{2}-ab^{2}-b^{3}$ | (36.21) | ||

$\displaystyle=a^{2}+(a^{2}b-ab^{2})+(ab^{2}-ab^{2})-b^{3}$ | (36.22) | |||

$\displaystyle=a^{3}-b^{3}$ | (36.23) |

For the second part, we want to apply this identity to the expression to rationalise. We have an expression in the form

And using the identity (Equation 36.20), we know that

Which implies (using the properties of reciprocals and fractions^{1}^{1}
1
See Section 4.1.1 if you’re unsure about
this) that

$\displaystyle\frac{1}{a-b}$ | $\displaystyle=\frac{a^{2}+ab+b^{2}}{a^{3}-b^{3}}$ | (36.26) |

Given that our $a=\sqrt[3]{3}$ and $b=\sqrt[3]{2}$, this is exactly what we want! As $a^{3}=3$ (which is rational) and $b^{3}=2$ which is rational, $a^{3}-b^{3}$ is also rational (specifically $a-b=1$). Therefore, we have that