Chapter 36 Assorted problems

Solutions to problems which have yet to be categorised.

36.0.1 Cones problem

Problem: Two similar cones, XX and YY, have surface areas 270cm2270\text{cm}^{2} and 120cm2120\text{cm}^{2} respectively.

The volume of cone XX is 1215cm31215\text{cm}^{3}.

Show that the volume of cone Y is 360cm3360\text{cm}^{3}.


We have two cones, let’s let XX be this one, with radius rr and height hh


Then we can draw YY (which is smaller than XX, but not drawn to scale here). As they are similar cones, the radius of YY is equal to arar (where aa is the scale factor between the two cones) and the height of YY is equal to ahah


First, the volume of a cone is πr2h3\frac{\pi r^{2}h}{3} and the surface area of a cone is πr(r+h2+r2)\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 XX to the value given in the question

πr(r+h2+r2)=270\pi r(r+\sqrt{h^{2}+r^{2}})=270 (36.1)

We can also do the same for YY

πar(ar+(ah)2+(ar)2)=120\pi ar(ar+\sqrt[]{(ah)^{2}+(ar)^{2}})=120 (36.2)

This can be simplified a bit by factoring the a2a^{2} from the square rooted part of the equation, giving

πar(ar+a2(h)2+(r)2)=120\pi ar(ar+\sqrt{a^{2}}\sqrt[]{(h)^{2}+(r)^{2}})=120 (36.3)

and thus that

πa2r(r+(h)2+(r)2)=120\pi a^{2}r(r+\sqrt{(h)^{2}+(r)^{2}})=120 (36.4)

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

πa2r(r+(h)2+(r)2)πr(r+h2+r2)=120270\frac{\pi a^{2}r(r+\sqrt{(h)^{2}+(r)^{2}})}{\pi r(r+\sqrt{h^{2}+r^{2}})}=\frac% {120}{270} (36.5)

and thus, after cancelling, that

a2=94a^{2}=\frac{9}{4} (36.6)

which means that

a=32a=\frac{3}{2} (36.7)

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

We know that the volume of XX is

πr2h=1215\pi r^{2}h=1215 (36.8)

and the formula for the volume of YY is equal to

π(ar)2(ah)=a3πr2h\pi(ar)^{2}(ah)=a^{3}\pi r^{2}h (36.9)

Thus, by multiplying Equation 36.8 through by a3a^{3}, we have

a3πr2h=a31215a^{3}\pi r^{2}h=a^{3}1215 (36.10)

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

a31215\displaystyle a^{3}1215 =(23)31215\displaystyle=\left(\frac{2}{3}\right)^{3}\cdot 1215 (36.11)
=360\displaystyle=360 (36.12)

Which is what we needed to show.

36.0.2 Divisibility of abc10abc_{10}

Example 36.0.1

When we represent 789 in decimal, we mean 7 hundreds plus 8 tens plus 9 units. If abc10abc_{10} is taken to represent a decimal number (not aa times bb times cc as in algebra), we mean the value 100a+10b+c100a+10b+c. Show that if a+b+ca+b+c is divisible by 9, then abc10abc_{10} is divisible by 9.

Solution: Anything that is divisible by 9 can be written as 9something9\cdot\text{something}. As this is algebra, we can use a letter (e.g. mm) to represent \saysomething. As we are assuming that a+b+ca+b+c is divisible by 9, we can write that

a+b+c=9ma+b+c=9m (36.13)

Where mm is some number (which depends on the value of aa, bb and cc). Then we can consider abc10abc_{10}. First, we can write this (as given in the question) as

abc10=100a+10b+cabc_{10}=100a+10b+c (36.14)

To show that this is divisible by 99 we need to write it as 9something9\cdot\text{something}. We need to apply the fact that a+b+c=9ma+b+c=9m here (because we are trying to show that, given this assumption, abc10abc_{10} is divisible by 99). We want to somehow extract an a+b+ca+b+c, here. Currently, we have a cc, which is great (because it’s what we’re after), and all we need is a bb and a cc, which we can get by breaking 100a100a into 99a+a99a+a and 10b10b into 9b+b9b+b. When we apply this to the expression as a whole, our desired result pops out fairly quickly.

abc10\displaystyle abc_{10} =100a+10b+c\displaystyle=100a+10b+c (36.15)
=99a+a+9b+b+c\displaystyle=99a+a+9b+b+c (36.16)
=9(11a+b)+a+b+c\displaystyle=9(11a+b)+a+b+c (36.17)
=9(11a+b)+9m\displaystyle=9(11a+b)+9m (36.18)
=9(11a+b+m)\displaystyle=9(11a+b+m) (36.19)

Therefore, if a+b+c=9ma+b+c=9m (i.e. a+b+ca+b+c is divisible by 99) abc10abc_{10} is also divisible by 99.

36.0.3 Rationalising

Example 36.0.2

Show that a3b3=(ab)(a2+ab+b2)a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2}), and hence rationalise the denominator of

13323\frac{1}{\sqrt[3]{3}-\sqrt[3]{2}} (36.20)


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

(ab)(a2+ab+b2)\displaystyle(a-b)(a^{2}+ab+b^{2}) =a2+a2b+ab2ba2ab2b3\displaystyle=a^{2}+a^{2}b+ab^{2}-ba^{2}-ab^{2}-b^{3} (36.21)
=a2+(a2bab2)+(ab2ab2)b3\displaystyle=a^{2}+(a^{2}b-ab^{2})+(ab^{2}-ab^{2})-b^{3} (36.22)
=a3b3\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

1ab\frac{1}{a-b} (36.24)

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

ab=a3b3a2+ab+b2a-b=\frac{a^{3}-b^{3}}{a^{2}+ab+b^{2}} (36.25)

Which implies (using the properties of reciprocals and fractions11 1 See Section 4.1.1 if you’re unsure about this) that

1ab\displaystyle\frac{1}{a-b} =a2+ab+b2a3b3\displaystyle=\frac{a^{2}+ab+b^{2}}{a^{3}-b^{3}} (36.26)

Given that our a=33a=\sqrt[3]{3} and b=23b=\sqrt[3]{2}, this is exactly what we want! As a3=3a^{3}=3 (which is rational) and b3=2b^{3}=2 which is rational, a3b3a^{3}-b^{3} is also rational (specifically ab=1a-b=1). Therefore, we have that

13323=(33)2+3323+(23)21\frac{1}{\sqrt[3]{3}-\sqrt[3]{2}}=\frac{(\sqrt[3]{3})^{2}+\sqrt[3]{3}\sqrt[3]{% 2}+(\sqrt[3]{2})^{2}}{1} (36.27)