### 23/03/2021

What is the perimeter of this rectangle?

-/|\-

### Solution to follow …

-oOo-

### 16/03/2021

Mean of Two Unknown Numbers

If

$$\frac{1}{x}+\frac{1}{y}=10$$and

$$xy=10$$What is the mean of $x$ and $y$ ?

-/|\-

### Solution

The mean $$\stackrel{-}{x}=\frac{x+y}{2}$$

Multiplying both parts of the quotient by $xy$ gives

$$\stackrel{-}{x}=\frac{\left(x+y\right)xy}{2xy}=\frac{1}{2}(\frac{1}{x}+\frac{1}{y})xy$$

Substituting the values given we get
$$\stackrel{-}{x}=\frac{10\times 10}{2}=50$$

-oOo-

### 09/03/2021

Surds and Combined Shapes - enough for even the hungriest mathematician

Given that the area of the collection of congruent rectangles is 480 cm². What are the lengths and widths of the congruent rectangles.

-/|\-

### Solution to follow …

-oOo-

### 02/03/2021

Avoiding the Overfilled Can

A production line fills 9.0 litres of paint into a cylindrical can of diameter 30cm.

What is the maximum (and minimum) height of the paint in the filled can?

-/|\-

### Solution

The paint can is cylindical and the formula for the volume, $V$, of cylinder is
$$V=\pi {r}^{2}h$$
where the radius, $r=15\mathrm{cm}$ and
$V=9.0\mathrm{lt}=9000{\mathrm{cm}}^{3}$.

Which leads to
$$h=\frac{9000}{\pi {\left(15\right)}^{2}}=\frac{10}{\pi}\mathrm{cm}=12.73\mathrm{cm}\text{(to 2 d.p.)}$$
But this question is really about bounding (upper and lower) because it asks for the maximum height of the paint. To get the maximum height we need to recalculate with the upper bound of the volume,
${V}_{\text{upper}}=9.05\mathrm{lt}=9050{\mathrm{cm}}^{3}$
and the lower bound of the radius, ${r}_{\text{lower}}=\frac{29.5}{2}=14.75\mathrm{cm}$.

Using these values we get
$${h}_{\text{upper}}=\frac{9050}{\pi {\left(14.75\right)}^{2}}=\mathrm{13.24081342...}\mathrm{cm}$$
Similarly for
$${h}_{\text{lower}}=\frac{8950}{\pi {\left(15.25\right)}^{2}}=\mathrm{12.24992628...}\mathrm{cm}$$
So $h=12.7\pm 0.5\mathrm{cm}$.

-oOo-

### 16/02/2021

Female Shinty Players

At a school students play one of three sports: Basketball; Cycling; Shinty.

The school has 50% more boys than girls.

Half the girls play shinty.

The same number of boys play basketball as girls play shinty.

There are 120 less female cyclists than male.

Each of the 310 bicycles are used when the boys and girls go cycling.

If a student is picked at random, what is the probability that they are a female shinty player?

-/|\-

### Solution to follow …

-oOo-

### 09/02/2021

Parallelogram or not?

OABC is a parallelogram, where $\stackrel{\rightharpoonup}{\mathrm{OJ}}=\frac{2}{3}\stackrel{\rightharpoonup}{\mathrm{OC}}$,
$\stackrel{\rightharpoonup}{\mathrm{OK}}=\frac{1}{4}\stackrel{\rightharpoonup}{\mathrm{OC}}$ and
$\stackrel{\rightharpoonup}{\mathrm{OL}}=\frac{8}{5}\stackrel{\rightharpoonup}{\mathrm{OA}}$

Show that $\mathrm{JKL}$ is a straight line. (note the figure is not to scale)

-/|\-

### Solution

To solve this one we need to show that $\stackrel{\rightharpoonup}{\mathrm{JK}}$ is parallel to $\stackrel{\rightharpoonup}{\mathrm{JL}}$. If this is true and as they start from the same point they must be on a straight line.

To make the algebra simpler let's make $\stackrel{\rightharpoonup}{\mathrm{OA}}=\mathbf{a}$ and $\stackrel{\rightharpoonup}{\mathrm{OC}}=\mathbf{c}$.

Let's continue with determining $\stackrel{\rightharpoonup}{\mathrm{JK}}$.

We chose $J$ as the common point but we could have chosen any of the three. $K$ would be interesting as we would have to show that $\stackrel{\rightharpoonup}{\mathrm{KJ}}$ and $\stackrel{\rightharpoonup}{\mathrm{KL}}$ were antiparallel.

to be continued ...

-oOo-

### 02/02/2021

Mean Rugby Scores

A rugby team has a mean score of 10.5 after eight games. The next game increases the mean to eleven. What was the score in that game?

-/|\-

### Solution

So, we are told that after eight games the mean is 10.5, i.e.,

$10.5=\frac{S}{8}$

where $S$ is the sum of all the scores and is

$S=84$

The ninth game adds an unknown amount to this,

$84+x$

therefore the new mean

$11=\frac{84+x}{9}$

So, $x=15$ the score of the ninth game.

-oOo-

### 26/01/2021

Is this Quadrilateral a Parallelogram?

The diagram show the quadrilateral ABCD (not to scale). Is it a parallelogram?

Given that

$A=3{x}^{2}+5x$,
$B={x}^{2}-x$,
$C=30x+50$ and
$D=11x-20$

-/|\-

### Solution

In a parallelogram, opposite angles are equal, therefore,

$30x+50=3{x}^{2}+5x$

and

$11x-20={x}^{2}-x$.

Simplifying, we get,

$3{x}^{2}-25x-50=0$

and

${x}^{2}-12x+20=0$.

These have solutions,

$x=\{10,-\frac{5}{3}\}$

and

$x=\{10,2\}$

So, if $x=10$ then the quadrilateral is a parallelogram.

-oOo-

### 19/01/2021

What is $\sqrt{44100}$ ?

Given that $4410=2\times {3}^{2}\times 5\times {7}^{2}$,

Tip: you don't need a calculator.

-/|\-

### Solution

This is a problem about factors and noticing that

$44100=4410\times 10$

$44100=2\times {3}^{2}\times 5\times {7}^{2}\times 2\times 5$

$44100={2}^{2}\times {3}^{2}\times {5}^{2}\times {7}^{2}$

So ...

$\sqrt{44100}=\sqrt{{2}^{2}\times {3}^{2}\times {5}^{2}\times {7}^{2}}$

$\sqrt{44100}=2\times 3\times 5\times 7$

$\sqrt{44100}=210$

-oOo-

### 12/01/2021

Was I Late?

Each work day, before COVID-19, the probability of my leaving before 0800 was 0.25. Leaving after 0830 had a probability of 0.10

The probability of arriving late for a tutoral was dependent on my time of leaving home.

These probabilities are given in the following table:

Time I left my house | Before 0800 |
Between 0800 - 0830 |
After 0830 |
---|---|---|---|

Probability of being late | 0.01 | 0.09 | 0.9 |

I tutored 220 days in 2018. How many times was I late?

-/|\-

### Solution to follow …

The solution is this is best found via a tree diagram.

$$P\left(\mathrm{late}\right)=0.25\times 0.01+0.65\times 0.09+0.10\times 0.90$$
$$P\left(\mathrm{late}\right)=0.151$$

So the number of times I was late equals $0.151\times 220=33.22$

Thirty three times late in a year is not good. I shall have to improve.

-oOo-

### 05/01/2021

Don't Panic!

Quadratics and Circle Theorems

Find the value of $x$

-/|\-

### Solution

Using the intersecting chords theorem states

$\left|\stackrel{\rightharpoonup}{\mathrm{AX}}\right|\left|\stackrel{\rightharpoonup}{\mathrm{BX}}\right|=\left|\stackrel{\rightharpoonup}{\mathrm{CX}}\right|\left|\stackrel{\rightharpoonup}{\mathrm{DX}}\right|$

where $A$, $B$, $C$ and $D$ are the point where the chords meet the circle and $X$ is where the chords cross.

From the diagram this simplifies to

$(2x+5)x=3$

$2{x}^{2}+5x-3=0$

${x}^{2}+\frac{5}{2}x-\frac{3}{2}=0$

${(x+\frac{5}{4})}^{2}-\frac{25}{16}=\frac{3}{2}$

$x+\frac{5}{4}=\pm \sqrt{\frac{25}{16}+\frac{24}{16}}$

$x=-\frac{5}{4}\pm \sqrt{\frac{49}{16}}$

$x=-\frac{5}{4}\pm \frac{7}{4}$

Giving $x=\frac{1}{2}$ or $x=-3$.

However, the context implies than $x>0$,

so, $x=\frac{1}{2}$.

-oOo-

### 29/12/2020

The New Year Problem

Find $x$.

${\left(2019\right)}^{0}+{\left(2020\right)}^{1}-{\left(2021\right)}^{x}=0$

-/|\-

### Solution

An easy one to see the new year in with.

The first term, ${\left(2019\right)}^{0}=1$ as any number raised to the zeroth power is one (if it's not zero), and

the second term, ${\left(2020\right)}^{1}=2020$ as any number raised to the power of one is that number.

So, the first two terms sum to

${\left(2019\right)}^{0}+{\left(2020\right)}^{1}=1+2020=2021$

Now our equation simplifies to $2021-{\left(2021\right)}^{x}=0$,

and further to ${\left(2021\right)}^{x}=2021$

Hence, $x=1$.

-oOo-

### 22/12/2020

Here's a Festive Message, but wait! It's Encrypted?

HZMMT XCMDNOHVN VIY V CVKKT IZR TZVM

-oOo-

### Solution

Decoding encrypted messages is part mathematics and part investigation.

There is a clue in the title of this problem: Festive.

This might inspire us to make a good guess as to the meaning of the message.

Continuing the investigation, we might conclude that this message contains seven words and that the forth is either an **A** or an **I**. These being the most common single letter words.

There are two three letter words. However, that doesn't narrow it down too much as there are 17576
$\left({26}^{3}\right)$combinations of three letters.

This however coupled with the idea that **V** is either **A** or **I**, we might conclude that **VIY** is **AND**.

Let's now see if we can make sense of this with some mathematics.

So far, we have:

V | ⇀ | A |

Y | ⇀ | D |

I | ⇀ | N |

Look at **I** ⇀ **N** we've moved five letters on from **I** to get **N**.

So let's try this idea with **V**. By moving five letters on from **V** we go one beyond the end of the alphabet.

By thinking of the alphabet as a continuous line of alphabets, one after the other, the next letter would be **A** which is correct.

So, it seems that we move five letters up the alphabet to get the real letter. If we drop of the end of the alphabet, we just wrap around to the beginning again.

Now we have a method (an algorithm) that we think might work, what does it decode the message to?

MERRY CHRISTMAS AND A HAPPY NEW YEAR

Definitely a festive message.

Here's the full decoding algorithm.

Encrypted Message Character | V | W | X | Y | Z | A | B | C | D | E | F | G | H | I | J | K | L | M | N | O | P | Q | R | S | T | U | $\alpha $ |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|

Maps to ... | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ | ↓ |

Real Message Character | A | B | C | D | E | F | G | H | I | J | K | L | M | N | O | P | Q | R | S | T | U | V | W | X | Y | Z | $\beta =\alpha +5$ |

The last column is the algorithm that gives the entire algorithm in one easy equation. In this case it is extremely easy and so not difficult to break.

In general, the more complex this algorithm is the more difficult it is to break the code,

but we must be careful not to make the algorithm too complex or it'll be too time consuming to for the intended reader to get the message.

Coding is also easy as it's the inverse action of the decoding algorithm.

Have some fun making your own code or see encryption to learn more.

-oOo-

### 15/12/2020

How well did Jean do?

Jean's mean score is 22% less than her score in biology,

her biology score was 10% higher than her chemistry,

her English was the same as her mathematics result,

and physics was half the mathematics score and 30% points lower than the mean.

What was Jean's score in each subject and her mean score?

-oOo-

### Solution

Let's work each statement in the question one after the other.

"Jean's mean score is 22% less than her score in biology" implies

$M=b-22$,

where $M$ is her mean score and $b$ is her score in biology.

"her biology score was 10% higher than her chemistry" implies

$b=c+10$,

where $c$ is her chemistry score.

"her English was the same as her mathematics result" implies

$e=m$,

where $e$ is english and $m$ is maths.

"and physics was half the mathematics score and 30% points lower than the mean" gived us two mathematical statements

$p=\frac{m}{2}$ and $p=M-30$,

where $p$ is her physics score.

Now, let's consider the mean score ...

To be continued...

-oOo-

### 08/12/2020

Sketching a Hill

Sketch the curve

$y=4x-3-{x}^{2}$.

Find all the intersections and the co-ordinates of the vertex.

-oOo-

### Solution to follow …

-oOo-

### 01/12/2020

Ratios within Ratios

If the ratio $a:b$ is $2:5$ and the ratio $b:c$ is $3:10$. What is the ratio $a:c$?

-oOo-

### Solution

The $b$ part of each ratio has a LCM of 15. Multiplying each ratio so that the $b$ part is 15, we get $6:15$ for the ratio $a:b$ and $15:50$ for the ratio $b:c$. So, the ratio $a:c$ is $6:50$.

-oOo-

### 24/11/2020

Rupees

In how many ways can 16 rupees be divided among beggars so that no beggar receives less than any other?

-oOo-

### Solution

The question doesn't say how many of the poor we are helping, so we could be giving 16 rupees to one beggar, i.e., 16 rupees x 1 beggar.

There might be two beggars, so 8 rupees x 2 beggars.

This is looking like a factors question.

The factors of 16 are 1,2,4,8,16.

So, the possibilities are:

16 rupees x 1 beggar;

8 rupees x 2 beggars;

4 rupees x 4 beggars;

2 rupees x 8 beggars;

1 rupee x 16 beggars.

So there are five ways we could divide the rupees.

-oOo-

### 17/11/2020

Is it a tangent to the circle?

Show that the straight line with equation $4x+3y=25$ is a tangent to the circle with equation ${x}^{2}+{y}^{2}=25$.

-oOo-

### Solution.

There are two ways in which a circle and a line might intersect.

They might also not intersect at all. A circle and a line might intersect once or twice.

A singular intersection can only occur if the line is a tangent, so for this problem we need only show that there is one intersection.

The equation for the line can be re-written as

$y=\frac{25-4x}{3}$

Substituting this for y in the equation of the circle,

${x}^{2}+{\left(\frac{25-4x}{3}\right)}^{2}=25$

Squaring out the bracket and multiplying by nine,

$9{x}^{2}+{(25-4x)}^{2}=225$

$9{x}^{2}+625-200x+16{x}^{2}=225$

$25{x}^{2}-200x+400=0$

${x}^{2}-8x+16=0$

Using the determinant on this quadratic,

${b}^{2}-4ac={8}^{2}-4\times 1\times \mathrm{16}=0$

This shows there is only one intersection, so the line is a tangent to the circle.

-oOo-

### 10/11/20

All you need to know is three things about a triangle: part II (it turns out to be untrue)

Triangle ABC has lengths $\mathrm{AB}=\mathrm{10\; cm}$, $\mathrm{BC}=\mathrm{6\; cm}$ and $\mathrm{\angle BAC}=\mathrm{30\xb0}$. Find the possible areas of this triangle.

-oOo-

### Solution to follow ...

-oOo-

### 03/11/2020

Prove

$0.\stackrel{.}{9}=1$?

-oOo-

### Solution

Let $x=0.\stackrel{.}{9}=\mathrm{0.999\dots}$

then $10x=\mathrm{9.999\dots}$

$10x-x=\mathrm{9.999\dots}-\mathrm{0.999\dots}$

$9x=9$

∴$x=1$

Q.E.D.

-oOo-

27/10/2020

All you need to know is three things about a triangle - part I?

A triangle has two angles of 60° and 90°, and an area of 10cm².

What is the perimeter?

Solution to follow ...

### 20/10/2020

Flowers

A species of plant can have one to five flowers. The probability of each number of flowers is given by:

Number |
1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|

Probability | 0.12 | 0.24 | 0.23 | 0.20 | ? |

Peter plants 96. How many of Peter's plants will have three, or more, flowers?

-oOo-

### Solution

Firstly, our table is missing the probability for five flowers. This can be found by knowing that the probabilities of all the outcomes sum to one.

$0.12+0.24+0.23+0.20+P\left(\text{5 flowers}\right)=1$

$P\left(\text{5 flowers}\right)=1-0.79$

∴ $P\left(\text{5 flowers}\right)=0.21$

The question asks for the number of plants with three, or more, flowers, i.e., the number of planted times the probability of three or more flowers,

$NP(n\ge 3)$

where $N$ is the number planted and $n$ is the number of flowers.

$N=96$ from the question;

and

$P(n\ge 3)=P(n=3)+P(n=4)+P(n=5)$

$P(n\ge 3)=0.64$

So, the number of plants with three, or more, flowers is $96\times 0.64=61.44$

This is a rather unusual count for the number of plants, because this is the mathematical answer. We now have to think of this number in the context of the problem. Rounding to the nearest integer we get 61.

-oOo-

### 13/10/2020

What is the smallest number of beans?

A bag contains some red, some yellow and some blue beans.

The probability of picking a red bean at random is 0.2.

It is known that the ratio of yellow to blue beans is $5:7$.

-oOo-

### Solution

From the questions we know $P(\mathrm{colour}=\mathrm{red})=0.2$,

but we also know that $P(\mathrm{colour}=\mathrm{red})=\frac{r}{n}$,

where $r$ is the number of red beans and $n$ is the total number of beans.

So $r=\frac{n}{5}$ (1).

We also know that $r+y+b=n$ (2),

where $y$ is the number of yellow beads and $b$ is the number of blues.

From the ratio given $\frac{y}{b}=\frac{5}{7}$,

so $y=\frac{5b}{7}$ (3).

Substituting equations (1) and (3) into (2) we get $\frac{n}{5}+\frac{5b}{7}+b=n$.

$b=\frac{7n}{15}$ (4),

and $y=\frac{n}{3}$ (5).

Using (1), (4) and (5) to create a ratio,

$\frac{n}{5}:\frac{n}{3}:\frac{7n}{15}$

$3:5:7$.

So, the smallest number of beans is $3+5+7=15$.

-oOo-

### 06/10/2020

Time to Fill a Pool

The pool can be filled by a big pump in 5 hours and by a small pump on 20 hours.

How long will both pumps take to fill the pool?

-oOo-

### Solution

This is a question about rates.

In this case the volume of water pumped pre hour.

For the big pump this is $\frac{V}{5}$ and $\frac{V}{20}$ for the little pump, where $V$ is the volume of the pool.

If both pumps are used together, then these rates are added to give the rate at which the pool is filled, i.e.,

$\frac{V}{\mathrm{t}}=\frac{V}{5}+\frac{V}{20}$,

where $t$ is the time to fill the pool with both pumps.

The $V$'s cancel,

$\frac{1}{\mathrm{t}}=\frac{1}{5}+\frac{1}{20}$.

Rearranging,

$t=\frac{5\times 20}{5+20}$

Giving,

$t=4\mathrm{hours}$

Is this answer reasonable?

Well, it took 5 hours to fill with the big pump, so if we add another no matter how slow it is it'll help fill the pool quicker.

Our calculated time is quicker, so we should be happy with that.

-oOo-

### 15/09/2020

Electronic Gadgets Everywhere

In a class of 24 students, 22 have a mobile phone and 10 have a portable DVD player. One student has neither.

How many students have:

a) a mobile phone only;

b) a portable DVD player only;

c) both?

-oOo-

### Solution

From the numbers of students given in the question we can see that some students definitely have more than one device; 23 students own devices and there 32 devices. Therefore, nine must own both types of device (that answers part c), 13 must own only a mobile phone (part a) and one has just a DVD player (part b).

Notice in this question I didn't answer the questions in order. Perfectly valid and sometimes a good technique to overcome difficulties with a question.

-oOo-

### 08/09/2020

Find the next three terms in this sequence

14, 24, 36, 50, 66, ..., ..., ...,

-oOo-

### Solution

As with any sequence we are given first let's look to see if it is not one we know. It's not one I know, so let's find the first difference between adjacent terms.

$n$ | 1 | 2 | 3 | 4 | 5 | |||||

${n}^{\mathrm{th}}$
term $\left({U}_{n}\right)$ |
14 | 24 | 36 | 50 | 66 | |||||

First difference
$({U}_{n+1}-{U}_{n})$ |
10 | 12 | 14 | 16 |

This first difference is an arithmetic sequence with a common difference of 2, which enables us to work out the next three first differences:

$n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | ||||||||

${n}^{\mathrm{th}}$ term $\left({U}_{n}\right)$ |
14 | 24 | 36 | 50 | 66 | ? | ? | ? | ||||||||

First difference $({U}_{n+1}-{U}_{n})$ |
10 | 12 | 14 | 16 | 18 | 20 | 22 |

Adding these first differences in term to the previous term in the sequence we get:

$n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | ||||||||

${n}^{\mathrm{th}}$ term $\left({U}_{n}\right)$ | 14 | 24 | 36 | 50 | 66 | 84 | 104 | 126 | ||||||||

First difference $({U}_{n+1}-{U}_{n})$ |
10 | 12 | 14 | 16 | 18 | 20 | 22 |

So the sequence is 14, 24, 36, 50, 66, 84, 104, 126,…

We can go further and find any term in the sequence. The first difference is an arithmetic sequence with a common difference of 2, implying that the original sequence is a quadratic sequence of the form ${U}_{n}=a{n}^{2}+bn+c$, where a is the common difference divided by 2, ∴ $a=1$.

$n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | ||||||||

${n}^{\mathrm{th}}$ term $\left({U}_{n}\right)$ |
14 | 24 | 36 | 50 | 66 | 84 | 104 | 126 | ||||||||

First difference $({U}_{n+1}-{U}_{n})$ |
10 | 12 | 14 | 16 | 18 | 20 | 22 | |||||||||

$a{n}^{2}$ | 1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | ||||||||

${U}_{n}-a{n}^{2}$ | 13 | 20 | 27 | 34 | 41 | 48 | 55 | 62 |

But ${U}_{n}-a{n}^{2}=bn+c$, so the last line is an arithmetic sequence with common difference $b=7$ and zeroth term $c=6$. Therefore, the ${n}^{\mathrm{th}}$ term of the sequence is given by ${U}_{n}={n}^{2}+7n+6$.

-oOo-

### 01/09/2020

Find $\mathrm{\angle ABC}$

$A$, $C$ and $D$ are point on a circle centred $O$;

$\mathrm{AB}$ and $\mathrm{CB}$ are tangents to the circle;

$\mathrm{AD}$ and $\mathrm{CD}$ are chords in the circle;

$\angle \mathrm{ADC}=\mathrm{115\xb0}$.

-oOo-

### Solution to follow …

-oOo-

### 25/07/2020

Solve Simultaneously

$\sqrt{x}+y=11$

$x+\sqrt{y}=7$

-oOo-

### Solution:

The equations imply that $x,y>0$.

Both equations have integers of the RHS

implying $x,y,\sqrt{x},\sqrt{y}$ are elements of the integers

moreover $x,y$ are perfect squares

so $0<x<7$ and $0<y<11$.

So $x\in \{1,4\}$ and $y\in \{1,4,9\}$.

Therefore $x=4$ and $y=9$

Are there any other solutions?

-/|\-