Maths Problem of the Week

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

Given that

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

-/|\-

-oOo-

19/01/2021What is $\sqrt{44100}$ ?

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

Tip: you don't need a calculator.

-/|\-

This is a problem about factors and noticing that

$44100=4410×10$
$44100=2×{3}^{2}×5×{7}^{2}×2×5$
$44100={2}^{2}×{3}^{2}×{5}^{2}×{7}^{2}$

So ...

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

$\sqrt{44100}=210$

-oOo-

12/01/2021Was 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?

-/|\-

-oOo-

Find the value of $x$

-/|\-

Solution

Using the intersecting chords theorem states

$|\stackrel{⇀}{\mathrm{AX}}||\stackrel{⇀}{\mathrm{BX}}|=|\stackrel{⇀}{\mathrm{CX}}||\stackrel{⇀}{\mathrm{DX}}|$

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

$\left(2x+5\right)x=3$

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

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

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

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

$x=-\frac{5}{4}±\sqrt{\frac{49}{16}}$

$x=-\frac{5}{4}±\frac{7}{4}$

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

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

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

-oOo-

29/12/2020The 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-

22/12/2020Here'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 commonest 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 IN 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 Maps to ... Real 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 $\alpha$ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ ↓ 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.

-oOo-

15/12/2020How 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-

08/12/2020Sketching a Hill

Sketch the curve

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

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

-oOo-

-oOo-

01/12/2020Ratios 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-

24/11/2020Rupees

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-

17/11/2020Is 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,

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

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

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

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

Using the determinant on this quadratic,

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

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

-oOo-

10/11/20All you need to know is three things about a triangle: part II (it turns out to be untrue)

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

-oOo-

-oOo-

03/11/2020Prove

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

-oOo-

Solution

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

then $10x=9.999\dots$

$10x-x=9.999\dots -0.999\dots$

$9x=9$

$x=1$

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?

20/10/2020Flowers

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

 Number of Flowers Probability 1 2 3 4 5 0.12 0.24 0.23 0.2 ?

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,

$NP\left(n\ge 3\right)$

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

$N=96$ from the question;

and

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

$P\left(n\ge 3\right)=0.64$

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

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-

13/10/2020What 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 $P\left(\mathrm{colour}=\mathrm{red}\right)=0.2$,
but we also know that $P\left(\mathrm{colour}=\mathrm{red}\right)=\frac{r}{n}$,
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-

06/10/2020Time 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.,

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

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

The $V$'s cancel,

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

Rearranging,

$t=\frac{5×20}{5+20}$

Giving,

$t=4\mathrm{hours}$

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-

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-

08/09/2020Find 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 $\left({U}_{n+1}-{U}_{n}\right)$ 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$\left({U}_{n+1}-{U}_{n}\right)$ 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$\left({U}_{n+1}-{U}_{n}\right)$ 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$\left({U}_{n+1}-{U}_{n}\right)$ 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-

01/09/2020Find $\mathrm{\angle ABC}$

$A$, $C$ 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;
$\angle \mathrm{ADC}=115°$.

-oOo-

-oOo-

25/07/2020Solve 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 and $0.
So $x\in \left\{1,4\right\}$ and $y\in \left\{1,4,9\right\}$.
Therefore $x=4$ and $y=9$
Are there any other solutions?

-/|\-

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