
A boy has twice as much money as his friend. If they have 84p together, how much does each have?

Hint: Simplify the problem, like this: if one boy has 1p then the friend has 2p. That's 3p altogether.Solution: They have 28p and 56p.Explanation: If one boy has twice as much as his friend, this can be thought of as being 3portions. If 3 portions amount to 84p then one portion will be 28p (84 ÷ 3). So one friend has 28p and the other must have 56p (28 × 2).


Along a certain street the lights are placed $30m$ apart. If there are $10$ lights, how far is it from the first to the last?

Hint:
How many gaps are there between $2$ lamp posts?
Solution:$270m$.
Explanation:Between each pair of lamp posts, there is one gap, so between $10$ lamp posts, there are $9$ gaps. $9 × 30m = 270m$.


$2$ icecreams are eaten by $2$ children in $2$ minutes. How long will it take $5$ children to eat $5$ icecreams?

Hint:
How long does it take for a child to eat an icecream (according to this puzzle)?
Solution:$2$ minutes
Explanation:If it takes $2$ minutes for the $2$ children to eat their icecreams, then it takes one of the children $2$ minutes to eat their icecream. So it will still take $2$ minutes if one million children are eating one million icecreams, if we assume that they eat one each, at exactly the same rate.


An electric train is traveling due north. If the wind is blowing from the north west, in which direction is the smoke blowing?

Hint: Read the question! Now read the question again! Now read the question really carefully!Solution: There is no smoke. It is an electric train.Explanation: This rather silly little puzzle shows how important it is to read problems very carefully indeed and not just to skim read the text.


Two farmers are taking their cows to market. On the way, one says to the other, "If you give me one of your cows, we would have the same number of cows." The other farmer replies, "True, but if you give me one of your cows, I will have twice as m any as you.". How many cows did the two farmers each have?

Hint: Think about how far apart the numbers must be. Also, do you think the numbers are odd or even?Solution:
One of the farmers has $5$ cows and the other has $7$ cows.
Explanation: If farmer 1 gives one cow to farmer 2 who then has twice as many as farmer 1, farmer 2 must have an even number of cows after the exchange, so he has an odd number before the exchange. If farmer 2 gives a cow to farmer 1 and they have the same number, then the number of cows each has must be 2 apart. It is reasonably obvious (at least, I think it is obvious) that the numbers must be fairly small, so try odd numbers which are 2 apart to see if they work.


A window frame in Salt's Mill, just outside Bradford in Yorkshire, consists of two equal semicircles and a circle inside a large semicircle with each touching the other three as shown. The width of the frame is $4m$.
What is the radius of the circle in metres?

Hint: 1. What lengths do you know?
2. Give variable names (letters) to lengths you don't know.
3. Remember Pythagoras' Theorem.Solution: $r={\frac{2}{3}}m$Explanation:The radius of the inner semicircles is $1m$.
Let the radius of the inner circle br $r$.
The height (radius) of the outer semicircle is $2$.
Length $AB = (1+r)m$.
Length $OA = \big(2r\big)m$.
Length $OB = 1m$


The diagram shows an annulus, which is the region between two circles with the same centre. Twelve equal touching semicircles are placed inside the annulus. The diameters of the semicircles lie along the diameters of the outer circle.
What fraction of the annulus is shaded?

Hint: 1. Can you identify the angle between each of the semicircles?
2. Try to find the radii of the various circles and semicircles.Solution:$\frac{1}{4}$ of the annulus is shaded.
Explanation:The angle between each semicircle must be $30^{\circ}$ as three identical semicircles make up a right angle ($90^{\circ}$).
Let the radius of the semicircles be $1 \ unit$.
Using trigonometry on triangle $OAB$, we can see that the hypotenuse $OB = 2 \ units$ long.
This tells us that the radius $OC$ of the outer circle is $3 \ units$.
The area of the outer circle is $\pi(3)^2 = 9 \pi$.
The area of the inner circle is $\pi(1)^2 =1 \pi$.
So the area of the annulus is $8\pi$.


Three squares are drawn on the outside of a right angled triangle, whose shorter sides have lengths $a$ and $2a$. The whole figure is surrounded by a rectangle, as shown.
What is the ratio of the area of the shaded region to the area of the outer rectangle?

Hint: 1. What are the areas of the shaded shapes?
2. Can you work out the length and width of the large rectangle?Solution: $5 : 9$Explanation:Find the areas of the two smaller squares. By Pythagoras' theorem, the area of the large square is $(2a)^2+a^2 = 5a^2$.
This also gives us the length of the hypotenuse, which is $a\sqrt{5}$.
Now examine the vertical length of the large rectangle, dividing it up into two triangles and a rectangle, as shown in the diagram. The triangles marked with an angle of $\theta$ are all similar and in each case, the hypotenuse is known.
By Pythagoras' theorem, using the central triangle, we can see that $\tan \theta = \frac{1}{2}$.
From here, using rightangled trigonometry, we can see that $\cos\theta=\frac{2}{\sqrt{5}} \ and \ \sin\theta = \frac{1}{\sqrt{5}}$.
This enables us to find the lengths we require. The vertical length of the large rectangle is $\frac{2a}{\sqrt{5}}+\frac{4a}{\sqrt{5}}\frac{5a}{\sqrt{5}}=\frac{11a}{\sqrt{5}}$ and the horizontal length is $\frac{2a}{\sqrt{5}}+\frac{5a}{\sqrt{5}}\frac{2a}{\sqrt{5}}=\frac{9a}{\sqrt{5}}$.
The area of the large rectangle is $\frac{1a}{\sqrt{5}}\times \frac{9a}{\sqrt{5}}=\frac{99a^2}{5}$.
The shaded area is $4a^2+a^2+a^2+5a^2 = 11a^2$.
So the ratio of the shaded area to the area of the large rectangle is \[\begin{aligned} 11a^2&:\frac{99a^2}{5}\\ 1&:\frac{9}{5}\\ 5&:9 \end{aligned}\]
