(Please note some solutions are dependent on images and are omitted here. If you need a

particular solution please e-mail).

**Volume ****7 (issues 19, 20 and 21)**

**CIRCA 19: Solutions**

**Front cover/The Pharaoh’s obelisk**

1/2 + 1/3 + ? = 1, 3/6 + 2/6 + 1/6 = 1 1/6 = 5 cubits.

5 cubits x 6 = 30 cubits. The Pharaoh’s obelisk is 30 cubits tall.

You may interpret the question as requiring the height of the obelisk above the surface of the

water, in which case the answer is 15 cubits.

**Pages 2 and 3/Finding fractions**

Sharing apples

The two apples could be cut into halves (4 pieces), one of which is then divided into three pieces

(sixth of an apple). Each receives 1/2 + 1/6

Sharing chocolate

1/2 + 1/4 would seem to be the best division. Others are possible.

This problem can be extended to how best to share 3 chocolate bars between 5 people and so on.

Fraction parade

The characters with false ID’s are (b) and (d).

(b)’s numerator is greater than her denominator. This makes her statement untrue.

She is an improper fraction.

(d) is not improper. His numerator is neither the same nor greater than his denominator.

(d) is a common or vulgar fraction**.**

**Pages 4 and 5/****Egyptian fractions**

Egyptian numbers

(a) 44500 (b) 1232143

Writing Egyptian fractions

(c) 1/537

Tohoser’s fraction problem

Tohoser’s initial idea for division of the canes would need 35 cuts

yielding 40 pieces. A better division needing less cuts is to halve four of the canes and divide

the remaining cane into eighths. This would need only 11 cuts with each child receiving

1/2 + 1/8.

Liquid fractions

1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 = 63/64

**Page 6 and 7/****Unit fractions**

(a) 1/7 = 1/8 + 1/56

(b) 1/10 = 1/11 + 1/110

(c) 1/4 = 1/5 + 1/20

(d) 1/20 = 1/21 + 1/420

**Pages 8 and 9****/Fraction Mansion Maze**

Requires visual solution. Please e-mail.

**Pages 10 and 11****/Roll up! Roll up!**

The big wheel that gives the ride of longest length is Tenpozan. The London Eye and the

proposed Voyager give the longest ride in time because Tenpozan stops for getting on and off.

particular solution please e-mail).

1/2 + 1/3 + ? = 1, 3/6 + 2/6 + 1/6 = 1 1/6 = 5 cubits.

5 cubits x 6 = 30 cubits. The Pharaoh’s obelisk is 30 cubits tall.

You may interpret the question as requiring the height of the obelisk above the surface of the

water, in which case the answer is 15 cubits.

The two apples could be cut into halves (4 pieces), one of which is then divided into three pieces

(sixth of an apple). Each receives 1/2 + 1/6

Sharing chocolate

1/2 + 1/4 would seem to be the best division. Others are possible.

This problem can be extended to how best to share 3 chocolate bars between 5 people and so on.

Fraction parade

The characters with false ID’s are (b) and (d).

(b)’s numerator is greater than her denominator. This makes her statement untrue.

She is an improper fraction.

(d) is not improper. His numerator is neither the same nor greater than his denominator.

(d) is a common or vulgar fraction

(a) 44500 (b) 1232143

Writing Egyptian fractions

(c) 1/537

Tohoser’s fraction problem

Tohoser’s initial idea for division of the canes would need 35 cuts

yielding 40 pieces. A better division needing less cuts is to halve four of the canes and divide

the remaining cane into eighths. This would need only 11 cuts with each child receiving

1/2 + 1/8.

Liquid fractions

1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 = 63/64

(a) 1/7 = 1/8 + 1/56

(b) 1/10 = 1/11 + 1/110

(c) 1/4 = 1/5 + 1/20

(d) 1/20 = 1/21 + 1/420

Requires visual solution. Please e-mail.

proposed Voyager give the longest ride in time because Tenpozan stops for getting on and off.

**Pages 12 and 13****/Fnd the percentages
**4 out of the sample of 10 are wearing roller skates.

4/10 = 40%. 40% of 2000 is 800 contestants

Guests 1, 2 and 3 all have half a cake in total on their plates already.

Ernest gives guest 4 the two quarters from plates 5 and 6 by swapping

the thirds with quarters.

Climbing pharaohs

A pharaoh takes 5 minutes to climb a tree. Seven pharaohs starting at the same

time would take 5 minutes each to climb a tree. A single pharaoh, assuming

he didn’t tire, would take 35 minutes to climb 7 trees one after the other.

Biggest total

The biggest total is 51.

Zasra’s mind reading trick

This trick involves manipulation of digital routes and some probability. No matter which number

(1 to 10) you choose, you get a multiple of 9. All digital routes of multiples of 9 are 9, so taking

away 5 will always leave 4, and hence, to D.

How many countries can you think of beginning with D? There are two, Denmark (the likely

choice) and Dominion Republic (less likely). Likely animals to come to mind beginning with E

are (in order of popularity of small sample) elephant (18), eel (0), emu (1). The Dominion

Republic was chosen by 1 person (out of 20) and the animal beginning with o was ostrich.

Other animals beginning with o are octopus, orang-utan, ostrich, otter, owl; there are others.

The life of Tut

Tut lived for 72 years. Start by adding all the fractions (baby = 1/36,

child = 1/6, school boy = 1/12, charioteer in the army = 1/18, wise pharaoh = 1/3).

They total 24—36 or 2–3. So, one third of Tut’s life is accounted for by the 24 years

he was a young pharaoh. 3 x 24 = 72.

5 jesters are wearing hats with two bells and 6 jesters are wearing hats with three bells.

This puzzle can be solved by trial and improvement. There is an explanation of how to do it

and a further puzzle on page 15 (see solutions for ‘Are you puzzled?’).

For the sharp-eyed...

1. In the bottom left there is a copy of a magic square (arrangement of consecutive numbers,

usually starting at one, where all the rows, columns and both diagonals sum to the same total)

that can be found on Du¨rer’s engraving of 1514, Melancolia. Some of the numbers are obstructed.

Which ones? What is the total? For further information on magic squares see CIRCA 11.

2. The banners hanging from the balcony provide a puzzle/sorting activity involving shape

on page 11 (see solution The family tree of the Quadrilaterals).

3. There are also various characters to spot who can also be found within the magazine,

e.g. Shelley from page 8, the ‘mathmagician’ from page 6, etc.

The jesters do a turn...

All but one of the jesters is holding the wrong card. This is how it should be:

Greg Grinning should be holding card D,

Joking Jane should be holding card B,

Laughing Laura is holding the correct card (C),

Stan Smile should be holding card E, and

Will Wisecracker should be holding card A.

A = 78° (90° + 12° + 78° = 180°)

B = 35° (90° + 55° + 35° = 180°)

C = 60° (90° + 30° + 60° = 180°)

An explanation of how the trick works is given on page 7.

A similar trick appeared in CIRCA 14 using dominoes.

It is an interesting exercise to work out how the two tricks are connected.

“Choose any domino from a double six set (6/6, 6/5, 6/4 ... 1/3, 1/2, 1/1, 1/0, 0/0).

Don’t tell me which!”

“Pick one of the numbers on the domino. Multiply it by 5”.

“Add 7 and then double the total.”

“Add the number from the other half of the domino.”

“Subtract 14”

“You are thinking of a 2-digit number. The first digit is the number of spots on one half of

the domino you chose. The other digit is the number of spots on the other half.”

(C) is the correct silhouette.

Starboard and port

Needs visual solution. Please e-mail.

Which door?

Door 2 has the biggest number (582). Door 1 was 155 and door 2 was 509.5.

Regular, light or Whatever

Only £1.00 is needed to be sure of getting a light cola, the next time you put a coin in.

Here’s why:

Because the buttons never give what their labels say, if you press the ‘Whatever’ button

and get a regular, you will know that this button will always provide regular cola. Pressing

the ‘Light’ button will therefore give you random colas (i.e. ‘Whatever’) and pressing ‘Regular’

will give you light. Alternatively, if the ‘Whatever’ button gives you a light, you can deduce

that this button will always provide light colas.

Trial and improvement

A knight at the theatre

The 77 pennies bought tickets for 13 adults and 4 children.

Blake postcard/keyring 3.48 euro

Courtney pen/snowscene 6.90 euro

Denise keyring/pen 4.98 euro

From the information in the cartoon we can find out which souvenirs Alice, Blake and Courtney

bought. By taking the amount of the change from the E20 note that Alice paid with we know how

much the group spent in total. Subtracting the spending of the other three gives us the amount

spent by Denise. Only two of the items on the souvenir stall sum to that amount. TOP.

After one second engine 1 on its own would have propelled the spaceship 1/20 of the distance

to Planet Zargo. In the same time engine 2 could have propelled the craft 1/30 of the distance.

Used together they would travel 1/20 + 1/30 of the distance in a second. Therefore the ship

would travel 1/12

( 1/20 + 1/30 = 5/60 = 1/12 ) the distance in a second.

So, it would take the aliens 12 seconds to reach Planet Zargo.

Ratio reckoning

(1) C (2) D (3) A

(4) E (5) F (6) B

The correct ratio of yellow paint to blue is 2:1.

Mixing half a litre of blue with the remaining litre of yellow would give the correct mix.

Half a litre of blue paint will be left.

Measuring the reproduced and reduced version can be slightly inaccurate. Pupils should find

(1) height 91.5mm and span 92mm making both almost equal.

(2) Head 13mm. The head divides into the height approximately 7 times.

(3) Since Leonardo’s time artists have generally accepted an average ratio of about 7:1 of

head lengths to height of grown men and women.

Lisa’s height

The length of Lisa’s head is 208mm. 208mm x 7 = 1456mm –

making Lisa 1.456 metres tall (4ft 9ins).

Changing proportions

Ratios that children find if they measure each other will vary from Vitruvian Man as the proportions

of a child change throughout childhood. A new born child will have short limbs and a relatively large

headsize. As the child gets older head growth slows (brain size growth more-or-less stops at

age 5) but limb growth speeds up.

Purple (4) goes to Zumbos.

Blue (7) goes to Zom.

Green (3) goes to Ziggus.

Yellow (6) goes to Zargo.

Longest and shortest chains

Count all the daisies in the chain including the starting pair but excluding the finishing repeated pair.

(0, 0) will be the shortest = 2 daisies

(0, 5) = 3 daisies

(2, 6) = 4 daisies

(0, 1) = 60 daisies

as does (1,0). 60 is the longest chain.

Starting pairs

There are 100 possible starting pairs.

First daisy has 10 possibilities 1, 2, 3, 4, 5, 6, 7, 8, 9, or 0.

Each of these has 10 possibilities e.g. (1,1), (1,2), (1, 3) and so on. Stress to the children that

they have to count ordered pairs because swapping around the numbers could give different results

[e.g. (2,1) produce a shorter chain than (1,2).] except in the case of repeated digits [e.g. (2,2)].

Other stories... other graphs

Graph Title y-axis

A 2 (iii)

B 1 (ii)

C 3 (i)

(d) is the cube made when the net is folded-up.

Which card is which?

(A) is the Jack of Clubs

(B) is the Queen of Hearts

(C) is the four of diamonds

Cistern problems

Tara and Brenda could consume the cola in 2 minutes.

(In one minute they will drink 1/3 + 1/6)

every other member).

Alice clinks with Louis, Gina, Howie and Rosie A= 4

Louis clinks with Gina, Howie and Rosie L= 3

Gina clinks with Howie and Rosie G= 2

Howie clinks with Rosie H= 1

Rosie doesn’t initiate any ‘clinks’ R = 0

TOP.