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Irish Times Puzzles 2013

Explanations of Irish Times Maths Week Puzzles

Irish Times Puzzle number 6

Saturday 19th October

Budget Shorts

With another austerity budget it is more important than ever to manage your money carefully. Test your money-maths with these five short questions.

(a) You pay €145.20 for an item and this price includes 21% VAT. How much VAT are you paying?

(b) Passing a window you see an item you would like priced at €100. The following week you see that the item has been marked up by 20%. A week later there is a sale and you see that all items are reduced by 20% on the marked prices. When was the best time to buy the item?

(c )If 3 coffees and a 2 muffins costs €9 and 2 coffees and 4 muffins costs €10, how much is a coffee and how much is a muffin?

(d) A coffee and a biscuit costs €2:50. If the coffee costs €2 more than the biscuit how

much do each cost?

(e) A 250ml bottle of shampoo costs €4.75. The same brand in a 375ml bottle costs €6:99. Which is the best value?


(a)   €25.20

To find out what the price was before VAT was added at 21%:

consider that €145.20 is 121% of that price. To find out the ex-VAT price divide 145.20 by 1.21

145.20/1.21 =120

If the original cost is €120 then you have paid €25.20 in VAT

(b)   At the sale. The original price is €100. When it is put up 20% the price becomes €120. The marked price then becomes €120. 20% of €120 is €24. So at the sale it will cost €96

(c)    A coffee costs €2 and a muffin costs €1.50. If 3 coffees and 2 muffins cost €9 then double that order 6 coffees and 4 muffins will cost €18. Imagine we bought that amount.

If we passed on 2 coffees and 4 muffins to someone costing €10 and got €10 from them

we would have 4 coffees left and it would have cost us €8.

Therefore a coffee costs €2 each.

Knowing this, it is simply a matter of saying that if 2 coffees and 4 muffins cost €10 and the two coffees cost €4 then 4 muffins would cost €10 - €4 = €6

If 4 muffins cost €6 then one muffin costs €1.50

This is more usually done by using the technique in algebra for simultaneous equations.

In this method we call the cost of a coffee X and the cost of a muffin Y.

We write our equations thus:

3X + 2Y = 9      equation 1

2X + 4Y = 10    equation 2

To preserve the integrity of an equation, whatever we do to one side we have to do to the other.

If we multiply both sides of equation 1 we get -

6X + 4Y = 18

Subtracting equation 2

6X + 4Y = 18

- 2X + 4Y = 10

= 4X + 0  = 8

Therefore X = 2

And if 2X and 4Y = 10 and X=2 then

4 + 4Y = 10    (subtracting 4 from each side)

4Y = 6

Y = 6/4           (dividing each side by 4)

Y = 1.5

(d)   A coffee costs €2.25 and the biscuit costs 25c.

(e)   The 375 ml bottle is marginally better value (costs 1.86c/ml as opposed to other bottle at 1.9 c/ml)


Irish Times Puzzle number 5

Friday 18th October

Master of the Rolls

I buy twin packs of kitchen towel and on the packet it says that there is 18m of paper towel on each roll. I measured an unused roll and found that the core measured 5 cm across and the whole roll was 15 cm across the diameter. The roll that I had been using in the kitchen measured 12cm across the diameter (with the same size core). What length of paper towel is left on it?

It is in the proportion of the (cross sectional) areas of the paper

Looking down from above I can see the area of paper.

Area of paper on the unused roll is the total area the roll minus the area of the core.

Area of circle diameter 15cm (r = 7.5) = 176.714 cm2

Area of core (r = 2.5) = 19.634 cm2

Area of paper = 157.079 cm2


Area of paper on used roll

= 113.097 – 19.634 = 93.463 cm2


So the proportion left on the partially used roll is 93.463/157.079 = 0.595

0.595of the 18m is left

0.595 x 18 = 10.71m


Irish Times Puzzle number 4

Thursday 17th October

Weighing Up

A shop needs to weigh items (in 1kg multiples) between 1 and 10kg. The money-conscious shopkeeper acquired an old balance scales with two trays but it had no metric weights. Weights he discovered cost €1 each plus €1 per kilogram. He gave his assistant €13 and sent him to buy weights. What should the assistant buy that could enable them to weigh out all required quantities from 1 to 10kg?

Answer 1, 3 and 6kg weights will do the job (or 1,2 & 7).

Firstly the weights will all have to add up to 10kg so that all together they can weigh the maximum amount. All the numbers between 1 and 10 can be made up from adding and subtracting combinations of these numbers.

10kg will cost €10 plus €1 for each weight. So that leaves €3 – therefore maximum number of weights allowed is 3.

With weights 1kg and 3 kg you can weigh 2 kg by putting one weight on each tray

Then it is 3-1 = 2kg

Other combinations get us all the way to 10


Left Balance

Right Balance











3 +1



6 + 1









6 +3






6 + 3 + 1



Irish Times Puzzle number 3

Wednesday 16th October - Hamilton Day

Harry and Bill were driving from Dublin to Waterford. Sean and Mary were driving down also. Harry (in the passenger seat) rang Sean and Mary to find out where they were. Mary also a passenger informed him that they were just passing exit 3 and they had set their cruise control to 100 km/h to economise on fuel. Harry noticed that they were just then passing a sign that said that exit 3 was 16 km away and that Waterford was 110 km.

In Bill sets their cruise control to 120 km/h how far from Waterford would they be when they overtake Sean and Mary?


Ans: 14km


The difference in speeds is 20km/h and they are 16km apart. That means that Bill and Harry are gaining on Sean and Mary at 20km/h. They would gain 20km in one hour. (That means after an hour they would be 4 km ahead.)

They would cover the 16km in 16/20 * 60 minutes and that would be 48 minutes.

So how far would they get in 48 minutes driving at 120 km/h?

They would get 120km in an hour (60 mins).

And in 48 mins they would get-

(48/60)* 120 = 96 km in 48 minutes.

And because they were 110 km from Waterford at the beginning they would be 110 – 96 = 14 km from Waterford when they overtake the others.


Irish Times Puzzle number 2

Tuesday 15 October

Pat and Mary’s ages add up to 76. Pat is now twice as old as Mary was when they got married. Mary is now one and a half times as old as Pat was when they got married.

How long are they married?

Answer is 16 years. It could be done by trial and retrial, but here is how to do it with algebra:

We have three unknowns, we will call Pat’s age X, Mary’s age Y and the time they are married, Z.

Therefore Pat’s age when they were married was X-Z

Mary’s Age when they were married was Y-Z


So what information do we have?

Pat and Mary’s ages add up to 76

X + Y = 76

Which can be rewritten as

X+ Y -76 = 0               we’ll call this     equation 1

Pat’s age is now twice what Mary’s age was when they got married.

We can write this:

X = 2(Y-Z)

X = 2Y – 2Z

X – 2Y + 2Z = 0                                  equation 2


Mary is now one and a half times what Pat’s age was when they got married.

We can write this:

Y = 3/2(X – Z)

2Y = 3(X-Z)

2Y = 3X – 3Z

3X – 2Y -3Z = 0                                  equation 3

We have three separate pieces of information now expressed in three equations. We have three unknowns. Our strategy is to combine equations 2 and 3 to eliminate Z and then we will combine the resultant equation with X and Y with equation 1.

To eliminate Z we must multiply equation 1 by 3 and equation 2 by 2.

3(X – 2Y + 2Z) = 3(0)

3X – 6Y + 6Z = 0                                equation 4


2(3X – 2Y – 3Z) = 2(0)

6X -4Y – 6Z = 0                                  equation 5


Adding equations 4 and 5 –


3X – 6Y + 6Z = 0

6X – 4Y – 6Z = 0

9X -10Y           = 0                              equation 6


Combining equations 6 and 1 we can get X on its own.

Firstly multiply equation 1 by 10 –


10(X + Y - 76) = 0

10X + 10Y -760 = 0


Adding equation 6-


10X + 10Y - 760 = 0

9X – 10Y = 0

19X – 760  = 0


Therefore X = 760/19 = 40


Therefore Mary was 20 when they got married.


And Mary’s age now is

Y = 76 – 40

Y = 36.


If Mary is 36 now and was married at 20 then they are 16 years married.

Irish Times Puzzle Number 1

Monday 14th October

Party Time

Not being very political myself, my friend was surprised when I told him that I had a meeting last night with a number of TDs from the three largest parties for some last minute lobbying about the budget. “How many of them came?” he asked me. Wanting to appear circumspect I replied “All were Fine Gael except two, all were Labour except two and all were Fianna Fail except two”. Can you figure out how many were at the meeting in total?

Answer - Four – Three TDs and me!

Three TDs

All were Fine Gale except 2 -   1 out of 3 was Fine Gael 2 were not

all were Labour except two - 1 was Labour, 2 were not

all were Fianna Fail except two 1 was fianna Fail and 2 were not.

That is three altogether.

This puzzle is a rewording of an "old chestnut" but to add a twist (or sting) to the tail I borrow from "as I was going to St Ives" and place myself in the picture.

This caught out a number of people who recognised the puzzle and who produced the answer from memory. Reading carefully, The "all were ... except" passage refers to the TDs and the question asks how many were at the meeting altogether!