Consider the following table of numbers and letters:

20, GC

15, IJ

45, AC

35, CG

70, JC

80, HF

25, EB

60, DG

All numbers in the first column have been increased by the same percentage, *P*, to give the results in the second column.

The resulting numbers are coded; each digit is replaced by letter. A given letter stands for the same digit every time it appears in that column.

What is the value of *P*, the percentage change?

This seems impossible but you can work through the logic to figure it out. Give it a try and then watch the video for a solution.

**Seemingly Impossible 7th Grade Math Problem – The Coded Table**

Or keep reading.

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**Answer To Seemingly Impossible 7th Grade Math Problem – The Coded Table**

Here is how I worked it out.

Since the second column has distinct whole numbers, we know the percentage increase corresponds to a positive fractional increase. So we can search for a common factor in the first column.

As 35 = 5 × 7 and 20 = 5 × 4, the common factor in the numbers is 5, giving the possibilities 1/5, 2/5, 3/5, 4/5, and 5/5. These correspond to percentage increases of 20%, 40%, 60%, 80%, and 100%.

Since 80(1 + 25%) = 100, but the resulting number for 80 has only two digits, the percentage increase has to be less than 25%.

Therefore, the percentage increase is 20%. And like magic we have found the answer!

We should check this does indeed work. For example:

20(1 + 20%) = 24 = GC

35(1 + 20%) = 42 = CG

The rest of the values also work out so each letter corresponds to a different digit.

]]>At the beautiful Three Geysers National Park, a placard explains that its three geysers erupt at precise intervals of time. Geyser A erupts exactly every 2 hours, geyser B erupts exactly every 4 hours, and geyser C erupts exactly every 6 hours.

However, you have no idea how the three eruptions are staggered. Assuming each geyser started erupting independently at a random point in history, what are the probabilities that each (A, B, and C) will be the first to erupt after your arrival?

Can you figure it out? Watch the video for a solution.

**Can You Solve The Three Erupting Geysers Riddle?**

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**Answer To The Three Geysers Probability Puzzle**

(This post was written quickly based on the video–please let me know if there are any typos/errors and I will correct them, thanks).

First I want to explain the incorrect trap answer.

Suppose the geysers erupted randomly, and their average rates of eruptions were every 2 hours (A), every 3 hours (B) and every 6 hours (C). This is akin to how a coin shows heads every 2 flips on average, or a standard dice shows the number 3 (or any number) every 6 rolls on average.

If this were the case, we could derive the answer as follows. Over a 12 hour period, we expect A to erupt 12/2 = 6 times, B to erupt 12/3 = 3 times, and C to erupt 12/6 = 2 times. In other words, the geysers A, B, C, respectively, erupt in proportion of 6:3:2. So the chance any one erupts first is the relative ratio:

Pr(A) = 6/(6 + 3 + 2) = 6/11

Pr(B) = 3/(6 + 3 + 2) = 3/11

Pr(C) = 2/(6 + 3 + 2) = 2/11

This is the correct answer if the geysers erupted on average every 2 hours (A), every 3 hours (B), and every 6 hours (C).

But the problem we want to solve is different. The geysers actually erupt on regular intervals. For example, suppose geyser C erupts at 12am. It then will erupt every 6 hours–12am, 6am, 12pm, 6pm, etc. In the average case, geyser C would erupt at 4 random times in a 24 hour period–say 12am, 4:23am, 11:12am, 5pm, etc.

So what is the proper way to solve the puzzle?

**Solving for regular intervals**

When we arrive, geyser A is equally likely to erupt at any time in the next 2 hours, geyser B in the next 4 hours, and geyser C in the next 6 hours.

Since geyser A erupts for sure in the next 2 hours, let us focus on the next 2 hours. What is the chance each geyser erupts in the next 2 hours? For geyser A there is a 100% = 1 chance, geyser B there is a 2/4 = 1/2 chance, and geyser C there is a 2/6 = 1/3 chance.

As each geyser B and C can either erupt or not in the next 2 hours, there are 4 possibilities (B erupts, C erupts), (B erupts, C does not), (B does not, C erupts), (B does not, C does not).

Along with A erupting for sure, we can enumerate the 4 possible cases for the next 2 hours.

(I) A, B, C all erupt

(II) A, B erupt, C does not

(III) A, C erupt, B does not

(IV) A erupts, B, C do not

In case (I), there is an equal 1/3 chance each is the first to erupt. Case I happens with chance Pr(A and B and C erupt) = (1)(1/2)(1/3) = 1/6.

In case (II), there is an equal 1/2 chance either A or B is first. Case II happens with chance Pr(A and B erupt and C does not) = (1)(1/2)(1 – 1/3) = 1/3.

In case (III), there is an equal 1/2 chance either A or C is first. Case III happens with chance Pr(A and C erupt and B does not) = (1)(1 – 1/2)( 1/3) = 1/6.

In case (IV), there a 100% chance A is first. Case IV happens with chance Pr(A erupts and B and C do not) = (1)(1 – 1/2)(1 – 1/3) = 1/3.

Now we can solve for the probability each geyser is the first to erupt as the weighted average of these cases, where x = I, II, III, IV.

Pr(A erupts first)

= Pr(A erupts first | case x)*Pr(case x)

= (1/3)(1/6) + (1/2)(1/3) + (1/2)(1/6) + (1)(1/3)

= 23/36

Pr(B erupts first)

= Pr(B erupts first | case x)*Pr(case x)

= (1/3)(1/6) + (1/2)(1/3) + (0)(1/6) + (0)(1/3)

= 2/9

Pr(C erupts first)

= Pr(C erupts first | case x)*Pr(case x)

= (1/3)(1/6) + (0)(1/3) + (1/2)(1/6) + (0)(1/3)

= 5/36

**Alternate geometric probability interpretation**

This is a creative approach, but I personally felt it was hard to visualize and calculate the volumes. But it is neat:

Brian Galebach created the problem and gave permission for me to post it in a video. It was previously shared on Oliver Roeder’s Riddler column on 538:

https://fivethirtyeight.com/features/which-geyser-gushes-first/.

Some adults felt this was too challenging for primary school students, as they themselves were confused about how to solve it! Here is the problem.

On the coast there are three lighthouses.

The first light shines for 3 seconds, then is off for 3 seconds.

The second light shines for 4 seconds, then is off for 4 seconds.

The third light shines for 5 seconds, then is off for 5 seconds.

All three lights have just come on together.

1) When is the first time all three lights will be off at the same time?

2) When is the next time all three lights will come on together at the same moment?

Watch the video for a solution.

Adults Baffled By Primary School Math Problem – The Lighthouse Puzzle

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**Answer The Lighthouse Puzzle**

(This was transcribed quickly after I made the video–please let me know if there are any typos/errors and I will correct them, thanks).

For **problem 1**, the third light is off after 5 seconds, and during this time the first and second are also off. All three lights will be off just after 5 seconds.

For **problem 2**, the first light turns on every 6 seconds, the second light turns on every 8 seconds, and the third light turns on every 10 seconds. They will all come on together at the least common multiple of these numbers.

LCM(6, 8, 10) = LCM(2×3, 2×2×2, 2×5) = 2×2×2×3×5 = 120

So the lights all turn on together at 120 seconds.

**Sources**

Mumsnet

https://www.mumsnet.com/Talk/_chat/3245037-Can-someone-help-with-this-KS2-homework

*The Sun*

**This primary school kid’s maths homework has left mums baffled…so can YOU solve it?**

https://www.thesun.co.uk/fabulous/6269845/primary-school-kids-maths-homework-mums-baffled-can-you-solve-it/

Daily Mail.

**‘Ridiculous’ maths problem intended for PRIMARY students leaves parents baffled – and no one can agree on an answer**

http://www.dailymail.co.uk/femail/article-5723087/Ridiculous-lighthouse-maths-problem-intended-PRIMARY-students-leaves-parents-baffled.html

The total amount of sand contained in Boxes A, B and C is 360 grams.

Each box starts with some unknown amount of sand, and a mathematician decides to play with the sand.

First the mathematician pours 1/6 of the sand from Box A into Box B. After that, the mathematician pours 1/3 of the sand in Box B into Box C. Finally, the mathematician pours 1/5 of the sand in Box C into Box A. At the end, each box has an equal amount of sand. How much sand did each box have at the beginning?

The problem was assigned to 10 year olds who have not yet studied algebra, so it is solvable without setting up a system of simultaneous equations.

Can you figure it out? Watch the video for a solution.

**Sand Mixing Puzzle – Math Homework 10 Year Olds In Singapore**

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**Tricky Sand Mixing Puzzle For 10 Year Olds In Singapore**

(This was transcribed quickly after I made the video–please let me know if there are any typos/errors and I will correct them, thanks).

Here’s how I solved it. I think it makes sense to start at the end. As each box has the same amount of sand at the end, each box ends with 360/3 = 120 grams (g) of sand.

End: (A, B, C) = (120, 120, 120)

Now we work one step backwards. In the very last step, 1/5 of Box C is poured into Box A, after which Box C ends with 120 g. If 4/5 of the sand in Box C is then equal to 120 g, then the total sand in Box C was equal to 5/4 of 120 g, which is 150 g.

Now we know Box A received 150 – 120 = 30 g from Box C, and it then ended up with 120 g of sand. So Box A had 90 g before this last mixing.

This step also had no effect on Box B which stays at 120.

Before C–>A: (A, B, C) = (90, 120, 150)

Now we go one step backwards. Box B ended up with 120 g after losing 1/3 of its contents to C. So if the amount in Box B, less 1/3 of itself, is equal to 120 g, then 2/3 of Box B was equal to 120 g, and so Box B had 3/2 of 120 g, which is 180 g, before being poured into Box C.

This also means Box C received 60 g from Box B, so it started with 90 before this step.

Box A is unaffected from this step.

Before B–>C: (A, B, C) = (90, 180, 90)

Finally, the 90 g in Box A is the result after pouring 1/6 of Box A into Box B. So if the original amount in Box A, less 1/6 of itself, is equal to 90 g, then 5/6 of Box A was equal to 90 g, and so Box A started with 6/5 of 90 g, which is 108 g.

This also means Box B received 18 g from Box A, so it started with 180 – 18 = 162 g before this step.

This step has no effect on Box C which stays at 90.

Before A–>B: (A, B, C) = (108, 162, 90)

So we know Box A started with 108 g, Box B started with 162 g, and so Box C started with 90 g. And that’s the answer.

Wow, that’s pretty hard for 10 year old homework!

Just for good measure I worked the problem out algebraically…it’s actually much harder!

**Source**

I received the problem by email from Tim.

The website kaisuparents shows how to solve it using a visual model–see the answer by Jmaths.SG:

https://www.kiasuparents.com/kiasu/question/27884-2/

Here is the problem:

Chris has an umbrella that is 85 centimeters (cm) long.

His locker is a cuboid with length 40 cm, width 40 cm, and height 70 cm.

Can the umbrella fit inside the locker diagonally between corners *P* and *M*.

Can you figure it out? Give it a try, and then watch the video for a solution.

**Tricky Umbrella Question Stumps Scottish Students**

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**Answer To Tricky Umbrella Question Stumps Scottish Students**

(This was transcribed quickly after I made the video–please let me know if there are any typos/errors and I will correct them, thanks).

**Method 1: distance formula**

Set up a coordinate system so *P* = (0, 0, 0) and then *M* = (40, -40, 70).

(Thanks to Max Krass for pointing out the early access video had incorrect coordinate axes)

Then we can find the length of PM using the distance formula:

√((40 – 0)^{2} + (-40 – 0)^{2} + (70 – 0)^{2})

= √(8100)

= 90

The umbrella can fit inside the locker’s diagonal because 90 cm is larger than 85 cm.

**Method 2: Pythagorean Theorem twice**

A very similar approach is to consider right triangles. Label point *N* as the opposite corner to *P* for the square base.

Then *PN* is the hypotenuse of a right triangle with legs of 40 and 40, meaning its length is:

√(40^{2} + 40^{2})

= 40√2

Then we can consider triangle *PNM*, where *PM* is the hypotenuse of the triangle with legs of 40√2 and 70, so its length is:

√((40√2)^{2} + 70^{2})

= √8100

= 90

So once again we can see the umbrella can fit inside the locker.

**Source**

The Scottish Sun:

https://www.thescottishsun.co.uk/news/2596833/scottish-pupils-national-5-maths-exam-petition/

A few years ago, the same exam had a crocodile problem that stumped Scottish students:

https://wp.me/p6aMk-4lL

There is also a similar problem from a Malaysian exam that is nice:

https://wp.me/p6aMk-5oj