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Each month, a new set of puzzles will be posted.  Come back next month for the solutions and a new set of puzzles, or subscribe to have them sent directly to you.

## Puzzle One

The board shown below has 32 cells, one of which is labelled S (Start) and another F (Finish). The shortest path starting at S and finishing at F involves exactly nine other cells and ten moves, where each move goes from cell to cell ‘horizontally’ or ‘vertically’ across an edge.

 From S, there are two starting paths: We can represent the number of paths to a particular square like this: There is only one way to travel along the edges and we can build up our diagram like this, noticing that each square not on an edge can only be reached from the left or from below. How many possible paths are there from S to F by using the instructions above?

## Puzzle Two

There are three circles with centre points A, B and C. Each circle touches the other two circles, and each centre lies outside the other two circles. The sides of the triangle, drawn connecting the 3 points, have lengths 13 cm, 16 cm and 20 cm. What are the radii of each of the three circles?

## Puzzle Three

A large square is split into four congruent squares, two of which are shaded. The other two squares have smaller shaded squares drawn in them whose vertices are the midpoints of the sides of the unshaded squares. What percentage of the diagram is shaded?

## Feedback

There are more than one way of doing these puzzles and may well be more than one answer.  Please let me and others know what alternatives you find by commenting below.  We also welcome general comments on the subject and any feedback you'd like to give.

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## Puzzle One

Four men are throwing darts at a garage door. Their first names are Andy, Brian, Carl and Danny, and their last names (not in the same order) are Anderson, Brown, Clinton, and Dickson. Their ages (not in the same order) are 31, 32, 33 and 34 and each man has one dart. The colours of the four darts (again not in the same order) are Blue, Red, Green, and White.

Danny’s dart sits a little to the right of the Red dart, 72 cm from Brown’s dart, and a little under Andy’s dart. Dickson’s dart sits a little right under the Red dart. Andy’s dart is 97 cm from Dickson’s dart.

Brian is the youngest and the one with the Red dart is older than Clinton, being 2 years younger than the White dart, who again is older than Carl.

For the four men, can you determine what their last names, ages and dart colours are? Also, the distance between the Green and Blue darts?

Solution:

## Puzzle Two

If Alex ate 1/2 of the plate of cookies he found on the counter, then his sister Lucy ate 1/3 of what was left, and then Charley ate 1/2 of those that were left, when their Mum came home and found 2 cookies, how many did the kids eat?

Solution:

Working backward:

1. 2 cookies were left. If Charley had eaten half of what she found, she must eaten the same amount.  Charley ate 2.
2. This means Lucy left 4 cookies, which was two-thirds of the cookies she found.  Lucy also ate 2.
3. So, Alex left 6 cookies.  He also ate half of what he found, so he must have found 12 cookies.

As a check, 12/2 = 6,  6/3*2 = 4, 4/2 = 2 cookies

## Puzzle Three

The ration of two numbers is 7:10 and their difference is 105.  What are the numbers?

Solution:

Given: The difference between numbers is 105 and the difference between ratio of numbers is 3.

Divide 105 by 3 = 35

Multiply both terms of the ratio by 35 gets the numbers

(35 * 10) – (35 * 7) = 350 – 245 = 105

Therefore, the numbers are 245 and 350.

## Puzzle Four

Jack and Jill went up the hill to fetch a pail of water. They started at the same time, but Jack arrived at the top a half of an hour before Jill. On the way down, Jill calculated that if she had walked 50% faster and Jack had walked 50% slower, then they would have arrived at the top of the hill at the same time.  How long did Jill take to walk up to the top of the hill?

Solution:

Let t equal the number of hours that Jill took to walk to the top of the hill.

So the time taken by Jack was (t – ½ ) hours. If Jack had walked 50% more slowly, he would have taken twice as long, i.e., 2(t – ½) = (2t – 1) hours.

If Jill had walked 50% faster, she would have taken 3/2 t hours.

Therefore, with Jack’s 50% slower equalling Jill’s 50% faster

2t – 1 = 3/2 t

1/2 t = 1

t = 2

As a check: (2t – 1) must = (3/2 t) or (2 * 2) – 1 = 3/2 * 2 and thus 3 equals 3.