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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

In two years time my age will be three times the age of my son was two years ago. Three years ago my age was twice the age of my same son will be in three years time. How old are we both?

## Puzzle Two

There are 10 points equally spaced around the circle with interconnecting lines drawn between them.
a) How many lines are needed to draw the 10 point pattern shown below?
b) How many lines would you need if you had 50 points on the circle?
c) How many lines would you need if you had 100 points on the circle?

## Puzzle Three

The missing numbers are between 0 & 9. The numbers in each row add up to the totals on the right, and the columns add up to the totals along the bottom. The sums of the diagonals are also given. Find the missing numbers? (Note: there may be more than one solution for each of these puzzles.)

## 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

The price of an item was increased by 25% and later decreased by 25%, which makes the price of the item \$375. What was the original price?

Solution:

Let p = the original price

Then, 1.25 x 0.75 x p = 375

p = £400

## Puzzle Two

If 3/4 is Peter’s salary is equal to 2/4 of Mary’s salary. What is Peter’s salary if Mary’s salary is £120,000?

Solution:

Let: Peter’s Salary = p

Then, ¾ x p = ½ x 120,000

p = 4/3 x 60,000 = £80,000

## Puzzle Three

418 marbles are initially shared among three kids A, B and C. A got half the amount of C’s share which is 3 times B’s share. If A now wins 35 marbles from B and loses 8 marbles to B in a game of marbles how many marbles do each now have?

Answer: A ends up with 141 marbles, B ends up with 84 marbles, and C has 193 marbles

Solution:

Let x = C’s share

Givens:

A’s share = 1/2 of C’s share, so A’s share = x/2
C’s share = x = 3 x B’s share, so B’s share = x/3
418 = A + B + C = x/2 + x/3 + x = (11/6)(x)

Therefore:

x = (418)(6/11) = (38)(6) = 228 (or C’s share)
B’s share = x/3 = 228/3 = 76 marbles
A’s share = x/2 = 228/2 = 114 marbles
A + B + C = 114 + 76 + 228 = 418

If A now wins 35 marbles from C, but loses 8 marbles to B, then
A now has 114 + 35 – 8 = 141 marbles
B now has 76 + 8 = 84 marbles and
C now has 228 – 35 = 193 marbles

Checking answers, A + B + C = 141 + 84 + 193 = 418

## Puzzle Four

There is a three-digit number. The second digit is four times as big as the third digit (no zeros) while the first digit is three less than the second digit. What is the three-digit number?

Solution:

Let:

x = 1st single digit number
y = 2nd single digit number
z = 3rd single digit number
n = total 3-digit number where no two digits are the same or zero.

Given:
a) 100x + 10y + z = n
b) y = 4z
c) x = y – 3
d) z <= 3 (or 0, 1, 2, or 3)

z can = 1 or 2 but not = 0 or 3 because of step b) and y must be a single digit non zero.

If z = 1, then y = 4 from step b) and x = 1 from step c); x & z can not be the same value or 1)

Therefore, z = 2, then y = 8 from step b), x = 5 from step c) and n = 582 per step a).