u3a - Maths Challenge - Week 2

Welcome to week 2 of our weekly maths challenge, with problems and puzzles posed by Gordon Burgin, Andrew Holt and the U3A Maths and Stats Subject Adviser - David Martin. Thank you so much for joining in Week One - the solutions are at the bottom of this page.

If you would like to share your ideas on how to solve these puzzles please join our learning forum or discuss within your U3A and interest group. Check back each week for the solutions and let us know how you get on by contacting This email address is being protected from spambots. You need JavaScript enabled to view it.. New maths puzzles will go up onto the website every Thursday.

Questions Week 2

Question 1

If the cold tap alone would fill a bath in 20 minutes and the hot tap alone would fill a bath in 30 minutes, how many minutes would it take to fill the bath with both the cold and hot taps on?

Question 2

The six angles of two different triangles are listed in decreasing order. The list starts with 115, 85, 75 and 35. What is the last angle in the list?

Question 3

Eight pens cost £10 and some pence while eleven more cost £26 and some pence. What is the cost of a pen?

Question 4

An octahedron is drawn by connecting the centres of the faces of a cube. What is the ratio of the volume of the octahedron to that of the cube?


Solutions for Week One

Question 1 Solution

By the time the faster car starts the slower car will have travelled one hour at 40 m.p.h. and so will be 40 miles ahead. The faster car will catch up at a rate of 10 m.p.h. and hence will take 4 hours to catch up these 40 miles. The distance travelled at 50 m.p.h. during these 4 hours before overtaking will be 200 miles.

Question 2 Solution

Classifying by side length there are:
Sixteen unit squares: 9 shaded in grey/brown; plus 2 in each of the second, third, and fifth rows and 1 in the fourth in white
Seven 2 x 2 squares: 3 starting in second row, 1 starting in third row, 3 starting in fourth row
Three 3 x 3 squares: 1 starting in top row, 2 starting in third row
Two 4 x 4 squares: Both starting in second row.
One 5 x 5 square: Enclosing the diagram
So, 16 + 7 + 3 + 2 + 1 = 29 in all.

Question 3 Solution

If n is a number such that (n ‐1) is divisible (assuming from the question it is
not 1) by a prime p then (n ‐1) = mp for some m. Applying this to the four first
primes we get
n = + 1. The only such number below 300 is 211 (for which m = 1

Question 4 Solution

Let the length of the square sides be x and the height be h.
Then the 10 cm2 will cover a surface area of 2x2 + 4xh
So 10 = 2x2 + 4xh and h = (10 – 2x2)/4x = ½ (5 – x2)/x
Hence V = x2h = x2.1/2. (5 – x2)/x = ½. (5x – x3) *
dV/dx = ½ (5 – 3x2) and d2V/dx2 = -3x
So V has a maximum at x = √5/3 (note d2V/dx2 < 0 here)
And hence from * maximum volume is ½. (5 – 5/3)√(5/3) = ½ .10/3.√(5/3)
= 2.1517 cm3 to 4 dec. places