# Week One Maths Challenge Solutions

Week One Maths Challenge Solutions

Solution 1
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.
Solution 2
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.
Solution 3
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 = 2.3.5.7.m + 1. The only such number below 300 is 211 (for which m = 1
Solution 4
Let the length of the square sides be x and the height be h.
Then the 10 cm2 will cover a surface area of 2×2 + 4xh
So 10 = 2×2 + 4xh and h = (10 – 2×2)/4x = ½ (5 – x2)/x
Hence V = x2h = x2.1/2. (5 – x2)/x = ½. (5x – x3) *
dV/dx = ½ (5 – 3×2) 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

Week One Maths Challenge Solutions

Did you get them all correct?  send me some feedback at…seahamdistrictu3a@mail.com

Week Two will be published this afternoon.