Solve the system of equations to find the values of x, y and z: xy/(x+y)=1/2, yz/(y+z)=1/3, zx/(z+x)=1/7
If all the faces of a tetrahedron have the same perimeter then show that they are all congruent.
A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £2 for women and 50p for children. How many men, women and children are there in the group?
Wow! We received loads of great solutions to this - thanks to everyone who took part, tried our problem and submitted a solution.
The answer, as given correctly by almost everyone who responded, is 21. Congratulations to everyone who got the right answer! We didn't ask for them, but in case you were wondering, the individual symbols have values as follows: the circle is 4; the triangle is 6; the hexagon is 7; the square is 8. Below are just a few of the solutions we were given.
Henry, Jack and Tobi from Junior King's Canterbury used a trial-and-improvement method, but spotted the following, which helped them find the right answer more quickly:
From the information in the third row down, we see that triangle has to be in the three times table. We tried 3, and that didn't work, but then 6 did work.
Simran, from SS Peter and Paul, also used trial-and-improvement, but spotted the following way of estimating the value of certain shapes quickly:
First I looked for a row (or column) that had most shapes the same. When I found a row (or column) like this, I divided the total of the row (or column) by the number of shapes which were the same in that row (or column).
Emma, from Walton High School, used that reasoning when she looked at the third row which contained three circles:
To begin with I looked at the third row and saw that the circle had to be 5 or under because if it was 6 or more then the triangle wouldn't be worth anything.
Leo and Jenni, from the Russell School, noticed the following way of speeding things up:
First of all we looked at the 2nd row: two squares and two hexagons. The total was 30 so we divided that by 2, which is 15. The whole numbers that are closest to half of 15 are 7 and 8. So we started by approximating the hexagon as 7 and the square as 8.
Many students followed some of the other methods we gave, and worked out exactly the values of the shapes without trial-and-improvement. For example, Laura from Beaconsfield High School used simultaneous equations:
Let T stand for Triangle, S for Square, etc.
Using row 1: 2T + 2S = 28, so T + S = 14.
Using column 4: T + S + 2C = 22, so 2C = 8, and so C = 4...
Huy, from Vietnam, wrote:
Comparing the two rows at the bottom of the table, I find out that the square is worth 2 more than the triangle.
Then looking at the first row, I can figure out that the value of the triangle is 6:
2 squares + 2 triangles = 4 triangles + 4 = 28,
so triangle = 6
Then I can figure out the value of the square: 6+2 = 8.
Eddie from Wilson's School gave a comparison of the various methods described:
Methods 1 and 5 just use trial and error.
Methods 2, 3, 4, 6 and 7 compare various rows and columns and use some more advanced logic.
Method 8 is the most straightforward, and yet probably the most overlooked.