Parallelogram 8 • Level 3 • 26 Oct 2023Line Rider

We work out the value of each expression in turn:

• A: 6 + 5 + 4 = 11 + 4 = 15.
• B: 6 + 5 − 4 = 11 − 4 = 7.
• C: 6 + 5 × 4 = 6 + 20 = 26.
• D: 6 − 5 × 4 = 6 − 20 = −14.
• E: 6 × 5 ÷ 4 = 6 × 1.25 = 7.5.

We see that expression B has the value that is closest to zero.

Suppose there were equal numbers of girls and boys at the school. Then there would be 300 girls and 300 boys. Each time we add one more girl and take away one boy, the difference between the number of girls and the number of boys goes up by 2, while the total number of pupils remains 600. So to have 30 more girls than boys, we would need to add 15 more girls and take away 15 boys. So a school with 600 pupils and 30 more girls than boys has 315 girls.

Alternatively, suppose that the school has g girls. Then, as there are 600 pupils altogether, there are 600−g boys.

Since there are 30 more girls than boys: g−600−g=30.

This gives g−600+g=30, that is, 2g−600=30, and hence 2g=630.

Therefore g=315.

Hence, there are 315 girls at Gill’s school.

One method is to do the following divisions, to find out in which case the answer is not an integer.

• A) 12 ÷ 2 = 6.
• B) 123 ÷ 3 = 41.
• C) 1,234 ÷ 4 = 308.5.
• D) 12,345 ÷ 5 = 2,469.
• E) 123,456 ÷ 6 = 20,576.

From these calculations we see that 1234 is not a multiple of 4. So the statement that is false is C.

Alternatively, we have the following tests for whether a positive integer is a multiple of n, for small values of n:

• n) Test for being a multiple of n
• 2. The units digit is even.
• 3. The sum of the digits is a multiple of 3.
• 4. The tens and units digits taken together make an integer which is a multiple of 4.
• 5. The units digit is 0 or 5.
• 6. A multiple of 2 and a multiple of 3.

We apply these tests to the statements given in the options:

• A) The units digit of 12 is 2, which is even. So 12 is a multiple of 2.
• B) The sum of the digits of 123 is 1 + 2 + 3 = 6. This is a multiple of 3. So 123 is a multiple of 3.
• C) The tens and units digits make 34. This is not a multiple of 4. So 1,234 is not a multiple of 4.
• D) The units digit is 5. So 12,345 is a multiple of 5.
• E) 123,456 has units digit 6 and the sum of its digits is 21. So it is a multiple of 2 and 3, and hence of 6.

Therefore it is statement C that is false.

We note first that the colour of the centre square does not affect the symmetries of the design. Since we are looking for the largest number of squares that can be coloured black, we need only consider colourings in which this square is black.

To have a colouring which has rotational symmetry of order 2, each pair of opposite squares, that is, the pairs labelled p,q,r and s in the diagram on the left, must be the same colour.

The second diagram above shows that, if all four of these pairs are black, then the design has four lines of symmetry. The third and fourth diagrams show that, if one of these pairs is coloured white and the other three pairs are black, then the design has two lines of symmetry.

The diagram on the right shows, however, that we can create a design with no lines of symmetry by colouring the appropriate two pairs white and the other two pairs black. It can be seen that the design in this diagram has rotational symmetry of order 2.

This shows that the largest number of squares that can be coloured black to make a design with rotational symmetry of order 2, but no lines of symmetry, is 5.

We assume that the small squares making up the grid have a side length of 1 unit.

The pentagon consists of a 6 × 6 square, with half of a 3 × 3 square removed. Hence the area of the pentagon, in square units, is:

6×6−123×3=36−92=632.

The area of a triangle is half the base multiplied by the height. The shaded triangle has a base of length 3 units and height 6 units. Therefore, its area, in square units, is:

123×6=9.

It follows that the fraction of the area of the pentagon that is shaded is:

9632=9×263=27.