Math Olympiad Problems for Grades 9 and 10: Problems with Solutions

Solution:
Let [tex]A[/tex] be the number of Alpha rovers and [tex]B[/tex] be the number of Beta rovers.
We set up a system of linear equations:
1. Total rovers: [tex]A + B = 45 \Rightarrow B = 45 - A[/tex]
2. Total wheels: [tex]4A + 6B = 210[/tex]
Substitute the first equation into the second: [tex]4A + 6(45 - A) = 210[/tex]
[tex]\Rightarrow 4A + 270 - 6A = 210[/tex]
[tex]\Rightarrow -2A = -60[/tex]
[tex]\Rightarrow A = 30[/tex].
There are [tex]30[/tex] Alpha rovers.

Solution:
The number ’[tex]1[/tex]’ appears [tex]1[/tex] time, ’[tex]2[/tex]’ appears [tex]2[/tex] times, ’[tex]3[/tex]’ appears [tex]3[/tex] times, and so on.
To find the 50th position, we find the sum of the first [tex]n[/tex] integers that gets closest to [tex]50[/tex].
The formula is [tex]\frac{n(n+1)}{2}[/tex].
For [tex]n = 9[/tex], the sum is [tex]\frac{9(10)}{2} = 45[/tex].
The last ’[tex]9[/tex]’ is at the 45th position.
The next [tex]10[/tex] positions (46th to 55th) will all be filled with the number ’[tex]10[/tex]’.
Therefore, the 50th pulse is [tex]10[/tex].

Solution:
The total sum of all available numbers from [tex]2[/tex] to [tex]10[/tex] is [tex]2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 54[/tex].
Since there are [tex]3[/tex] rows and they all must have the exact same sum, the sum of any single row (the ”magic constant”) must be [tex]54 \div 3 = 18[/tex].
In any [tex]3 \times 3[/tex] magic square, the central cell is exactly one-third of the magic constant: [tex]18 \div 3 = 6[/tex].

Solution:
Drawing an auxiliary line parallel to the panels through node [tex]P[/tex] splits angle [tex]x[/tex] into two smaller alternate interior angles.
The top part equals the [tex]42^\circ[/tex] angle, and the bottom part equals the [tex]36^\circ[/tex] angle.
Total angle [tex]x = 42^\circ + 36^\circ = 78^\circ[/tex].

Solution:
The total number of ways to arrange [tex]5[/tex] modules without restrictions is [tex]5! = 120[/tex].
Next, calculate the number of arrangements where [tex]P[/tex] and [tex]R[/tex] are together by treating them as a single block.
There are [tex]4! = 24[/tex] ways to arrange the [tex]4[/tex] units.
Inside the block, [tex]P[/tex] and [tex]R[/tex] can switch places ([tex]PR[/tex] or [tex]RP[/tex]), which adds a factor of [tex]2! = 2[/tex].
Ways they are together [tex]= 24 \times 2 = 48[/tex].
To find the number of ways they are separated, subtract the grouped ways from the total: [tex]120 - 48 = 72[/tex] ways.