Solution

Suppose that in the class there are n boys, then the number of ways to choose from them a pair is equal to n $($n – 1$)$, the number of ways to choose a pair from the whole class is $25 \times 24$. Therefore, the probability that two boys will be chosen is $n (n-1)/(25 \times 24)$ = 3/25. Hence n $($n – 1$)$ = 72 = $9 \times 8$. Consequently, n = 9, that is, in the class there are 9 boys and 16 girls.

Answer

16 girls.

Hint

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Solution

We can divide the rectangle into 5 sections as shown on the diagram. One of these sections will contain at least two points, but the distance between any two points within each of these sections does not exceed $\sqrt5$.

Hint

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Solution

Split the regular hexagon into 6 equilateral triangles of side 1. Then at least one of the triangles will contain two points. The distance between them will not be greater than the side length of the triangle.

Hint

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Solution

Divide the 1m large square into 25 small squares of side 20cm. Using the pigeon hole principle we know that at least one small square must contain at least three dots.

Hint

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Solution

Each of the smaller triangles cannot cover more than one corner of the larger triangle. Therefore one corner will always remain uncovered, so the larger triangle will not be completely covered.

Hint

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