A

9000139708

Level: 
A
The shelf contains \(15\) books. From this amount, \(9\) books are in English and \(6\) books in other languages. Find the number of possibilities how to rearrange the books on the shelf, if all English books have to be on the left and the other on the right.
\(9!\, 6!=261\:273\:600\)
\(9^{6}=531\:441\)
\(\frac{9!} {6!}=504\)
\(\frac{9!} {6!\, 3!}=84\)

9000140003

Level: 
A
Solve the following equation with unknown \(x\) and a real parameter \(a\in\mathbb{R}\setminus\{0\}\). \[ax - \frac{2} {a^{2}} = \frac{4x+1} {a} \]
\(\begin{array}{cc} \hline \text{Parameter} & \text{Solution set}\\ \hline a=-2 & \mathbb{R} \\ a=2 & \emptyset \\ a\notin\{-2,0,2\} & \left\{\frac1{a(a-2)}\right\} \\\hline \end{array}\)
\(\begin{array}{cc} \hline \text{Parameter} & \text{Solution set}\\ \hline a=-2 & \mathbb{R}\setminus\{1\} \\ a=2 & \emptyset \\ a\notin\{-2,0,2\} & \left\{\frac1{a(a-2)}\right\} \\\hline \end{array}\)
\(\begin{array}{cc} \hline \text{Parameter} & \text{Solution set}\\ \hline a=-2 & \emptyset \\ a=2 & \mathbb{R} \\ a\notin\{-2,0,2\} & \left\{\frac1{a(a-2)}\right\} \\\hline \end{array}\)

9000139507

Level: 
A
The average mass of five melons is \(2\: 400\, \mathrm{g}\). We have to add another melon such that the new average value of all six melons will be \(2\: 420\, \mathrm{g}\). Find the mass of the sixth melon.
\(2\: 520\, \mathrm{g}\)
\(2\: 540\, \mathrm{g}\)
\(2\: 480\, \mathrm{g}\)
\(2\: 460\, \mathrm{g}\)

9000139501

Level: 
A
Ten apples in a box have average mass \(200\, \mathrm{g}\). We remove one apple of the mass \(200\, \mathrm{g}\) from the box. What is the change in the average mass of the apples from the box?
The average mass of the apples does not change.
The average mass of the apples decreases by \(20\, \mathrm{g}\).
The average mass of the apples increases by \(20\, \mathrm{g}\).
There is not enough information to solve this problem.

9000139305

Level: 
A
There are five rooms with three beds and one room with five beds in a hotel. A group of \(20\) people booked rooms at this hotel. Determine the number of possible choices for the people to the five-bed room.
\(\frac{20!} {5!\, 15!}=15\:504\)
\(20\cdot 3\cdot 5=300\)
\(\frac{20!} {3!\, 5!}=3\:379\:030\:566\:912\:000\)
\(20^{5}=3\:200\:000\)