Yes, your solution is correct! 🎉

Let's break down the path the children took and count the animals they saw.

1. Goats:
   Nele mentions they went straight past the goats, so they didn’t stop at the goat enclosure.
   • Animals seen: 4 goats.
2. Donkeys:
   They didn't see any donkeys, so we don’t need to count them.
3. Roe Deer:
   They walked past the roe deer, meaning they didn’t stop at this enclosure either.
   • Animals seen: 4 roe deer.
4. Deer:
   They stopped at the deer enclosure.
   • Animals seen: 4 deer.
5. Forester’s Lodge:
   They took a break at the forester's lodge but didn’t mention seeing any more animals after this point. We don’t need to count any animals here.

Total animals seen:

• Goats: 4
• Roe deer: 4
• Deer: 4

So, the total number of animals they saw on the way to the forester’s lodge is: 4 + 4 + 4 = 12 animals.

Yes, your solution is correct! 🎉

To solve this problem, we add up the number of dots cut out by each friend:

• Jan cuts out 7 dots.
• Nele cuts out 5 dots.
• Paula cuts out 5 dots.
• Tim cuts out 3 dots.

Now, sum them up:
7+5+5+3=20

Final Answer:

The 4 friends cut out a total of 20 dots for the ladybirds. 

Yes, your solution is correct! 🎉

1. There are 5 dogs, each with a jumper that has dots, checks, or stripes as the pattern.
2. The Klein family's dog:
   • Has a jumper with a hood.
   • Does not have stripes.

Only the Schnauzer and the Great Dane have sweaters with hoods. However, the Schnauzer's sweater has stripes, which means it cannot belong to the Klein family, as their dog does not have stripes. Therefore, the Great Dane (next to a heart) must be the Klein family dog.

Yes, your solution is correct! 🎉

The images illustrate six different ways to read the word "HASE."

Yes, your solution is correct! 🎉

We completed the drawing and confirmed that each of the four sections of the new enclosure is the same size as that of the old one.

This indicates that the old enclosure can fit into the new enclosure four times.

Yes, your solution is correct! 🎉

Let’s analyze the problem systematically by defining the quantities of each type of chocolate animal:

• R = Number of reindeer
• O = Number of owls
• S = Number of squirrels
• H = Number of hedgehogs
• B = Number of rabbits

Given clues:

1. Fewer hedgehogs than reindeer:
   H<R
2. The reindeer and the owls together are as many as the rabbits:
   R+O=B
3. Fewer squirrels than owls:
   S<O

Goal: Determine which type has the most chocolate animals.

Solution:

From clue 2, rabbits (B) are equal to the sum of reindeer (R) and owls (O):

B=R+O

Since R>H and O>S, it follows that rabbits (B) are at least as large as any other type of chocolate animal.

Conclusion:

The rabbits are the type of chocolate animals with the most in total.

Yes, your solution is correct! 🎉

We follow Nele's idea from the task and calculate the total number of votes for each of the four remaining animals.

• The stork received a total of 4 + 4 + 3 + 5 = 16 votes.
• The hedgehog received a total of 5 + 4 + 3 + 5 = 17 votes.
• The raven received a total of 5 + 4 + 5 + 5 = 19 votes.
• The fox received a total of 3 + 5 + 3 + 2 = 13 votes.

The smallest total is 13, indicating that the fox received the fewest votes.

Yes, your solution is correct! 🎉

Two conditions must be met for Nele’s plate:

1. The number of kangaroos is exactly twice the number of hedgehogs.
2. The number of rabbits is one more than the number of kangaroos.

Let’s analyze the options:

• Heart: There is 1 kangaroo and 3 hedgehogs. The first condition is not fulfilled.
• Candle: There are 4 kangaroos and no hedgehogs. The first condition is not fulfilled.
• Star: There are 2 kangaroos and 1 hedgehog, and there are 3 rabbits (one more than the kangaroos). Both conditions are fulfilled.
• Tree: There are 2 kangaroos and 2 rabbits. The second condition is not fulfilled.
• Bell: There are 3 kangaroos and 1 hedgehog. The first condition is not fulfilled.

Conclusion: The correct plate is “Star.”

Yes, your solution is correct! 🎉

With its first two steps, the robot reaches the small rear wall. There, it turns to the left. Next, it takes two steps to the field between the flower and the bone. It then turns left again and, with one final step, reaches the bone.

Yes, your solution is correct! 🎉

Given the details provided:

• Two parrots only have red and blue feathers: the one at "candle" and the one at "heart".
• Of these, only the parrot at "heart" has a dark beak.

Conclusion:

The parrot Paula named Jojo is the one at "heart", as it uniquely satisfies the conditions of having red and blue feathers and a dark beak.

Yes, your solution is correct! 🎉

In the missing section, there are four honeycombs in a row. At the left end, one honeycomb connects to the row diagonally upwards, while at the right end, it connects diagonally downwards. The picture illustrates the missing section:

Yes, your solution is correct! 🎉

We can describe the necklace as follows: 'M' represents mussels, and 'S' represents snails. The sequence is:

MMS S SMMS S SMMS S SMM

In total, 8 mussels and 9 snails have been used. However, neither type is sufficient to create a longer chain. This leaves Paula with 1 mussel and 2 snails for her crafting projects.

Yes, your solution is correct! 🎉

The three cats at 🎄, 🌟, and ♥  all have brown fur. Among them, the plush cats at 🎄 and 🌟 have white paws. Of these two, only the cat at 🎄 also has a white-tipped tail.

Yes, your solution is correct! 🎉

When you look in the mirror, you see the dachshund on the left and the husky on the right. This applies to the pictures at 🛎, 🕯 , and 🌟. The picture is not the mirror image, because the husky is standing there.
The dachshund's dog leash is in exactly the same direction as the dachshund is standing.
This is drawn incorrectly in the picture 🛎. The husky's lead also points in the wrong direction in this picture. The 🌟 is a mirror image.

Yes, your solution is correct! 🎉

We connect each molehills in the picture to each other and count the connections. There are 6.

Yes, your answer is correct! 🎉

We followed the path outlined in the text from Thick lime to the forester's lodge, counting the animal enclosures along the way.
In total, there are six enclosures: wild boar, donkeys, ferrets, deer, geese, and roe deer.
However, the donkeys were hidden, so we only observed animals in five of the enclosures.
Since there were exactly three animals visible in each enclosure, we saw a total of 5 x 3 = 15 animals along the path.

Yes, your answer is correct! 🎉

Since Marlene remembers there were as many 2-point ladybirds as 7-point ladybirds, let's assume there are x ladybirds of each type.

• The total points from the 2-point ladybirds is 2x.
• The total points from the 7-point ladybirds is 7x.

Adding these together:

2x+7x=9x

So, the total number of points is 9 times the number of ladybird pairs.

For example:

• If there is 1 pair (1 of each type), the total is 9 points.
• If there are 2 pairs, the total is 18 points.
• If there are 3 pairs, the total is 27 points.

Answer: The total points depend on the number of pairs, and they are always a multiple of 9 (e.g., 9, 18, 27, 36, 45 etc.).

Yes, your answer is correct! 🎉

The only striped sweater is at position (U), belonging to Daisy. The single-colored sweater, an orange one, is at position (E) and worn by Kalle. The only remaining sweater with more than two colors is at position (K), belonging to Einstein. The sweater with ducks, at position (N), is worn by Maggie. Therefore, Jack is the dachshund at position (D).

Yes, your answer is correct! 🎉

Here you can see where SCHAF, OCHSE, DACHS, and FISCH are hiding:

The FROSCH cannot be located in the specified manner.

Yes, your answer is correct! 🎉

The approach of dividing the new fence into parts to directly calculate the difference in length is a great way to simplify the problem.

Here's how that works:

Explanation:

• The extra length comes from the 2 m extension on each of the 8 sections (4 sides, each split into two ends).
• For each section, the added length is 2m, and there are 8 sections in total.

Thus, the additional length is: 8×2=16m

Yes, your answer is correct! 🎉

With the first three clues, three possibilities remain regarding which chocolate animal each person has in the boot:

- With clue (X), possibilities 1 and 2 remain.

- The aunt cannot clearly determine who found which chocolate animal.

- With clue (W), possibilities 1, 2, and 3 remain.

- With clue (O), possibilities 1 and 3 remain.

- With clue (B), possibilities 1 and 2 remain.

Even with these clues, the aunt still cannot clearly identify who received which animal. However, with clue (Y), only one possibility remains, specifically possibility 2. Therefore, clue (Y) is the one we are looking for.

Yes, your answer is correct! 🎉

According to the votes counted so far, the Beaver is currently in the lead. If all 12 children from class 4b vote for the Beaver, it will still remain ahead and win the election. Therefore, the Beaver's victory is possible.

The Hedgehog has the second most votes, standing at 22. If all 12 children from class 4b vote for the Hedgehog, he would then have a total of 22 + 12 = 34 votes. This would allow the Hedgehog to overtake the Beaver, who would then have only 31 votes. Thus, the Hedgehog could potentially win the election.

The Stork is in third place, with 16 votes so far. Even if all 12 remaining votes go to the Stork, he would only reach 16 + 12 = 28 votes, which is still fewer than the Beaver's total. Therefore, the Stork cannot win the election.

It is also clear that the two animals with fewer votes than the Stork cannot win.

Consequently, only the Hedgehog or the Beaver can become the school mascot.

Yes, your answer is correct! 🎉

We count the cookies on the plates and observe that plate (S) has the most cookies—6 cookies. Thus, plate (S) belongs to Jule. The cookies on Jule's plate weigh as much as 6 rabbits.

Next, we analyze the weights of the cookies on the other plates:

• Plate (P): Weighs as much as 6 rabbits.
• Plate (D): Weighs as much as 5 rabbits.
• Plate (A): Weighs as much as 7 rabbits.
• Plate (N): Weighs as much as 5 rabbits.

Since Milan's plate must weigh the same as Jule's plate (6 rabbits) but have fewer cookies, plate (P) is Milan's plate.

Conclusion: Milan's plate is plate (P).

Yes, your answer is correct! 🎉

We illustrate Robbie's paths using a pen or by tracing them with a finger. Robbie can only reach the bone by pressing the (J) key.

First, Robbie moves forward 2 spaces and then turns to the right. Next, he runs backward 4 spaces between the walls. After that, he turns to the left, runs forward 2 spaces, and finally lands on the bone.

Yes, your answer is correct! 🎉

We first count the number of bleached feathers in each of the bearded vultures: (M) has 5, (T) has 6, (N) has 5, (R) has 5, and (S) has 4.

It is established that more feathers were bleached in Bavaria than in Recka, but fewer than in Dagmar. Therefore, we conclude that Dagmar must have 6 bleached feathers, Bavaria has 5, and Recka has 4. This indicates that Dagmar is represented by (T) and Recka by (S).

Among the remaining three bearded vultures, only those at (M) and (N) do not have bleached tail feathers, which means they are Wally and Herculis.

Additionally, the bearded vulture at (R) is identified as Bavaria.

Yes, your answer is correct! 🎉

We fill in the edges of the missing honeycombs.

Now, the gap can be easily compared with the possible answers. The piece (Q) fits into the gap.

Yes, your answer is correct! 🎉

For this task, both the direction of the circle's arcs and the sizes of the paper squares must be considered.

Counting the missing squares in the snail, we find that their side lengths must be 3 boxes and 8 boxes wide. Only answer option (B) meets these requirements, as all other options include at least one square with an incorrect side length.

Additionally, the direction of the circular arcs in (B) matches perfectly. Thus, these squares complete the snail spiral accurately.

Theoretically, the paper squares could also be rotated. However, only the squares in (B) have the correct dimensions, making the solution uniquely determined.

The side lengths of the squares in the spiral follow the Fibonacci sequence, where each number is the sum of the two preceding ones. This characteristic further confirms the accuracy of the solution.

Yes, your answer is correct! 🎉

Try combinations of packs to get 12 cans

Case 1: Use two 5-packs and two single cans

• Two 5-packs = 2×€4.10=€8.20
• Two single cans = 2×€1.00=€2.00
• Total = €8.20 + €2.00 = €10.20

Case 2: Use one 8-pack and one 4-pack

• One 8-pack = €6.80
• One 4-pack = €3.60
• Total = €6.80 + €3.60 = €10.40

Case 3: Use one one 5-pack, one 4-pack and three single can

• One 5-pack = €4.10
• One 4-pack = €3.60
• Three single can = €3.00
• Total = €4.10 + €3.60 + €3.00 = €10.70

Case 4: Use three 4-packs

• Three 4-packs = 3×€3.60=€10.80
• Total = €10.80

Step 3: Choose the lowest cost

The lowest cost is achieved in Case 1, where Pina buys two 5-packs and two single cans for a total of €10.20.

Final Answer: (B) €10.20

Yes, your answer is correct! 🎉

When Amadeus looks into the mirror, turning to the left shows items positioned farther forward in the cage, while turning to the right reveals items positioned farther back.

As a result, in the mirror, the food bowl appears to the right of Krümel. This observation is true only for options (V) and (F). Since Krümel is facing the food bowl, the correct mirror image is option (F).

Yes, your answer is correct! 🎉

To solve this, we calculate how many ways Gregor and Mila can each occupy a molehill. There are 4 molehills, and both Gregor and Mila must choose different ones.

1. Step 1: Gregor chooses a molehill.
   • He has 4 choices (any of the 4 molehills).
2. Step 2: Mila chooses a molehill.
   • Mila cannot choose the same molehill as Gregor, so she has 3 choices (the remaining molehills).
3. Step 3: Multiply the choices.
   • The total number of possibilities is 4×3=12.

Final Answer: 12