Multiple answers are possible, such as calendar, new year / X’mas / other greeting cards etc.

One possible solution is 14. There can be other solutions, based on different logic applied.

Water Freezes at 32 Degrees Fahrenheit

Present Labels:

Bag 1: Silver
Bag 2: Gold
Bag 3: Silver and Gold

It is known that all three bags are labelled wrongly. Hence:

Open the first bag (labelled ‘Silver’). As we know that the label is wrong, it may contain either (i) gold or (ii) gold and silver

(i) If it contains gold, bag 2 may contain (a) silver OR (b) gold & silver. But if it contains silver, then the third bag must contain gold & silver, which makes its label correct, contradicting the given fact that all three bags are labelled wrong. Thus, it follows that the second bag must contain gold & silver, and the third bag must contain silver. Thus, the bags are identified as Bag 1 => Gold, Bag 2 => Gold & Silver and Bag 3 => Silver, and hence can be labelled correctly.

(ii) If Bag 1 contained gold and silver, then bag 2 may contain (a) gold or (b) silver. Here again, by similar logic as above, bag 2 can’t contain gold (because it would imply that the label is correct), and hence must contain silver, thereby reaching the conclusion as Bag 1 => Gold & Silver, Bag 2 => Silver and Bag 3 => Gold, and hence the bags can be labelled accordingly.

By applying similar logic, the correct labelling can be derived by opening ANY one of the bags.

Multiple solutions are possible, such as:

2184
4263
5163
5442
6342
7242
7521
8421
9321

Mr. Chef has an ‘impossible task’ at hand!

Let the qty of butter be X. Then, by condition 2, the qty of sugar must be 2X, whereas by condition 1, the amount of flour must be 4X, and by condition 3, the total of all 3 must be 600, which means X + 2X + 4X = 7X = 600 => X = 600/7 = 85.714285714285714285714285714286… (and counting) Happy searching for a balance that can weigh this quantity exactly!

Multiple answers are possible, such as:

195, 285, 375, 465, 555, 690, 780, 870, 960

Since the first tree is Apple (A), the second can be either Apple or Orange (O). In both scenarios, the remaining 3 trees can have 8 combinations each. Thus, total 16 combinations are possible.  However, since an A followed by an O must be followed by another A only, 8 of the possibilities get ruled out, leaving 8 possible arrangements.

In the list below, highlighted arrangements are NOT possible.

Just draw another line longer than the first, so that the first line becomes shorter compared to the second.