Solution:

Mushroom

Solution:

Align the toothpicks to form TEN. Use one toothpick for each of the lines.

Solution:

The password is U472BMT.

Read aloud the line and you will sound something like "U Four Seven 2 B M T".

Solution:

Set the first switches on for abt 10min, and then switch on the second switch and then enter the room.
Three cases are possible
1.Bulb is on => second switch is the ans
2.Bulb is off and on touching bulb , you will find bulb to be warm
=>1st switch is the ans.
3.Bulb is off and on touching second bulb , you will find bulb to be normal(not warm)
=>3rd bulb is the ans.

Solution:

Four

three placed flat on the table in a triangle(touching each other) and put the fourth one on top of them in the middle.

Solution:

'five'

Its the count of alphabet

Solution:

prinka is 10, in a system that awards five for each vowel.

Solution:

It has been clearly stated that none of the two students gave same number of correct answers which leads us to two different possibilities of correct answers:
Either (1 + 2 +3 + 4 + 5 = 15) or (0 + 1 + 2 + 3 + 4 = 10) stands true.

After analyzing the question, the maximum number of correct answers possible with the answers Gerry, Billy, Clark, Peeta and Jonathan gave:
Question 1 = 2 (b or c)
Question 2 = 2 (b or c)
Question 3= 4 (True)
Question 4 = 4 (True)
Question 5 = 3 (True)

Let us first assume with the maximum number of correct answers possible according to our data which are 15 (2+2+4+4+3). According to it, Billy must have given all correct answers as he was the only one who answered True for questions 3, 4 and 5. But in that case, Gerry and Clark must have given exactly three correct answers. Furthermore, Peeta and Jonathan must have given two correct answers. Therefore none of them got all the five correct answers which is wrong according to what we have analyzed.

Thus we move on to our next assumption according to which, the total number of correct answers are 10 (0+1+2+3+4). Now we must acknowledge that Questions 3 as well as Question 4 cannot be False as it will mean that the number of correct answers would not be 10. So the students giving wrong answers cannot be Gerry, Billy and Peeta.

Assume that Jonathan got all wrong, it will suggest that Gerry, Clark and Peeta each would have at least two correct answers. Then, Billy would have to be the person with only one correct answer and in that situation, the correct answers for questions 1 and 2 would be a and a respectively. Since that is not possible as then the total number of correct answers would be 1 (a) + 1 (a) + 1 (False) + 4 (True) + 2 (Flase) = 9.

Thus it must be Clark who has given all wrong answers. The correct answers are then b, a, True, False and False. Also, the scores are Clark (0), Billy (1), Gerry (2), Jonathan (3) and Peeta (4).

Solution:

TIF56

Explanation:
The password is formed as:
T (5th letter of Shruti)
I (5th letter of Arshika)
F(1st letter of Friday)
5(5th day of week)
6(Length of Friday)

Similarly, we will pick the 1st letter for Monday and 3rd letter for Wednesday.

Solution:

The driver is actually the wife and mother.

Solution:

So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.

There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3

(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.