Solution:

Ruskin Stole Chihuahua
Sam stole Hyena
Terry Stole Deer

Explanation:
Let us consider the first two statements again; they are conflicting. However both can be true together. Also both of them cannot be incorrect which means that Terry stole Chihuahua and his statement will be true. But it has been already inferred that the one who stole Chihuahua was telling a lie thus it can’t be possible.

Focus on the given facts again. Terry might have stolen a hyena or a deer. Let us say that he stole hyena. In such an event the statement of Terry will be untruthful. But we know that the person who stole hyena told the truth, thus it is controverting our hypothesis and so cannot be true.

Considering everything, now we can be sure of the fact that Terry stole the deer.

It means that the declaration by Ruskin is totally false and the declaration by Sam is true which shows that the Chihuahua and hyena are stolen by these two. Also we know that the one who stole hyena is correct and the one stealing Chihuahua is a falsifier. Thus we now know that Ruskin stole Chihuahua and Sam stole hyena.

Solution:

987,654

Let us assume with a smaller number and then increase the number by extrapolating. If there are two citizens, the number of hairs on their heads will be zero and one. Therefore for ‘n’ number of citizens, the hair distribution will range from 0 to (n-1).

If you take up a case of a number greater than 987,654 let’s say 987,655; and consider the fact that none of the citizens have 987,654 hairs and they must be having different number of hairs, one must be having more than 987,654 hairs which is bluntly violating the fact of the town that there are more citizens than the hairs on any individual.

Solution:

B

sideways clockwise 1/4 turn

Solution:

From left to right
#1 Mr. Reed Richard – Blue Marbles
#2 Ms. Sue Storm – Red Marbles
#3 Mr. Johnny Storm – White Marbles
#4 Mr. Ben Grimm – Green Marbles.

Let us label the houses from left to right as #1, #2, #3, #4. Now let us split the hints in different category.
a) Mr. Reed Richard’s house is to the left of the green marbles one
b) The third one is made of white marbles
c) Ms. Sue Storm owns a red marbles house
d) Mr. Johnny Storm does not live on either ends
e) Mr. Johnny Storm lives right of the blue marbles house
f) Mr. Ben Grimm lives in the fourth house
g) The first house is not made up of red marbles

By seeking all the (g) and (b), we are sure that the #1 is not made up of red marbles and nor is the #3. Now with (f) we know that Mr. Ben Grimm lives in #4 made up of green marbles and keeping in mind (c) #2 must be made of Red Marbles and Ms. Sue Storm lives there. Now moving to the hint (d) Mr. Johnny Storm must be living in #3 and adding up with hint (b), it must be made from white marbles. Now by hint (a) #4 must be made up of green marbles (or Mr. Reed Richard’s house can’t be on its left). Also by (f) Mr. Ben Grimm lives there. All of it leaves Mr. Reed Richards living in #1 with blue marbles construction.

Solution:

Answer
Kid4 and Kid6

Solution: According to given information, sitting arrangement of six people is -

Kid4 Kid5 Kid1 Kid3 Kid2 Kid6
or
Kid6 Kid1 Kid3 Kid2 Kid5 Kid4
or
Kid4 Kid5 Kid2 Kid3 Kid1 Kid6
or
Kid6 Kid2 Kid3 Kid1 Kid5 Kid4

So, in all the situations, Kid6 and Kid4 are at corners of the row.

Solution:

Only one(dead one)

other flew away :--)

Solution:

Step1 -> Start running both the hourglasses.
Step2 -> When 4 hours hourglass runs out, turn it over.
Step3 -> When 7 hours hourglass runs out, turn it over.
Step4 -> Again when 4 hours hourglass runs out, turn 7 minute hourglass over (it was running for exactly one minute.)
final step -> 8 minute already passed and allow one minute of 7 hour glass to get over. ==>9 minute

Solution:

Perhaps surprisingly, the apples now weigh 50 kilograms. The sack of apples originally weighed 100 kilograms, and 1 percent of its weight, or 1 pound, was non-water. If that 1 pound now makes up 2 percent of the total weight, then the total weight must be 50 kilograms.

Solution:

C

VIII has taken place of VII.

Solution:

Man1 is carrying 5 bags and the Man2 is carrying 7 bags.

Let's assume that the Man1 was carrying M1 bags and the Man2 was carrying M2 bags.
M2 + 1 = 2 * (X-1)
M2 + 1 = 2M1 - 2
M2 = 2M1 - 3

Also,
M2 - 1 = M1 + 1
M2 = M1 + 2

=>
2M1 - 3 = M1 + 2
M1 = 5

=>
M2 = 7