Solution:

Seven

Dividing the horses in a group of five, we will have five different groups of horses. A race will be conducted separately for each group which will help us find the fastest three horses in every group. We have used five different races till now.

Now we can have a race among the winners from every race which will help us determine the fastest of them all. We have used the sixth race.

The second fastest horse can be the one who came second in the sixth race or the second fastest horse in the finals winners' first race.

If we look at the first case, the third fastest horse can be the third fastest horse in the sixth race or the second fastest horse in either the fastest or second fastest horse first race.
If we look at the other case, the third fastest horse may be the second fastest in the sixth race or the third fastest in the final winners' first race.

If you calculate our assumptions, we have five more horses to race (seventh race) and determine the second fastest and the third fastest horse.

Solution:

Keep one coin from the first set and place it on the scale along with the two from the second set etc. If the weight is off by one hundredth of an ounce, then you will know that it is the first set that is faulty and if the weight is off by two hundred of an ounce, then the second set is faulty and so on.

Solution:

This one definitely sounds too unusual. But there is an easy way to look at things.

Let us say that there is one unfaithful husband in the town. In such a case, 99 women will know about him. The wife of that man will be the only one who will not know about him and would have always thought that his husband is truly faithful to her. But since the mayor announced that there is at least one cheater among them, she will come to a realization that her husband must be cheating on her. So her husband will be executed on the first day of the announcement.

Now let us take up a case that there are two cheating husbands in the town. Now 98 women of the town must be knowing about them both. The two wives of those men will think that there is only one cheater in the town. Both of them believes that their respective husbands are loyal and this would not report about them

But when both of them will find out on the next day that no one was executed, they will realize that it could only mean that both of their husbands are cheaters. Thus both the husbands will be executed on the second day.

Thus if we have X number of cheating husbands, they all will die on the Xth day after the announcement is being made.

Solution:

One thing you must take note of is that in the question they are talking about pair of socks not individual socks.

Thus, total socks = (9x2) + (8x2) = 18 + 16 = 34
Total possibilities = 34*33*32/3*2*1 = 5984

The next part can be a little tricky. Read the question again. You are asked to calculate the probability that she will pick up a pair of blue socks. That also includes picking up three blue socks.

Thus our favorable events = select 2 blue + all 3 blue
= (18*17/2*1)(16/1)+ (18*17*16)/(3*2*1)
= 2448 + 816
= 3264

Hence the Probability = 3264/5984 = 6/11 = 0.54545

Solution:

Original are of the rectangle = l*b where l= length and b = breadth
After the sides are increased by 100%, the area will increase by
= (((2l*2b)-(a*b))*100) / (a*b) = 300%

Solution:

Total games won by the team = 30 percent of 60 = 18
Now, the team wins continuous matches and raises winning % to 50
Thus, ((18 + x) / (60 + x)) * 100 = 50
(18 + x) / (60 + x) = 1/2
x = 24
Thus the team must have won 24 matches to raise the winning percentage to 50 percent.

Solution:

The ant climbs 6 meters in one minute and then falls down 3 meters in the second minute, the distance that it covers in 2 minutes is 3 meters.

Thus for climbing 54 meters, the ant will require time = 52*2/3 = 36 minutes.

We have counted only for 54 meters as it will climb the remaining six meters in one minute.

Thus the total time required to climb the 60 meters wall = 36 + 1 = 37 minutes.

Solution:

Mike Cooper is the one.

We can exclude Martin from our investigations as he was the victim.

Terry Singer possesses an orange car. The thief possesses a blue car, thus Terry Singer cannot be the murderer.

The suspect who owns a blue car was wearing purple shoes.

The suspect who weighs 190 pounds was wearing purple shoes.

The suspect who weighs 190 pounds is not the one who has black hair. Therefore, one suspect was wearing purple shoes, possesses a blue car, weighs 190 pounds, and is not the one who has black hair.

But Jim Condon was wearing brown shoes, Sylvian Bogard has black hair, Jonah Jameson weighs 210 pounds, and Terry Singer possesses an orange car.

Therefore, the only suspect who could be this suspect is Mike Cooper.

As Mike Cooper, owns a blue car and the thief owns a blue car, he is definitely the thief.

Solution:

Calculate the LCM of 2, 4, 6, 8, 10 and 12 seconds.
The outcome will be 120 seconds.

Thus at 2 minutes, it rings once.

Which means that the bells will ring 15 times in the 30 minutes.

But wait that will be an incorrect answer as the bells will also ring together in the 0th second.

Which means the bells toll together 15 + 1 = 16 times.

Solution:

What we will do first is we will subtract the driving time of twenty minutes from the 2 hours 30 minutes difference and we are left with 2 hours 10 minutes.

Now Tia left in morning, thus for the mirror effect, the hands must be pointing up and down i.e. at 6 o'clock. (It could have been at 12 o'clock as well as the hands are straight up but we know she left in the morning).

Now we have to divide 2 hours and 10 minutes by 2 in order to find a center point that will be 65 minutes in this case for compensating for the mirror.

Now if we add 65 minutes to 6 o'clock, we get the time when she left the home which will be 7:05 and the time which she saw in the mirror was 4:55.

Now if we add the 20 minutes we had subtracted previously again to the time she left her home, we will get the time she reached at the party which will be 7:25.

Thus the time when she reached the party was 7:25.