It will require some extra tosses, but you can get a fair result this way:
Always use two throws to determine the answer, a HH or TT throw is discarded.HT == Heads and TH == Tails

If we start dropping the egg from 1st floor and try till 100 till the egg doesn’t break .We will require 100 eggs in worst case. Lets try to drop the egg from every 2nd floor .This will require 51 drops in worst case .Here’s how (100/2)=50 , let’s say the egg broke on 100 the floor now drop it from 99th floor it should not break. Similarly for dropping from every 3rd floor gives 35 drops [100/3 +3-1].Let’s say minimum number of steps required when we drops egg from every xth floor.

There fore minimum drops required would be [(100/10)+10-1]=19

The answer is 7 .Here’s how :
Divide the horses into group of 5 (A,B,C,D,E) and conduct 5 races to find 5 fastest horses .
Now conduct a race among these to find the fastest horse (Let’s say it’s from group E) .and let’s also assume that fastest horse from group D came 2nd and C came 3rd. Total 6 races done. Now the second fastest horse can be the one who came second in the group E which gave us the fastest horse or the one which came second in 6th race. Similarly the 3rd fastest horse can be
the either fastest horse from group D and C ,or the horses which came 2nd and 3rd in group E or the horse which came second in group D. Conduct a race among these 5 horses and you will get the result in final and 7th race.

This solution is calculated programatically. I will highly appreciate if someone can provide mathematical explanation of the points calculated here.

1. Initially 3000 bananas
   • 1st mile the camel carries 1000 bananas forward and eats 1, it eats 1 banana on the way back.
   • Again it carries 1000 bananas forward and eats 1, eats 1 going back.
   • 1000 bananas forward and eats 1.

   After 1 mile: (1000 – 1 – 1) + (1000 – 1 – 1) + (1000 – 1) = 2995 bananas left. So for every 1 mile camel eats 5 bananas. So the camel has to travel 200 miles, when the number of bananas left will be 3000 – 200*5 = 2000

2. Now the camel has to make only 2 trips for every 1 mile
   • 1st mile the camel carries 1000 bananas forward and eats 1, it eats 1 banana on the way back.
   • 1000 bananas forward and eats 1.

   After 1 mile: (1000 – 1 – 1) + (1000 – 1) = 1997 bananas left. So for every 1 mile camel eats 3 bananas.
   So to consume next 1000 bananas, the camel has to travel 1000/3 = 333.33 miles.

3. Now the camel is left with 1000 bananas at 200 + 333.33 = 533.33 miles mark. Now
   • The camel has to travel 1000 bananas forward and eat 1 per mile.
   • The camel still has to travel 1000 – 533.33 = 466.66 miles. So the camel is going to consume another 466.66 bananas.

Therefore the final number of bananas left will be 1000 – 466.66 = 533.33 or 533 (who wants to eat a camel eaten banana anyway

Open the Orange/Apple basket. If we get Apple then basket labeled as Orange will be “Orange/Apple” basket.

Send 1 and 2 first, 1 comes back, send 5 and 10, 2 comes back, send 2 and 1 to the other side.

Let G, S, and F represent the ages in years of the grandson, the son, and the father, respectively. Since a year has 365 days, or 52 weeks, or 12 months, the problem can be represented by three equations with three unknowns:

365 G = 52 S (The grandson is about as many days old as the son is in weeks)
12 G = F (The grandson is approximately as many months old as the father is in years)
G + S + F = 120 (The ages of the grandson, the son, and the father add up to 120 years)

Because the grandson’s age is G = F/12, the son’s age can be represented in terms of F by substituting:

S = (365/52)×G
S = (365/52)×(F/12)
The third equation can now be represented with only the father’s age as an unknown:

F/12 + (365/52)×(F/12) + F = 120

Multiplying by 12, we get: 

F + (365/52)×F + 12 F = 120×12
20 F = 1440
F = 72
G = F/12 = 6
S = (365/52)×6 = 42

The father is 72, the son is 42, and the grandson is 6 years old.
You may have noticed that the statement of the problem is somewhat ambiguous: “The grandson is approximately as many months old as the father is in years” indicates that there may be some variance. A year really has 365 1/4 days instead of 365, and 52 weeks times 7 days gives only 364 days in a year. These inconsistencies produce fractional results that need to be rounded.

Doing the division in the following equation:
F + (365/52)×F + 12 F = 120×12

produces 20.019 F = 1440, which is rounded to 20 F = 1440

Pierre = apple,
Michelle = muffin,
Irene = raisin,
Jean = chocolate,
Sarah = orange

Start with 2 children crossing.
One child gets out of boat, other returns.
Child gets out, soldier crosses.
Soldier gets out, child returns.
Both children cross…etc…