61. TRIANGLE PROBABILITY.
A stick of length 1 is divided randomly into 3 parts. What is the probability that a triangle can be made with those three parts?
Answer:
The probability, that a triangle can be made by randomly dividing a stick of length 1 into 3 parts, is 25%. A triangle can be made, if and only if, sum of two sides is greater than the third side. Thus, X1 <> X1 + X2 + X3 = 1. From above equations: X1 < 1/2, X2 < 1/2, X3 < 1/2. Thus, a triangle can be formed, if all three sides are less than 1/2 and sum is 1. Now, let's find the probability that one of X1, X2, X3 is greater than or equal to 1/2. Note that to divide stick randomly into 3 parts, we need to choose two numbers P and Q, both are between 0 & 1 and P. Now, X1 will be greater than or equal to 1/2, if and only if both the numbers, P & Q , are greater than or equal to 1/2. Thus, probability of X1 being greater than or equal to 1/2 is = (1/2) * (1/2) = 1/4. Similarly, X3 will be greater than or equal to 1/2, if and only if both the numbers, P & Q, are less than or equal to 1/2. Thus, probability of X3 being greater than or equal to 1/2 is = (1/2) *(1/2) = 1/4. Also, probability of X2 being greater than or equal to 1/2 is = (1/2)* (1/2) = 1/4. The probability that a triangle can not be made = (1/4) + (1/4) + (1/4) = (3/4). Thus, the probability that a triangle can be made = 1 - (3/4) = (1/4) = 25 %. Thus, the probability that a triangle can be made by randomly dividing a stick of length 1 into 3 parts is 25%.
ADDITIONAL INFORMATION FOR INTERESTED PUZZLE LOVERS:
Let's generalize the problem. What is the probability that a polygon with (N+1) sides can be made from (N+1) segments obtained by randomly dividing a stick of length l into (N+1) parts?The probability is = 1 - (N+1)*(1/2)^N. The probability tends to 1 as N grows. Thus, it is easier to make a N-sided polygon than it is to make a triangle!!!
62. MEASURE 45 MINUTES.
There are two ropes. Each one can burn in exactly one hour. They are not necessarily of the same length or width as each other. They also are not of uniform width (may be thiner/wider in middle than on the end), thus burning half of the rope is not necessarily 1/2 hour. By burning the ropes, how do you measure exactly 45 minutes worth oftime?
Answer:
If you light both ends of one rope simultaneously, it will burn in exactly a 1/2 hour. Thus, burn one rope from both ends and the other rope from only one end. Once the one rope (which is burning from both ends) finally burns out (and you know a 1/2 hour has elapsed), you also know that the other rope (which is burning from only one end) has exactly 1/2 hour left to burn. Since you only want 45 minutes, light the second end of the rope. This remaining piece will burn in15 minutes. Thus, totaling 45 minutes.
63. ODD BALL.
You are given 12 balls and one of them has a weight defect - either heavier or lighter. We don't know which one it is. You are allowed to use the balance only three times. Find out which one of the twelve is the defective ball?
Answer:
It is always possible to find the odd ball in 3 weighings and to tell whether the odd ball is heavier or lighter.
Step: 1) Take 8 balls and weigh 4 against 4. If both are not equal, goto step 2. If both are equal, goto step 3.
Step: 2) One of these 8 balls is the odd one. Name the balls on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the balls on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is one ball from the remaining 4 balls inintial weighing. If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3. If both are equal, L4 is the odd ball and is lighter. If L2 is light, L2 is the odd ball and is lighter. If L3 is light, L3 is the odd ball and is lighter. If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2. If both are equal, there is some error in weighing. If H1 is heavy, H1 is the odd ball and is heavier. If H2 is heavy, H2 is the odd ball and is heavier. If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4. If both are equal, L1 is the odd ball and is lighter. If H3 is heavy, H3 is the odd ball and is heavier. If H4 is heavy, H4 is the odd ball and is heavier.
Step: 3) One of the remaining 4 balls is the odd one. Name the balls as C1, C2, C3, C4. Weight (C1, C2) against (C3, X) where X is anyball from the first weighing of 8 balls. If both are equal, C4 is the odd ball. Weigh C4 with X. If both are equal, there is some error in weighing. If C4 is heavy, C4 is the odd ball and is heavier. If C4 is light, C4 is the odd ball and is lighter. If (C1, C2) is heavier side, one of C1, C2 is heavier or C3 is lighter. Weigh C1 and C2. If both are equal, C3 is the odd ball and is lighter. If C1 is heavy, C1 is the odd ball and is heavier. If C2 is heavy, C2 is the odd ball and is heavier. If (C1, C2) is lighter side, one of C1, C2 is lighter or C3 is heavier. Weigh C1 and C2. If both are equal, C3 is the odd ball and is heavier. If C1 is light, C1 is the odd ball and is lighter. If C2 is light, C2 is the odd ball and is lighter.
64. HOW MANY PEOPLE?
In the middle of the confounded desert, there is the lost city of "Mystery". To reach it, I will have to travel overland by foot from the coast. On a trek like this, each person can only carry enough rations for five days and the farthest we can travel in one day is 30 miles. Also, the city is 120 miles from the starting point. What I am trying to figure out is the fewest number of persons, including myself, that I will need in our Group so that I can reach the city, stay overnight, and then return to the coast without running out of supplies. How many persons (including myself) will I need to accomplish this mission?
Answer:
Total 4 persons (including you) required. It is given that each person can only carry enough rations for fivedays. And there are 4 persons. Hence, total of 20 days rations is available. First Day : 4 days of rations are used up. One person goes back using one day of rations for the return trip. The rations remaining for the further trek is for 15 days.
Second Day : The remaining three people use up 3 days of rations. One person goes back using 2 days of rations for the return trip. The rations remaining for the further trek is for 10 days.
Third Day : The remaining two people use up 2 days of rations. One person goes back using 3 days of rations for the return trip. The rations remaining for the further trek is for 5 days.
Fourth Day : The remaining person uses up one day of rations. He stays overnight. The next day he returns to the coast using 4 days ofrations. Thus, total 4 persons, including you are required.
65. TYING THE EARTH.
Assume for a moment that the earth is a perfectly uniform sphere of radius 6400 km. Suppose a thread equal to the length of the circumference of the earth was placed along the equator, and drawn to a tight fit. Now suppose that the length of the thread is increased by 12 cm, and that it is pulled away uniformly in all directions. By how many cm, will the thread be separated from the earth's surface?
Answer:
1.908 cm.
Let the outer circle have radius as 6400+h cm and therefore its circumference is 2*Pi*(6400+h) Earth's circumference is 2*Pi*6400. Hence the difference = 2*Pi*h = 12 cm. Therefore h=1.908 cm. It is worthwhile to note that this factor does not involve the radius of the sphere and is independent of it. This means that irrespective of the size of the sphere, i.e. irrespective of the radius of it, whether it is a giant globe or a small tennis ball, it holds good.
66. HOW MANY DIAGONALS?
A polygon has 1325 diagonals. How many vertices does it have?
Answer:
The formula to find number of diagonals (D) given total number of vertices or sides (N) is D = N * (N - 3)/2. Using the formula, we get 1325 * 2 = N * (N - 3). N**2 - 3N - 2650 = 0. Solving the quadratic equation, we get N = 53 or -50. It is obvious that answer is 53 as number of vertices can not be negative. Q.E.D.
67. HOW FAR POLES APART?
A cable, 16 meters in length, hangs between two pillars that are both 15 meters high. The ends of the cable are attached to the tops of the pillars. At its lowest point, the cable hangs 7 meters above the ground. How far are the two pillars apart?
Answer:
Zero meters apart. Since the lowest point is 7 metres from the ground, the distance of the lowest point from the top of the pillar is 8 metres which is half of 16 metre (lenth of cable). So cable should be in folded position. So distance between the pillars is 0 meters.
68. OPTIMUM CHANCE.
Long long ago, a young prince wanted to marry a kings's daughter. The king decided to test the prince. He gave the prince two empty boxes, 100 white balls, and 100 black balls. "You must put all the balls in the boxes", he told the prince. "After this, I will call my daughter from the room next door. She will take a random ball from one of the two boxes. If this ball is a black one, you are allowed to marry my daughter. The Question is What was the best way in which the prince could divide the balls over the boxes?
Answer:
The best way is to put one black ball in the first box, and all other balls in the second box. Then, the probability of grabbing a blackball from the first box is 1, and the probability of grabbing a black ball from the second box is 99/199. The total probability of grabbing a black ball is 0.5 � 1 + 0.5 � 99/199 = 298/398(approximately 74.9%).
69. FLYING DISTANCE.
Consider 4 (dimensionless) flies. They are situated at the corners of a square whose side is 1 meter. Each fly tries to reach the another fly in front of it. Since the flies are flying towards another, they will meet each other at a certain time in the center of the square. The Question is what is the length of the path they have travelled at the moment they reach each other?
Answer:
Because all flies constantly fly perpendicular to another fly, they all travel the shortest distance to each other, which is 1 meter (all flies make a kind of spiral flight to the center of the square, and during this flight, the flies constantly form a square until theymeet in the center). Therefore the flies all travel 1 meter each.
70. FRIGHTENING FRACTION.
Exactly 63.41463414 % of the people asked if they used a certain product, replied "Yes". What is the smallest number of people who could have been asked thequestion?
Answer:
As a fraction, 63.41463414% = 63414/99999 = 7046/11111 = 26/4141 people were asked, 26 replied yes.
