To Think or Not to Think
- Thread starter adilsaleem
- Start date
Divide the balls in three groups of 6,6, 1.
Weigh the two groups of six balls(I).If equal, the last one is heavier.
If unequal, take the heavier group and devide in group of three's and weight them (II),Whichever is heavier take that group of three balls and divide them in three groups of one ball each. Weigh two groups (III), if equal , last ball else the ball which tilted the balance, is the heaviest one.
Pankaj
Ok so we change the name of Traderji to Teaserji now
Weigh the two groups of six balls(I).If equal, the last one is heavier.
If unequal, take the heavier group and devide in group of three's and weight them (II),Whichever is heavier take that group of three balls and divide them in three groups of one ball each. Weigh two groups (III), if equal , last ball else the ball which tilted the balance, is the heaviest one.
Pankaj
Ok so we change the name of Traderji to Teaserji now
Yes !! If teacher wins the case (I don't know how), court will compel student to give the fees and if teacher loses the case, the student, having won his 1st case, is liable to pay the fee to his teacher.
So... chit bhi meri, pat bhi meri !!!
Abhay (AAD)
So... chit bhi meri, pat bhi meri !!!
Abhay (AAD)
I guess there is something more to this ....
The answer to the 13 balls was perfect.
If 6 x 4 = 12
8 x 6 = 32
11 x 8 = 66
then 10 x 10 = ??
Submitted by : Nick Chimenti
Answer
80
The pattern is multiply the first number by second after reducing second number by 2. One can view is as follow:
6 x (4 - 2) = 12
8 x (6 - 2) = 32
11 x (8 - 2) = 66
10 x (10 - 2) = 80
Few years ago, a Law teacher came across a student who was willing to learn but was unable to pay the fee. The student struck a deal saying, "I would pay your fee the day I win my first case in the court". Teacher agreed and proceeded with the law course.
When the course was finished and teacher started pestering the student to pay up the fee, student reminded the deal and pushed days. Fed up with this, the teacher decided to sue the student in the court of law and both of them decided to argue for themselves.
Did the teacher make the right decision? Will he get his fee from the student?
Answer
This is one of the greatest paradoxes ever recorded in history.
The teacher can put forward his argument saying: "If I win this case, as per the court of law, student has to pay me. And if I lose the case, student will still pay me because he would have won his first case. So either way i will have to get the money".
Equally student can argued back saying: "If I win the case, as per the court of law, I don't have to pay anything to the teacher. And if I lose the case, I don't have to pay him because I haven't won my first case yet. So either way, I am not going to pay the teacher anything".
You have 13 balls which all look identical. All the balls are the same weight except for one. Using only a balance scale, can find the odd one out with only 3 weighings?
Is it possible to always tell if the odd one out is heavier or lighter than the other balls?
Submitted by : Brett Hurrell
Answer
It is always possible to find odd ball in 3 weighings and in most of the cases it is possible to tell whether the odd ball is heavier or lighter. Only in one case, it is not possible to tell the odd ball is whether heavier or lighter.
Take 8 balls and weigh 4 against 4.
If both are not equal, goto step 2
If both are equal, goto step 3
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 5 balls in intial 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.
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.
One of the remaining 5 balls is the odd one. Name the balls as C1, C2, C3, C4, C5. Weight (C1, C2, C3) against (X1, X2, X3) where X1, X2, X3 are any three balls from the first weighing of 8 balls.
If both are equal, one of remaining 2 balls is the odd i.e. either C4 or C5. Weigh C4 with X1
If both are equal, C5 is the odd ball. But you can not tell whether it is heavier or lighter.
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, C3) is heavier side, one of C1, C2, C3 is the odd ball and is heavier. Weigh C1 and C2.
If both are equal, C3 is the odd ball and is heavier.
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, C3) is lighter side, one of C1, C2, C3 is the odd ball and is lighter. 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.
In Laloo's family, each son has the same number of sisters and brothers. Also, each daughter has twice the number of brothers than sisters.
How many sons and daughters does Laloo have?
Answer
4 sons and 3 daughters
Laloo must be having one more son then daughter, as each son has same number of sisters and brothers. Using this and little trial-and-error, we can get the result i.e. 4 sons and 3 daughters.
Each brother has 3 sisters and 3 brothers.
Each sister has 2 sisters and 4 brothers.
A fly is flying between two trains, each travelling towards each other on the same track at 60 km/h. The fly reaches one engine, reverses itself immediately, and flies back to the other engine, repeating the process each time.
The fly is flying at 90 km/h. If the fly flies 180 km before the trains meet, how far apart were the trains initially?
Answer
Initially, the trains were 240 km apart.
The fly is flying at the speed of 90 km/h and covers 180 km. Hence, the fly flies for 2 hours after trains started.
It's obvious that trains met 2 hours after they started travelling towards each other. Also, trains were travelling at the speed of 60 km/h. So, each train travelled 120 km before they met.
Hence, the trains were 240 km apart initially.
8 x 6 = 32
11 x 8 = 66
then 10 x 10 = ??
Submitted by : Nick Chimenti
Answer
80
The pattern is multiply the first number by second after reducing second number by 2. One can view is as follow:
6 x (4 - 2) = 12
8 x (6 - 2) = 32
11 x (8 - 2) = 66
10 x (10 - 2) = 80
Few years ago, a Law teacher came across a student who was willing to learn but was unable to pay the fee. The student struck a deal saying, "I would pay your fee the day I win my first case in the court". Teacher agreed and proceeded with the law course.
When the course was finished and teacher started pestering the student to pay up the fee, student reminded the deal and pushed days. Fed up with this, the teacher decided to sue the student in the court of law and both of them decided to argue for themselves.
Did the teacher make the right decision? Will he get his fee from the student?
Answer
This is one of the greatest paradoxes ever recorded in history.
The teacher can put forward his argument saying: "If I win this case, as per the court of law, student has to pay me. And if I lose the case, student will still pay me because he would have won his first case. So either way i will have to get the money".
Equally student can argued back saying: "If I win the case, as per the court of law, I don't have to pay anything to the teacher. And if I lose the case, I don't have to pay him because I haven't won my first case yet. So either way, I am not going to pay the teacher anything".
You have 13 balls which all look identical. All the balls are the same weight except for one. Using only a balance scale, can find the odd one out with only 3 weighings?
Is it possible to always tell if the odd one out is heavier or lighter than the other balls?
Submitted by : Brett Hurrell
Answer
It is always possible to find odd ball in 3 weighings and in most of the cases it is possible to tell whether the odd ball is heavier or lighter. Only in one case, it is not possible to tell the odd ball is whether heavier or lighter.
Take 8 balls and weigh 4 against 4.
If both are not equal, goto step 2
If both are equal, goto step 3
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 5 balls in intial 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.
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.
One of the remaining 5 balls is the odd one. Name the balls as C1, C2, C3, C4, C5. Weight (C1, C2, C3) against (X1, X2, X3) where X1, X2, X3 are any three balls from the first weighing of 8 balls.
If both are equal, one of remaining 2 balls is the odd i.e. either C4 or C5. Weigh C4 with X1
If both are equal, C5 is the odd ball. But you can not tell whether it is heavier or lighter.
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, C3) is heavier side, one of C1, C2, C3 is the odd ball and is heavier. Weigh C1 and C2.
If both are equal, C3 is the odd ball and is heavier.
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, C3) is lighter side, one of C1, C2, C3 is the odd ball and is lighter. 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.
In Laloo's family, each son has the same number of sisters and brothers. Also, each daughter has twice the number of brothers than sisters.
How many sons and daughters does Laloo have?
Answer
4 sons and 3 daughters
Laloo must be having one more son then daughter, as each son has same number of sisters and brothers. Using this and little trial-and-error, we can get the result i.e. 4 sons and 3 daughters.
Each brother has 3 sisters and 3 brothers.
Each sister has 2 sisters and 4 brothers.
A fly is flying between two trains, each travelling towards each other on the same track at 60 km/h. The fly reaches one engine, reverses itself immediately, and flies back to the other engine, repeating the process each time.
The fly is flying at 90 km/h. If the fly flies 180 km before the trains meet, how far apart were the trains initially?
Answer
Initially, the trains were 240 km apart.
The fly is flying at the speed of 90 km/h and covers 180 km. Hence, the fly flies for 2 hours after trains started.
It's obvious that trains met 2 hours after they started travelling towards each other. Also, trains were travelling at the speed of 60 km/h. So, each train travelled 120 km before they met.
Hence, the trains were 240 km apart initially.