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# Monthly Archives: January 2011

# Four People on a Rickety Bridge

There are four people who need to cross a rickety bridge in 17 minutes. Because of its dilapidated condition, the bridge can only support up to 2 people at any given time. In addition, since it is pitch black out and there is only one lantern between the four people, anybody crossing the bridge must carry the lantern (or if two people are crossing the bridge at the same time, then one of them must be carrying the lantern). Someone must always bring the lantern back to the people who haven’t crossed the bridge yet, so that when they cross, they don’t fall through the holes in the bridge. You must take into account the additional time it will take for them to cross back.

The four people can cross the bridge at the rate of 1, 2, 5, and 10 minutes, each way.

There are no tricks: you can’t carry anyone, or throw the lantern or drop it on the bridge, or swim across.

# Gold Bar For 7 Days of Work

A worker is to perform work for you for seven straight days. In return for his work, you will pay him 1/7th of a bar of gold per day. The worker requires a daily payment of 1/7th of the bar of gold. What and where are the fewest number of cuts to the bar of gold that will allow you to pay him 1/7th each day?

Day 1: Give worker 1 bar

Day 2: Take back 1 bar and give 2 piece bar

Day 3: Give 1 bar [Making 2 + 1 = 3]

Day 4: Take back 1 & 2 piece bar and give 4 piece bar

Day 5: Give 1 piece bar [Making 4 + 1 = 5]

Day 6: Take back 1 piece and give 2 piece

[Making 5 – 1 + 2 = 6]

Day 7: Give 1 piece bar

Day 1: Give worker 1 bar

Day 2: Take back 1 bar and give 2 piece bar

Day 3: Give 1 bar [Making 2 + 1 = 3]

Day 4: Take back 1 & 2 piece bar and give 4 piece bar

Day 5: Give 1 piece bar [Making 4 + 1 = 5]

Day 6: Take back 1 piece and give 2 piece

[Making 5 – 1 + 2 = 6]

Day 7: Give 1 piece bar

# A Box of Defective Balls (1 Box in total of 10 boxes)

You have 10 boxes of balls (each ball weighing exactly10 gm) with one box with defective balls (each one of the defective balls weigh 9 gm). You are given an electronic weighing machine and only one chance at it. How will find out which box has the defective balls?

Mark the boxes as box1,box2,box3………box10

then from 1st box we have to take 1 ball,from 2nd box 2 ball……….from 10th box we kept 10 balls. and weight the total ball….

Mark the boxes as box1,box2,box3………box10

then from 1st box we have to take 1 ball,from 2nd box 2 ball……….from 10th box we kept 10 balls. and weight the total ball….

so if total nos ball is :n(n+1)/2 (n= total nos of box that is 10)=55

if all the ball is 10 gm.then total weight will be=550gm.

suppose total weight is 546gm.

so the difference is(550-546=4gm).

then the d defective ball is in box no:4.

similarly we can try for all possible box difference gives us the box number which contains the defective balls

# Defective Ball in Identical Balls – Optimum Weighing times

12 balls are given. All but one are of equal weight. You do not know whether the defective ball is lighter or heavier than the normal balls. You are given a comparison balance. You can use the balance only 3 times. How would you find out which is the defective ball? Is it heavier or lighter?

**Solution :**

There may be more than one solution, however, I am describing here the one which struck me. We will divide balls in groups of four. Thus we will have three groups, A, B and C. Without loss of generality , put group A and B on comparison balance.

**case 1 :**A = B

In this case, we know that defective ball is in group C. Put (C1.C2,C3) and ( A1,A2,A3) on comparison balance.

**case 1a :**(C1,C2,C3) = ( A1,A2,A3)

Clearly, C4 is the defective ball. Compare it with A1 and find out whether it is heavier or lighter.

**case 1b :**(C1,C2,C3) < ( A1,A2,A3)

Now, we know that defective ball is lighter. Compare C1 and C2. if they are unequal, we know which one is defective depending upon which one is lighter. If they are equal, we know that C3 is the lighter defective ball.

**case 1c :**(C1,C2,C3) > (A1,A2,A3)

Now, it is obvious that defective ball is heavier. Carry out the same procedure as described in case 1b, to find out which one is heavier defective ball.

**case 2 :**A > BFrom this observation we can deduce that one of the ball in group A is heavier or one of the ball in group B is lighter. Now, compare ( A4, B3,B4) and ( B1,B2,C1).

**case 2a :**( A4,B3,B4) > ( B1,B2,C1)

This tells us that either A4 is heavier or one of B1-B2 is lighter. Compare B1 and B2. If they are equal then A4 is the heavier defective ball. If not, we know which one is defective depending on which one is lighter.

**case 2b :**(A4,B3,B4) < ( B1,B2,C1)

This tells us that one of B3 and B4 is lighter defective ball.Compare B3 and B4 and find out which one is defective.

**case 2c :**(A4,B3,B4) = (B1,B2,C1)

This is an easier case. Now we know that one of A1,A2,A3 is defective and heavier. Compare A1 and A2 if they are equal A3 is defective, if not, the heavier of the two is defective.

**case 3 :**A < B

Line of reasoning is exactly similar to case 2. ( replace heavier by lighter and vice versa ).