Challenge your brains! Can you solve this problem?

[Poseidon]

Answer= Rs.800

Let the cost price of the product be X.
On selling for 10 gain : 1.1*x
Selling for Rs.80 less : x-80 0.9*x
0.1*x 80
and so x Rs.800

[Fan from VT]

I understand that thinking, but look at it this way.  The prize will be behind each door 1/3rd of the time, and the door that's eliminated is always an incorrect choice.  So here's all the possible outcomes if you always choose to switch:

I pick A, prize is actually behind A.  B or C is eliminated.  I switch, I lose.

I pick A, prize is behind B.  C is eliminated, I switch, I win.

I pick A, prize is behind C.  B is eliminated, I switch, I win.

The guesses aren't independent because only wrong answers are eliminated.  All else being equal, you win 2/3rds of the time when switching, and 1/3rd if you don't.

Actually, your logic there is flawed.  You're considering the eliminations of B or C in that first case to be combined incidences.  However, they are separate, just like your last two scenarios.  it should be:

I pick A, prize is actually behind A.  B is eliminated.  I switch, I lose.

I pick A, prize is actually behind A.  C is eliminated.  I switch, I lose.

I pick A, prize is behind B.  C is eliminated, I switch, I win.

I pick A, prize is behind C.  B is eliminated, I switch, I win.

So, no matter what, at the end, you have a 50/50 shot, regarding the selection with the last two doors as a separate step.  

EDIT:  Or maybe my logic is flawed.  The simulator sure seems to support your theory.

Yeah - the problem there is you're assuming your first two options occur the same % of the time, but in actuality A is only right 1/3rd of the time.  Your setup would require A to be right 50% of the time. Whether B or C is eliminated is random within that category, but both events are still a subset of the 1/3rd of the time that A has the prize.  They are combined incidences in that sense.

I've always thought that the easiest way to think about this problem is kind of backwards.

-If you switch every time, when do you lose?

-Well, you only lose if on your first pick you pick the door that has the prize. This happens 1/3 of the time, so you have a 1/3 chance of losing.  Ergo, you have a 2/3 chance of winning if you switch every time.



Or think of it this way:

If you pick a door NOT covering the prize on your first pick, then the other non-prize door is eliminated and if you switch, you win (because if you pick a non-winning door first, only 2 doors remain: winning and non-winning; the door that gets thrown away can only be non-winning, so the only one left is winning). Well, your odds of picking a non-winning door first is 2/3, and you win every time you pick a non-winning door, so you have a 2/3 chance of winning if you switch every time.

[Fan from VT]

I understand that thinking, but look at it this way.  The prize will be behind each door 1/3rd of the time, and the door that's eliminated is always an incorrect choice.  So here's all the possible outcomes if you always choose to switch:

I pick A, prize is actually behind A.  B or C is eliminated.  I switch, I lose.

I pick A, prize is behind B.  C is eliminated, I switch, I win.

I pick A, prize is behind C.  B is eliminated, I switch, I win.

The guesses aren't independent because only wrong answers are eliminated.  All else being equal, you win 2/3rds of the time when switching, and 1/3rd if you don't.

Actually, your logic there is flawed.  You're considering the eliminations of B or C in that first case to be combined incidences.  However, they are separate, just like your last two scenarios.  it should be:

I pick A, prize is actually behind A.  B is eliminated.  I switch, I lose.

I pick A, prize is actually behind A.  C is eliminated.  I switch, I lose.

I pick A, prize is behind B.  C is eliminated, I switch, I win.

I pick A, prize is behind C.  B is eliminated, I switch, I win.

So, no matter what, at the end, you have a 50/50 shot, regarding the selection with the last two doors as a separate step.  

EDIT:  Or maybe my logic is flawed.  The simulator sure seems to support your theory.

Yeah - the problem there is you're assuming your first two options occur the same % of the time, but in actuality A is only right 1/3rd of the time.  Your setup would require A to be right 50% of the time. Whether B or C is eliminated is random within that category, but both events are still a subset of the 1/3rd of the time that A has the prize.  They are combined incidences in that sense.

Right. If you pick A, there's a 1/3 chance the prize is there, A, THEN there's 50% shot B is eliminated and a 50% shot C is eliminated; doesn't matter because both doors are empty. So the odds of occurrence for you situation is:
1/6 lose
1/6 lose
1/3 win
1/3 win

1/3 lose, 2/3 win

[wjb3000]

I understand that thinking, but look at it this way.  The prize will be behind each door 1/3rd of the time, and the door that's eliminated is always an incorrect choice.  So here's all the possible outcomes if you always choose to switch:

I pick A, prize is actually behind A.  B or C is eliminated.  I switch, I lose.

I pick A, prize is behind B.  C is eliminated, I switch, I win.

I pick A, prize is behind C.  B is eliminated, I switch, I win.

The guesses aren't independent because only wrong answers are eliminated.  All else being equal, you win 2/3rds of the time when switching, and 1/3rd if you don't.

Actually, your logic there is flawed.  You're considering the eliminations of B or C in that first case to be combined incidences.  However, they are separate, just like your last two scenarios.  it should be:

I pick A, prize is actually behind A.  B is eliminated.  I switch, I lose.

I pick A, prize is actually behind A.  C is eliminated.  I switch, I lose.

I pick A, prize is behind B.  C is eliminated, I switch, I win.

I pick A, prize is behind C.  B is eliminated, I switch, I win.

So, no matter what, at the end, you have a 50/50 shot, regarding the selection with the last two doors as a separate step.  

EDIT:  Or maybe my logic is flawed.  The simulator sure seems to support your theory.

Yeah - the problem there is you're assuming your first two options occur the same % of the time, but in actuality A is only right 1/3rd of the time.  Your setup would require A to be right 50% of the time. Whether B or C is eliminated is random within that category, but both events are still a subset of the 1/3rd of the time that A has the prize.  They are combined incidences in that sense.

I've always thought that the easiest way to think about this problem is kind of backwards.

-If you switch every time, when do you lose?

-Well, you only lose if on your first pick you pick the door that has the prize. This happens 1/3 of the time, so you have a 1/3 chance of losing.  Ergo, you have a 2/3 chance of winning if you switch every time.



Or think of it this way:

If you pick a door NOT covering the prize on your first pick, then the other non-prize door is eliminated and if you switch, you win (because if you pick a non-winning door first, only 2 doors remain: winning and non-winning; the door that gets thrown away can only be non-winning, so the only one left is winning). Well, your odds of picking a non-winning door first is 2/3, and you win every time you pick a non-winning door, so you have a 2/3 chance of winning if you switch every time.

Cool, I finally came to that realization after reading everyone elses responses, thanks for posting.

Good thread!

[chicagogreen]

Answer= Rs.800

Let the cost price of the product be X.
On selling for 10 gain : 1.1*x
Selling for Rs.80 less : x-80 0.9*x
0.1*x 80
and so x Rs.800

I got something close to this.
But, gotta say, the use of language is really weak in the original question.  Assuming I understand what it is saying, try this.

selling at 80 represents a 10% loss. so,
.9x=80
x(the cost of the thing for the seller) = 80/.9x-
x=88.88

but the question is what is the price the seller is selling at that represents a 10% profit.

so cost x (88.88) plus 10% of x
88.88 + 8.88 =
97.6
rupees?

thats what i get

[MBz]

Suppose you had eight billiard balls, the recruiter began. One of them is slightly heavier, but the only way to tell is by putting it on a scale against the others. What's the fewest number of times you'd have to use the scale to find the heavier ball?

Twice.  I'll name each ball a number from 1-8.  Start by putting balls 1-3 on one side and balls 4-6 on the other.  Leave 7 and 8 off.  If the first weighing is of equal weight,  put balls 7 and 8 on seperate sides and see which one weighs more.  If in the first weighing, they are not equal, take the side that weighs more.  Pick any 2 of those 3 balls to put on the scale, if one is heavier, that is your heaviest ball, if they are equal, the ball which was put to the side is the heaviest.

[Fan from VT]

Suppose you had eight billiard balls, the recruiter began. One of them is slightly heavier, but the only way to tell is by putting it on a scale against the others. What's the fewest number of times you'd have to use the scale to find the heavier ball?

Twice.  I'll name each ball a number from 1-8.  Start by putting balls 1-3 on one side and balls 4-6 on the other.  Leave 7 and 8 off.  If the first weighing is of equal weight,  put balls 7 and 8 on seperate sides and see which one weighs more.  If in the first weighing, they are not equal, take the side that weighs more.  Pick any 2 of those 3 balls to put on the scale, if one is heavier, that is your heaviest ball, if they are equal, the ball which was put to the side is the heaviest.

Like it.

[hwangjini_1]

Suppose you had eight billiard balls, the recruiter began. One of them is slightly heavier, but the only way to tell is by putting it on a scale against the others. What's the fewest number of times you'd have to use the scale to find the heavier ball?

Twice.  I'll name each ball a number from 1-8.  Start by putting balls 1-3 on one side and balls 4-6 on the other.  Leave 7 and 8 off.  If the first weighing is of equal weight,  put balls 7 and 8 on seperate sides and see which one weighs more.  If in the first weighing, they are not equal, take the side that weighs more.  Pick any 2 of those 3 balls to put on the scale, if one is heavier, that is your heaviest ball, if they are equal, the ball which was put to the side is the heaviest.

Like it.

ditto. nice play on that one.

[bstnsportsfan3]

i just read this entire thread...and all i can say is my brain now hurts

[rondo987]

heres one...

One machine can shred a truck load of paper in 4 hours, and another machine can shred a truck load of paper in 2 hours. If the machines were to shred together, how long would it take to shred a truck load of paper?

[rondo987]

I couldn't figure it out to the second, partly because I'm too lazy to think. But im pretty sure its somewhere around 1 hour and 20 minutes

[GKC]

The ball problem reminded me of something the Dallas Mavericks stats guy said. You always shoot a 3 to win especially when away, when down 2. Reason being, is that you have say 35% to hit a 3. 50% to hit a 2, but usually 50% chance to win overtime (home teams win slightly more). So if you take a 2, you only have about 25% chance to win the game.

Obviously foul trouble and momentum should be looked at, but all things equal, go for the win.

heres one...

One machine can shred a truck load of paper in 4 hours, and another machine can shred a truck load of paper in 2 hours. If the machines were to shred together, how long would it take to shred a truck load of paper?

Truck A = X
Truck B = Y
Z = Minutes
A = How many truckloads of paper is shredded
X shreds 1/240 truck loads of paper per minute (240 minutes in 4 hours)
Y shreds 1/120 truck loads of paper per minute (120 minutes in 2 hours)

So A = (1/240)z + (1/120)z
If Z = 1, A = 0.0125
If A = 1, Z = 80

So 80 minutes.

---

This one's one I found in class:

You have 2 pieces of string.
Each, when lit, will burn from one end to another in an hour.
However, the interval in which it burns is random (ie it could burn faster at certain points, slower at others. Though total it will take an hour).

How do you use the 2 pieces of string and a lighter to measure 45 minutes?

[Montana]

Here's one: The professor gave one of the students a 40 pound brick and told him to cut it up into 4 separate pieces so that it would be possible to weigh any amount up to the 40 pounds. The student came back to class only to tell the professor that he couldn't figure out how to do it and then accidentally dropped it on the floor and low and behold it broke into the 4 exact pieces. What was the weight of each piece?

[mgent]

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

[Fan from VT]

The ball problem reminded me of something the Dallas Mavericks stats guy said. You always shoot a 3 to win especially when away, when down 2. Reason being, is that you have say 35% to hit a 3. 50% to hit a 2, but usually 50% chance to win overtime (home teams win slightly more). So if you take a 2, you only have about 25% chance to win the game.

Obviously foul trouble and momentum should be looked at, but all things equal, go for the win.

heres one...

One machine can shred a truck load of paper in 4 hours, and another machine can shred a truck load of paper in 2 hours. If the machines were to shred together, how long would it take to shred a truck load of paper?

Truck A = X
Truck B = Y
Z = Minutes
A = How many truckloads of paper is shredded
X shreds 1/240 truck loads of paper per minute (240 minutes in 4 hours)
Y shreds 1/120 truck loads of paper per minute (120 minutes in 2 hours)

So A = (1/240)z + (1/120)z
If Z = 1, A = 0.0125
If A = 1, Z = 80

So 80 minutes.

---

This one's one I found in class:

You have 2 pieces of string.
Each, when lit, will burn from one end to another in an hour.
However, the interval in which it burns is random (ie it could burn faster at certain points, slower at others. Though total it will take an hour).

How do you use the 2 pieces of string and a lighter to measure 45 minutes?

Light one string at one end and at the same time light the other string at both ends.

When the string that is doubly lit completely burns out (i.e. the flames meet), it's been 30 minutes. At that moment, light the unlit end of the other string, the one burning at only one end. When those two flames meet and extinguish the second string, it's been 45 minutes