## Welcome to BrainDen.com - Brain Teasers Forum

 Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account. As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends. Of course, you can also enjoy our collection of amazing optical illusions and cool math games. If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top. If you have a website, we would appreciate a little link to BrainDen. Thanks and enjoy the Den :-)
Guest Message by DevFuse

# negative*negative=positive

### #1 kkhebbar

kkhebbar

Newbie

• Members
• 2 posts

Posted 18 September 2007 - 11:32 PM

Here is logical querry. A man is in the intersection of a unmarked Y junction. One road goes to a place say "Heaven" and the other road goes to a place called "Hell". This man wants to head in the direction of Heaven but does not know which direction to head.

Each road has a gatekeeper who are look alikes and one is from Heaven and the other is from Hell. They both are standing at the junction who could help this man for direction. The only problem is the person from Heaven always tells the truth and person from Hell always tells a lie. This fact is known to the stranger man too.

What could be the possible question this stanger man should ask the gatekeeper and can be sure that leads to Heaven? Given a criterion that he can ask only one question and to only one person. The person can be either from Heaven or Hell but the stranger man does not know about it. (HINT = in the subject of the puzzle)

Spoiler for Solution

• 0

### #2 Martini

Martini

Senior Member

• Members
• 770 posts

Posted 19 September 2007 - 12:47 AM

This is a variant of Honestants and Swindlecants III.. Any further discussion can be had in the existing thread.