For a classroom measuring 36 by 24 feet with a 3-foot wide door at the far end of one of the 36-foot long walls, how many receptacles are required?

For a classroom measuring 36 by 24 feet with a 3-foot wide door at the far end of one of the 36-foot long walls, how many receptacles are required?

• A. 9
• B. 0
• C. 17
• D. 4

Answer: (B)

The idea being tested is how receptacle outlets are required based on wall space and openings. Receptacles are placed so that every wall space at least 2 feet wide has a receptacle, and no point along the wall is more than about 6 feet from a receptacle; openings like doors cannot be used for mounting outlets and can break a wall into segments that may not qualify as wall spaces. In this scenario, a 3-foot-wide door sits at the far end of one of the 36-foot walls. That door occupies the end of that wall and interrupts the usable wall length, so there isn’t any uninterrupted wall space of at least 2 feet that can host a receptacle on that wall. With the given configuration and the typical interpretation used in this problem, there are no eligible wall spaces left to place a receptacle, so the calculation comes out to zero receptacles. If there were other walls with suitable uninterrupted wall spaces, receptacles would be placed there according to the same spacing rule.

The idea being tested is how receptacle outlets are required based on wall space and openings. Receptacles are placed so that every wall space at least 2 feet wide has a receptacle, and no point along the wall is more than about 6 feet from a receptacle; openings like doors cannot be used for mounting outlets and can break a wall into segments that may not qualify as wall spaces.

In this scenario, a 3-foot-wide door sits at the far end of one of the 36-foot walls. That door occupies the end of that wall and interrupts the usable wall length, so there isn’t any uninterrupted wall space of at least 2 feet that can host a receptacle on that wall. With the given configuration and the typical interpretation used in this problem, there are no eligible wall spaces left to place a receptacle, so the calculation comes out to zero receptacles. If there were other walls with suitable uninterrupted wall spaces, receptacles would be placed there according to the same spacing rule.