hard · Elliott Wave Theory
An impulse advances from a wave-1 origin where wave 1 spans 100 points. Wave 4 ends at the 0.382 retracement of the entire wave-1-through-3 distance, and waves 1 and 3 together cover exactly 261.8 points. An analyst must project the most common Fibonacci target for wave 5 measured from the wave-4 low.
Using the standard convention that wave 5 frequently equals the net traveled distance of waves 1 through 3 multiplied by 0.618 when wave 3 is extended, what is the projected length of wave 5?
- Approximately 161.7 points, taking 0.618 of the 261.8-point waves-1-through-3 net distance
- Approximately 100 points, on the basis that wave 5 equals wave 1 when wave 3 is the extended wave
- Approximately 38.5 points, applying 0.382 to the 100-point wave 1 as the wave-4 retracement implies
- Approximately 161.8 points, taking the 0.618 expansion of wave 1 plus the wave-4 retracement depth
Sign up free to see the explanation and track your rank →
More Elliott Wave Theory practice
- In a five-wave advance, Wave 1 is 10 points long, Wave 3 is… — How should this count be co
- A commodity price moves from $80 to $96, pullbacks to $88, t… — If an analyst identifies t
- Which is more likely?
- According to the Swing Count Validation technique, what should the trader conclude?
- Based on common Fibonacci relationships, how far might Wave C drop from the end of Wave B?
- An analyst sees a 'Close-Below-Prior-Swing Test' fire when p… — What does this likely sign
- According to the 'Fourth-Wave Target Zone' guideline, where is a correction most likely to
- According to the Guideline of Alternation, what should you expect for Wave 4?