Afroditee
5 collectors have some rare collection, then comes the \(6^{th}\) collector with 50 pieces of collection, and then the average piece of collection is 5 more than the average before. Then comes the \(7^{th}\) collector with some collection, and the average is 5 more than the average of the first 6 collectors. How many pieces of collection does the \(7^{th}\) collector have?
A. 54
B. 60
C. 64
D. 68
E. 70
Using a reasoning and logical approach:
The arithmetic mean is a measure of what the “typical element” in a set should be.
(avg) (count) = total sum —— we can use the Arithmetic Mean as a “proxy” for each person’s number by apportioning this total sum evenly to each person (i.e., giving each person his or her own “average representation” to stand for the actual data point)
Say the original average of the 5 people = X per person
C1 - C2 - C3 - C4 - C5
X —-X — X —- X —X
Before the 6th collector comes in, each person has an apportioned average = X
The 6th person comes in with 50 items.
These 50 items are enough to raise the average by +5 points —— meaning each person gets an extra +5 points out of the 50 brought in
C1 — C2 — C3 — C4 — C5 — 6th person
X —— X —- X —- -X —— X —- 50
+5 —- +5 — +5 —- +5 —-+5 —-> — 25
So - 25 is taken out of the 50 brought in, leaving 25 for the sixth person’s “average representation element”
This means the new average must be 25.
Original average was 20.
The 7th collector now comes in and brings an unknown amount of items: call it N
(1st) N must be enough to raise the prior 6 people’s “average representation” by +5 points
AND
(2nd) still be enough so that the new, 7th person can also have his or her own “average representation”.
(1st) +5 points given to prior 6 people: (+5) (6) = 30
AND
(2nd) Since the new average of the 7 people is 30, the 7th person needs to bring in another 30 for his own average representation.
30 + 30 = 60
7th person must have 60 items.
Answer: 60
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