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13 Apr 2026, 13:22
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
15% (04:13) correct
85%
(02:57)
wrong
based on 26
sessions
History
Date
Time
Result
Not Attempted Yet
Stefanos and his five colleagues have a total of 350 sales leads. No two people have the same number of leads, and each person has at least one lead. The median number of leads among the six people is 75.
If Stefanos has 20 fewer leads than one of his colleagues, which of the following could be the number of leads Stefanos has?
I. 55
II. 70
III. 90
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III
If Stefanos has 20 fewer leads than one of his colleagues, which of the following could be the number of leads Stefanos has?
I. 55
II. 70
III. 90
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III
16 Apr 2026, 05:49
Kudos
Bookmarks
Stefanos and his five colleagues have a total of 350 sales leads. No two people have the same number of leads, and each person has at least one lead. The median number of leads among the six people is 75.
If Stefanos has 20 fewer leads than one of his colleagues, which of the following could be the number of leads Stefanos has?
I. 55
Stefanos: 55
Colleague: 75
There are six values, an even number of values. So, for the median to be 75, the middle two values have to average to 75. So, at least one other has to be 75. However, no two can be the same.
Eliminate.
II. 70
Stefanos: 70
Colleague: 90
Total for other four: 190
An 80 would make the median 70 + 80 = 75
Leaves 110 for the other three, of which two must be below 70 and one must be above 80.
Could be 107, 1, and 2.
70 could be the number of leads Stefanos has.
III. 90
We already know that I is out and II is in. So, the correct answer must be (B) II only.
We can check this one just to see how it works though.
Stefanos: 90
Colleague: 110
Total for other four: 150
For the median to be 75, the middle two have to average to 75. So the minimum for the other four is 74 + 76 + 1 + 2.
150 < 74 + 76 + 1 + 2
Eliminate.
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III
Correct answer: B
If Stefanos has 20 fewer leads than one of his colleagues, which of the following could be the number of leads Stefanos has?
I. 55
Stefanos: 55
Colleague: 75
There are six values, an even number of values. So, for the median to be 75, the middle two values have to average to 75. So, at least one other has to be 75. However, no two can be the same.
Eliminate.
II. 70
Stefanos: 70
Colleague: 90
Total for other four: 190
An 80 would make the median 70 + 80 = 75
Leaves 110 for the other three, of which two must be below 70 and one must be above 80.
Could be 107, 1, and 2.
70 could be the number of leads Stefanos has.
III. 90
We already know that I is out and II is in. So, the correct answer must be (B) II only.
We can check this one just to see how it works though.
Stefanos: 90
Colleague: 110
Total for other four: 150
For the median to be 75, the middle two have to average to 75. So the minimum for the other four is 74 + 76 + 1 + 2.
150 < 74 + 76 + 1 + 2
Eliminate.
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III
Correct answer: B
16 Apr 2026, 08:20
Kudos
Bookmarks
This is best to be done via POE.
First one is not possible because if you have 55 then 55+20=75. But 75 does not exist because 75 is the median itself and number of people ar 6.
You cant have a+b/2 = 75.
Thus not possible.
Second one:
I didnt how it fits into the picture but maybe with some combination of 3rd and 4th values they fit in. Doubtful but lets move on as we have no convincing way to prove it does not fit as yet.
Third one:
90 is cheeky because if you have 90 then 110 is the other number and both are right side of median and they add up to 200.
the 3rd and 4th value adds up to 150.
200+150=350 and you have no room for 2 more values to exist which have to be 0,0 but that is not possible based on the stem.
Thus eliminated.
Only 2 remains and hence is the answer.
First one is not possible because if you have 55 then 55+20=75. But 75 does not exist because 75 is the median itself and number of people ar 6.
You cant have a+b/2 = 75.
Thus not possible.
Second one:
I didnt how it fits into the picture but maybe with some combination of 3rd and 4th values they fit in. Doubtful but lets move on as we have no convincing way to prove it does not fit as yet.
Third one:
90 is cheeky because if you have 90 then 110 is the other number and both are right side of median and they add up to 200.
the 3rd and 4th value adds up to 150.
200+150=350 and you have no room for 2 more values to exist which have to be 0,0 but that is not possible based on the stem.
Thus eliminated.
Only 2 remains and hence is the answer.
kevincan
