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08 May 2026, 01:03
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
29% (02:43) correct
71%
(03:51)
wrong
based on 7
sessions
History
Date
Time
Result
Not Attempted Yet
The arithmetic mean of five positive integers A, B, C, D, E is 48. If A < B < C < D < E and E = 100, what is the difference between the greatest and least possible values of the median of the five integers?
A) 48
B) 50
C) 53
D) 55
E) 60
A) 48
B) 50
C) 53
D) 55
E) 60
08 May 2026, 03:34
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Hello diva, Can you explain why the answer is 53, but not 30?
Here is how I solve the problem
Max C
Want C is the greatest
Make A, B, D is have to be smallest
- A = 1
- B = 2
- D = C + 1
So that:
- 1+2+C+(C+1)=140
- 2C+4=140→2C=136→C(Max)=68
Min C
Want C is smallest
- A = 1
- B = 2
- C = C
- D = lớn nhất có thể (<100)
1+2+C+D=140→C+D=137
Cuz D<100 ⇒ max D = 99
C=137−99=38
So Difference=68−38=30
Here is how I solve the problem
Max C
Want C is the greatest
Make A, B, D is have to be smallest
- A = 1
- B = 2
- D = C + 1
So that:
- 1+2+C+(C+1)=140
- 2C+4=140→2C=136→C(Max)=68
Min C
Want C is smallest
- A = 1
- B = 2
- C = C
- D = lớn nhất có thể (<100)
1+2+C+D=140→C+D=137
Cuz D<100 ⇒ max D = 99
C=137−99=38
So Difference=68−38=30
09 May 2026, 01:08
Kudos
Bookmarks
Hello,
Your approach for the maximum value of C is completely correct.
My logic to find the minimum value case is below :
You have:
A+B+C+D=140
and
A<B<C<D
We want the smallest possible C.
You said:
1,2,C,99
Then:
C=38
This is valid.
BUT...
If we ask ourself this:
Can we make C even smaller?
Yes.
How?
By increasing A and B.
Why increasing A and B helps
Look at this set:
1,2,38,99
Total = 140.
Now suppose we move some value from 38 into the smaller numbers:
12,14,15,99
Total is still 140.
But now:
C=15
which is much smaller than 38.
The core idea which worked for me:
If A and B are too tiny, then C is forced to become big to make the total reach 140.
So making A,B, super small actually hurts our intention of finding minimum value of C.
Hence the difference :
68−15=53
Answer: C) 53
Your approach for the maximum value of C is completely correct.
My logic to find the minimum value case is below :
You have:
A+B+C+D=140
and
A<B<C<D
We want the smallest possible C.
You said:
- make A,B smallest
- make D biggest
1,2,C,99
Then:
C=38
This is valid.
BUT...
If we ask ourself this:
Can we make C even smaller?
Yes.
How?
By increasing A and B.
Why increasing A and B helps
Look at this set:
1,2,38,99
Total = 140.
Now suppose we move some value from 38 into the smaller numbers:
12,14,15,99
Total is still 140.
But now:
C=15
which is much smaller than 38.
The core idea which worked for me:
If A and B are too tiny, then C is forced to become big to make the total reach 140.
So making A,B, super small actually hurts our intention of finding minimum value of C.
Hence the difference :
68−15=53
Answer: C) 53
xmi
