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Answer:E
3=Average after correct list - Average before correct list of scores
3= (78+x)/3-78/3
x=9
New list of scores satisfying median remain same
21,25,32+x
21,25,41
Range=41-21=20
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Since the average score increases by 3 points after correcting the error, the impact on the set would be= 3*3= 9
Since the student found that his score was less by 'x', we need to add 9 to one of the values without changing the median.
The only value to which 9 can be added without altering the median is 32. So 32+9= 41.
And the range would be 41-21=20
Answer: E
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If the average of 3 scores increases by 3 points then total score of a student increase by 9 points. As the median remains same so student having 32 points will increase by 9 points and his new score will become 41 points.
So range =41-21=20
Option E is the answer.

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Option E,

The difference between the old and new total is 3*(29-26) = 9, thus one score increased by 9. Now if 21 were to increase, 25 will no longer be the median thus 32 increased to 41. Range then will be 41-21 I.e 20

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for given three scores 21,25,32 the median is 25 and avg is 78/3 ; 26
later when score error is know the avg score increase by 3 points or say to 29 so total sum = 29*3 ; 87
net change in points of one of scores ; 87-78 ; 9
given that median remains same so the +9 gets added to 32 ; new score ; 41
21,25,41 ; range = 41-21 ; 20
OPTION E

The list of scores obtained by 3 students in a test is as follows: 21, 25, 32. One of the students found that his score was less by ‘x’ points due to a calculation error. If the average of three score increases by 3 points after correcting this error and the median score remains the same before and after correction, which of the following could be the range of list of corrected scores?

A 2
B 5
C 7
D 11
E 20
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Average of 3 students = 21+25+32/3 = 78/3 = 26
If the student's score is increased by X no. The average increase by 3 point.
Therefore 78+x/3 = 26+3
78+x/3 = 29
78+x = 87
X = 9
The student's marks has to be increase by 9 marks such that median = 25 remain same.

If we add 9 to 21 marks of student 1, median will shift to 30(21,25,32 --> 21+9,25,32 ->>> 25,30,32)

If we add 9 to 25 marks of student 2, median will shift to 32(21,25,32 --> 21,25+9,32 ->>> 21,32,34)

If we add 9 to 32 marks of student 3, median will remain same 25(21,25,32 --> 21,25,32+9 ->>> 21,25,41)

The range of marks is (highest-lowest) => 41-21 =20
Answer is supposedly E

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E.
No which will be increased is 32 by 9 points(avg change).
Range will be 41-21=20

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The list of scores obtained by 3 students in a test is as follows: 21, 25, 32. One of the students found that his score was less by ‘x’ points due to a calculation error. If the average of three score increases by 3 points after correcting this error and the median score remains the same before and after correction, which of the following could be the range of list of corrected scores?

A 2
B 5
C 7
D 11
E 20

Since the average of the scores increases by 3, the total score will increase by 9.

If the median does not change, 21 and 25 cannot increase by 9, So 32 will increase by 9 to 41.

The range of the new score is 41-21 = 20.

Answer E is correct
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Given:
1. The list of scores obtained by 3 students in a test is as follows: 21, 25, 32.
2, One of the students found that his score was less by ‘x’ points due to a calculation error.
3. The average of three score increases by 3 points after correcting this error and the median score remains the same before and after correction,

Asked: Which of the following could be the range of list of corrected scores?

1. The list of scores obtained by 3 students in a test is as follows: 21, 25, 32.
Scores = {21,25,32}

2, One of the students found that his score was less by ‘x’ points due to a calculation error.
3. The average of three score increases by 3 points after correcting this error and the median score remains the same before and after correction,
Median = 25 before and after correction
Sum increased by 3*3 = 9 points after correcting the error
x = 9

21 + 9 = 30; Not feasible since median will change
25 + 9 = 34; Not feasible since median will change
32+9 = 41; Feasible since median will remain 25 before and after correction

Corrected Scores = {21,25,41}; Range = 41 - 21 = 20

IMO E
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Quote:
The list of scores obtained by 3 students in a test is as follows: 21, 25, 32. One of the students found that his score was less by ‘x’ points due to a calculation error. If the average of three score increases by 3 points after correcting this error and the median score remains the same before and after correction, which of the following could be the range of list of corrected scores?

A 2
B 5
C 7
D 11
E 20

E , IMO

median = 25 .
score increased by 3*3 = 9 points .
so earlier 32 will become now 32+9=41
so range = 41-21= 20 .
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Let C be the corrected score & I be the incorrect score.
The student found his score is less by x points. So his Correct score would be greater.
\(\therefore C = I +x\)

Sum of Scores = 21+25+32 = 78
\(Average = \frac{78}{3} = 26\)

Since after correction, the average increased by 3,
\(Average+3 = \frac{SumOfCorrectedScores}{3}\)
\(26+ 3= \frac{(78+x)}{3}\)
\(\therefore x=9\)

Adding 9 to 21 results in the scores: 25,30,32 ---> Median Changed
Adding 9 to 25 results in the scores: 21,32, 34---> Median Changed
Adding 9 to 32 results in the scores: 21,25,41 ----> Median remains the same as stated in question stem.

Range = 41-21 = 20. Answer (E) 20
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Three scores 21, 25, 32
Now some incorrect calculation happened.
Total score 21+25+32 = 78
Average is 26.
After correction avg score increased by 3;
Therefore 26+3=29
Total score is 87.

But median remains same i.e., 25.
So 87-25=62 remains for two person

Additional amount 62-(21+32)=9 will add to only one person.

Let, 21+9= 30 so 30, 25, 32 this cannot happen here median changes.
If 32+9= 41 so 21, 25, 41 here median is same as required.

Thus the new score list will be 21, 25, 41
Therefore the range is 20.

IMO ans is E
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