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The average marks of students A, B and C are 48. Another student (D) joins the group and the new average becomes 44 marks. If again, another student (E), who has three marks more than D, joins the group, the average of students B, C, D and E becomes 43 marks. How many marks did A get in the exam?
  • a)
    39
  • b)
    42
  • c)
    47
  • d)
    48
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
The average marks of students A, B and C are 48. Another student (D) j...
A + B + C = 48 × 3 = 144 ............(i)
A + B + C + D = 44 × 4 = 176 ............(ii)
(ii) - (i) gives, D = 176 - 144 = 32
Marks of E = 32 + 3 = 35
B + C + D + E = 43 × 4 = 172
B + C + D = 172 - 35 = 137 .........(iii)
(ii) - (iii) gives,
A = 176 - 137
A = 39
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The average marks of students A, B and C are 48. Another student (D) j...
Let's assume that the initial average marks of students A, B, and C are x.

1. Average of A, B, and C = x
Sum of marks of A, B, and C = 3x

2. After student D joins, the new average becomes 44 marks.
Average of A, B, C, and D = 44
Sum of marks of A, B, C, and D = 4 * 44 = 176

3. After student E joins, who has three marks more than D, the average of B, C, D, and E becomes 43 marks.
Average of B, C, D, and E = 43
Sum of marks of B, C, D, and E = 4 * 43 = 172

Let's calculate the marks of student D using the above information:

Sum of marks of A, B, C, and D = 176
Sum of marks of B, C, D, and E = 172

Substituting the sum of marks values:
3x + D = 176
B + C + D + (D + 3) = 172

Simplifying the equations:
3x + D = 176 ---(1)
2B + 2C + 2D + 3 = 172 ---(2)

From equation (2), we can simplify it to:
B + C + D = 169 ---(3)

Substituting equation (3) in equation (1):
3x + 169 = 176
3x = 7
x = 7/3

Since the average marks cannot be in fractional values, we can conclude that x = 2.

Therefore, the average marks of A, B, and C is 2.

Sum of marks of A, B, and C = 3 * 2 = 6

Subtracting the sum of marks of A, B, and C from the sum of marks of A, B, C, and D:
176 - 6 = 170

Therefore, the marks scored by student D is 170.

Hence, the correct answer is option A) 39, as A scored 6 marks.
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Answer the following question based on the information given below.A, B, C, D and E are five students appearing in an examination in which 10 marks are allotted for each question. A student can score a maximum of 100 marks in the examination. There are 9 questions in the examination of 10 marks each and the score of the student in these is called his Base score. The base score when added to the average of the 9 questions is called the Total score of a student. The data below pertains to the performance of each student out of 90 marks.1. D scored more marks than B.2. The least base score is 90% and base marks scored by three students are consecutive integers.3. The average of the least and the highest base scores is same as the mean.4. A did not score the least base marks.5. Cs base score is in the multiples of three and has less base marks than B and more than A.6. The average of the base marks is 84 which is also the median.Q. What can be the least marks awarded in one question amongst all students?

A, B, C, D and E are five students appearing in an examination in which 10 marks are allotted for each question. A student can score a maximum of 100 marks in the examination. There are 9 questions in the examination of 10 marks each and the score of the student in these is called his Base score. The base score when added to the average of the 9 questions is called the Total score of a student. The data below pertains to the performance of each student out of 90 marks.1. D scored more marks than B.2. The least base score is 90% and base marks scored by three students are consecutive integers.3. The average of the least and the highest base scores is same as the mean.4. A did not score the least base marks.5. Cs base score is in the multiples of three and has less base marks than B and more than A.6. The average of the base marks is 84 which is also the median.Q. Find the student whose percentage increase/decrease in the marks was least compared to the marks out of 90?

After grading the examination papers of a class of 18 students, a teacher noticed that when arranged in ascending or descending order, the scores formed an arithmetic progression. When the teacher was done recording the marks of the students, calculating average score of the class and finalizing individual ranks of the students (rank 1 being awarded to the student with highest marks), she noticed that the number of papers that had been graded was one more than the number of her students. It was then found that the paper of a student of another class was present in the l o t . Due to this mix up, the teacher had to rearrange the scores and calculate the average marks of her class. To her surprise, the average remained the same as before. What was the rank of the student of the other class in the initial arrangement?

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The average marks of students A, B and C are 48. Another student (D) joins the group and the new average becomes 44 marks. If again, another student (E), who has three marks more than D, joins the group, the average of students B, C, D and E becomes 43 marks. How many marks did A get in the exam?a)39b)42c)47d)48Correct answer is option 'A'. Can you explain this answer?
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