Directions: Refer the data given below to answer the following questions.
The following table gives the scores of 10 students of a class who have opted for various credits (subjects). The credits are labeled A through F and are rated on a scale of 1 to 5, where students can score integral values only. Not all credits are necessarily taken by a student, and not all students necessarily take a credit. The range of scores indicates the minimum and maximum scores in that credit by students who may have chosen that credit. The average is the average of the scores in a credit of students who may have chosen a particular credit. The number of credit takers out of the 10 students is given in the last column.
Explanation: Students scoring less than 2 (i.e. 1) is only possible in A, C and D.
In A, total score of 6 students is 6 x 4.5 = 21.
If minimum is 1 and maximum is 4, then the remaining 4 students have to score 21 – 4 – 1 = 16.
It means all students have scored 4, except one.
In C, total score of 7 students is 7 x 4 = 28. If 2 of then score 1 and 5, then the remaining 5
Students score 22. It is not possible for any of these 5 to score 1.
In D, total score of 3 students is 3 x 1.33 = 4. If 2 of them score 1 and 2, the third has to score 1.
But the students who have scored 1 in A, C and D could be the same. So atleast 2 of them have scored 1.
Directions: Refer the data given below to answer the following questions.
The following table gives the scores of 10 students of a class who have opted for various credits (subjects). The credits are labeled A through F and are rated on a scale of 1 to 5, where students can score integral values only. Not all credits are necessarily taken by a student, and not all students necessarily take a credit. The range of scores indicates the minimum and maximum scores in that credit by students who may have chosen that credit. The average is the average of the scores in a credit of students who may have chosen a particular credit. The number of credit takers out of the 10 students is given in the last column.
Explanation: Using similar approach as in the previous question, we get A→5, B→1, C→5, D, E→ cannot be maximum and F→2. Maximum is 5.
For e.g. For A, using the above logic, we know that five students have scored 4, and one has scored 1.
For C, the total is 28, and two scores are 1 and 5. The balance 22 is scored by 5 students. It is now possible for 1 of them to score 2 and the others to score 4 each. Hence, maximum = 5.
Now these 5 could be the same across all credits.
Directions: Refer the following data to answer the questions given below.
The following grid show the marks scored with ranks of a student in 4 tests I, II, III, and IV. Initially the marks were entered in a grid such as the one shown below with marks and rank of subject 1 on the top left, II on the top right, etc.
The data is to be read as follows. In test I, the person scored 40 and secured the 1st rank, in test II the person scored 30 and secured 3rd rank, etc.
Using a concept similar to the one above, marks of 4 students A, B, C and D were entered in 4 grids as shown below. However, in the following grids, the subjects for B, C and D have been erased and these 3 grids have been turned clockwise by 90º, 180º and 270º, not necessarily in that order. There are 10 students in the class, and in no subjects do any two students score the same marks. The more marks a person scores, the higher his rank in the subject. A scored the first rank in 2 subjects, while C and D scored the first rank in 1 subject each.
Let us assume that grid of A is fixed. So in the subject that A scores 60 and secures rank 1, others must have scored less than 60. For B it could be marks 40, 40 or 50, for C it could be 40 or 30, and for D the corresponding subject mark could be 20, 50 ro 50.
Similarly, in the subject that C scored the 1st rank, it is clear that A should have scored less than 60 i.e. A has scored 90 in that subject. So the mark of A aligned to mark of C would be as follows.
Now, we know that in the subject that a scores 60 and secures 5th rank, C is not the 1st rank. Hence D must be the 1st rank in that subject. We can now align all subjects of all students accordingly.
Directions: Refer the following data to answer the questions given below.
The following grid show the marks scored with ranks of a student in 4 tests I, II, III, and IV. Initially the marks were entered in a grid such as the one shown below with marks and rank of subject 1 on the top left, II on the top right, etc.
The data is to be read as follows. In test I, the person scored 40 and secured the 1st rank, in test II the person scored 30 and secured 3rd rank, etc.
Using a concept similar to the one above, marks of 4 students A, B, C and D were entered in 4 grids as shown below. However, in the following grids, the subjects for B, C and D have been erased and these 3 grids have been turned clockwise by 90º, 180º and 270º, not necessarily in that order. There are 10 students in the class, and in no subjects do any two students score the same marks. The more marks a person scores, the higher his rank in the subject. A scored the first rank in 2 subjects, while C and D scored the first rank in 1 subject each.
Let us assume that grid of A is fixed. So in the subject that A scores 60 and secures rank 1, others must have scored less than 60. For B it could be marks 40, 40 or 50, for C it could be 40 or 30, and for D the corresponding subject mark could be 20, 50 ro 50.
Similarly, in the subject that C scored the 1st rank, it is clear that A should have scored less than 60 i.e. A has scored 90 in that subject. So the mark of A aligned to mark of C would be as follows.
Now, we know that in the subject that a scores 60 and secures 5th rank, C is not the 1st rank. Hence D must be the 1st rank in that subject. We can now align all subjects of all students accordingly.
Directions: Refer the following data to answer the questions given below.
The following grid show the marks scored with ranks of a student in 4 tests I, II, III, and IV. Initially the marks were entered in a grid such as the one shown below with marks and rank of subject 1 on the top left, II on the top right, etc.
The data is to be read as follows. In test I, the person scored 40 and secured the 1st rank, in test II the person scored 30 and secured 3rd rank, etc.
Using a concept similar to the one above, marks of 4 students A, B, C and D were entered in 4 grids as shown below. However, in the following grids, the subjects for B, C and D have been erased and these 3 grids have been turned clockwise by 90º, 180º and 270º, not necessarily in that order. There are 10 students in the class, and in no subjects do any two students score the same marks. The more marks a person scores, the higher his rank in the subject. A scored the first rank in 2 subjects, while C and D scored the first rank in 1 subject each.
Let us assume that grid of A is fixed. So in the subject that A scores 60 and secures rank 1, others must have scored less than 60. For B it could be marks 40, 40 or 50, for C it could be 40 or 30, and for D the corresponding subject mark could be 20, 50 ro 50.
Similarly, in the subject that C scored the 1st rank, it is clear that A should have scored less than 60 i.e. A has scored 90 in that subject. So the mark of A aligned to mark of C would be as follows.
Now, we know that in the subject that a scores 60 and secures 5th rank, C is not the 1st rank. Hence D must be the 1st rank in that subject. We can now align all subjects of all students accordingly.