Test: Line Charts (March 13) - CAT MCQ

The number of students who failed in all the six subjects in all the three terms would be limited by the highest average among all sections in all terms.

The highest average is for Hindi in term I which is 98.

The total marks of 25 students for Hindi in term I = 98 x 25 = 2450

If we assume 1 student fails, then maximum marks of the other 24 students = 2450 – 34 = 2416
which is not possible.

Thus, no student can fail in all six subjects in all three terms.

The number of students failed will be minimum for the subject with the maximum average marks.
For the II term, Hindi has the maximum average marks.
The total marks scored by all the 25 students in Hindi = 25 x 94 = 2350.
If we assume 3 students failed, then the maximum total marks of the remaining 22 students =2350 – 34×3 = 2248.
But since maximum marks for a subject is only 100, this is not possible.
If we assume 2 students failed, then the maximum total marks of the remaining 23 students = 2350 – 34×2 = 2282 which is possible. Thus, the maximum number of students who have failed in all the subjects in the second term = 2

The number of students that scored more than 80 in all subjects in all the terms will be limited by the subject having the lowest average which is 60 for Biology in the first term.
The total score of all the sixty students = 60 x 25 = 1500
Thus, the maximum number of students who scored greater than 80 in all the six subjects for all the terms = 1500/80 = 18.75 i.e 18 students.

For ease of calculation let’s assume the number of people in Boselvania in 2005 is 100.

In Graph II, the percentage increase in 2006 is 20%.

Thus, the population would be 100×1.2 = 120 in 2006.

Also, the percentage crimes in Boselvania in 2006 is 30%.

Thus, the number of crimes in Boselvania is 120x.3 = 36.

In this way, the table can be formed for all the given years.

From the table, we can see that the number of people committing crimes was second lowest in 2008

For ease of calculation let’s assume the number of people in Boselvania in 2005 is 100.

In Graph II, the percentage increase in 2006 is 20%.

Thus, the population would be 100×1.2 = 120 in 2006.

Also, the percentage crimes in Boselvania in 2006 is 30%.

Thus, the number of crimes in Boselvania is 120x.3 = 36.

In this way, the table can be formed for all the given years.

From the table, we can see that the population was third highest in the year 2006.

The imports are minimum proportionate to the exports implies that the ratio of the value of imports to exports has the minimum value.

Now, this ratio has a minimum value 0.35 in 1997, i.e., the imports are minimum proportionate to the exports in 1997.

The graph gives only the ratio of imports to exports for different years. To find the percentage increase in imports from 1997 to 1998, we require more details such as the value of imports or exports during these years.

Hence, the data is inadequate to answer this question.

The exports are more than the imports imply that the ratio of value of imports to exports is less than 1.

Now, this ratio is less than 1 in years 1995, 1996, 1997 and 2000.

Thus, there are four such years.