Table Charts - 2 - Free MCQ Practice Test with solutions, SSC CGL Quant

Let the total population of State R be x million.

Then, population of State R above poverty line

    = [(100 - 24)% of x] million

And so, male population of State R above poverty line

But, it is given that male population of State R above poverty line = 1.9 million.

∴ Total population of State R = 6.25 million.

Female population below poverty line for State P = 2.1 million

Let the male population below poverty line for State P be x million.

∴ Population below poverty line for State P = (2.1 + 1.75) million = 3.85 million.

Let the population above poverty line for State P by y million.

Since, 35% of the total population of State P is below poverty line, therefore, 65% of the total population of State P is above poverty line i.e., the ratio of population below poverty line to that above poverty line for State P is 35 : 65.

∴ Population above poverty line for State P = 7.15 million and so, male population above poverty line for State P

For State Q:

Male population below poverty line = 2.4 million.

Let the female population below poverty line be x million.

∴ Total population below poverty line = (2.4 + 4) = 6.4 million.

If N_q be the total population of State Q, then,

= 25.6 million.

For State T:

Male population below poverty line = 6 million.

Let the female population below poverty line be y million.

∴ Total population below poverty line = (6 + 3.6) = 9.6 million.

If N_t be the total population of State T, then,

Total population of State S = 7 million.

∴ Population above poverty line

    = [(100 - 19)% of 7] million

    = (81% of 7) million

    = 5.67 million.

And so, the number of females above poverty line in State S

= 2.43 million.

Total number of students passed in annual exams in a section

= [ (No. of students failed in half-yearly but passed in annual exams) + (No. of students passed in both exams)] in that section

∴ Success rate in annual exams in Section A

Similarly, success rate in annual exams in:

Section B

Clearly, the success rate in annual examination is maximum for Section A.

Total number of failures in half-yearly exams in a section

= [ (Number of students failed in both exams) + (Number of students failed in half-yearly but passed in Annual exams)] in that section

Therefore Failure rate in half-yearly exams in Section A

Similarly, failure rate in half-yearly exams in:

Clearly, the failure rate is minimum for Section D.

Since the classification of the students on the basis of their results and sections form independent groups, so the total number of students in the class:

= (28 + 23 + 17 + 27 + 14 + 12 + 8 + 13 + 6 + 17 + 9 + 15 + 64 + 55 + 46 + 76)

= 430.

Number of students who passed half-yearly exams in the school

= (Number of students passed in half-yearly but failed in annual exams)
+ (Number of students passed in both exams)

= (6 + 17 + 9 + 15) + (64 + 55 + 46 + 76)

= 288.

Also, Number of students who passed annual exams in the school

= (Number of students failed in half-yearly but passed in annual exams)
+ (Number of students passed in both exams)

= (14 + 12 + 8 + 13) + (64 + 55 + 46 + 76)

= 288.

Since, the number of students passed in half-yearly = the number of students passed in annual exams. Therefore, it can be inferred that both the examinations had almost the same difficulty level.

Thus Statements (a), (b) and (d) are false and Statement (c) is true.

Pass percentages in at least one of the two examinations for different sections are:

Clearly, the pass percentage is maximum for Section D.

Required percentage

Required percentage

Required average

The percentages of candidates qualified to candidates appeared from State P during different years are:

∴ Maximum percentage is for the year 2001.

Required percentage

Required percentage