Tabular DI Quiz 1

Answer: 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)

= (4+12+10+13) + (65+50+57+66)

= 39+ 238

= 277

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

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

= (16+10+12+9) +(65+50+57+66)

= 47 +238

= 285

The students passing the half yearly exams are less in comparison to the students passing the exams in Annual Exams. So this concludes that Half-yearly examinations were more difficult.

Answer: Since students are classified based on their grades and sections form individual classes, the total number of students in the class is:

= 12+13+27+21+16+10+12+9+4+12+10+13+65+50+57+66

= 397

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

For Section A = \frac{16+4+65}{12+16+4+65} = \frac{85}{97}\times 100 \% = 82.45\%

For Section B = \frac{10+12+50}{13+10+12+50} \frac{72}{85}\times 100 \% = 61.2\%

For Section C = \frac{12+10+57}{27+12+10+57} \frac{79}{106}\times 100 \% = 83.74\%

For Section D = \frac{9+13+66}{21+9+13+66} \frac{88}{109}\times 100 \% = 95.92\%

Clearly, the pass percentage is maximum for Section D

Answer: 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

The success rate in annual exams in

Section A = \frac{16+65}{97}\times 100\% =78.57\%

Similarly, the success rate in annual exams in:

Section B = \frac{10+50}}{85}\times 100\% =  51\%

Section C = \frac{12+57}{106}\times 100\% = 73.14\%

Section D = \frac{9+66}{109}\times 100\% = 81.75\%

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

Answer: 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

= \left[\frac{\text{ Number of students of Section A failed in half-yearly }\times 100}{\text{ Total number of students in Section A }} \right]\%

= \left[\frac{(12+16)\times 100}{(12+16+4+65)} \right]\%

=\left[ \frac{28\times 100}{97} \right]\%

= 28.86%.

Similarly, failure rate in half-yearly exams in:

Section B

\left[ \frac{(13+10)\times 100}{(13+10+12<br /> <br /> =50)} \right]\% = \left[ \frac{23\times 100}{85} \right]\% = 27.05\%.

Section C

\left[ \frac{(27+12)\times 100}{(27+12+10+57)} \right]\% = \left[ \frac{39\times 100}{106} \right]\% = 36.79\%.

Section D

\left[ \frac{(21+9)\times 100}{(21+9+13+66)} \right]\% = \left[ \frac{30\times 100}{109} \right]\% = 27.52\%.

Clearly, the failure rate is minimum for Section B.

Answer:

Total number of covid positive children staying in all the cities together

= 18% of 16000 + 13% of 13000 + 12% of 12500 + 15% of 14200

= 2880 + 1690 + 1500 + 2130

= 8200

Solution:

The two cities with minimum number of cases are Muzaffarpur and Bhagalpur

Number of covid positive women in Muzaffarpur  = 42% of 13000 = 5460

Number of covid positive women in Bhagalpur  = 42% of 12500 = 5250

∴ Total number of covid positive women in this two cities = 5460 + 5250 = 10710

Hence, option D is correct.

Answer: Option a

The number of men below the age group of 50 years = 30% of 14200 = 4260

∴ The number of men not eligible for Vaccination = 52% of 14200 – 4260

= 7384 – 4260 = 3124

Answer: Option b

Number of Covid positive women and children in Patna = 62% of 16000 = 9920

Number of Covid positive women and children in Muzaffarpur = 55% of 13000 = 7150

Number of Covid positive women and children in Bhagalpur= 54% of 12500 = 6750

Number of Covid positive women and children in Hazipur= 48% of 14200 = 6816

Answer: Option c

Total number of positive girls in Patna = 55% of 18% of 16000 = 1584

Total number of positive women in Patna = 44% of 16000 = 7040

∴The required total = 7040 + 1584 = 8624