In the same time, they cover 110 km and 90 km respectively.

Therefore, Ratio of their speeds = 110 : 90 = 11 : 9

Relative speed of the thief and policeman  =  (11 – 10) km/hr = 1 km/hr 

Distance covered in 6 minutes  = (1/60) x 6 km   = 1/10 km = 100 m

Therefore, Distance between the thief and policeman = (200 – 100) m = 100 m.

speed    = (5x5/18)m/sec
             = 25/18 m/sec.

Distance covered in 15 minutes    = (25/18 x 15 x 60)m
                                                 = 1250 m.

Total distance travelled in 12 hours    =(35+37+39+.....upto 12 terms)
This is an A.P with first term, a=35, number of terms,
n= 12,d=2.
Required distance    = 12/2[2 x 35+{12-1) x 2]
=6(70+23)
= 552 kms.

1. ANSWER : C 

 Explanation -  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. 

 2. ANSWER : D 

 Explanation -  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. 

 3. ANSWER : D 

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

  For Section A = 14+6+6428+14+6+64×100 = 84112×100% = 75%

 For Section B =12+17+5523+12+17+55×100 % = 78.5%

 For Section C = 8+9+4617+8+9+46×100%= 78.75%

 For Section D = 13+15+7627+13+15+76×100%= 79.39%

  Clearly ,the pass percentage is maximum for Section D 

 4. ANSWER : A

 Explanation - 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= 14+64112 × 100% = 69.64%

 Similarly, success rate in annual exams in:

 Section B = 12+55107×100% =  62.62% 

 Section C = 8+4680×100% = 67.5% 

 Section D = 89131×100% = 67.94% 

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

 5. ANSWER : D 

 Explanation - 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

 Failure rate in half-yearly exams in Section A %= 37.5 %

 Similarly, failure rate in half-yearly exams in:

  Section B = 32.71%

  Section C = 31.25%

  Section D = 30.53%

  Clearly, the failure rate is minimum for Section D.

 = (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.

1. ANSWER : D 

Explanation - The ratio of imports to exports for the years 1998 and 1999 are 1.25 and 1.40 respectively. 

Let the exports in the year 1999 = Rs. x crores.  

Then, the exports in the year 1998 = Rs. (500 - x) crores. 

 250500-x = 1.25 => x = 300 crores      [ Using ratio for 1998 ]  

Thus, the exports in the year 1999 = Rs. 300 crores. 

Let the imports in the year 1999 = Rs. y crores.  

Then, Imports in the year 1999 = y300=1.4 => y = 420= Rs. 420 crores.

2. ANSWER : C

Explanation - 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. 

3. ANSWER : D 

Explanation - 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. 

4. ANSWER : B 

Explanation -  Ratio of imports to exports in the year 1996 = 0.85. 

Let the exports in 1996 = Rs. x crores. 

Then , 272x = 0.85 => x = 320  

Exports in 1996 = Rs. 320 crores.  

5. ANSWER : D  

Explanation - 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.

1. ANSWER : 2  

Explanation -  Compare the respective pass percentage for three years : 2000, 2001 and 2002   

= (140 x 100)/170 < (150 x 100)/180 and (150 x 100)/180 > (160 x 100)/200

 = 82.35% < 83.33% and 83.33% > 80%

2. ANSWER : 3  

Explanation -  Total percentage of students who succeeded in at least one of three exams in 2000 = (140 x 100)/170= 82.35 %   

3. ANSWER : 2  

Explanation -  Total percentage increase in the number of students in 2002 over 2000 is = (30 x 100)/170 =17.64 %  

4. ANSWER : 4  

Explanation -  Total percentage of students who cleared CAT in 2000 =(20 x 100)/170 = 11.76 %