Table chart


Direction: The following table shows the number of students of 7 colleges A, B, C, D, E, F, G participating in extra-curricular activities :

  1. The difference of the range of number of students in activity IV and the average is of number of students in activity III per college is :









  1. View Hint View Answer Discuss in Forum

    As per given table , we can see that the
    Number of students in all given seven colleges
    A = 65 , B = 130 , C = 420 , D = 75 , E = 540 , F = 220 , G = 153
    Total numbers of colleges = 7
    Number of students in activity IV in college C = 438
    Number of students in activity IV in college D = 105

    Average number of students in activity III = sum of number of students in all seven colleges
    total numbers of colleges

    ∴ Difference = Range of number of students in activity IV - Average number of students in activity III

    Correct Option: C

    As per given table , we can see that the
    Number of students in all given seven colleges
    A = 65 , B = 130 , C = 420 , D = 75 , E = 540 , F = 220 , G = 153
    Total numbers of colleges = 7
    Number of students in activity IV in college C = 438
    Number of students in activity IV in college D = 105
    Range of number of students in activity IV = Number of Highest Students in activity IV - Number of lowest Students in activity IV
    Range of number of students in activity IV = 438 − 105 = 333

    Average number of students in activity III = sum of number of students in all seven colleges
    total numbers of colleges

    ∴ Average number of students in activity III =65 + 130 + 420 + 75 + 540 + 220 + 153
    7

    ∴ Average number of students in activity III =1603 = 229
    7

    Difference = Range of number of students in activity IV - Average number of students in activity III
    ∴ Difference = 333 − 229 = 104


  1. The median of data pertaining to activity III is:









  1. View Hint View Answer Discuss in Forum

    Required Median = n + 1th observation
    2

    Correct Option: C

    Arranging the observations of activity III in ascending order:
    65, 75, 130, 153, 220, 420, 540
    Number of observations ( n )= 7(odd)

    Required Median = n + 1th observation
    2

    ∴ Median = 7 + 1th observation
    2

    = fourth observation = 153



  1. The college in which minimum number of students participate in extra-curricular activities is :









  1. View Hint View Answer Discuss in Forum

    As per given table , we can see that
    The minimum number of student participated in extra curriculum in All 7 collages is 65 student.

    Correct Option: D

    As per given table , we can see that
    The minimum number of student participated in extra curriculum in All 7 collages is collage A and 65 student.


Direction: Read the following information carefully and answer the questions based on it.

  1. Number of students leaving school C from the year 1990 to 1995 is approximately what percentage of number of students taking admission in the same school and in the same year?











  1. View Hint View Answer Discuss in Forum

    Number of students leaving school C from 1990 to 1995
    = 130 + 150 + 125 + 140 + 180 = 725
    Number of students admitted during the period
    = 1100 + 320 + 300 + 260 + 240 + 310 = 2530 =2530

    Correct Option: E

    Number of students leaving school C from 1990 to 1995
    = 130 + 150 + 125 + 140 + 180 = 725
    Number of students admitted during the period
    = 1100 + 320 + 300 + 260 + 240 + 310 = 2530 =2530

    ∴ Required Percentage =725X100≈ 29%
    2530



  1. In which of the following schools, percentage increase in the number of students from the year 1990 to 1995 is maximum?











  1. View Hint View Answer Discuss in Forum

    Increase in number of students in school A = (230 − 120) + (190 − 110) + (245 − 100) + (280 − 150) + (250 − 130) = 585

    Correct Option: B

    Increase in number of students in school A = (230 − 120) + (190 − 110) + (245 − 100) + (280 − 150) + (250 − 130) = 585
    ∴ % increase from 1990 (1025) to 1995

    =585x 100= 57.07 %
    1025

    Similarly, we can calculate for other schools.