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Aptitude miscellaneous
  1. In an examination, the average of marks was found to be 50. For deducting marks for computational errors, the marks of 100 candidates had to be changed from 90 to 60 each and so the average of marks came down to 45. The total number of candidates, who appeared at the examination, was








  1. View Hint View Answer Discuss in Forum

    Here , The average of marks was found to be 50
    Let total number of candidates be n.
    According to question,
    ∴ 50n – 30 × 100 = 45n
    ⇒ 50n - 45n = 30 × 100

    Correct Option: A

    Here , The average of marks was found to be 50
    Let total number of candidates be n.
    According to question,
    ∴ 50n – 30 × 100 = 45n
    ⇒ 50n - 45n = 30 × 100
    ⇒ 5n = 3000
    ⇒ n = 3000 ÷ 5 = 600

  1. In an exam, the average marks obtained by the students was found to be 60. After omission of computational errors, the average marks of 100 candidates had to be changed from 60 to 30 and the average with respect to all the examinees came down to 45 marks. The total number of candidates who took the exam, was








  1. View Hint View Answer Discuss in Forum

    Let the number of candidates be n, then
    According to question ,
    60n – 45n = 30 × 100
    ⇒ 15n = 3000

    Correct Option: A

    Let the number of candidates be n, then
    According to question ,
    60n – 45n = 30 × 100
    ⇒ 15n = 3000

    ⇒ n =
    3000
    		 = 200
    15

  1. The average of 18 observations is recorded as 124. Later it was found that two observations with values 64 and 28 were entered wrongly as 46 and 82. Find the correct average of the 18 observations.








  1. View Hint View Answer Discuss in Forum

    As per the given details in question ,
    Difference in observations = 64 + 28 – 46 – 82 = – 36

    ∴ Correct average = 124 -
    36
    18

    Correct Option: B

    As per the given details in question ,
    Difference in observations = 64 + 28 – 46 – 82 = – 36

    ∴ Correct average = 124 -
    36
    		 = 122
    18

  1. Out of 4 numbers, whose average is 60, the first one is onefourth of the sum of the last three. The first number is








  1. View Hint View Answer Discuss in Forum

    Let the first number be p,

    then, p =
    240 - x
    		 = 48
    4

    ⇒ 4p = 240 – p

    Correct Option: C

    Let the first number be p,

    then, p =
    240 - x
    		 = 48
    4

    ⇒ 4p = 240 – p
    ⇒ 5p = 240
    ⇒ p = 240 ÷ 5 = 48

  1. The average marks of a class of 35 children is 35. The marks of one of the students, who got 35, was incorrectly entered as 65. What is the correct average of the class?








  1. View Hint View Answer Discuss in Forum

    According to question ,
    Total correct marks of 35 children = 35 × 35 + 35 – 65 = 1225 – 30 = 1195

    Required average =
    1195
    		 = 34.14
    35

    Second method to find the required average ,
    Diffference = – 65 + 35 = – 30

    Correct Option: B

    According to question ,
    Total correct marks of 35 children = 35 × 35 + 35 – 65 = 1225 – 30 = 1195

    Required average =
    1195
    		 = 34.14
    35

    Second method to find the required average ,
    Diffference = – 65 + 35 = – 30
    Required average = 35 –
    30
    35

    Required average = 35 – 0.857 = 34.143