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Aptitude miscellaneous
  1. The mean of 50 numbers is 30. Later it was discovered that two entries were wrongly entered as 82 and 13 instead of 28 and 31. Find the correct mean.








  1. View Hint View Answer Discuss in Forum

    Required average = 30 +
    (28 + 31 - 82 - 13)
    50

    Required average = 30 + -
    36
    50

    Required average = 30 – 0.72 = 29.28
    Second method to solve this question with the help of given formula :
    Here, n = 50, m = 30 , a = 28, b = 31 , p = 82, q = 13
    Correct Average = m +
    (a + b - p - q)
    n

    Correct Option: C

    Required average = 30 +
    (28 + 31 - 82 - 13)
    50

    Required average = 30 + -
    36
    50

    Required average = 30 – 0.72 = 29.28
    Second method to solve this question with the help of given formula :
    Here, n = 50, m = 30 , a = 28, b = 31 , p = 82, q = 13
    Correct Average = m +
    (a + b - p - q)
    n

    Correct Average = 30 +
    (28 + 31 - 82 - 13)
    50

    Correct Average = 30 + -
    59 - 95
    50

    Correct Average = 30 -
    36
    50

    Correct Average = 30 – 0.72 = 29.28

  1. The average marks of 100 students were found to be 40. Later on it was discovered that a score of 53 was misread as 83. Find the correct average corresponding to the correct score








  1. View Hint View Answer Discuss in Forum

    On the basis of given details in question ,
    Total of correct marks = 100 × 40 – 83 + 53 = 3970
    ∴ Correct average marks = 3970 ÷ 100 = 39.70
    Second method to solve this question with the help of given formula :
    Here, n = 100, m = 40 a = 53, b = 83

    Correct Average = m +
    (a -b)
    n

    Correct Option: C

    On the basis of given details in question ,
    Total of correct marks = 100 × 40 – 83 + 53 = 3970
    ∴ Correct average marks = 3970 ÷ 100 = 39.70
    Second method to solve this question with the help of given formula :
    Here, n = 100, m = 40 a = 53, b = 83

    Correct Average = m +
    (a -b)
    n

    Correct Average = 40 +
    53 - 83
    100

    Correct Average = 40 -
    30
    		 = 39.70
    100

  1. The average of marks in Mathematics for 5 students was found to be 50. Later, it was discovered that in the case of one student the marks 48 were misread as 84. The correct average is :








  1. View Hint View Answer Discuss in Forum

    Here , The average of marks in Mathematics for 5 students was found to be 50
    Total marks obtained by 5 students = 50 × 5 = 250.
    Now, in this total marks, 84 is included instead of 48.
    ∴ Correct total marks = 250 – 84 + 48 = 214
    ∴ Correct average = 214 ÷ 5 = 42.8
    Second method to solve this question with the help of given formula :
    Here, n = 5, m = 50 , a = 48 , b = 84

    Correct Average = m +
    (a - b)
    n

    Correct Option: C

    Here , The average of marks in Mathematics for 5 students was found to be 50
    Total marks obtained by 5 students = 50 × 5 = 250.
    Now, in this total marks, 84 is included instead of 48.
    ∴ Correct total marks = 250 – 84 + 48 = 214
    ∴ Correct average = 214 ÷ 5 = 42.8
    Second method to solve this question with the help of given formula :
    Here, n = 5, m = 50 , a = 48 , b = 84

    Correct Average = m +
    (a - b)
    n

    Correct Average = 50 +
    (48 - 84)
    5

    Correct Average = 50 -
    36
    5

    Correct Average = 50 – 7.2 = 42.8

  1. The average of a collection of 20 measurements was calculated to be 56 cm. But later it was found that a mistake had occurred in one of the measurements which was recorded as 64 cm., but should have been 61 cm. The correct average must be








  1. View Hint View Answer Discuss in Forum

    Total length of 20 measurements = 56 × 20 = 1120 cm
    Correct length of 20 measurements = 1120 – 64 + 61 = 1117
    Correct average = 1117 ÷ 20 = 55.85 cm
    Second method to solve this question with the help of given formula :
    Here, n = 20, m = 56 , a = 61, b = 64

    Correct Average = m +
    (a - b)
    n

    Correct Option: C

    Total length of 20 measurements = 56 × 20 = 1120 cm
    Correct length of 20 measurements = 1120 – 64 + 61 = 1117
    Correct average = 1117 ÷ 20 = 55.85 cm
    Second method to solve this question with the help of given formula :
    Here, n = 20, m = 56 , a = 61, b = 64

    Correct Average = m +
    (a - b)
    n

    Correct Average = 56 +
    61 - 64
    20

    Correct Average = 56 -
    3
    20

    Correct Average = 56 – 0.15 = 55.85 cm

  1. Among three numbers, second is twice the first and also thrice the third. If the average of the three numbers is 33, then the largest number is :








  1. View Hint View Answer Discuss in Forum

    Let the first number be p.
    ∴ Second number = 2p

    And Third number =
    2p
    3

    According to the question, p + 2p +
    2p
    		 = 33 × 3
    3

    ⇒ 3p +
    2p
    		 = 99
    3

    ⇒
    9p + 2p
    		 = 99
    3

    Correct Option: B

    Let the first number be p.
    ∴ Second number = 2p

    And Third number =
    2p
    3

    According to the question, p + 2p +
    2p
    		 = 33 × 3
    3

    ⇒ 3p +
    2p
    		 = 99
    3

    ⇒
    9p + 2p
    		 = 99
    3

    ⇒ 11p = 99 × 3
    ⇒
    99× 3
    		 = 27
    11

    ∴ Largest number = 2p = 2 × 27 = 54