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Aptitude miscellaneous
  1. A cricketer whose bowling average is 12.4 runs per wicket, takes 5 wickets for 26 runs and thereby decreases his average by 0.4. The number of wickets taken by him till the last match was








  1. View Hint View Answer Discuss in Forum

    Let required number of wickets = n
    According to question,
    ⇒ 12.4 × n + 26 = (n + 5) (12.4 – 0.4)
    ⇒ 12.4 × n + 26 = (n + 5) × 12
    ⇒ 12.4n + 26 = 12n + 60
    ⇒ 12.4n – 12n = 60 – 26
    ⇒ 0.4n = 34

    Correct Option: D

    Let required number of wickets = n
    According to question,
    ⇒ 12.4 × n + 26 = (n + 5) (12.4 – 0.4)
    ⇒ 12.4 × n + 26 = (n + 5) × 12
    ⇒ 12.4n + 26 = 12n + 60
    ⇒ 12.4n – 12n = 60 – 26
    ⇒ 0.4n = 34
    ⇒ n = 34 ÷ 0.4

    ∴ n =
    340
    		 = 85
    4

  1. The average of 9 observations was found to be 35. Later on, it was detected that an observation 81 was misread as 18. The correct average of the observations is :








  1. View Hint View Answer Discuss in Forum

    Given that , The average of 9 observations = 35
    Later on, it was detected that an observation 81 was misread as 18 .
    Correct sum of 9 observations = 9 × 35 – 18 + 81
    Correct sum of 9 observations = 315 + 63 = 378

    Correct Option: B

    Given that , The average of 9 observations = 35
    Later on, it was detected that an observation 81 was misread as 18 .
    Correct sum of 9 observations = 9 × 35 – 18 + 81
    Correct sum of 9 observations = 315 + 63 = 378
    ∴ Required correct average = 378 ÷ 9 = 42

  1. The average of runs of a cricket player of 10 innings was 32. How many runs must he make in his next inning so as to increase his average of runs by 4 ?








  1. View Hint View Answer Discuss in Forum

    Let the batsman make p runs.
    Total runs in 10 innings = 10 × 32 = 320

    ∴
    320 + p
    		 = 32 + 4
    11

    ⇒ 320 + p = 36 × 11
    ⇒ p = 396 – 320 = 76
    Second method to solve this question with the help of given formula :
    Here, T = 32, N = 11, t = 4

    Correct Option: A

    Let the batsman make p runs.
    Total runs in 10 innings = 10 × 32 = 320

    ∴
    320 + p
    		 = 32 + 4
    11

    ⇒ 320 + p = 36 × 11
    ⇒ p = 396 – 320 = 76
    Second method to solve this question with the help of given formula :
    Here, T = 32, N = 11, t = 4
    Required Run = T + Nt [Here N is taken as (n + 1)]
    Required Run = 32 + 11 × 4
    Required Run = 32 + 44 = 76

  1. A cricket player after playing 10 tests scored 100 runs in the 11th test. As a result, the average of his runs is increased by 5. The present average of runs is








  1. View Hint View Answer Discuss in Forum

    Here , A cricket player after playing 10 tests scored in the 11th match = 100 runs
    If the average in 10 tests be n , then
    According to question ,
    ⇒ n × 10 + 100 = (n + 5) × 11
    ⇒ 11n – 10n = 100 – 55

    Correct Option: C

    Here , A cricket player after playing 10 tests scored in the 11th match = 100 runs
    If the average in 10 tests be n , then
    According to question ,
    ⇒ n × 10 + 100 = (n + 5) × 11
    ⇒ 11n – 10n = 100 – 55
    ⇒ n = 45
    ∴ Required average = 50

  1. A student, by mistake, wrote 64 in place of 46 as a number at the time of finding the average of 10 given numbers and got the average as 50. The correct average of the numbers is :








  1. View Hint View Answer Discuss in Forum

    Here , Average of 10 given numbers = 50
    A student, by mistake, wrote 64 in place of 46 , then
    Correct sum of numbers = 10 × 50 – 64 + 46
    Correct sum of numbers = 500 – 18 = 482

    Correct Option: A

    Here , Average of 10 given numbers = 50
    A student, by mistake, wrote 64 in place of 46 , then
    Correct sum of numbers = 10 × 50 – 64 + 46
    Correct sum of numbers = 500 – 18 = 482
    ∴ Correct average = 482 ÷ 10 = 48.2