Average


  1. The average height of 40 students is 163 cm. On a particular day, three students namely A, B, C were absent and the average of the remaining 37 students was found to be 162 cm. If A and B have equal height and the height of C be 2 cm less than that of A, find the height of A.









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    Let the height of A, B, C be y cm, y cm and (y – 2) cm respectively.
    Now, sum of height of 40 students = 163 × 40 = 6520 cm
    Sum of height of 37 students excluding A, B and C = 162 × 37 = 5994 cm.
    ∴ Sum of heights of A, B and C = (6520 – 5994) = 526 cm
    According to question ,
    ⇒ y + y + y –2 = 526
    ⇒ 3y = 526 + 2 = 528

    Correct Option: C

    Let the height of A, B, C be y cm, y cm and (y – 2) cm respectively.
    Now, sum of height of 40 students = 163 × 40 = 6520 cm
    Sum of height of 37 students excluding A, B and C = 162 × 37 = 5994 cm.
    ∴ Sum of heights of A, B and C = (6520 – 5994) = 526 cm
    According to question ,
    ⇒ y + y + y –2 = 526
    ⇒ 3y = 526 + 2 = 528
    ⇒ y = 528 ÷ 3 = 176 cm
    ∴ Height of A = 176 cm


  1. The average of 6 observations is 12. A new seventh observation is included and the new average is decreased by 1. Find the seventh observation.









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    The average of 6 observations = 12
    Sum total of 6 observations = 6 × 12
    The average of 7 observations = 11
    Sum of 7 observations = 7 × 11
    Seventh observation = Sum of 7 observations – Sum total of 6 observations

    Correct Option: A

    The average of 6 observations = 12
    Sum total of 6 observations = 6 × 12
    The average of 7 observations = 11
    Sum of 7 observations = 7 × 11
    Seventh observation = Sum of 7 observations – Sum total of 6 observations
    ∴ Seventh observation = (7 × 11) – (6 × 12) = 77 – 72 = 5



  1. The average of ten numbers is calculated as 15. It was discovered later on that while calculating the average one number namely 36 was wrongly read as 26. Find the correct average.









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    Since 36 was misread as 26, i.e., 26 was counted while calculating average.
    Incorrect average = 15
    ∴ Incorrect sum of ten numbers = 10 × 15 = 150
    ∴ Correct sum total = 150 + 36 – 26 = 160

    Correct Option: A

    Since 36 was misread as 26, i.e., 26 was counted while calculating average.
    Incorrect average = 15
    ∴ Incorrect sum of ten numbers = 10 × 15 = 150
    ∴ Correct sum total = 150 + 36 – 26 = 160
    ∴ Correct average = 160 10 = 16


  1. The average of 50 numbers is 38. If two numbers, namely 45 and 55 are discarded, what is the average of the remaining numbers?









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    The average of 50 numbers = 38
    Sum total of 50 numbers = 38 × 50 = 1900
    Sum total of remaining 48 numbers = 1900 – (45 + 55) = 1800

    Correct Option: D

    The average of 50 numbers = 38
    Sum total of 50 numbers = 38 × 50 = 1900
    Sum total of remaining 48 numbers = 1900 – (45 + 55) = 1800
    and their average = 1800 ÷ 48 = 37. 5.



  1. A team of 8 persons joins in a shooting competition. The best marksman scored 85 points. If he had scored 92 points, the average score for the team would have been 84. The number of points the team scored was









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    Here , Average score for 8 persons team = 84
    Total score for 8 persons team = 84 × 8
    Difference = ( 92 - 85 ) points
    Required total points scored = 84 × 8 – (92 – 85)

    Correct Option: B

    Here , Average score for 8 persons team = 84
    Total score for 8 persons team = 84 × 8
    Difference = ( 92 - 85 ) points
    Required total points scored = 84 × 8 – (92 – 85)
    Required total points scored = 672 – 7 = 665