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Aptitude miscellaneous
  1. The average weight of 12 parcels is 1.8 kg. Addition of another new parcel reduces the average weight by 50 g. What is the weight of the new parcel ?








  1. View Hint View Answer Discuss in Forum

    The average weight of 12 parcels = 1.8 kg
    Total weight 12 parcels = 12 × 1.8 = 21.6 kg.
    when addition of another new parcel reduces the average weight by 50 g.
    ∴ New average of 13 parcels = 1.8 – 0.05 = 1.75 kg.
    Total weight of 13 parcels = 13 × 1.75 = 22.75 kg.
    ∴ Weight of new parcel = 22.75 – 21.6 = 1.15 kg.
    Second method to solve this question with the help of given formula :

    Here, T = 1.8, N= (12 + 1) = 13 ,t =
    50
    		 = 0.05
    1000

    Correct Option: C

    The average weight of 12 parcels = 1.8 kg
    Total weight 12 parcels = 12 × 1.8 = 21.6 kg.
    when addition of another new parcel reduces the average weight by 50 g.
    ∴ New average of 13 parcels = 1.8 – 0.05 = 1.75 kg.
    Total weight of 13 parcels = 13 × 1.75 = 22.75 kg.
    ∴ Weight of new parcel = 22.75 – 21.6 = 1.15 kg.
    Second method to solve this question with the help of given formula :

    Here, T = 1.8, N= (12 + 1) = 13 ,t =
    50
    		 = 0.05
    1000

    Weight of new parcel = T – Nt
    Weight of new parcel = 1.8 – 13 × 0.05
    Weight of new parcel = 1.8 – 0.65 = 1.15 kg

  1. A cricketer, whose bowling average was 12.4 runs/wicket takes 5 wickets for 22 runs in a match, thereby decreases his average by 0.4. The number of wickets, taken by him before this match was :








  1. View Hint View Answer Discuss in Forum

    Let the number of wickets before the last match be y.
    According to the question,
    ⇒ 12.4y + 22 = (y + 5 ) × 12
    ⇒ 12.4y + 22 = 12y + 60
    ⇒ 12.4y – 12y = 60 – 22
    ⇒ 0.4y = 38

    Correct Option: C

    Let the number of wickets before the last match be y.
    According to the question,
    ⇒ 12.4y + 22 = (y + 5 ) × 12
    ⇒ 12.4y + 22 = 12y + 60
    ⇒ 12.4y – 12y = 60 – 22
    ⇒ 0.4y = 38
    ⇒ y = 38 ÷ 0.4

    ⇒ y =
    380
    		 = 95
    4

  1. In the first 10 overs of a cricket game, the run rate was only 3.2. The run rate in the remaining 40 overs to reach the target of 282 runs is








  1. View Hint View Answer Discuss in Forum

    According to question ,
    Total runs scored in first 10 overs = 3.2 × 10 = 32
    Runs to be scored in remaining 40 overs = 282 – 32 = 250

    Correct Option: C

    According to question ,
    Total runs scored in first 10 overs = 3.2 × 10 = 32
    Runs to be scored in remaining 40 overs = 282 – 32 = 250
    ∴ Required run–rate = 250 ÷ 40 = 6.25

  1. The average of runs scored by a cricketer in his 99 innings is 99. How many runs will he have to score in his 100th innings so that his average of runs in 100 innings may be 100?








  1. View Hint View Answer Discuss in Forum

    Here , The average of runs scored by a cricketer in his 99 innings = 99
    Sum of runs scored by a cricketer in his 99 innings = 99 × 99
    Number of runs scored in 100th innings = 100 × 100 – 99 × 99
    Number of runs scored in 100th innings = 10000 – 9801 = 199
    Second method to solve this question with the help of given formula :

    Correct Option: C

    Here , The average of runs scored by a cricketer in his 99 innings = 99
    Sum of runs scored by a cricketer in his 99 innings = 99 × 99
    Number of runs scored in 100th innings = 100 × 100 – 99 × 99
    Number of runs scored in 100th innings = 10000 – 9801 = 199
    Second method to solve this question with the help of given formula :
    Increase in average = 1 run
    ∴ Runs scored in 100th innings = 100 + 99 = 199

  1. The averages of runs scored by a cricket player in 11 innings is 63 and the average of his first six innings is 60 and the average of last six innings is 65. Find the runs scored by him in the sixth innings.








  1. View Hint View Answer Discuss in Forum

    Here , The averages of runs scored by a cricket player in 11 innings = 63
    Sum of runs scored in 11 innings = 11 × 63
    The average of his first six innings = 60
    Sum of runs scored in first six innings = 6 × 60
    and the average of last six innings = 65
    Sum of runs scored in last six innings = 6 × 65
    ∴ Runs scored by the cricketer in the 6th innings = 6 × 60 + 6 × 65 – 11 × 63

    Correct Option: D

    Here , The averages of runs scored by a cricket player in 11 innings = 63
    Sum of runs scored in 11 innings = 11 × 63
    The average of his first six innings = 60
    Sum of runs scored in first six innings = 6 × 60
    and the average of last six innings = 65
    Sum of runs scored in last six innings = 6 × 65
    ∴ Runs scored by the cricketer in the 6th innings = 6 × 60 + 6 × 65 – 11 × 63
    Runs scored by the cricketer in the 6th innings = 360 + 390 – 693 = 57