///Worded Average Problems
Worded Average Problems2018-12-08T06:57:23+10:00

Maths Made Easy Forums Standard Data & Statistics Worded Average Problems

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    The mean height of the 5 Sydney storm basketball players on the court at the start of a game was 1.90m. During the game, a player who is 1.85m tall was injured and was replaced by a player who is 1.95m tall. What was the mean height of the 5 players after the replacement?
    5 × 1.9 = 9.5m first we find the total of the 5 players: If they have a mean of 1.9, then the total heights would be 5 lots of 1.9
    9.5 – 1.85 = 7.65 remove the 1.85 player
    7.65 + 1.95 = 9.6 now add in the replacement player at 1.95
    9.6 ÷ 5 = 1.92 to find the new mean, divide the new total by 5, since there are still only 5 players on the court
    ∴ mean = 1.92m  

     

  • AdminAdmin

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    The average temperature in June for the past 30 years in a city has been 65.2. This year the average June temperature was 71.5. What is the new 31 year average?
    65.2 × 30 find the total for the last 30 years if the average was 65.2, that means that each year can be considered as 65.2 and since there were 30 years, to find the total you multiply 65.2 by 30
    =1956 the total for the 30 years is 1956
    1956 + 71.5 = 2027.5 to find the new total for the 31 years, add the 30 year total to the new temperature
    2027.5 ÷ 31 = 65.4 to find the new average for the 31 years, divide the new total by 31

     

  • AdminAdmin

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    The total rainfall for the month of November was 97.2mm. There were 21 dry days in November. The average daily rainfall for November correct to 1 decimal place is:
    a. 3.2mm    b. 4.6mm     c. 10.8mm    d. unknown

    There are 30 days in November, so average rainfall per day would be for the entire month, not just for the rainy days
    97.2 ÷ 30 = 3.24
    ∴ a

  • AdminAdmin

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    Isabella has completed four assessment tasks, each marked out of 100, giving her a mean of 71%. She wants to increase her mean to 75% after the trial. Which of the following below will calculate the mark (x) she needs to achieve in the next assessment task?
    a. $$x=\frac{71+75}{2}$$

    b. $$x=\frac{71+75}{2}$$

    c. $$75=\frac{71\times4+x}{2}$$

    d. $$75=\frac{71\times4+x}{5}$$

    the total of 4 tests with an average of 71 will be 71 × 4
    then the 5th test is x
    so the total of all five tests will be 71 × 4 + x

    to find the average of these 5 tests, divide by 5

    since the average is to be 75 then $$75=\frac{71\times4+x}{5}$$
    D

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