9 thoughts on “Weighted Averages – 1

  1. Dear J, I came across a question which can be solved using weighted avg but I want able to. Kindly have a look.
    In a laboratory , three containers A, B, C have equal volumes of different mixtures of liquid oxygen and liquid nitrogen. The concentration(by vol) of liquid Oxygen in different mixtures is 20% , 40% and 80% resp. First 1/5th of the contents of A are poured into B, then 2/5th of the contents of B are poured into C. The final concentration of liquid Oxygen in C is approximately:
    a) 62.4 %
    b) 61.7 %
    c) 68.6 %
    d) 65.9 %

    I did it this way:
    20% (wt : 1/5 ) ————- 40% (wt: 1)
    5 : 1 . so avg conc = 40 – (40-20)*1/6 = 40 – 10/3 = 110/3

    then:
    10/3% (wt : (2/5)*(1+1/5) ) ————- 80% (wt: 1)
    (1+ 1/5) because the volume of the resultant mixture has increased.
    so ratio becomes: 25 : 12
    and the final conc of C comes out to be 80 – 20.8 = 59.6 BUT THE ANSWER IS 65.9
    Even if I take the Wt of 10/3% mixture as only 2/5 ,
    then also the answer is coming out to be: 67.6
    Please help me out.

    • The mistake is towards the end – you have taken
      “so ratio becomes: 25 : 12
      and the final conc of C comes out to be 80 – 20.8 = 59.6 ”
      The 20.8 is wrong….you have taken 12/25 of the difference but you should have taken 12/(12+25) = 12/37 of the difference….12/37 * (80 – 110/3) = 14.1x which when subtracted from 80 leaves 65.8x ~65.9%.

      regards
      J

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  6. I did not understand the last two example of weighted average 1 .as how do we get 15 and 9 on dividing 24 into the ratio of 5;3,and so on in last question . plz help sir as soon as possible.

    • 5x + 3x = 24. So 8x = 24. So x = 3. So 5x and 3x are 15 and 9. Knowledge of ratios is a basic prerequisite for applying this logic! So get comfortable with them first.

      regards
      J

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