11 thoughts on “Weighted Averages – 3

  1. Hi Sir,

    I have a confusion in the last example where line AB is divided into 5:1 ratio.
    As per my understanding when a line segment AB is internally divided into 5:1 ratio, then the point which divides AB should be 5/6th distance away from A and 1/6th distance away from B.
    To solve the last question you have taken ratio as 1:5 not 5:1. Answer comes out to be (3,7)

    Please correct me if my understanding is wrong.


      • Dear Sir,

        If we consider the ratio as 1:5 in the question, then you have not taken the “inverse ratio” as mentioned in the 1st post of weighted averages.
        As per my understanding, this situation is slightly different because 1:5 is not the ratio of the weightages but the ratio in which AB is actually divided. Hence, in this case inverse ratio is not to be taken.
        Please confirm if my understanding is correct.

  2. Dear J,
    I have a question. In problem 2, we could have calculated the desired answer by applying the weighted avg technique only on C+H and Sheep, we get S/(C+H) =7/4 so S/(C+H+S) = 7/(7+4) = 7/11. Why do we need to use the previous condition? Does that matter ?

    • You’re perfectly right, Kartik. I just explained the full detail (including finding the ratio of all three) to demonstrate how to apply multiple “seesaws”; what you said would be quicker and just as correct. Your thought process is absolutely correct and hopefully that would spell good things for you in the exam 🙂 Good work!


  3. Dearest J,
    Can u please tell me how to solve this one: (please share multiple approaches, if there are. and if it’s using weighted averages then please tell me what should be the weights and avg values.) PFA the question in pic.

    • No weighted average needed here. If the times are x and y then 2x : 3y = 1 : 2 which gives x : y as 3 : 4. Or just plug in the answer choices into the above equation and it will instantly give you the answer!


  4. sir in this first one you took the the ratio as 1:6,and not 6:1..but previously in every problem, you used to take the inverse values..and here you haven’t..just want to know the basic why you are taking inverse somewhere and not here..:)

  5. Sir in second question how can you take average when units are not same…42 is area grazed not cows…m unable to get sir…

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