-3
$\begingroup$

İn a class average weight of students without mary is 51 and average weight of students without john is 50 kg.

Without both john and mary the other students average weight is 50 kg.what is the average weight of john and mary?

  • 1
    The question might make more sense if it asked about median or mode, rather than mean. – 2017-01-27
  • 0
    Why the many downvotes? I agree that it could have been better formulated and the OP could have explained his/her line of thought, but other than that I don't see anything wrong. – 2017-01-27
  • 0
    @temizsaclihippie If you had $5$ students with weights $49,49,50,53,54$ what is the number that you call "*average weight*"? – 2017-01-27
  • 0
    İ mean like Arithmetical average.like collect all the number and divide the result with number of the digits.english is my 3rd language so i cant explain myself very much .sorry about that.and thank you for your answers – 2017-01-27
  • 0
    @temizsaclihippie In that case the question cannot be answered with just the information provided. – 2017-01-27

1 Answers 1

1

Let's say there are $n$ ($n\geq 3$) students in total, and denote there weights by $x_1, \dots ,x_n$ with $x_1=\text{weight of Mary}$ and $x_{2}=\text{weight of John}$. Then $(x_1+\sum_{i=3}^nx_i)/(n-1)=50$ and $(x_2+\sum_{i=3}^nx_i)/(n-1)=51$. Thus $x_1=(n-1)50-\sum_{i=3}^nx_i$ and $x_2=(n-1)51-\sum_{i=3}^nx_i$. Moreover, we know that $(\sum_{i=3}^nx_i)/(n-2)=50$. It follows that $$\frac{x_1+x_2}{2}=\frac{(n-1)(50+51)-2(n-2)50}{2}=\frac{n+99}{2}.$$

Unless I made a mistake, the answer seems to depend on the number of students in the class.

  • 0
    Thank you for your answer.i can find the weight of mary but i cant find the weight of john.its 3 am at where i live and i am still trying to solve the question. – 2017-01-27
  • 0
    How did you find the weight of Mary? In my answer this also depends on $n$ and it looks like I used the three pieces of information that were given. – 2017-01-27
  • 0
    X= number of the students.Y=total weight of the students without john and mary.M= weight of mary..y/x=50. Y+m/x+1=50. y+m=50x + 50. Y=50x. So mary is 50 kg – 2017-01-27
  • 0
    Very simply, if adding a sample to a set doesn't change its average, then that sample must have the average value (in this case, 50kg). – 2017-01-27
  • 0
    Yeah, I noticed that too, indeed very straightforward, but still the average weight of John and Mary seems to depend on the number of students. – 2017-01-27
  • 0
    Yes mary is the easy part .but i cant find the john s weight so i can find their average weight. – 2017-01-27
  • 0
    Let me rephrase that. Since we know the weight of Mary, knowing the average weight of John and Mary is equivalent to knowing the weight of John, but by my answer the weight of John is $n+49$, which depends on $n$. Thus you cannot give his weight without knowing something extra. – 2017-01-27
  • 0
    İsnt it possible that there is a different solution? – 2017-01-27
  • 0
    Unless I made mistakes, no. – 2017-01-27