The number of students in each of 10 different classes is a different

S-2) Given Students aregreater than 63. So possible range is from 64 to 88. Avg of given set of original numbers is 73. So to keep this average, we need to eliminate numbers 84 to 88. So we have elements of set {64 to 82}. Now we can find the numbers and SD. Sufficient.

S-1) I thought Mean = Median so equidistance elements. So the sequence of elements can be starting 58 or any number SD will remain same. This is because adding constant C, Standard Deviation doesnot change. But the answer says only B. Cant I infer the set as Equidistance ? Bunnel or Any one kindly help ?

Math experts please provide an explanation on how to tackle such questions.

Hi,

There are 16 numbers with mean in middle of 72&74 or 73..
Let's see the statements..

1. the Mean number of students is equal to the median number of students in the the 10 classes.
mean=median does not necessarily mean equidistant..
Example..
Five numbers with average 74..
1) 70,72,74,76,78.... Equidistant
2) 66,72,74,78,80... Now these are not Equidistant..
So the standard deviation will be different for both above examples
Hence insufficient

2. number of students in any class is more than 63 and the avg. number of students in 10 classes is same as average of the above list
This tells us that the lowest number can be 64 and average is 73..
Only possibility is the numbers are 64,66,68,70,72 and corresponding numbers will be 74,76,78,80,82
Hence the standard deviation will always be same
Sufficient

B

Sir can you explain how you solved for statement 2? It is said that the number of students in each class is not equal to the numbers in the above list......and since possibilities are it can start from 65 and end at 87........it is impossible to distribute 10 numbers between 65 and 87 with mean 73........Least possible 10 numbers are...65,67,69,71,73,75,77,79,81 and 83......in this case average is 74??

Sir, please see.....am i doing something wrong?

Statement 1 : Theory :yes, we know an evenly spread set or a set where all the elements are equal will have mean=median but the reverse is not necessarily true
this means we either have an evenly spaced set of 10 integers like this 64,66,68,70,72,74,76,78,80,82 or 66,68,70,72,74,76,78,80,82,84. Either of these two will have the same SD.However if the set is formed like this : 60,64,70,72,74,76,78,82,86,88 we have the mean=median but it will have a different SD from either of the 2 sets above because the spreads from the mean are different.
Insufficient

Statement 2 : We can only choose from 64 onwards and we must choose such that the mean is 73=mean of the entire set in the question prompt.
We can do this in only way : 64,66,68,70,72,74,76,78,80,82 which will have a particular mean of 73.If we attempt to form any other set of 10 numbers from the listed numbers above such that mean=73 ,we will not be able to. Sufficient

Answer is B

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