AnkitK wrote:

Dear Fluke could you throw some light your answer.Also y hase you taken -7 when it 7 in the set ?

{-10,-8,0,6,7}

Typo. It is -10,0,7.

The stem says two elements are taken out. Find out the mean.

1. Median remains unaltered even after those two elements are taken out.

Original Median = 0 for {-10,-8,0,6,7}

If median remains unchanged, we are sure that the two elements that are taken out must be one each from pair {-10,-8} and {6,7}

Possible sets will be:

{-10,0,6},{-10,0,7},{-8,0,6},{-8,0,7} each having different means. Thus, we can't find the exact mean after two element are removed.

Not Sufficient.

2. The range will decrease by 3 after two elements are removed.

What is the original range: maximum-minimum= 7-(-10) = 17. It should be reduced by 3 making the new range 14.

For what combination of the set of the three will make the range=14. Only when minimum=-8 and maximum=6;

Range=6-(-8)=14.

Meaning thereby, the two elements that were removed were indeed {-10,7} and the set becomes {-8,0,6} making the mean = -2/3.

We found the exact mean here.

Sufficient.

If you want the brute force method, it is going to take some time.

In how many ways can you remove two elements from a set of 5. \(C^{5}_{2}=10 ways\)

{-10,-8,0,6,7}

Total possible ways to remove two elements:

{-10,-8} leaves {0,6,7}. Range= (7-0) = 7 != 14

{-10,0} leaves {-8,6,7}. Range= (7-(-8)) = 15 != 14

{-10,6} leaves {-8,0,7}. Range= (7-(-8)) = 15 != 14

{-10,7} leaves {-8,0,6}. Range= (6-(-8)) = 14 = 14. Possible{-8,0} leaves {-10,6,7}. Range= (7-(-10)) = 17 != 14

{-8,6} leaves {-10,0,7}. Range= (7-(-10)) = 17 != 14

{-8,7} leaves {-10,0,6}. Range= (6-(-10)) = 16 != 14

{0,6} leaves {-10,-8,7}. Range= (7-(-10)) = 17 != 14

{0,7} leaves {-10,-8,6}. Range= (6-(-10)) = 16 != 14

{6,7} leaves {-10,-8,0}. Range= (0-(-10)) = 10 != 14

You can see that there is just one possible way.

_________________

~fluke

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