S is a set containing 9 different numbers. T is a set containing

Question

S is a set containing 9 different numbers. T is a set containing 8 different numbers, all of which are members of S. which of the following statements cannot be true?

A. The mean of S is equal to the mean of T
B. The median of S is equal to the median of T
C. The range of S is equal to the range of T
D. The mean of S is greater than the mean of T
E. The range of S is less than the range of T

my question : what if the extra number is zero ?? ???

Bunuel · Accepted Answer

S is a set containing 9 different numbers. T is a set containing 8 different numbers, all of which are members of S. which of the following statements cannot be true?

The range of a set is the difference between the largest and smallest elements of a set.

Consider the set S to be {-4, -3, -2, -1, 0, 1, 2, 3, 4} -->  mean=median=0  and  range=8 .

A. Mean of S = mean of T --> remove 0 from set S, then the mean of T still would be 0;

B. Median of S = Median of T --> again remove 0 from set S, then the median of T still would be 0;

C. Range of S = range of T --> again remove 0 from set S, then the range of T still would be 8;

D. Mean of S > mean of T --> remove 4, then the mean of T would be negative -0.5 so less than 0;

E. Range of S < range of T --> the range of a subset cannot be more than the range of a whole set: how can the difference between the largest and smallest elements of a subset be  more  than the difference between the largest and smallest elements of a whole set.

Answer: E.

Hope it helps.

Abhii46 · Answer

Mean of both the sets can be equal. Let us suppose the mean of any 8 number is 10 then the 9th number could also be 10 and mean remains the same.

In the same way median can also be same. In set of 9 numbers median will be the 5th number when arranged in ascending order and in set T it will be the mean of 4th and 5th number.

If we take S = { 1, 2, 3, 4 ,5, 6, 7, 8, 9 } and T as { 1, 2, 3, 4, 6, 7, 8, 9 } then median in both the cases will be 5.

From the above eg range is same in both the cases.

If the number which is not the part of set T is greater than mean of T then the mean of set S will be greater than that of set T

Range of S will always be greater than or equal to range of T because of an additional number. If that number is greater than the greatest number in set T the range will be more, if that number is smaller than the smallest number in set T again the range will be more since now the new number is the smallest number. If that number lies in between then the range will be equal.

fozzzy · Answer

So this is a property of sets?

Bunuel · Answer

Sure. The range of a subset cannot be greater than the range of the whole set.

Alexmsi · Answer

@Bunuel:

I agree with you that the range of subsequent set is less than the whole set, but it can also be the same ? For exmaple:

100 14 13 2 whole set.
100 14 2 subsequent set.

All mebers of the subsequent set are also members of the whole set. But the range are in both cases the same. So it must be as follows: The range of the subsequent set can be equal or less than that of the whole set ?

Bunuel · Answer

The way it's written in my post is the same:  the range of a subset cannot be greater than the range of a whole set . This means that the range of a subset is always less than or equal to the range of the whole set.

superpus07 · Answer

Solve for an "easier" problem and make up an example to see what's going on in the problem.

Set S = {1,2,3}
Set T = {2,3}

Okay so if we can get an answer that is true, we can eliminate that answer choice.

a) hmm, how to get the mean equal each other? Oh, just remove the 2 instead of 1. 
b) same
c) same
d) remove 3 instead of 1

At this point we can choose e) and move on, but to be sure just test some numbers again.

e)
range set S = 3-1 = 2
range set T = 3-2 = 1
range set T = 3-1 = 1

So, this can never be true. --> Bingo

This is basically the same method as Bunuel posted above, but for me it sometimes works better if I have a simpler set to work with.

akadmin · Answer

The range of a set is the difference between the largest and smallest elements of a set.

Consider the set S to be {-4, -3, -2, -1, 0, 1, 2, 3, 4} -->  mean=median=0  and  range=8 .

A. Mean of S = mean of T --> remove 0 from set S, then the mean of T still would be 0;
B. Median of S = Median of T --> again remove 0 from set S, then the median of T still would be 0;
C. Range of S = range of T --> again remove 0 from set S, then the range of T still would be 8;
D. Mean of S > mean of T --> remove 4, then the mean of T would be negative -0.5 so less than 0;
E. Range of S < range of T --> the range of a subset cannot be more than the range of a whole set: how can the difference between the largest and smallest elements of a subset be  more  than the difference between the largest and smallest elements of a whole set.

Answer: E.

Hope it helps.

ganand · Answer

Hi  akadmin ,

Yes, it is still possible.

Consider following examples:

T = {2, 4, 6, 8, 10, 12, 21} Mean of T = 63/7 = 9

S = {2, 4, 6, 8, 9, 10, 12, 21} Mean of S = 9

Key idea is that create a set with 7 elements such that the mean is not a member of the set.

For larger set (i.e. set S) you can always add mean as an additional element and mean of the larger set will still be same as smaller set (set T).

Hope it helps.

dave13 · Answer

Bunuel but if i take this set of numbers S - -\frac{0 + 1 +2 +3 +4+5 +6 +7 +8}{9}= 4 T --- \frac{0 +1 + 2+ 3 +4 +5 + 6 +7}{8} = 4.5 mean of set S is 4 and mean of Set T 3.5 median of set S is 4 and median of set T is 3.5 then how should i solve this question

pushpitkc

pushpitkc · Answer

Hi  dave13

Though you have written the values down correctly, but while calculating the mean
for T you seem to have made a mistake. Anyways, now that you have the values you
just start by substituting values and you will find out that Option A,B,C,E are incorrect

For instance, We know Mean(S) = 4 and Mean(T) = 3.5

So, Option D is possible and cannot be the answer. You need to choose values for the
two sets, S and T and such that we cab eliminate all but 4 options. The answer to this
question is Option E because it is never going to be possible

Hope that helps!

KarishmaB · Answer

S has 9 different numbers. T has 8 of these different numbers in it.

Which of the following cannot be true?

A. The mean of S is equal to the mean of T
This is possible. Let's say the mean of set S is a number 20 and one of the numbers in S is 20. If T has all numbers in it except 20, its mean will still be 20.

B. The median of S is equal to the median of T
This is possible. Let's say the median of set S is the middle number 20 and set T does not have 20 but has 19 and 21 in the middle. The median of T will still be 20.

C. The range of S is equal to the range of T
When we read range, think of numbers arranged from lowest to highest. Range is highest - lowest. T could include both the lowest number and the highest number and hence its range will be same as the range of S.

D. The mean of S is greater than the mean of T
Mean of T could be less than or higher than that of S. Again, arrange all elements of S in ascending order. If you pick just the greatest 8 numbers to have in T, mean of T will be more than that of S. If you pick the smallest 8 elements, mean of T will be less than mean of S.

E. The range of S is less than the range of T
Can the range of T be greater than that of S? Arrange all numbers of S in ascending order. Which 8 numbers will you pick to have a higher range? For maximum range, you can pick the smallest and greatest number of S. But that will give just the same range as that of S. T cannot have a range greater than the range of S.

Answer (E)

thunderbolt27 · Answer

Regarding option B:
 Let S = {n1, n2, n3, n4, n5, n6, n7, n8, n9}  
 T = {n1, n2, n3, n4, n5, n6, n7, n8} where n1 < n2 < n3 < n4 < n5 < n6 < n7 < n8< n9  
 Median of S = n5; Median of T = 1/2 * (n4 + n5)  
 If Median S = Median T, (2* n5 = n4 + n5) i.e, (n4 = n5). But it is given that all numbers are different. How can it be ?

Krunaal · Answer

Taking your example,

Let's suppose n4 = 9, n5 = 10, and n6 = 11

Set S is as you've mentioned, median n5 = 10

Now for set T, we remove n5 and instead include n9, so the median becomes 1/2 * (n4 + n6) => 1/2 * (9 + 11) = 10 = n5

So we have a case that is possible, hence statement B can hold true and is not the right answer.

siddhantvarma · Answer

KarishmaB   Bunuel 
I approached this by writing down equations and I'm running into conflict with the premise in the question when I try to solve it algebraically. 
We're given that: S is a set containing 9 different numbers. T is a set containing 8 different numbers, all of which are members of S.
All numbers of set S are different and so are the numbers in set T. 
Let S = {a1,a2,a3,...,a9}
And T = {a1,a2,...,a8}

B. The median of S is equal to the median of T
Median of Set S : a5
Median of Set T : (a4+a5)/2
a5 = (a4+a5)/2 => a4 = a5, but we know a4 cannot be equal to a5 since all the numbers are given to be different. So how can statement (B) be true? I know taking examples as others have pointed out makes it clear that it can be true, so then where am I wrong in trying to reason it algebraically?

KarishmaB · Answer

You are assuming that set T has the 8 smallest numbers of set S. That is not necessary.

S = 1, 2, 3, 4, 5, 6, 7, 8, 9
T = 1, 2, 3, 4, 6, 7, 8, 9 (this is a possibility)

Hence it is possible that medians are the same. Not necessary of course, but possible.

egmat · Answer

Hi siddhantvarma,

Great question! Your algebra is actually correct for the specific case you set up — but you accidentally locked yourself into just one case.

Your mistake:  You wrote T = {a 1 , a 2 , ..., a 8 }, which means you assumed the element REMOVED from S to create T is always a 9  (the largest element). But the problem only says T is missing one element from S — and that missing element could be ANY of the  9 .

Let S = {a 1  < a 2  < a 3  < ... < a 9 } in sorted order. The median of S is a 5 .

Now, what if we remove a 5  itself?

T = {a 1 , a 2 , a 3 , a 4 , a 6 , a 7 , a 8 , a 9 }

Median of T = (a 4  + a 6 ) /  2

Setting this equal to a 5 : a 4  + a 6  =  2  * a 5

This requires a 5  to be exactly the average of a 4  and a 6  — which is totally possible with all different numbers!

Example:  S = { 1 ,  2 ,  3 ,  4 ,  5 ,  6 ,  7 ,  8 ,  9 }. Remove  5 . T = { 1 ,  2 ,  3 ,  4 ,  6 ,  7 ,  8 ,  9 }. Median of T = ( 4  +  6 ) /  2  =  5 . Median of S =  5 . Equal! No contradiction.

General principle:   When a problem says T is a subset of S with one element missing,  never assume WHICH element is missing . You need to consider all possibilities before concluding something is impossible. Your algebraic approach was sound — you just needed to test all  9  cases for which element gets removed, not just one.

Answer: E

S is a set containing 9 different numbers. T is a set containing

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S is a set containing 9 different numbers. T is a set containing

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