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chetan86
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When one new number is included in an existing set of 6 numbers 12 Sep 2014, 07:24
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64% (01:46) correct 36% (01:49) wrong based on 557 sessions

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When one new number is included in an existing set of 6 numbers, does the median of the set change?

(1) The mean of the original 6 numbers is 50.
(2) At least 2 of the numbers in the original set were 50.
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When one new number is included in an existing set of 6 numbers 12 Sep 2014, 08:01
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chetan86
When one new number is included in an existing set of 6 numbers, does the median of the set change?

(1) The mean of the original 6 numbers is 50.
(2) At least 2 of the numbers in the original set were 50.
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If the set is {50, 50, 50, 50, 50, 50}, then no matter what number we add the median (50) will not change.
If the set is {0, 0, 0, 50, 50, 200}, the median = (0 + 50)/2 = 25, and if new number is between 0 and 50, but not 25, for example 10, then the median will change.

Answer: E.
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When one new number is included in an existing set of 6 numbers 29 Mar 2016, 19:26
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goshh, i misread the first statement.

Thought it was median = 50

If the first statement actually states that median = 50

Will C correct?
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When one new number is included in an existing set of 6 numbers 29 Mar 2016, 19:35
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goshh, i misread the first statement.

Thought it was median = 50

If the first statement actually states that median = 50

Will C correct?
Show more

Yes C will be correct..
from 1 we will know that 3rd and 4th numers are 50..
when we add the number smaller than median or bigger than median, Median will shift to 3rd or 4th number..
As both numbers are 50, answer will be 50..
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When one new number is included in an existing set of 6 numbers 19 Nov 2018, 01:13
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Hello!

The main concept being tested in this problem is the concept of Median and how it is calculated. A challenging aspect of this problem is that there are no values given and hence it entails trying out multiple cases. Of course, your knowledge of the Mean is also being tested in this problem, since you have to use it while analysing Statement 1.

Since there are 6 numbers, the median which is the middle value will be the average of the two values in the middle i.e. the median will be the average of the 3rd and the 4th numbers. But, it has to be borne in mind that the median can be calculated only after arranging the given numbers in ascending/descending order. Hence, we can say that the 1st and the 2nd numbers should be smaller than the 3rd and the 4th; similarly, the 5th and the 6th numbers are greater than the 3rd and the 4th. However, the exception to this case is when some or all of the six numbers are equal to each other.

Remember that the given DS question is a Yes or No type of question - which means that your answer should be a definite Yes or a definite No. In any case, if you get a Yes and a No, the data would not be sufficient to answer the question.

Now, when we add a new number to this data set of 6 numbers, the median will be the middle number i.e. the 4th number. So, giving a definite Yes or No depends on the magnitude of the number added and also on its position in the data set.

Let us analyze Statement I now. As per statement I alone, the mean of the six numbers is 50. Therefore, the sum of the six numbers should be 50 x 6 = 300 ( since Mean = Sum of values / Number of values). Let's now see how we can get the sum of 6 numbers to be 300. If the numbers are a, b, c, d, e and f,

a+b+c+d+e+f = 300.

This can happen if a=b=c=d=e=f=50. Since the numbers are equally spaced, Mean = Median = 50. If the seventh number is less than 50, Median =50; if the seventh number is 50, Median = 50; and if the seventh number is more than 50, Median =50. In this case, regardless of the number that you add, the median will remain 50. Hence, we can answer the main question with a NO.

However, if a=10, b=20, c=30, d=40, e=50, f=150 and the seventh number is also 150, the median becomes 40. hence, we can answer the main question with a YES.

Since the data given in statement I alone is not giving us a conclusive answer, statement I alone is insufficient!

Let us now analyse statement II. At least 2 of the six numbers are 50 - 2 0r 3 or 4 or 5 or all 6 numbers can be 50.

If 2 numbers are 50, the other numbers can be anything and we cannot find out the median. And hence, when we add the seventh number, we will be essentially looking at the same situation where we get a Yes and a No (it is very easy to get carried away at this stage and say that the other four numbers should give me 200 more since the total is 300 - remember, NOW, you don't know that the total is 300; that is given in Statement I and we have not combined the statements yet!)

We can conclude similarly for cases when there are 3/4/5 numbers having values of 50 each.

If we assume all 6 numbers to be 50, then we get a case similar to the one which we saw while analysing Statement I.

Hence, the second statement alone is also insufficient.

Combining statements I and II, we can still look at 2 cases:

Case 1: a=b=c=d=e=f = 50 and here, the median does not change on adding a new number.

Case 2: a=25, b=25, c=25, d=50, e=50, f=125 for which the median is 37.5. Now, adding a new number = 50 to this set will change the median to 50.

Hence, even on combining the two statements, we do not get a definite Yes or No. Therefore, E is the answer.

The thing to remember about such problems is that the absence of values is actually something you can use to your advantage since you can take multiple sets of values and thereby eliminate options. But while picking values, it is very easy to go overboard by taking complex sets of values, whereas, simple numbers (like the ones we have demonstrated) are enough to disprove most statements. However, remember to try values from different ranges to ensure that you have covered all bases.

Hope this helps!
Cheers,

CrackVerbal Academics Team
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When one new number is included in an existing set of 6 numbers 13 Dec 2024, 07:43
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combined together we got atleast 2 50s
now it could be that originally all are 50 - so 50,50,50,50,50,50 now in such case whatever number you add, median (mid no. will be 50 only)

OR it if we say 2 nums are 50 then remaining 4 will have sum 200, now we can have something like 10, 20, 50, 50, 70, 90 so again avg = 40 and median = 50, now if we add any number in b/w them say 49 or 51 then median will be 50 again BUT we can also have something like 0,0,0,50,50,200 then current median = 25 but if we add another 0 then median will change to 0 so E is the answer.
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