Hey all,
Couldn't figure this one out. I took this from Zuleron's 198 question doc. A brief explanation would be very much appreciated. Thanks!
26) If the average of five numbers, x, 7, 2, 16, and 11 = the median, what is x?
a. 7 < x > 11
b. x is median of the five numbers
GMAT PREP I MEDIAN/AVERAGE
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I took this from zuleron's 198 gmat prep questions and it obviously didn't look right to me at first either.
I think a is supposed to be 7 < x < 11.
The answer is D and it makes sense.
the median=mean, therefore, x=9.
I think a is supposed to be 7 < x < 11.
The answer is D and it makes sense.
the median=mean, therefore, x=9.
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yep the answer is D.
the 5 numbers can be arranged as follows
2, 7, 11,16 and x.
Stmt 1 says x is between 7 and 11.
so the arrangement becomes
2, 7, x, 11, 16.
If u look at this seq carefully , u will notice 2 and 7, 11 and 16 are evenly spaced (diff = 5).
So we can ignore them in calculating the value of x.
so x cna be calculated simply as 7+11/2. = 9.
So sufficient.
Stmt 2 says the same
Hence ans is D.
HTHelps
-V
the 5 numbers can be arranged as follows
2, 7, 11,16 and x.
Stmt 1 says x is between 7 and 11.
so the arrangement becomes
2, 7, x, 11, 16.
If u look at this seq carefully , u will notice 2 and 7, 11 and 16 are evenly spaced (diff = 5).
So we can ignore them in calculating the value of x.
so x cna be calculated simply as 7+11/2. = 9.
So sufficient.
Stmt 2 says the same
Hence ans is D.
HTHelps
-V
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Well, if the st 1: is considered to be 7<x>11 which boils down to x> 11 then statement I is insufficient
and the answer will be "B"
as opposed to considering 7<x<11, where answer will be D.
and the answer will be "B"
as opposed to considering 7<x<11, where answer will be D.
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In consecutive nos the Mean = Median. So can one conclude that if the no are arranged in ascending order 2 7 11 16 , x has to be the middle value ie.between 7 and 11. As mentioned before in this thread the diff between 7 & 2 = 11 & 16 = 5. Also the average of 2 7 11 and 16 is 9. So x cannot be in the beginning or end of the series.
Is 1 and 2 necessary as they are stating the obvious. Please point out the error in my reasoning.
Is 1 and 2 necessary as they are stating the obvious. Please point out the error in my reasoning.
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Dear kstv,
you are right and wrong.
In consecutive nos the Mean = Median but the converse does not have to be true.
If mean = median, they don't have to be consecutive numbers.
Consider set A = 1, 9, 16, 25, 29 Clearly, 16 is the median
mean = (1+9+16+25+29)/5 = 16
Thus, mean = median and the numbers are not consecutive (or the set is not evenly spaced set, to be precise). Consecutive integers is special case of evenly spaced set.
you are right and wrong.
In consecutive nos the Mean = Median but the converse does not have to be true.
If mean = median, they don't have to be consecutive numbers.
Consider set A = 1, 9, 16, 25, 29 Clearly, 16 is the median
mean = (1+9+16+25+29)/5 = 16
Thus, mean = median and the numbers are not consecutive (or the set is not evenly spaced set, to be precise). Consecutive integers is special case of evenly spaced set.
kstv wrote:In consecutive nos the Mean = Median. So can one conclude that if the no are arranged in ascending order 2 7 11 16 , x has to be the middle value ie.between 7 and 11. As mentioned before in this thread the diff between 7 & 2 = 11 & 16 = 5. Also the average of 2 7 11 and 16 is 9. So x cannot be in the beginning or end of the series.
Is 1 and 2 necessary as they are stating the obvious. Please point out the error in my reasoning.
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26) If the average of five numbers, x, 7, 2, 16, and 11 = the median, what is x?
a. 7 < x < 11
b. x is median of the five numbers
Arranging terms in ascending order, we have,
2, 7, 11, 16 x can be any number.
It is given that mean = median. Depending on the value of x,
if x < 7, median = 7
7 < x < 11, median = x
x > 11, median = 11
Stmt. I
7 < x < 11
which gives us, median = mean = x
Hence, 2+7+11+16+x = 5*x
Solving, x = 9
Suff.
Stmt. II
x is the median of the five number. Essentially, this is the same info. as given in Stmt. I
Hence, Suff.
Answer is "D"
kstv,
This is a value question, which means we must find a definite value for "x". The theory for consecutive numbers does not really come into play here.
a. 7 < x < 11
b. x is median of the five numbers
Arranging terms in ascending order, we have,
2, 7, 11, 16 x can be any number.
It is given that mean = median. Depending on the value of x,
if x < 7, median = 7
7 < x < 11, median = x
x > 11, median = 11
Stmt. I
7 < x < 11
which gives us, median = mean = x
Hence, 2+7+11+16+x = 5*x
Solving, x = 9
Suff.
Stmt. II
x is the median of the five number. Essentially, this is the same info. as given in Stmt. I
Hence, Suff.
Answer is "D"
kstv,
This is a value question, which means we must find a definite value for "x". The theory for consecutive numbers does not really come into play here.
kstv wrote:Thanks, using this can U disprove/elaborate what I have concluded about the options not being necessary.
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was looking for the part which you explained
Arranging terms in ascending order, we have,
2, 7, 11, 16 x can be any number.
It is given that mean = median. Depending on the value of x,
if x < 7, median = 7
7 < x < 11, median = x
x > 11, median = 11
If x is 19 Mean = Median = 11.
so x does not necessarily lie between 7 and 11 if Mean = Median.
My doubt was whether the rule of cosecutive no applies universally to a ascending series. Thanks.
Arranging terms in ascending order, we have,
2, 7, 11, 16 x can be any number.
It is given that mean = median. Depending on the value of x,
if x < 7, median = 7
7 < x < 11, median = x
x > 11, median = 11
If x is 19 Mean = Median = 11.
so x does not necessarily lie between 7 and 11 if Mean = Median.
My doubt was whether the rule of cosecutive no applies universally to a ascending series. Thanks.
- ganesh prasath
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but the first stmt says that 7<x>11 so doesnt it mean that 7<11<x ? how have u written as 7<x<11 ??
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even if we use 7<x>11
this means x>11
and thus arranging in ascending order,
2,7,11,x,16 or 2,7,11,16,x (depends on whether x <16)
Median is still 11
and thus (2+7+11+x+6)/5 =11
and we get a single value of x.
Thus A is sufficient
also B is sufficient on its own.
Thus D
this means x>11
and thus arranging in ascending order,
2,7,11,x,16 or 2,7,11,16,x (depends on whether x <16)
Median is still 11
and thus (2+7+11+x+6)/5 =11
and we get a single value of x.
Thus A is sufficient
also B is sufficient on its own.
Thus D