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# Tom, Jane, and Sue each purchased a new house. The average

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Tom, Jane, and Sue each purchased a new house. The average [#permalink]

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09 Oct 2009, 22:27
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Tom, Jane, and Sue each purchased a new house. The average (arithmetic mean) price of the three houses was \$120,000. What was the median price of the three houses?

(1) The price of Tom’s house was \$110,000.
(2) The price of Jane’s house was \$120,000.
[Reveal] Spoiler: OA

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09 Oct 2009, 22:53
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Tom, Jane, and Sue each purchased a new house. The average (arithmetic mean) price of the three houses was \$120,000. What was the median price of the three houses?

(1) The price of Tom’s house was \$110,000.
(2) The price of Jane’s house was \$120,000.

[Reveal] Spoiler:
OA B IMO C

We have three prices: a, b and c. (a+b+c)/3=120
The median price would be: the second biggest.
a<=b<=c --> median price b.

(1) One of the prices is 110, less than average of 120. It's possible 110 to be a or b price, so insufficient.

(2) One of the prices is 120 equals to average. It must be the b price, as it's not possible this price to be lowest or highest because it's equals to the average, only 2 cases a<120<c or a=b=c=120

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04 Dec 2012, 18:55
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By defintion, Median divides the distribution of values such that exactly half lie below the median and half above the median. example, given 2,5,9,50 (notice that to calculate median, the values must first be arranged in ascending order), the median is (5+9)/2=7. meaning below 7 like two values and above 7 lie two values.

By contrast, Mean or average doesn't necessarily do that. It is affected by the magnitude of each value. if one value is extreme, the mean or average shifts towards that extremity. In the above example, the average is about 16. half of the numbers aren't less than this 16, there are three numbers less than 16. similarly, half aren't above 16, only one i.e. 50 is above 16.

When you are told that there are only three values and one of them is actually the mean or the average, what's that really saying? if all three numbers are different, you know that the average has to fall somewhere between the smallest and the biggest number. and we are given that this middle number is 120,000 and also happens to be the mean. it doesn't matter now what the other two numbers are. this becomes an evenly spaced set of three numbers. the smaller the smallest number, the larger the largest number has to be to keep the average 120,000 constant.

So as a rule, you can remember that for any evenly spaced set, the mean is always equal to the median. example, 2,4,6 or 10,20,30,40.

eaakbari wrote:
Mean and one of the values out of 3 values are same, hence median has to be equal to mean.

Ans. is B.

Is the above a rule general to all sets?

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10 Oct 2009, 07:50
good explanation ,thaank you very much

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05 Nov 2009, 22:29
kirankp wrote:
Tom, Jane, and Sue each purchased a new house. The average (arithmetic mean) price of the three houses was \$120,000. What was the median price of the three houses?

(1) The price of Tom’s house was \$110,000.
(2) The price of Jane’s house was \$120,000.

Did not you get B?
If one of the house is equal to mean, then it is the median because other 2 houses (both) cannot be > 120,000 or < 120,000. The wrost case, is one is < 120,000 and the other is >120,000.

So B is suff....
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05 Nov 2009, 22:29
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Mean and one of the values out of 3 values are same, hence median has to be equal to mean.

Ans. is B.

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21 Apr 2012, 00:53
Bunuel wrote:
We have three prices: a, b and c. (a+b+c)/3=120
The median price would be: the second biggest.
a<=b<=c --> median price b.

(1) One of the prices is 110, less than average of 120. It's possible 110 to be the a or b price, so insufficient.

(2) One of the prices is 120 equals to average. It must be the b price, as it's not possible this price to be lowest or highest because it's equals to the average, only 2 cases a<120<c or a=b=c=120

if one of the price is 110, aka not avg 120, wouldn't it be "a"? Since there are 3 numbers, average is 120, anything less than average will be "a"? How can 110 possibly be b?

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21 Apr 2012, 01:04
catty2004 wrote:
Bunuel wrote:
We have three prices: a, b and c. (a+b+c)/3=120
The median price would be: the second biggest.
a<=b<=c --> median price b.

(1) One of the prices is 110, less than average of 120. It's possible 110 to be the a or b price, so insufficient.

(2) One of the prices is 120 equals to average. It must be the b price, as it's not possible this price to be lowest or highest because it's equals to the average, only 2 cases a<120<c or a=b=c=120

if one of the price is 110, aka not avg 120, wouldn't it be "a"? Since there are 3 numbers, average is 120, anything less than average will be "a"? How can 110 possibly be b?

Try to construct different scenarios, you'll see that it's not that hard:

Sum is 120*3=360.
110+120+130=360;
100+110+150=360.
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06 Sep 2012, 10:38
Bunuel wrote:
Tom, Jane, and Sue each purchased a new house. The average (arithmetic mean) price of the three houses was \$120,000. What was the median price of the three houses?

(1) The price of Tom’s house was \$110,000.
(2) The price of Jane’s house was \$120,000.

[Reveal] Spoiler:
OA B IMO C

We have three prices: a, b and c. (a+b+c)/3=120
The median price would be: the second biggest.
a<=b<=c --> median price b.

(1) One of the prices is 110, less than average of 120. It's possible 110 to be a or b price, so insufficient.

(2) One of the prices is 120 equals to average. It must be the b price, as it's not possible this price to be lowest or highest because it's equals to the average, only 2 cases a<120<c or a=b=c=120

You have no idea how long I've been looking for a more well explained answer, thank you so much!!! Everyone else justs calculates the 360-250 and shows examples, this Issa much more intuitive when you pair it with a bell curve, thank you thank you thank uou

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04 Dec 2012, 10:53
Mean and one of the values out of 3 values are same, hence median has to be equal to mean.

Ans. is B.

Is the above a rule general to all sets?

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04 Dec 2012, 21:31
koisun wrote:
By defintion, Median divides the distribution of values such that exactly half lie below the median and half above the median. example, given 2,5,9,50 (notice that to calculate median, the values must first be arranged in ascending order), the median is (5+9)/2=7. meaning below 7 like two values and above 7 lie two values.

By contrast, Mean or average doesn't necessarily do that. It is affected by the magnitude of each value. if one value is extreme, the mean or average shifts towards that extremity. In the above example, the average is about 16. half of the numbers aren't less than this 16, there are three numbers less than 16. similarly, half aren't above 16, only one i.e. 50 is above 16.

When you are told that there are only three values and one of them is actually the mean or the average, what's that really saying? if all three numbers are different, you know that the average has to fall somewhere between the smallest and the biggest number. and we are given that this middle number is 120,000 and also happens to be the mean. it doesn't matter now what the other two numbers are. this becomes an evenly spaced set of three numbers. the smaller the smallest number, the larger the largest number has to be to keep the average 120,000 constant.

So as a rule, you can remember that for any evenly spaced set, the mean is always equal to the median. example, 2,4,6 or 10,20,30,40.

eaakbari wrote:
Mean and one of the values out of 3 values are same, hence median has to be equal to mean.

Ans. is B.

Is the above a rule general to all sets?

Thanks for the explanation.

So I can generalize that

if a number in a set is equal to the mean, the set is an evenly spaced set and hence the mean = median.

???

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27 Dec 2012, 07:25
Yes it is given that 120 is the mean. In statement 2 it says one of the numbers is 120. You can memorize it as a rule or test it on any 3 nos. The median has to be 120. Read bunuel's post too.
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Re: Tom, Jane, and Sue each purchased a new house. The average [#permalink]

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17 Jun 2013, 22:55
Tom, Jane, and Sue each purchased a new house. The average (arithmetic mean) price of the three houses was \$120,000. What was the median price of the three houses?

(1) The price of Tom’s house was \$110,000.
(2) The price of Jane’s house was \$120,000.

[Reveal] Spoiler:
OA B IMO C

here we go, the price of T is 110 this doesn't tell about other two so its not sufficient.

2. the price is 120 also we know mean is 120 so the other two have one > 120 other is < 120 and the median is the middle one which is 120 hence B is fine

first i too marked C but when have seen the QA then got to realize

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07 Mar 2014, 16:50
rajathpanta wrote:
Yes it is given that 120 is the mean. In statement 2 it says one of the numbers is 120. You can memorize it as a rule or test it on any 3 nos. The median has to be 120. Read bunuel's post too.

I think, this ruel (if a number in a set is equal to the mean, the set is an evenly spaced set and hence the mean = median.) is only true if the set has distinct values. For e.g. Mean = Median when set has same values (120, 120, 120) and this set is not evenly spaced.

Can the rule be - "For set of distinct values, if a number in a set is equal to the mean, then it is evenly spaced set and the mean = median."

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Re: Tom, Jane, and Sue each purchased a new house. The average [#permalink]

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02 Jun 2014, 00:59
Tom, Jane, and Sue each purchased a new house. The average (arithmetic mean) price of the three houses was \$120,000. What was the median price of the three houses?

(1) The price of Tom’s house was \$110,000.
(2) The price of Jane’s house was \$120,000.

They are asking us for the median house price, out of the group of three houses.

Given information:
(T+J+S)/3= 120,000.

Statement 1 is telling us that T = 110,000. This means that the other two houses has to have a mean of 130,000. This could be any two prices that deviate with the same amount from 130,000.

Statement one is therefore insufficient. We're looking for the house with the median price and that could be 110,000 ->130,000. We're not sure which one it is currently.

Statement 2 gives us that Jane's house is 120,000. This is the same as the mean and even though we might think that we need info on another house, we're actually already done.

Together with the information given in the question stem we've got the following:

(T +120,000+ S)/3=120,000.

(T+S)/2 = 120,000.

If T is above 120,000 then S has to be equally far from 120,000 in the negative direction. This means that either all of the houses has the mean price of 120,000 or:

T=120,000+x
J=120,000
S=120,000-x

Whatever x is(zero included) we still know that the median will be 120,000.

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Re: Tom, Jane, and Sue each purchased a new house. The average [#permalink]

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29 Aug 2015, 18:26
Since the list of numbers is odd, would this apply to all odd listed #? Exp. if the average of a,b,c,d,e is 20 and d=20, does that mean that d is the median?

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Re: Tom, Jane, and Sue each purchased a new house. The average [#permalink]

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29 Aug 2015, 18:34
kedusei wrote:
Since the list of numbers is odd, would this apply to all odd listed #? Exp. if the average of a,b,c,d,e is 20 and d=20, does that mean that d is the median?

Median is defined as the "middle most term" which in the case of odd numbered set mentioned by you will be 'c' (I am assuming that you wanted to mention this!).

For an even numbered set, the median will be the average of the 2 middle most terms. Example, median of a,b,c,d will be (b+c)/2

For more theory on sets (mean, median, mode etc.) look at: statistics-made-easy-all-in-one-topic-203966.html

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Re: Tom, Jane, and Sue each purchased a new house. The average [#permalink]

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01 Sep 2015, 18:12
Mean = 120, what is the Median?
T+J+S = 120/3
--> T+J+S = 360

1.) T = 110
Test Extreme Cases:
If T = 110, J = 1, S = 249 .... Median = 110
If T = 110, J = 125, S = 125 .... Median = 125
So, Insuff.

2.) J = 120
Test Extreme Cases:
If J = 120, T = 120, S = 120 .... Median = 120
If J = 120, T = 100, S = 140 .... Median = 120
If J = 120, T = 15, S = 125 .... Median = 120

As you can see, the Median will always equal 120 when one of the values is equal to 120.

So, Suff.

Answer = B, only statement 2.

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Re: Tom, Jane, and Sue each purchased a new house. The average [#permalink]

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22 Jul 2017, 02:06
Hi,

Concept involved :

Median of a no. is
For odd no. of digits in the series the middle term will be the median of the no.

e.g.
3, 13, 7, 5, 21, 23, 39, 23, 40, 23, 14, 12, 56, 23, 29

When we put those numbers in order we have:

3, 5, 7, 12, 13, 14, 21, 23, 23, 23, 23, 29, 39, 40, 56

There are fifteen numbers. Our middle is the eighth number:

3, 5, 7, 12, 13, 14, 21, 23, 23, 23, 23, 29, 39, 40, 56

The median value of this set of numbers is 23

for the even set of no.s the median is the average of the middle terms.

3, 13, 7, 5, 21, 23, 23, 40, 23, 14, 12, 56, 23, 29

When we put those numbers in order we have:

3, 5, 7, 12, 13, 14, 21, 23, 23, 23, 23, 29, 40, 56

There are now fourteen numbers and so we don't have just one middle number, we have a pair of middle numbers:

3, 5, 7, 12, 13, 14, 21, 23, 23, 23, 23, 29, 40, 56

In this example the middle numbers are 21 and 23.

To find the value halfway between them, add them together and divide by 2:

21 + 23 = 44
then 44 ÷ 2 = 22

So the Median in this example is 22.

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Re: Tom, Jane, and Sue each purchased a new house. The average [#permalink]

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17 Aug 2017, 02:33
1) toms house is 110,000 \$

mean is 120,000

sum/N = mean
sum =360,000 as n=3

now toms is 110,000 which means jane and sue has total of 250,000 value
either J or S can be 120,000 or 130,000 or vice versa which changes median..

A and D gone

2) jane is 120,000
T+S =240,000

max can be 120,000 in this case all will be 120,000
and in other cases one is less than 120,000 and other is more than 120,000
in that case also median is same 120,000

hence B is sufficient.
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Re: Tom, Jane, and Sue each purchased a new house. The average   [#permalink] 17 Aug 2017, 02:33
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