Tom, Jane, and Sue each purchased a new house. The average : GMAT Data Sufficiency (DS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 27 Feb 2017, 08:14

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Director
Joined: 12 Oct 2008
Posts: 554
Followers: 3

Kudos [?]: 439 [1] , given: 2

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

### Show Tags

09 Oct 2009, 22:27
1
KUDOS
27
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

59% (02:06) correct 41% (00:53) wrong based on 1141 sessions

### HideShow timer Statistics

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 Math Expert Joined: 02 Sep 2009 Posts: 37144 Followers: 7266 Kudos [?]: 96779 [7] , given: 10784 Re: Average [#permalink] ### Show Tags 09 Oct 2009, 22:53 7 This post received KUDOS Expert's post 10 This post was BOOKMARKED reply2spg 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

_________________
Senior Manager
Joined: 18 Sep 2009
Posts: 360
Followers: 3

Kudos [?]: 451 [0], given: 2

### Show Tags

10 Oct 2009, 07:50
good explanation ,thaank you very much
SVP
Joined: 29 Aug 2007
Posts: 2492
Followers: 69

Kudos [?]: 748 [0], given: 19

### Show Tags

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.... _________________ Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html GT Intern Affiliations: CA - India Joined: 27 Oct 2009 Posts: 45 Location: India Schools: ISB - Hyderabad, NSU - Singapore Followers: 20 Kudos [?]: 703 [0], given: 5 Re: mean n median [#permalink] ### Show Tags 05 Nov 2009, 22:29 1 This post was BOOKMARKED Mean and one of the values out of 3 values are same, hence median has to be equal to mean. Ans. is B. Manager Joined: 30 May 2008 Posts: 76 Followers: 1 Kudos [?]: 96 [0], given: 26 Re: Average [#permalink] ### Show Tags 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 Answer B. 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? Math Expert Joined: 02 Sep 2009 Posts: 37144 Followers: 7266 Kudos [?]: 96779 [0], given: 10784 Re: Average [#permalink] ### Show Tags 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 Answer B. 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. _________________ Intern Joined: 12 Sep 2011 Posts: 13 Followers: 0 Kudos [?]: 0 [0], given: 3 Re: Average [#permalink] ### Show Tags 06 Sep 2012, 10:38 Bunuel wrote: reply2spg 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
Manager
Joined: 24 Mar 2010
Posts: 81
Followers: 1

Kudos [?]: 62 [0], given: 134

### Show Tags

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?

_________________

- Stay Hungry, stay Foolish -

Intern
Joined: 07 May 2011
Posts: 42
GMAT 1: Q V
GMAT 2: Q V
Followers: 0

Kudos [?]: 18 [2] , given: 11

### Show Tags

04 Dec 2012, 18:55
2
KUDOS
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?

Manager
Joined: 24 Mar 2010
Posts: 81
Followers: 1

Kudos [?]: 62 [0], given: 134

### Show Tags

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.

???

_________________

- Stay Hungry, stay Foolish -

Senior Manager
Status: Prevent and prepare. Not repent and repair!!
Joined: 13 Feb 2010
Posts: 274
Location: India
Concentration: Technology, General Management
GPA: 3.75
WE: Sales (Telecommunications)
Followers: 9

Kudos [?]: 90 [0], given: 282

### Show Tags

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.
_________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan
Kudos drives a person to better himself every single time. So Pls give it generously
Wont give up till i hit a 700+

Manager
Joined: 03 Mar 2013
Posts: 91
Location: India
Concentration: General Management, Marketing
GPA: 3.49
WE: Web Development (Computer Software)
Followers: 0

Kudos [?]: 8 [0], given: 6

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

### Show Tags

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 Intern Joined: 21 Dec 2013 Posts: 24 Concentration: General Management, Technology WE: Project Management (Computer Software) Followers: 1 Kudos [?]: 18 [0], given: 21 Re: mean n median [#permalink] ### Show Tags 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." Current Student Joined: 25 Sep 2012 Posts: 300 Location: India Concentration: Strategy, Marketing GMAT 1: 660 Q49 V31 GMAT 2: 680 Q48 V34 Followers: 2 Kudos [?]: 136 [1] , given: 242 Re: Tom, Jane, and Sue each purchased a new house. The average [#permalink] ### Show Tags 31 May 2014, 13:37 1 This post received KUDOS Hey mods! the answer has not been registered here... As confirmed by earlier posts, the answer is B Math Expert Joined: 02 Sep 2009 Posts: 37144 Followers: 7266 Kudos [?]: 96779 [0], given: 10784 Re: Tom, Jane, and Sue each purchased a new house. The average [#permalink] ### Show Tags 31 May 2014, 15:00 b2bt wrote: Hey mods! the answer has not been registered here... As confirmed by earlier posts, the answer is B Added the OA. Thank you. _________________ Intern Joined: 28 May 2014 Posts: 16 Followers: 0 Kudos [?]: 2 [0], given: 0 Re: Tom, Jane, and Sue each purchased a new house. The average [#permalink] ### Show Tags 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.

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13990
Followers: 592

Kudos [?]: 168 [0], given: 0

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

### Show Tags

06 Jul 2015, 23:09
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Intern
Joined: 04 Feb 2014
Posts: 25
Followers: 0

Kudos [?]: 1 [0], given: 1

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

### Show Tags

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?
Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2651
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Followers: 120

Kudos [?]: 1373 [0], given: 789

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

### Show Tags

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
_________________

Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515
Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html
Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html

Re: Tom, Jane, and Sue each purchased a new house. The average   [#permalink] 29 Aug 2015, 18:34

Go to page    1   2    Next  [ 22 posts ]

Similar topics Replies Last post
Similar
Topics:
John and Tom leave the school at the same time to each of their houses 2 13 Feb 2017, 00:00
1 Jane and Tom spent a total of \$160 on clothes. How much did Tom spend? 5 16 Nov 2015, 21:49
2 Susan, Jordan, and Mitchell each purchased a used car. The average (ar 7 13 Feb 2015, 07:51
2 Sue purchased several identical brooms at Gigamart. 2 21 Feb 2014, 05:52
Tom and Jack are in a line to purchase iphones. How many 1 09 Jul 2011, 02:43
Display posts from previous: Sort by