Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 27 Aug 2016, 08:46

### 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

# In a village of 100 households, 75 have at least one DVD

Author Message
TAGS:

### Hide Tags

Retired Moderator
Status: The last round
Joined: 18 Jun 2009
Posts: 1309
Concentration: Strategy, General Management
GMAT 1: 680 Q48 V34
Followers: 78

Kudos [?]: 881 [9] , given: 157

In a village of 100 households, 75 have at least one DVD [#permalink]

### Show Tags

30 Jul 2010, 05:45
9
KUDOS
84
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

34% (03:21) correct 66% (02:12) wrong based on 1567 sessions

### HideShow timer Statistics

In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x – y is:

A. 65
B. 55
C. 45
D. 35
E. 25
[Reveal] Spoiler: OA

_________________
Intern
Joined: 04 Jun 2011
Posts: 36
Followers: 0

Kudos [?]: 40 [28] , given: 61

Re: 800 Score: Overlapping Set Problem [#permalink]

### Show Tags

24 Jun 2011, 13:58
28
KUDOS
10
This post was
BOOKMARKED
Am I missing something here??? it seems straightforward......

The obvious maximum that have all 3 is 55, because you are limited by the SMALLEST number.

The minimum is simply the sum of the max of each people who DONT have the product, so:

100-80 = 20 don't have Cell
100-75 = 25 don't have DVD
and 100-55 = 45 don't have MP3

SO a total of 20+25+45 = 90 combined who might NOT have SOME combination of the 3 products. So subtract that from 100, to give you the minimum of the people who COULD have all 3 and you get 100-90 = 10.

55-10 = 45
Math Expert
Joined: 02 Sep 2009
Posts: 34456
Followers: 6277

Kudos [?]: 79641 [19] , given: 10022

Re: 800 Score: Overlapping Set Problem [#permalink]

### Show Tags

30 Jul 2010, 06:25
19
KUDOS
Expert's post
8
This post was
BOOKMARKED
Hussain15 wrote:
In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x – y is:

A. 65

B. 55

C. 45

D. 35

E. 25

For me the best way to solve this problem is not use Venn diagram or formulas but to draw simple bars (note: each dash is 5):

Min overlap is 10:
-------------------- 80 phones;
-------------------- 75 DVD's;
-------------------- 55 MP3.

Max overlap is 55:
-------------------- 80 phones;
-------------------- 75 DVD's;
-------------------- 55 MP3.

55-10=45.

_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6830
Location: Pune, India
Followers: 1926

Kudos [?]: 11956 [18] , given: 221

Re: 800 Score: Overlapping Set Problem [#permalink]

### Show Tags

01 Dec 2010, 09:37
18
KUDOS
Expert's post
10
This post was
BOOKMARKED
hiredhanak: I am assuming you are looking for a venn diagram solution to this question..

It is pretty simple.
First of all maximum number of households: We want to bring the circles to overlap as much as possible.
80 - Cell phone
75 - DVD
55 - MP3
Lets take Cell phone and DVD circles since they will have maximum overlap. They must overlap in 55 households so that total number of households is 100. Now put the MP3 households in a way to maximize all three overlap.
Attachment:

Ques1.jpg [ 12.66 KiB | Viewed 29718 times ]

So at most 55 households can have all 3.

Now, minimum number of households: We want to take the circles as far apart from each other as possible. Now put the MP3 households in a way to minimize all three overlap. So make the MP3 households occupy the shaded region i.e. region occupied by DVD players alone and cell phone alone. You will be able to adjust 45 MP3s outside the common area but you will need to put 10 of the MP3 households in the common area. So minimum overlap is 10.
Attachment:

Ques2.jpg [ 16.98 KiB | Viewed 29712 times ]

x - y = 55 - 10 = 45
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Last edited by VeritasPrepKarishma on 24 Nov 2011, 19:49, edited 1 time in total. Intern Joined: 07 Jul 2010 Posts: 21 Followers: 1 Kudos [?]: 12 [9] , given: 6 Re: 800 Score: Overlapping Set Problem [#permalink] ### Show Tags 14 Aug 2010, 01:38 9 This post received KUDOS 2 This post was BOOKMARKED For mainhoon and others who still didn't understand, you have to create the situation of minimum possible overlap. This is done in the way Bunuel explains it: After the first step of distributing phones and DVDs among the 100, you get 55 as the minimum possible overlap for these 2. Now you bring in the MP3s, of which there are 55. Remember, you are trying to make sure as few people as possible get all 3 - if possible, none. So, what do you do when you have 55 MP3s, and 45 people left who don't have an DVD and a phone? Distribute it among them first, naturally! After doing that, you still have 10 left over. You are left with no choice but to distribute these among those who already have a DVD and phone. Hence the minimum overlap you can possibly achieve is 10. This gives you the answer of 55 - 10 = 45. Last edited by rednblack89 on 14 Aug 2010, 01:46, edited 1 time in total. Math Expert Joined: 02 Sep 2009 Posts: 34456 Followers: 6277 Kudos [?]: 79641 [8] , given: 10022 Re: 800 Score: Overlapping Set Problem [#permalink] ### Show Tags 08 Aug 2010, 03:24 8 This post received KUDOS Expert's post 4 This post was BOOKMARKED nitishmahajan wrote: Hi Bunuel, Could you please explain how did you get the minimum over lap as 10 ? Cheers, seekmba wrote: Hey Bunuel, Can you please explain the min overlap part where you got 10? Thanks Well everything is on the diagram but I'll try to elaborate: -----(-----------)---- 80 phones (black dashes represents # of people who don't have phone, so 20 people don't have phone); -----(-----------)---- 75 DVD's (black dashes represent # of people who don't have DVD, so 25 people don't have DVD); So you see that overlap of phone and DVD owners is 55 and 45 people don't have either phones or DVD's. To have min overlap of 3 let 20 people who don't have phone to have MP3 and also 25 people who don't have DVD's to have MP3. So we distributed 45 MP3 and there is no overlap of 3 yet. But 10 DVD's are still left to distribute and only people to have them is the group which has both MP3 and phone. -----(--)-------------; -----(--)-------------; -----(--)-------------. Hope it's clear. _________________ Current Student Status: Everyone is a leader. Just stop listening to others. Joined: 22 Mar 2013 Posts: 993 Location: India GPA: 3.51 WE: Information Technology (Computer Software) Followers: 159 Kudos [?]: 1259 [8] , given: 226 Re: In a village of 100 households, 75 have at least one DVD [#permalink] ### Show Tags 15 Jul 2013, 11:03 8 This post received KUDOS My imagination about minimum overlapping percentage. Attachment: overlapp.jpg [ 70.48 KiB | Viewed 24894 times ] _________________ Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use? | 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction". Intern Joined: 04 May 2013 Posts: 12 Followers: 2 Kudos [?]: 3 [7] , given: 47 Re: 800 Score: Overlapping Set Problem [#permalink] ### Show Tags 20 Jul 2013, 10:19 7 This post received KUDOS 1 This post was BOOKMARKED shs0145 wrote: Am I missing something here??? it seems straightforward...... The obvious maximum that have all 3 is 55, because you are limited by the SMALLEST number. The minimum is simply the sum of the max of each people who DONT have the product, so: 100-80 = 20 don't have Cell 100-75 = 25 don't have DVD and 100-55 = 45 don't have MP3 SO a total of 20+25+45 = 90 combined who might NOT have SOME combination of the 3 products. So subtract that from 100, to give you the minimum of the people who COULD have all 3 and you get 100-90 = 10. 55-10 = 45 i think this is the simplest way to solve this, even i did it the same way. CEO Status: Nothing comes easy: neither do I want. Joined: 12 Oct 2009 Posts: 2795 Location: Malaysia Concentration: Technology, Entrepreneurship Schools: ISB '15 (M) GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Followers: 220 Kudos [?]: 1510 [4] , given: 235 Re: 800 Score: Overlapping Set Problem [#permalink] ### Show Tags 30 Jul 2010, 07:59 4 This post received KUDOS 1 This post was BOOKMARKED Nice approach by Bunnel. Even if we have to use Venn diagram. Attachment: a.png [ 17.44 KiB | Viewed 30974 times ] For minimum overlap, 80 and 75 will have maximum overlap and 55 will be distributed among three sections : namely 80,75 and all three Reason : If 55 have elements with no overlapping then 75 and 80 can not have max overlap. To have minimum overlap, 75 and 80 must overlap to the max. So those elements which are not part of intersection of all the three can be shared between 75 and 80. a+b+c = 80 .............1 b+c+d = 75...............2 a+c+d = 55...............3 => 55 is distributed in 3 sections.... 80 intersection 55 ; 74 intersection 55; 55 intersection 75 intersection 80. a+b+c+d = 100............4 using 4 and 1 we get d = 20 using 4 and 2 we get a = 25 using the above values of a,d and equation 3 we get c = 55 -a-d = 55- 20-25 = 10 Hence minimum is 10. For max : 75 will have 55 inside 80 will have the intersection of 75 and 55 i.e. 55 thus total becomes 55 + 20 + 25 = 100 where 55 = intersection of all the three 20 = 75 outside the intersection of 75 and 55 25 = 80 outside the intersection of 75,55,80 thus 55 is the max. x-y = 55-10 = 45 _________________ Fight for your dreams :For all those who fear from Verbal- lets give it a fight Money Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook GMAT Club Premium Membership - big benefits and savings Gmat test review : http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html SVP Joined: 16 Jul 2009 Posts: 1629 Schools: CBS WE 1: 4 years (Consulting) Followers: 40 Kudos [?]: 884 [4] , given: 2 Re: 800 Score: Overlapping Set Problem [#permalink] ### Show Tags 10 Jan 2011, 10:12 4 This post received KUDOS I have done it in 1:41 doing this: x= (75+55+80-100-z)/2 for z=0 ->55 y=the same for z=100 ->10 Therefore x-y=45 PS: z is obviously the people who have 2 devices. _________________ The sky is the limit 800 is the limit GMAT Club Premium Membership - big benefits and savings Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6830 Location: Pune, India Followers: 1926 Kudos [?]: 11956 [3] , given: 221 Re: 800 Score: Overlapping Set Problem [#permalink] ### Show Tags 09 Jan 2011, 19:24 3 This post received KUDOS Expert's post 1 This post was BOOKMARKED suchoudh wrote: VeritasPrepKarishma wrote: Now, minimum number of households: We want to take the circles as far apart from each other as possible. Now put the MP3 households in a way to minimize all three overlap. So put it in the shaded region. You will need to put 10 of the MP3 households in the common area. So minimum overlap is 10. Ok, and how did you get to the number 10? (25 + 20 =) 45 households have either only Cell or only DVD Player. Out of the 55 households who have MP3 Players, put 45 in these areas so that all three do not overlap. But the rest of the (55 - 45 =)10 households that have MP3 players need to be put in the common region consisting of 55 households that have both Cell and DVD Player. Hence overlap of all three will be 10. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)
Followers: 90

Kudos [?]: 819 [2] , given: 43

Re: In a village of 100 households, 75 have at least one DVD [#permalink]

### Show Tags

14 Aug 2012, 06:41
2
KUDOS
Hussain15 wrote:
In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x – y is:

A. 65
B. 55
C. 45
D. 35
E. 25

A more algebraic approach:

If we denote by $$N_2$$ the number of households with exactly two types of devices, and by $$N_3$$ the number of those with all three types, than we can write the following equation:

$$100=75+80+55-N_2-2N_3,$$ from which we get that $$N_2+2N_3=110.$$
Households with exactly two types of devices we counted twice, so we have to subtract them once. Households with exactly three types of devices we counted three times, therefore we have to remove them twice.

Maximum for $$N_3$$ is 55, in which case $$N_2=0$$, meaning there are no households with exactly two types of devices, but only either with just one type or all three of them. Everybody who has an MP3 player, also has a DVD player and a cell phone.

In order to determine the minimum number of households with exactly three types of devices, let's take a look at the possible combinations of two devices. If $$N_3$$ is minimum, there should be households with exactly two types of devices but not three.

All types of devices, but also those having exactly two types, explicitly DVD players and Cell Phones - If $$N_{DC}$$ is the number of households that have exactly these two types of devices, then $$N_{DC}+N_3\geq75+80-100=55$$.
All types of devices, but also those having exactly two types, explicitly DVD players and MP3 players - If $$N_{DM}$$ is the number of households that have exactly these two types of devices, then $$N_{DM}+N_3\geq75+55-100=30$$.
All types of devices, but also those having exactly two types, explicitly Cell phone and MP3 players - If $$N_{CM}$$ is the number of households that have exactly these two types of devices, then $$N_{CM}+N_3\geq80+55-100=35$$.

Adding the above three inequalities, we obtain that $$N_{DC}+N_{DM}+N_{CM}+3N_3\geq120.$$ Since $$N_{DC}+N_{DM}+N_{CM}=N_2$$, we get that $$N_2+3N_3\geq120.$$ Taking into account that $$N_2+2N_3=110$$, we obtain $$110+N_3\geq{120}$$ or $$N_3\geq{10}.$$

$$55-10=45$$

_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Math Expert
Joined: 02 Sep 2009
Posts: 34456
Followers: 6277

Kudos [?]: 79641 [2] , given: 10022

Re: In a village of 100 households, 75 have at least one DVD [#permalink]

### Show Tags

11 Jun 2013, 08:24
2
KUDOS
Expert's post
1
This post was
BOOKMARKED
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Overlapping Sets:
how-to-draw-a-venn-diagram-for-problems-98036.html

All DS Overlapping Sets Problems to practice: search.php?search_id=tag&tag_id=45
All PS Overlapping Sets Problems to practice: search.php?search_id=tag&tag_id=65

_________________
Senior Manager
Joined: 17 Dec 2012
Posts: 442
Location: India
Followers: 23

Kudos [?]: 361 [2] , given: 14

Re: In a village of 100 households, 75 have at least one DVD [#permalink]

### Show Tags

21 Jul 2013, 01:04
2
KUDOS
I have given the solution to a similar problem. Here's the solution for this problem.

1. Let C, D and M be who have cell only, dvd only and mp3 only and CD be who have both cell and dvd only and so on.
2 CDM is minimum when C+D+M=0 and maximum when CD+CM+DM=0.
3. C+D+M+CD+DM+CM+CDM=100 -- (1) because the total of all the mutually exclusive has to sum up to 100 which is the number of households.
4. (C+CD+CM+CDM) + (D+CD+DM+CDM) +(M+CM+DM+CDM) = 80+75+55 or ,
C+D+M+2CD+2CM+2DM+3CDM =210 ----(2)
5. (2) - (1) = CD+CM+DM+2CDM=110 --- (3)
6. (3) - (1) => CDM-(C+D+M) =10 ---(4)
We can find the minimum value when we put (C+D+M) = 0 above giving minimum CDM=10
7. To find maximun value equation (1) can be changed as
C+D+M+CDM=100 -- (5) by putting CD+CM+DM=0
From (4) and (5) we have CDM=55 which is maximun CDM
8.The answer is max CDM - min CDM i.e., 55-10=45

As a shortcut for this type of problems where there are 3 items , the minimum possible for all the 3 can be found out by adding the percentages of the 3 i.e., 55+75+80=210 . and subtracting 200 i.e, 210-200=10 . the maximum can be found out by adding the minimum to 100 and dividing by 2 i.e., (100+10)/2 = 55. Both the answers are percentages.
_________________

Srinivasan Vaidyaraman
Sravna
http://www.sravnatestprep.com

Classroom and Online Coaching

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6830
Location: Pune, India
Followers: 1926

Kudos [?]: 11956 [2] , given: 221

Re: In a village of 100 households, 75 have at least one DVD [#permalink]

### Show Tags

22 Jun 2015, 20:29
2
KUDOS
Expert's post
wrightvijay wrote:

Where does it say that every household in the village have at least one of these three devices?
There could be none of these three devices in some households.

It's not given. It is something we have derived using logic.

Take a simpler case:

Say you have 3 siblings (A,B and C) and 5 chocolates which you want to distribute among them in any way you wish. Now you want to minimise the number of your siblings who get 3 chocolates. No one gets more than 3. What do you do?

Will you leave one sibling without any chocolates (even if he did rat you out to your folks!)? No. Because if one sibling gets no chocolates, the other siblings get more chocolates and then more of them will get 3 chocolates. So instead you give 1 to each and then give the leftover 2 to 2 of them. This way, no sibling gets 3 chocolates and you have successfully minimised the number of siblings who get 3 chocolates. Basically, you spread out the goodies to ensure that minimum people get too many of them.

This is the same concept. When you want to minimise the overlap, you basically want to spread the goodies around. You want minimum people to have all 3. So you give atleast one to all of them.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Intern
Joined: 29 Mar 2014
Posts: 14
Location: United States
Concentration: Entrepreneurship, Finance
GMAT 1: 720 Q50 V39
GPA: 3
Followers: 0

Kudos [?]: 19 [1] , given: 4

Re: In a village of 100 households, 75 have at least one DVD [#permalink]

### Show Tags

05 Apr 2014, 00:31
1
KUDOS
Stay away venn diagram for problems of this kind..
The maximum number is 55, you don't need any computing for this.

For finding the minimum possible household, just find out the number of households that would not have these gadgets

Number of houses that don't have a cellphone, DVD and MP3 are, 20, 25 and 45 respectively

When there is minimum overlap the number of households that cannot have all the three gadgets together is the sum of these three numbers which is 90. So the minimum possible number of households that can have all three gadgets is 10

55-10 = 45
VP
Joined: 08 Jul 2010
Posts: 1220
Location: India
GMAT: INSIGHT
WE: Education (Education)
Followers: 47

Kudos [?]: 1100 [1] , given: 40

Re: In a village of 100 households, 75 have at least one DVD [#permalink]

### Show Tags

22 Jun 2015, 05:29
1
KUDOS
Expert's post
wrightvijay wrote:
________________________________

Where does it say that every household in the village have at least one of these three devices?
There could be none of these three devices in some households.

Hi wrightvijay,

It doesn't matter whether they have given the highlighted fact above or not because it's the question of maximizing and minimizing the overlapping portion which will be maximized when everything else is minimized and vice versa.
_________________

Prosper!!!

GMATinsight

Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com
Call us : +91-9999687183 / 9891333772
http://www.GMATinsight.com/testimonials.html

Contact for One-on-One LIVE ONLINE (SKYPE Based) or CLASSROOM Quant/Verbal FREE Demo Class

GMATinsight
107, 1st Floor, Krishna Mall, Sector-12 (Main market), Dwarka, New Delhi-110075

______________________________________________________
Please press the if you appreciate this post !!

Senior Manager
Joined: 03 Nov 2005
Posts: 395
Location: Chicago, IL
Followers: 3

Kudos [?]: 42 [0], given: 17

Re: 800 Score: Overlapping Set Problem [#permalink]

### Show Tags

02 Aug 2010, 18:14
a+b+c+d = 100............4
using 4 and 1 we get d = 20
using 4 and 2 we get a = 25
using the above values of a,d and equation 3
we get c = 55 -a-d = 55- 20-25 = 10

Hence minimum is 10.

For max :
75 will have 55 inside
80 will have the intersection of 75 and 55 i.e. 55

thus total becomes 55 + 20 + 25 = 100 where 55 = intersection of all the three
20 = 75 outside the intersection of 75 and 55
25 = 80 outside the intersection of 75,55,80

thus 55 is the max.

x-y = 55-10 = 45[/quote]

gurpreetsingh,

Is it possible to demonstrate your approach using 3 overlapping sets. That's the reason why I can't make that much sence out of your explanation because here we have 3 overlapping sets. I cant figure out why you are using just 2.

All the help is highly appreciated.
_________________

Hard work is the main determinant of success

Director
Status: Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Joined: 18 Jul 2010
Posts: 690
Schools: Wharton, Sloan, Chicago, Haas
WE 1: 8 years in Oil&Gas
Followers: 15

Kudos [?]: 136 [0], given: 15

Re: 800 Score: Overlapping Set Problem [#permalink]

### Show Tags

08 Aug 2010, 10:24
Bunuel,
Your answer is elegant. But I must admit I would have jumped into the Venn diagram first. So can you please explain how to solve this in that approach using the usually established formulae? It would be great to think out of the box, but sadly my mind is not such

Also in your explanation above, you mention overlap of "3", did you mean "2" = 2x5=10?
_________________

Consider kudos, they are good for health

Senior Manager
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 408
Location: Milky way
Schools: ISB, Tepper - CMU, Chicago Booth, LSB
Followers: 8

Kudos [?]: 185 [0], given: 50

Re: 800 Score: Overlapping Set Problem [#permalink]

### Show Tags

14 Aug 2010, 18:04
Bunuel wrote:
nitishmahajan wrote:
Hi Bunuel,

Could you please explain how did you get the minimum over lap as 10 ?

Cheers,

seekmba wrote:
Hey Bunuel,

Can you please explain the min overlap part where you got 10?

Thanks

Well everything is on the diagram but I'll try to elaborate:

-----(-----------)---- 80 phones (black dashes represents # of people who don't have phone, so 20 people don't have phone);
-----(-----------)---- 75 DVD's (black dashes represent # of people who don't have DVD, so 25 people don't have DVD);

So you see that overlap of phone and DVD owners is 55 and 45 people don't have either phones or DVD's.

To have min overlap of 3 let 20 people who don't have phone to have MP3 and also 25 people who don't have DVD's to have MP3. So we distributed 45 MP3 and there is no overlap of 3 yet. But 10 DVD's are still left to distribute and only people to have them is the group which has both MP3 and phone.

-----(--)-------------;
-----(--)-------------;
-----(--)-------------.

Hope it's clear.

Fantastic explanation. Thanks Bunuel for your explanation and Hussain15 for posting this question.
_________________

Support GMAT Club by putting a GMAT Club badge on your blog

Re: 800 Score: Overlapping Set Problem   [#permalink] 14 Aug 2010, 18:04

Go to page    1   2   3    Next  [ 44 posts ]

Similar topics Replies Last post
Similar
Topics:
1 In the youth summer village there are 150 people, 75 of them are not w 1 19 Jun 2016, 11:03
1 In a certain village, m litres of water are required per household per 3 13 May 2016, 04:52
2 In how many no. between 100 and 1000 at least one of their digits is 7 2 14 Oct 2015, 21:23
3 In a neighborhood having 90 households, 11 did not have 11 23 May 2012, 12:24
11 In a village of 100 households, 75 have at least one DVD pla 18 26 Jul 2008, 10:53
Display posts from previous: Sort by