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Venn Diagrams- Basics, Problems, Maxima and Minima

Venn Diagrams- Basics, Problems, Maxima and Minima

byTotal Gadha- Tuesday, 18 September 2007, 04:49 PM

For all the CAT aspirants taking CAT in 2009 or 2010, this chapter should provide

some insight into Venn diagrams and methods for solving the problems. This

chapter comes on the demand of some high octane TG users who are responsible

for my lack of sleep and excessive intake of caffeine last night. I hope this resolves

many of their problems in Venn diagrams.

Venn diagrams are pictorial representations used to display mathematical or logical relationships betweentwo or more given sets (groups of things). The drawing consists of two or more circles, each representinga specific group. Each Venn diagram begins with a rectangle representing the universal set. Then eachset in the problem is represented by a circle. Any values that belong to more than one set will be placedin the sections where the circles overlap. A typical venn diagram is shown in the figure below:
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In the figure, set A contains the multiples of 2 which are less than 30 and set B contains multiples of 3which are less than 25. Therefore, A = {2, 4, 6, 8, 10, 12... 26, 28} and B = {3, 6, 9, 12... 21, 24}. Thevarious areas in the above diagram depict the following relationships:

Intersection (AB)- Denotes the set of elements that are shared by two or more given sets. In

the figuregiven below, the intersection of the two sets is shown.

A B = {6, 12, 18, 24}

Only A or Only B- The part of set A, or set B, which is not shared by any other set is known as"only A," or "only B." In the figure given below, the two parts are shown:

Only A = {2, 4, 8, 10, 14, 16, 20, 22, 26, 28}, only B = {3, 9, 15, 21}

Union (AUB)- Denotes all the elements of the given sets taken once.

A U B = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 26, 28}

It can be seen that


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The venn diagram for three sets is shown below:

It can be shown that

Problem-solving through Venn diagrams:

I use the following method to solve problems through Venn diagrams:

Solved Examples:

Of all the users on Totalgadha.com, 80% spend time in CAT Quant-DI forum whereas 60% spend time inCAT verbal forum. If only those users will crack CAT who spend time in both the forums, what percentageof users of TotalGadha

will crack CAT? will not crack CAT?

Answer: n(AUB) = n(A) + n(B) n(AB) 100% = 80% + 60% n(AB) n(AB) = 40%

Therefore, 40% users of TG will crack CAT. And 60% of users (only A + only B) will not crack CAT.

NOTE: See that the surplus (superfluous part) can only be adjusted inside the area denoted for theintersection of the sets, a fact we will use in maxima- minima type of questions.

A survey on a sample of 25 new cars being sold at a local auto dealer was conducted to see which of thethree popular options air conditioning, radio and power windows were already installed. Thesurvey found:

15 had air conditioning2 had air conditioning and power windows but no radios12 had radio6 had air conditioning and radio but no power windows11 had power windows4 had radio and power windows3 had all three options.

What is the number of cars that had none of the options? (CAT 2003)1. 4 2. 3 3. 1 4. 2


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Answer: We make the Venn diagram and start filling the areas as shown:

Total Number of cars according to the diagram = 2 + 6 + 3 + 1 + 5 + 2 + 4 = 23.Therefore, number of cars having none of the given options = 25 23 = 2.

New Age Consultants have three consultants Gyani, Medha and Buddhi. The sum of the number ofprojects handled by Gyani and Buddhi individually is equal to the number of projects in which Medha isinvolved. All three consultants are involved together in 6 projects. Gyani works with Medha in 14projects. Buddhi has 2 projects with Medha but without Gyani, and 3 projects with Gyani but withoutMedha. The total number of projects for New Age Consultants is one less than twice the number ofprojects in which more than one consultant is involved. (CAT 2003- Leaked)

What is the number of projects in which Gyani alone is involved?1. 02. 1.3. 4.4. cannot be determined

What is the number of projects in which Medha alone is involved?1. 02. 1.3. 4.4. cannot be determined

Answer: The Venn diagram for the three consultants is shown below:

Total Number of projects = 2 number of projects in which more than one consultant is involved 1 = 2 19 1 = 37.Therefore, X + 8 + 6 + 3 + Y + 2 + X + Y 16 = 37 X + Y = 17. The values of X or Y cannot be

uniquely determined. Medha alone is involved in X + Y 16 = 17 16 = 1 project.

Concept of Maxima and Minima:

1. When the total number of elements is fixed


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Let's have a look at the Venn diagram of two sets again:

Imagine that in the beginning, the number of elements in all the areas is zero, as shown above. All thesets are empty right now.

Let's see what happens if I insert one element inside AB:

We can see that adding 1 element to AB increases the number of elements in both A and B by 1. The

total number of elements in all areas combined is 1 only (0 + 1 + 0) but if you add the number ofelements in A and B (A + B), the addition will come up to 2. Therefore, adding 1 element to AB

gives an extra 1 element. Hence, for every surplus of 1 element we can add 1 element toAB.

Lets see the Venn diagram for 3 sets:

In diagram 1, we have added 1 element to intersection of only two sets (A and B but not C). We can seethat A and B both increase by 1 and therefore we get a surplus of 1 element.

In diagram 2, we have added 1 element to intersection of all the three sets (A and B and C). We can seethat A, B and C all three increase by 1 element each and therefore we get a surplus of 2 elements.

Therefore, in case of three sets, we can accommodate the surplus by

adding elements to intersection of only two sets in which case a surplus of 1 element can be

accommodated by increase of 1 element in the intersection of only two sets.

adding elements to intersection of three sets in which case a surplus of 2 elements can be


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accommodated by increase of 1 element in the intersection of three sets.

How is this related to maxima and minima?

Let's see:

According to a survey, at least 70% of people like apples, at least 75% like bananas and at least 80%like cherries. What is the minimum percentage of people who like all three?

Answer: Let's first calculate the surplus:

percentage of people who like apples + percentage of people who like bananas + percentage of peoplewho like cherries = 70% + 75% + 80% = 225% a surplus of 125%.

Now this surplus can be accommodated by adding elements to either intersection of only two sets or tointersection of only three sets. As the intersection of only two sets can accommodate only a surplus of100%, the surplus of 25% will still be left. This surplus of 25% can be accommodated by adding elementsto intersection of three sets. For that we have to take 25% out of the intersection of only two sets andadd it to intersection of three sets. Therefore, the minimum percentage of people who like all three =25%

The question can be solved mathematically also. Let the elements added to intersection of only two setsand intersection of three sets be x and y, respectively. These elements will have to cover the surplus.

x + 2y = 125%, where x + y 100%. For minimum value of y, we need maximum value of x. x = 75%, y = 25%.

In a college, where every student follows at least one of the three activities- drama, sports, or arts- 65%follow drama, 86% follow sports, and 57% follow arts. What can be the maximum and minimumpercentage of students who follow all three activities exactly two activities

Answer: Let us again see the surplus:

Percentage of students who follow drama + Percentage of students who follow sports + Percentage ofstudents who follow arts = 65% + 86% + 57% = 208% surplus = 108%. This surplus can be

accommodated through adding elements either to intersection of only two sets or to intersection of onlythree sets. As the intersection of only two sets can accommodate only a surplus of 100%, the surplus of8% will still be left. This surplus of 8% can be accommodated by adding elements to intersection of threesets. For that we have to take 8% out of the intersection of only two sets and add it to intersection ofthree sets. Therefore, the minimum percentage of people who like all three = 8%. In this casethe percentage of students who follow exactly two activities will be maximum = 92% .

The surplus of 108% can also be accommodated through adding elements to only intersection of threesets. As adding 1 element to intersection of three sets give a surplus of 2 sets, adding 54% tointersection of three sets will give a surplus of 108%. Therefore, the maximum value of studentswho follow all three activities is 54%. In this case the percentage of students who follow

exactly two activities will be minimum = 0%.

We can also solve it mathematically x + 2y = 108%, where x + y 100%. The maximum value of x

will give minimum value of y, whereas minimum value of x will give maximum value of y.

2. When the total number of elements is NOT fixed

In this case we assign the variables to every area of the Venn diagram and form the conditions keepingtwo things in mind: try to express the areas in the Venn diagram through least number of variables. all the numbers will be zero or positive. No number can be negative.

Out of 210 interviews of IIM- Ahmedabad, 105 CAT crackers were offered tea by the interview panel, 50

were offered biscuits, and 56 were offered toffees. 32 CAT crackers were offered tea and biscuits, 30were offered biscuits and toffees, and 45 were offered toffees and tea. What is the maximum and minimum number of CAT crackers who were offered all three snacks?


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maximum and minimum number of CAT crackers who were offered at least one snack?

Answer: Lets make the Venn diagram for this question. Since we want to assume least number ofvariables, we can see that assuming a variable for the number of students who were offered all threesnacks will help us express all the other areas. Let the number of students who were offered all threesnacks = x.

In the above diagram, we have expressed all the areas in terms of x. To decide maximum value of x, wenote that 32 x, 45 x and 30 x will be zero or positive. Therefore, the maximum value of x will be 30.(30 is the lowest among 30, 32 and 45). To decide minimum value of x, we note than x 19 and x 12 will be zero or positive. Therefore, x cannot be less than 19 (19 is the higher number between 19 and12).

Therefore, maximum and minimum number of CAT crackers who were offered all three snacks = 30 and19.

The number of CAT crackers who were offered at least one snack = Total number of CAT crackers in theVenn diagram = x + 28 + 32 x + x + 45 x + x 19 + 30 x + x 12 = 104 + x.

As the maximum and minimum values of x are 30 and 19, respectively, the maximum and minimumvalue of 104 + x will be 134 and 123, respectively.

Maximum and minimum number of CAT crackers who were offered at least one snack = 134 and 123.

I am afraid I shall have to end here and leave the rest of it for myCBT Club

students. I shall cover some problems based on this in theCBT Clubthis week.

Reply

Re: Venn Diagrams- Basics, Problems, Maxima and Minima

bySmall Wonder- Tuesday, 18 September 2007, 05:02 PM

TG Sir ki jai hoThanks a ton. God Bless You!Small Wonder!Show parent|Reply
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byVaidyanathan Ganesan- Tuesday, 18 September 2007, 05:11 PM

TG,

I was always on the look for your articles on the homepage and this time i was notdisappointed.. . A new article!!!!

Keep up your good work.

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byGul Gul- Tuesday, 18 September 2007, 05:11 PM

YAhooooooooo!!!!! Thnx TG Muahhhhhhhhh.......

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bykishore ayyadevara- Tuesday, 18 September 2007, 05:24 PM

Thanks a lot TG

i could not get the following point..

Percentage of students who follow drama + Percentage of students who follow sports + Percentage of

students who follow arts = 65% + 86% + 57% = 208% surplus = 108%. This surplus can be

accommodated through adding elements either to intersection of only two sets or to intersection of onlythree sets. As the intersection of only two sets can accommodate only a surplus of 100%, the

surplus of 8% will still be left. This surplus of 8% can be accommodated by adding elements to

intersection of three sets. For that we have to take 8% out of the intersection of only two sets and add

it to intersection of three sets

but here we are considering 3 sets...so how come u took only 2?

the other 2 intersection are 0 then?

Regards,

Kishore.

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bySri KLR- Tuesday, 18 September 2007, 05:36 PM

TG,

Good method to solve max and min.....

x+2y and x+y is too good...I like equations....no thinking needed from now on..

So far I used hit and trial.....but NO more ...feel relieved now....

Please give us some problems on four dimensionals also....

Also give some more prob on venns.. as exercise on Quant-DI forum...

I have kinda premonition that CAT might rear it's ugly head again....CAT(We ) is (are)

going thru Testing times...

I have seen one prob in cat 2 years back....one tough venn prob..don remember...willget back with it..

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bySoftware Engineer- Tuesday, 18 September 2007, 06:31 PM

Here, the quality of content is best, relally ! Thank You.

Software Engineer

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byKing Kong- Tuesday, 18 September 2007, 07:01 PM

TG == Too Good !!?? [ ] Awesome stuff...

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Hi Kishore,
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"intersection of only two sets" means area containing intersection of two sets but not the

third set. In the figure given below, the area in red is the intersection of only two sets.

Total Gadha

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bypadmaselvan lakshman- Tuesday, 18 September 2007, 07:38 PM

hi TG,

Thanx a lot for such a wonderful article.. U have made maxima and minima look a lot more easier through this article..

as suggested earlier, pls give us a quiz on Venn Diag for practice..

Show parent|Reply



hi TG,

the thing is...u said that the intersection can take a maximum surplus of 100%

but, why isnt it more than 100?

if the 3 shaded portions are a,b,c . then what if a+b = 100%(i.e., all the contents of the first

circle are in the shaded part) and c>0?

Regards,

Kishore.

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bythe underdog- Tuesday, 18 September 2007, 08:12 PM

"we are not worthy" "we are not worthy"*

* from the movie "Wayne's World"
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11/73

hehe.. thanks TG!

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byAkon Convict- Tuesday, 18 September 2007, 08:32 PM

Thank U very much TG.....

God Bless You

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Hi Kishore,

How can something be more than 100%? If I have given you 10 boys at most can you have

more than 10 boys combining all the areas?

Total Gadha

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byCrazy CAT- Tuesday, 18 September 2007, 08:49 PM

Thanx a tonne Tg ,

It was worth waiting for such a long time for this article.

one request Sir ,plz give some tips like this on functions and graph article also.

Regards

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12/73

a topic on cubes also please.....

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bythe underdog- Tuesday, 18 September 2007, 09:25 PM

+1Gullz Golu

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Help me in the following problem TGAll students in a class of 100 attended a summer camp. Each student had the option ofenrolling for coaching in atmost 3 namely Football, Cricket and hockey.

a students had enrolled for hockey

b for football

c for cricket

Also d students had enrolled for eactly 1 sport, e for exactly 2, f for exactly 3 and gstudents had not enrolled for any of the three sports.

Q1 If d>e>f and c is less than a as well as b, what is the maximum possible value of c?

Q2 If a,b,c,d,e,f and g are all distinct the minimum possible no. of students whoenrolled for atleast 1 sport is ?

Q2 If a is less than b as well as c, the no. of students who enrolled only for hockey is ?

Please Help....

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TG

seems i made some horrible interpretations then

thought A ,B,C are 100% each
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13/73

what i find is the thrice intersected part(part common to all the circles) is counted thrice and

(AnB) (BnC)(CnA)

are counted twice

coming to the problem,

so the surplus 108% should be distributed in the following way

x+2y = 108% ( where x is the intersection of 2 circles n y is common to three)

my problem is , i could not get how x+y = 100%...do u mean to say that the elements

belonging to ONLY A, B,C are all zero and every element lies in the intersection parts?

Regards,

Kishore.

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Kishore,

See if this is clear:

Say all students = 100 % in terms of equation a1+a2+a3=100 and we havea1+2a2+3a3=208. Subtract the above 2 equations we haave a2+2a3=108 now a3 shouldbe minimum 8 and a2 maximum 92 i.e. 92+2*8=108 see now a1+a2+a3 = 100 ismaintained. Also a3 max= 54 in this case a2=0 and a1 will take 46. Is it a bit clearnow?

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byTop CAT- Tuesday, 18 September 2007, 11:09 PM

Exhaustive yet Simple

that,s ur trademark.....

that's what we all like about this site...

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byAbhi S- Wednesday, 19 September 2007, 02:21 AM


14/73

nw iv gone mad...i thot u r lost somewhere...bt u...bt u came wid a bang....

thnx TG sir.....

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byAbhi S- Wednesday, 19 September 2007, 02:23 AM

nw plz...u may hav a rest....let us hav a sip f d same caffiene....take care TG sir...

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byanand pani- Wednesday, 19 September 2007, 08:17 AM

great

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bykishore ayyadevara- Wednesday, 19 September 2007, 10:40 AM

Thank You Gullz

it's clear now...i was having problems with interpreting the assumptions

and thanks TG for this useful article

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byGul Gul- Wednesday, 19 September 2007, 10:46 AM

Thanks to TG....

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bydinesh munna- Wednesday, 19 September 2007, 12:41 PM

Sir...I am sorry but this article was not exactly what i expected.I was expecting you to put sum

fundaes on solving problems like the one just posted here by sum guy.Pls luk into it and fit ur

solution into that.thanq

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15/73


byTotal Gadha- Wednesday, 19 September 2007, 01:22 PM

Hi Dinesh,

Unfortunately, I am not preparing you for a Math Olympiad but CAT only. Have a look at the

Venn diagram problem that came in CAT 2006:

A survey was conducted of 100 people to find out whether they had read recent issues of

Golmal, a monthly magazine. The summarized information regarding readership in 3 months

is given below: Only September: 18, September but not August: 23, September and July: 28,

September: 28, July: 48, July and August: 10, None of the three months: 24.

What is the number of surveyed people who have read exactly two consecutive issues (out of

the three)?

(1) 7 (2) 9 (3) 12 (4) 14 (5) 17

The article already has question that came in previous CAT papers. There has been no

mention of even maxima or minima in CAT.

I sincerely believe that this is all that you need to know to tackle Venn diagram problems in

CAT.

As for the fundas for solving problems like the one posted here, I can solve it through thefundas mentioned in the article only.

Total Gadha

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byTom Goel- Wednesday, 19 September 2007, 01:30 PM

I know what we can say in respect of TG falls very short of what he deserves...a wordof thanks for TG not from the keyboard but from heart.

Thanks a lot and keep the good work going.......

+1 regrading article on cubes ;)

Amit Goel

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16/73


byTotal Gadha- Wednesday, 19 September 2007, 01:46 PM

Hi Tom,

Let me try. I can certainly imagine the hours on 'paint' I will have to spend to draw images

of cubes

Total Gadha

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byGul Gul- Wednesday, 19 September 2007, 02:06 PM

TG

PLZ help out in the question posted i was able to crack half questions in that set but therest half posted is posing some problem. Plz help so that i can be completely confidentin this topic..

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byravi kumar- Wednesday, 19 September 2007, 03:06 PM

TG

I feel ashamed to learn so much from you for free. Give me your address I want to send you

some gurudakshina. Also feel sorry for ppl who are not gadhas yet

best regards

Ravi Kumar

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byjitendra havaldar- Wednesday, 19 September 2007, 04:37 PM

Hi TG,

In the problem of cars having ACs,power windows(PW),radio(R) i didn't understandthe way you have splitted the radio and the power windows part by taking (4-3) = 1 in
http://totalgadha.com/mod/forum/post.php?reply=11942http://totalgadha.com/mod/forum/post.php?reply=11942http://totalgadha.com/mod/forum/post.php?reply=11942http://totalgadha.com/user/view.php?id=2&course=1http://totalgadha.com/user/view.php?id=2&course=1http://totalgadha.com/user/view.php?id=2&course=1http://totalgadha.com/mod/forum/discuss.php?d=1262#11942http://totalgadha.com/mod/forum/discuss.php?d=1262#11942http://totalgadha.com/mod/forum/post.php?reply=11946http://totalgadha.com/mod/forum/post.php?reply=11946http://totalgadha.com/mod/forum/post.php?reply=11946http://totalgadha.com/user/view.php?id=4009&course=1http://totalgadha.com/user/view.php?id=4009&course=1http://totalgadha.com/user/view.php?id=4009&course=1http://totalgadha.com/mod/forum/discuss.php?d=1262#11946http://totalgadha.com/mod/forum/discuss.php?d=1262#11946http://totalgadha.com/mod/forum/post.php?reply=11951http://totalgadha.com/mod/forum/post.php?reply=11951http://totalgadha.com/mod/forum/post.php?reply=11951http://totalgadha.com/user/view.php?id=10991&course=1http://totalgadha.com/user/view.php?id=10991&course=1http://totalgadha.com/user/view.php?id=10991&course=1http://totalgadha.com/mod/forum/discuss.php?d=1262#11828http://totalgadha.com/mod/forum/discuss.php?d=1262#11828http://totalgadha.com/mod/forum/post.php?reply=11970http://totalgadha.com/mod/forum/post.php?reply=11970http://totalgadha.com/mod/forum/post.php?reply=11970http://totalgadha.com/user/view.php?id=1782&course=1http://totalgadha.com/user/view.php?id=1782&course=1http://totalgadha.com/user/view.php?id=1782&course=1http://totalgadha.com/user/view.php?id=1782&course=1http://totalgadha.com/user/view.php?id=10991&course=1http://totalgadha.com/user/view.php?id=4009&course=1http://totalgadha.com/user/view.php?id=2&course=1http://totalgadha.com/user/view.php?id=1782&course=1http://totalgadha.com/user/view.php?id=10991&course=1http://totalgadha.com/user/view.php?id=4009&course=1http://totalgadha.com/user/view.php?id=2&course=1http://totalgadha.com/user/view.php?id=1782&course=1http://totalgadha.com/user/view.php?id=10991&course=1http://totalgadha.com/user/view.php?id=4009&course=1http://totalgadha.com/user/view.php?id=2&course=1http://totalgadha.com/user/view.php?id=1782&course=1http://totalgadha.com/user/view.php?id=10991&course=1http://totalgadha.com/user/view.php?id=4009&course=1http://totalgadha.com/user/view.php?id=2&course=1http://totalgadha.com/user/view.php?id=1782&course=1http://totalgadha.com/user/view.php?id=10991&course=1http://totalgadha.com/user/view.php?id=4009&course=1http://totalgadha.com/user/view.php?id=2&course=1http://totalgadha.com/user/view.php?id=1782&course=1http://totalgadha.com/user/view.php?id=10991&course=1http://totalgadha.com/user/view.php?id=4009&course=1http://totalgadha.com/user/view.php?id=2&course=1http://totalgadha.com/user/view.php?id=1782&course=1http://totalgadha.com/user/view.php?id=10991&course=1http://totalgadha.com/user/view.php?id=4009&course=1http://totalgadha.com/user/view.php?id=2&course=1http://totalgadha.com/user/view.php?id=1782&course=1http://totalgadha.com/mod/forum/post.php?reply=11970http://totalgadha.com/mod/forum/discuss.php?d=1262#11828http://totalgadha.com/user/view.php?id=10991&course=1http://totalgadha.com/mod/forum/post.php?reply=11951http://totalgadha.com/mod/forum/discuss.php?d=1262#11946http://totalgadha.com/user/view.php?id=4009&course=1http://totalgadha.com/mod/forum/post.php?reply=11946http://totalgadha.com/mod/forum/discuss.php?d=1262#11942http://totalgadha.com/user/view.php?id=2&course=1


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the area common to both R & PW ...

It would really be great if you take some time to explain this ..any help from anyone isalso welcome on this....

Thanks.

Jitendra

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byAbhishek Deb- Wednesday, 19 September 2007, 04:49 PM

Hi TG,

I wanted to know one thing that u said that "minimum values of the s tudents following three activitieswill be 54% max in drama, arts ques.. I wanted to know then what will be the min of all the ppl havingtwo activities interest.... will it be 54%....?If not then how to solve it......

Abhishek.

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byNitin Jain- Wednesday, 19 September 2007, 05:14 PM

Hello TG

Please clarify my doubt. You said that there are two cases for solving maxima andminima problems. In the former the total no of elements are fixed and in the latter thetotal no of elements are not fixed. I guess total no of elements imply AUBUC? Because

in the last question of CAT interviewers , total no of elements are fixed i.e 210.Thenhow can we classify the question to fall under fixed elements or not?

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bysun is here- Wednesday, 19 September 2007, 09:04 PM


18/73

HI TG ,

A survey was conducted of 100 people to find out whether they had read recent

issues of Golmal, a monthly magazine. The summarized information regarding readership in 3

months is given below: Only September: 18, September but not August: 23, September and

July: 28, September: 28, July: 48, July and August: 10, None of the three months: 24.

What is the number of surveyed people who have read exactly two consecutive issues (out of

the three)?

(1) 7 (2) 9 (3) 12 (4) 14 (5) 17

iam not able to solve this problem please throw some light on it

Thanks

Sun

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byTotal Gadha- Thursday, 20 September 2007, 01:02 AM

Hi Gullz,

lease give me the options. I calculated highest value of c as 48% assuming some values but I

do not know if you need answer in terms of a, b, d, e, f or otherwise.

Total Gadha

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byMou Sukoshi- Thursday, 20 September 2007, 01:07 AM

A survey was conducted of 100 people to find out whether they had read recent issuesof Golmal, a monthly magazine. The summarized information regarding readership in 3months is given below: Only September: 18, September but not August: 23, Septemberand July: 28, September: 28, July: 48, July and August: 10, None of the three months:24.

What is the number of surveyed people who have read exactly two consecutive issues(out of the three)?(1) 7 (2) 9 (3) 12 (4) 14 (5) 17
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19/73

are these figures correct??

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byR V- Thursday, 20 September 2007, 07:47 AM

Hi Gullz how did you get the equation a1+2a2+3a3=108?? Could you please explain.

Thank you

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byGul Gul- Thursday, 20 September 2007, 09:37 AM

TG m so sorry i didnt write the options....I m in office and i forgot 2 bring the paper inwhich the question was given i will definitely post it in the evening... But i rememberthe options for following question

Q2 If a,b,c,d,e,f and g are all distinct the minimum possible no. of students whoenrolled for atleast 1 sport is ? a) 9 b)12 c) 15 d) none

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byGul Gul- Thursday, 20 September 2007, 09:41 AM

R V

a1 only 1

a2 exactly 2

a3 exactly 3

when you add all a2 is added twice and a3 thrice see a venn diagram for 3 circles u willget it..

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byTotal Gadha- Thursday, 20 September 2007, 10:17 AM

Hi Sun is here,


20/73

The options to this CAT 2006 question are NOT correct. This was one of the two questions in

CAT 2006 that was wrong.

Total Gadha

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byMohit Goyal- Thursday, 20 September 2007, 10:31 AM

Hi TG.

Wonderful Attempt to unearth the Concepts of venn Dig !! I have some very interesting sets

on Venn dig and will shortly post them but only if you promise to reply me after solving cause i

have posted 2 queries earlier n dint get ne reply....I am not a regular user of TG.com but ireally believe that you are doing a Noble Job Sir....Hats Off ...

Mohit

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bySS VV- Thursday, 20 September 2007, 10:57 AM

knew you were taking a while since u had to write it up to your standard which isexcellent :>:>

you rock!!

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bygourav kalra- Thursday, 20 September 2007, 01:24 PM

Hey..lovely article...any word of praise wud be less...

u really rock...am impressed... )

god bless u!!

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byTotal Gadha- Thursday, 20 September 2007, 02:43 PM


21/73

Hi Mohit,

I will certainly reply to you. Do post your problems.

Total Gadha

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byNitin Jain- Thursday, 20 September 2007, 03:12 PM

Hi TG

Pls reply to my post also given above.

Thanks

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byArun Prasad G- Thursday, 20 September 2007, 03:18 PM

TG, it is regarding this problem of apples, bananas and cherries.

x + 2y = 125%, where x + y 100%. For minimum value of y, we need maximum value of x. x = 75%, y = 25%.

I understand this. But after solving this, I wanted to fit this back into the venn diagrams and I amtotally confused. For example,

I wrote AnBnC as 25, and AnB excluding C as 75, BnC excluding A as 75 and so on. Now I am trying to

find only apples, only banana's and only cherries. I am not able to get it fully as I am not getting 100%after writing individual percentages. Please explain this to me.

Also I have posted a query on your "How to find the Units Digit of a Number?" . Please respond tp thatalso.

THANK YOU VERY MUCH FOR YOUR WONDERFUL CONTRIBUTION TO ALL ASPIRANTS. ITS FANTASTICTO READ EACH AND EVERY ARTICLE OF YOURS.

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byrishi sawla- Thursday, 20 September 2007, 03:41 PM

Hey Arun,

I think Sum of AnBexcluding C , BnC excluding A and AnC excluding B is 75.

Not individually.

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byravi teja- Thursday, 20 September 2007, 05:03 PM

hai gulz,

i was not able to solve even a single question in the set u mentioned.can u pass theanswer to me too if u get it please.if needed i can provide the options too.

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byneelakanta siva- Thursday, 20 September 2007, 05:20 PM

for GOLU'S question ,

only hockey+only football+only cricket > both H & C and not F +both C & Fand notH+both F &

Hand not C > ALL THREE

means H only + F only + C only > intersection of 2 only > intersection of 3

we can take intersection of 3 as 0

now intersection of 2 games only can take a max value of 49, then only the otherremaining part can be 51 and be greater.

49 can be alloted to the intersection of H and C only as we have to max cricket (C)

then intersection of F&C Xonly= 0

intersection of F&H Xonly=0

the remaining part can take 51 which should be divided as 16+17+18 and the 16 should
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23/73

be alloted to cricket(C) and 17 and 18 can be alloted to only F and only H. This is beczof the condition (C has to be less than A and B)

SO 0+0+49+16 = 65 IS THE MAX VALUE CRICKET CAN TAKE

PART 2 &3 OF THE Q IS NOT COMPLETE

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byTotal Gadha- Thursday, 20 September 2007, 07:46 PM

Hi Nitin,

In the previous sets, it has been mentioned that every student follows at least one activity.

This kind of statement is critical to determine that AUBUC is fixed. Even if the total number of

students/people etc. are given, if it is not given that every person belongs to at least one

category, AUBUC is not fixed.

Total Gadha

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byR V- Friday, 21 September 2007, 08:09 AM

hi Gullz, thanks for the explanation. i understood it.

RV

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byGul Gul- Friday, 21 September 2007, 09:04 AM

TG so sorry yesterday net was nt working at my place...,here r d options

If d>e>f and c is less than a as well as b, what is the maximum possible value of c?Options a) 63 b)64 c)65 d)66 e) none

If a,b,c,d,e,f and g are all distinct the minimum possible no. of students who enrolled


24/73

for atleast 1 sport is ? a)9 b)10 c)11 d) 13 e) none

If a is less than b as well as c, the no. of students who enrolled only for hockey is ? a)32 b) 48 c) 49 d) 50 e) none

Answers are c, e, c respectively......I hav 2 go 4 a full day training

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bysonal singh- Friday, 21 September 2007, 11:21 AM

hello Tg

Thanks for such a wonderful article.

Please explain how did u get 4-3 =1 for the following.and 15 had air conditioning specify all theelements in B or Only B

survey on a sample of 25 new cars being sold at a local auto dealer was conducted to see which of thethree popular options air conditioning, radio and power windows were already installed. Thesurvey found:

15 had air conditioning2 had air conditioning and power windows but no radios12 had radio6 had air conditioning and radio but no power windows11 had power windows4 had radio and power windows3 had all three options.

What is the number of cars that had none of the options? (CAT 2003)1. 4 2. 3 3. 1 4.2

Answer: We make the Venn diagram and start filling the areas as shown:
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25/73

Total Number of cars according to the diagram = 2 + 6 + 3 + 1 + 5 + 2 + 4 = 23.Therefore, number of cars having none of the given options = 25 23 = 2.

thanks

sonal

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bySumit Singla- Friday, 21 September 2007, 09:04 PM

Thanks for the brilliant article TG... Something that I solved mostly by hit and trial has been

made simple enough to be solved verbally! YOU ROCK!!

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bySaurabh Goyal- Friday, 21 September 2007, 11:46 PM

Hi Tg,

Found ur Venn diagram topic very helpful, but still getting ques with more difficulty...

I am facing problem in getting the logic of the question mentioned below..

Q. In a town 70% of the persons suffer from disease A, 80% from disease B, 75% from C and

85% from disease D and "P%" from all the four diseases. Find the minimum value of P.

i tried this ques by the above mentioned funda in your post, but not getting any where with

the logic. will it be 200% of surplus now?as the common 1 between all four fields will be

increased by 1 in comparison to 3 fields???

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byTotal Gadha- Saturday, 22 September 2007, 11:50 AM

Hi Saurabh,

A + B + C + D = 310%. Therefore, surplus = 210%. Now, the intersection of 4 sets can

accommodate a surplus of 3 per element. The intersection of 3 sets can accommodate a

surplus of 2 per element. The maximum surplus it can accommodate is 2 100% = 200%.

The surplus of 10% would be left. Now you can decrease intersection of two sets and shift the

decreased amount in the intersection of 3 sets. Going by this manner, you can quickly obtain

that if you keep 90% in intersection of 2 sets and 10% in the intersection of 3 sets, the surplus

accommodated = 90 2 + 10 3 = 210%.
http://totalgadha.com/mod/forum/discuss.php?d=1262#12229http://totalgadha.com/mod/forum/discuss.php?d=1262#12229http://totalgadha.com/mod/forum/post.php?reply=12249http://totalgadha.com/mod/forum/post.php?reply=12249http://totalgadha.com/mod/forum/post.php?reply=12249http://totalgadha.com/user/view.php?id=10061&course=1http://totalgadha.com/user/view.php?id=10061&course=1http://totalgadha.com/user/view.php?id=10061&course=1http://totalgadha.com/mod/forum/discuss.php?d=1262#11828http://totalgadha.com/mod/forum/discuss.php?d=1262#11828http://totalgadha.com/mod/forum/post.php?reply=12304http://totalgadha.com/mod/forum/post.php?reply=12304http://totalgadha.com/mod/forum/post.php?reply=12304http://totalgadha.com/user/view.php?id=10368&course=1http://totalgadha.com/user/view.php?id=10368&course=1http://totalgadha.com/user/view.php?id=10368&course=1http://totalgadha.com/mod/forum/discuss.php?d=1262#11828http://totalgadha.com/mod/forum/discuss.php?d=1262#11828http://totalgadha.com/mod/forum/post.php?reply=12315http://totalgadha.com/mod/forum/post.php?reply=12315http://totalgadha.com/mod/forum/post.php?reply=12315http://totalgadha.com/user/view.php?id=2&course=1http://totalgadha.com/user/view.php?id=2&course=1http://totalgadha.com/user/view.php?id=2&course=1http://totalgadha.com/user/view.php?id=2&course=1http://totalgadha.com/user/view.php?id=10368&course=1http://totalgadha.com/user/view.php?id=10061&course=1http://totalgadha.com/user/view.php?id=2&course=1http://totalgadha.com/user/view.php?id=10368&course=1http://totalgadha.com/user/view.php?id=10061&course=1http://totalgadha.com/user/view.php?id=2&course=1http://totalgadha.com/user/view.php?id=10368&course=1http://totalgadha.com/user/view.php?id=10061&course=1http://totalgadha.com/user/view.php?id=2&course=1http://totalgadha.com/mod/forum/post.php?reply=12315http://totalgadha.com/mod/forum/discuss.php?d=1262#11828http://totalgadha.com/user/view.php?id=10368&course=1http://totalgadha.com/mod/forum/post.php?reply=12304http://totalgadha.com/mod/forum/discuss.php?d=1262#11828http://totalgadha.com/user/view.php?id=10061&course=1http://totalgadha.com/mod/forum/post.php?reply=12249http://totalgadha.com/mod/forum/discuss.php?d=1262#12229


26/73

Therefore, the minimum value of P = 10%. The maximum will be when P accommodates all

the surplus. Therefore, maximum value = 210/3 = 70%

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