{"id":213093,"date":"2022-07-29T17:31:13","date_gmt":"2022-07-29T12:01:13","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=213093"},"modified":"2023-05-24T15:50:55","modified_gmt":"2023-05-24T10:20:55","slug":"cat-venn-diagrams-questions-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/cat-venn-diagrams-questions-pdf\/","title":{"rendered":"CAT Venn Diagrams Questions PDF [Most Important]"},"content":{"rendered":"

Venn Diagrams <\/b>is one of the most important topics in the CAT LRDI<\/strong> section. Sets on Venn Diagrams have appeared very frequently in the past CAT exams. <\/span>These sets are usually not very tough, and hence you should not miss out on the questions from Venn Diagrams. Firstly, understand the data given in the Venn Diagrams (2-set, 3-set, 4-set Venn Diagrams).<\/strong><\/span> Also, learn the <\/span>ways to Represent the data in Venn Diagrams<\/span><\/strong><\/span>. <\/span>You can check out these CAT Venn Diagrams questions<\/span> from the CAT Previous year’s papers<\/strong><\/span><\/a>.<\/span> This post will look at some important Venn Diagrams Sets in the CAT LRDI section. These are a good source of practice for CAT preparation; If you want to practice these questions, you can download this Venn Diagrams Questions<\/strong> for CAT<\/strong> LRDI<\/strong> PDF <\/strong>along with the detailed solutions (and video solutions) below, which is completely Free.<\/p>\n

Download Venn Diagrams Questions for CAT<\/a><\/p>\n

Enroll for CAT Complete Course<\/a><\/p>\n

Question 1:\u00a0<\/b>Shyam visited Ram during his brief vacation. In the mornings they both would go for yoga. In the evenings they would play tennis. To have more fun, they indulge only in one activity per day, i.e. either they went for yoga or played tennis each day. There were days when they were lazy and stayed home all day long. There were 24 mornings when they did nothing, 14 evenings when they stayed at home, and a total of 22 days when they did yoga or played tennis. For how many days Shyam stayed with Ram?<\/p>\n

a)\u00a032<\/p>\n

b)\u00a024<\/p>\n

c)\u00a030<\/p>\n

d)\u00a0None of these<\/p>\n

1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

Let the number of total days=N
\nThey played tennis for=N-14\u00a0 days
\nThey did yoga for =N-24 days<\/p>\n

<\/figure>\n

And the question says that total days when they did yoga or played tennis are 22<\/p>\n

which means<\/p>\n

N-14 + N-24 = 22<\/p>\n

2N – 38 = 22<\/p>\n

2N = 60<\/p>\n

N = 30<\/p>\n

Hence total days they stayed together were 30<\/p>\n

Instructions<\/b><\/p>\n

There were a hundred schools in a town. Of these, the number of schools having a play – ground was 30, and these schools had neither a library nor a laboratory. The number of schools having a laboratory alone was twice the number of those having a library only. The number of schools having a laboratory as well as a library was one-fourth the number of those having a laboratory alone. The number of schools having either a laboratory or a library or both was 35.<\/p>\n

Question 2:\u00a0<\/b>How many schools had none of the three viz., laboratory, library or play – ground?<\/p>\n

a)\u00a020<\/p>\n

b)\u00a05<\/p>\n

c)\u00a030<\/p>\n

d)\u00a035<\/p>\n

2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b>
\n<\/p>\n

The diagram for this question has been shown:<\/p>\n

Total number of schools having either or LAB or LIB or both = a+b+x\/2 – y + y + 3x = 7x\/2 + a + b = 35<\/p>\n

Here a = b = y = 0<\/p>\n

7x\/2 = 35<\/p>\n

x = 10<\/p>\n

Total number of schools having at least one of PG, LIB or LAB = 30+2x+x+x\/2 = 30+3x+x\/2 = 30+30+5 = 65<\/p>\n

Number of schools having neither of the three = 100-65 = 35<\/p>\n

Question 3:\u00a0<\/b>What was the ratio of schools having laboratory those having library?<\/p>\n

a)\u00a01 : 2<\/p>\n

b)\u00a05 : 3<\/p>\n

c)\u00a02 : 1<\/p>\n

d)\u00a02 : 3<\/p>\n

3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b>
\n<\/p>\n

The diagram for this question has been shown:<\/p>\n

Total number of schools having either or LAB or LIB or both = a+b+x\/2 – y + y + 3x = 7x\/2 + a + b = 35<\/p>\n

It has been given that the schools having playground don’t have a Library or Laboratory.<\/p>\n

Hence\u00a0 a = b = y = 0<\/p>\n

7x\/2 = 35<\/p>\n

x = 10<\/p>\n

Required ratio = 25:15 = 5:3<\/p>\n

Instructions<\/b>A survey of 200 people in a community who watched at least one of the three channels \u2014 BBC, CNN and
\nDD \u2014 showed that 80% of the people watched DD, 22% watched BBC, and 15% watched CNN.<\/p>\n

Question 4:\u00a0<\/b>What is the maximum percentage of people who can watch all the three channels?<\/p>\n

a)\u00a012.5%<\/p>\n

b)\u00a08.5%<\/p>\n

c)\u00a015%<\/p>\n

d)\u00a0Data insufficient<\/p>\n

4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

Let a be the number who watch only one channel, b be the number who watch only 2 channels and c be the number who watch all channels.<\/p>\n

a+b+c = 100<\/p>\n

a+2b+3c = 80+22+15 \u00a0=117<\/p>\n

Subtracting both equations,<\/p>\n

b+2c = 117-100 = 17<\/p>\n

Maximum c occurs when b = 0<\/p>\n

2c = 17<\/p>\n

c = 8.5<\/p>\n

Question 5:\u00a0<\/b>If 5% of people watched DD and CNN, 10% watched DD and BBC, then what percentage of people watched BBC and CNN only?<\/p>\n

a)\u00a02%<\/p>\n

b)\u00a05%<\/p>\n

c)\u00a08.5%<\/p>\n

d)\u00a0Cannot be determined<\/p>\n

5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

Applying AUBUC formula<\/p>\n

Let x be the number who watch BBC and CNN and y be the number who watch all three channels.<\/p>\n

100 = 80+22+15-(10+5+x)+y<\/p>\n

x-y = 2<\/p>\n

Hence only 2% people watch BBC and CNN only.<\/p>\n

Checkout: CAT Free Practice Questions and Videos<\/strong><\/a><\/em><\/p>\n

Question 6:\u00a0<\/b>Referring to the previous question, what percentage of people watched all the three channels?<\/p>\n

a)\u00a03.5%<\/p>\n

b)\u00a00%<\/p>\n

c)\u00a08.5%<\/p>\n

d)\u00a0Cannot be determined<\/p>\n

6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

Applying AUBUC formula<\/p>\n

Let x be the number who watch BBC and CNN and y be the number who watch all three channels.<\/p>\n

100 = 80+22+15-(10+5+x)+y<\/p>\n

x-y = 2<\/p>\n

We cannot find the exact value of y.<\/p>\n

Hence, the answer is “cannot be determined”.<\/p>\n

Question 7:\u00a0<\/b>Each of 74 students in a class studies at least one of the three subjects H, E and P. Ten students study all three subjects, while twenty study H and E, but not P. Every student who studies P also studies H or E or both. If the number of students studying H equals that studying E, then the number of students studying H is<\/p>\n

7)\u00a0Answer:\u00a052<\/b><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

Let us draw a Venn diagram using the information present in the question.<\/p>\n

<\/figure>\n

It is given that\u00a0the number of students studying H equals that studying E.<\/p>\n

Let ‘x’ be the total number of students who studied H, and H and P but mot E.We can also say that the same will be the number of students who studied E, and E and P but not H.Therefore,<\/p>\n

x +\u00a020 + 10 + x = 74<\/p>\n

x = 22<\/p>\n

Hence,\u00a0the number of students studying H\u00a0= 22 + 10+ 20 = 52<\/p>\n

Question 8:\u00a0<\/b>If among 200 students, 105 like pizza and 134 like burger, then the number of students who like only burger can possibly be<\/p>\n

a)\u00a023<\/p>\n

b)\u00a026<\/p>\n

c)\u00a096<\/p>\n

d)\u00a093<\/p>\n

8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

It has been given that among 200 students,\u00a0105 like pizza and 134 like burger.
\nThe question asks us to find out the number of students who can be liking only burgers among the given values.<\/p>\n

The least number of students who like only burger will be obtained when everyone who likes pizza likes burger too.
\nIn this case, 105 students will like pizza and burger and 134-105 = 29 students will like only burger. Therefore, the number of students who like only burger cannot be less than 29.<\/p>\n

The maximum number of students who like only burger will be obtained when we try to separate the 2 sets as much as possible.
\nThere are 200 students in total. 105 of them like pizza. Therefore, the remaining 95 students can like only burger and 134-95 = 39 students can like both pizza and burger. As we can see, the number of students who like burger cannot exceed 95.<\/p>\n

The number of students who like only burger should lie between 29 and 95 (both the values are included).
\n93 is the only value among the given options that satisfies this condition and hence, option\u00a0D is the right answer.<\/p>\n

Question 9:\u00a0<\/b>For two sets A and B, let A\u0394B denote the set of elements which belong to A or B but not both. If P = {1,2,3,4}, Q = {2,3,5,6,}, R = {1,3,7,8,9}, S = {2,4,9,10}, then the number of elements in (P\u0394Q)\u0394(R\u0394S) is<\/p>\n

9)\u00a0Answer:\u00a07<\/b><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

P = {1,2,3,4} and\u00a0 Q = {2,3,5,6,}
\nP\u0394Q = {1, 4, 5, 6}
\nR = {1,3,7,8,9} and S = {2,4,9,10}
\nR\u0394S = {1, 2, 3, 4, 7, 8, 10}
\n(P\u0394Q)\u0394(R\u0394S) = {2, 3, 5, 6, 7, 8, 10}
\nThus, there are 7 elements in\u00a0(P\u0394Q)\u0394(R\u0394S)\u00a0.
\nhence, 7 is the correct answer.<\/p>\n

Question 10:\u00a0<\/b>If A = {$6^{2n} -35n – 1$}, where $n$ = 1,2,3,… and B = {35($n$-1)}, where $n$ = 1,2,3,… then which of the following is true?<\/p>\n

a)\u00a0Every member of A is in B and at least one member of B is not in A<\/p>\n

b)\u00a0Neither every member of A is in B nor every member of B is in A<\/p>\n

c)\u00a0Every member of B is in A.<\/p>\n

d)\u00a0At least one member of A is not in B<\/p>\n

10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

If we carefully observe set A, then we find that $6^{2n} -35n – 1$ is divisible by 35. So, set A contains multiples of 35. However, not all the multiples of 35 are there in set A, for different values of $n$.
\nFor $n = 1$, the value is 0, for $n = 2$, the value is 1225 which is the 35th multiple of 3.
\nIf we observe set B, it consists of all the multiples of 35 including 0.
\nSo, we can say that every member of set A will be in B while every member of set B will not necessarily be in set A.
\nHence, option\u00a0A is the correct answer.<\/p>\n

Question 11:\u00a0<\/b>A club has 256 members of whom 144 can play football, 123 can play tennis, and 132 can play cricket. Moreover, 58 members can play both football and tennis, 25 can play both cricket and tennis, while 63 can play both football and cricket. If every member can play at least one game, then the number of members who can play only tennis is<\/p>\n

a)\u00a038<\/p>\n

b)\u00a032<\/p>\n

c)\u00a045<\/p>\n

d)\u00a043<\/p>\n

11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

Assume the number of members who can play exactly 1 game = I<\/p>\n

The number of members who can play exactly 1 game = II<\/p>\n

The number of members who can play exactly 1 game = III<\/p>\n

I+2II+3III=144+123+132=399….(1)<\/p>\n

I+II+III=256……(2)<\/p>\n

=> II+2III=143…..(3)<\/p>\n

Also, II+3III=58+25+63=146 ……(4)<\/p>\n

=> III = 3 (From 3 and 4)<\/p>\n

=> II =137<\/p>\n

=> I = 116<\/p>\n

The members who play only tennis = 123-58-25+3 = 43<\/p>\n

Question 12:\u00a0<\/b>Students in a college have to choose at least two subjects from chemistry,\u00a0mathematics and physics. The number of students choosing all three subjects is 18,\u00a0choosing mathematics as one of their subjects is 23 and choosing physics as one of\u00a0their subjects is 25. The smallest possible number of students who could choose\u00a0chemistry as one of their subjects is<\/p>\n

a)\u00a022<\/p>\n

b)\u00a021<\/p>\n

c)\u00a020<\/p>\n

d)\u00a019<\/p>\n

12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

View Video Solution<\/a><\/p>\n

Solution:<\/b><\/p>\n

Now 23 students choose maths as one of their subject.<\/p>\n

This means (MPC)+ (MC) + (PC)=23 where MPC denotes students who choose all the three subjects maths, physics and chemistry and so on.<\/p>\n

So MC + PM =5 Similarly we have PC+ MP =7<\/p>\n

We have to find the smallest number of students choosing chemistry<\/p>\n

For that\u00a0in the first equation let PM=5 and MC=0. In the second equation this PC=2<\/p>\n

Hence minimum number of students choosing chemistry will be (18+2)=20 Since 18 students chose all the three subjects.<\/p>\n

Enroll for CAT 2022 Complete Course <\/a><\/p><\/div>\n

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Download CAT Quant Formulas PDF<\/a><\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Venn Diagrams is one of the most important topics in the CAT LRDI section. Sets on Venn Diagrams have appeared very frequently in the past CAT exams. These sets are usually not very tough, and hence you should not miss out on the questions from Venn Diagrams. Firstly, understand the data given in the Venn […]<\/p>\n","protected":false},"author":32,"featured_media":213105,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3],"tags":[5119,5798],"class_list":{"0":"post-213093","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-cat","8":"tag-cat-2022","9":"tag-venn-diagrams-pdf-for-cat"},"better_featured_image":{"id":213105,"alt_text":"CAT Venn Diagrams Questions PDF","caption":"CAT Venn Diagrams Questions PDF","description":"CAT Venn Diagrams Questions 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