CAT DI Venn Diagrams Questions & Solutions

A sample of real CAT DI Venn Diagrams past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 48 DI Venn Diagrams questions in all — sign up free to practise them timed.

Passage
There are 15 girls and some boys among the graduating students in a class. They are planning a get-together, which can be either a 1-day event, or a 2-day event, or a 3-day event. There are 6 singers in the class, 4 of them are boys. There are 10 dancers in the class, 4 of them are girls. No dancer in the class is a singer.

Some students are not interested in attending the get-together. Those students who are interested in attending a 3-day event are also interested in attending a 2-day event; those who are interested in attending a 2-day event are also interested in attending a 1-day event.

The following facts are also known:
1. All the girls and 80% of the boys are interested in attending a 1-day event. 60% of the boys are interested in attending a 2-day event.
2. Some of the girls are interested in attending a 1-day event, but not a 2-day event; some of the other girls are interested in attending both.
3. 70% of the boys who are interested in attending a 2-day event are neither singers nor dancers. 60% of the girls who are interested in attending a 2-day event are neither singers nor dancers.
4. No girl is interested in attending a 3-day event. All male singers and 2 of the dancers are interested in attending a 3-day event.
5. The number of singers interested in attending a 2-day event is one more than the number of dancers interested in attending a 2-day event.
Q1.CAT 2022
How many boys are there in the class?
Answer: 50
Show solution
No. of girls = 15

Let the no. of boys be x

No. of singers = 6

No of boys who are singers = 4

Therefore, no of girls who are singers = 2

No of dancers = 10

No of boys who are dancers = 6

Therefore, no. of girls who are dancers = 4

No. of boys who are neither singers nor dancers = x-10

No. of girls who are neither singers nor dancers = 9

Now we fill the above table,

using statements 1 and 2, we get the following table

Let the number of girls who are interested in attending a 2-day event be a and the number of girls who are dancers and are interested in 2-day event be b.

Now using statements 3 and 4, we get

$2\le0.18x-4\le6$

$6\le0.18x\le10$

0.18x should be integer for which x should be a multiple of 50, and 0.18x lies between 6 and 10; therefore, the only possible value of x is 50.

From statement 5, we can say that,

4+ 0.4a - b = 5+b +1

or, 0.4a = 2+2b

or, a = 5(1+b)

a should be a multiple of 5 as b is a whole number. So possible values of a can be 5, 10 or 15. Now, as the maximum value of b can be 4 and the maximum value of 0.4a-b can be 2, so the only possible value of a satisfying the conditions above is 5. If a= 5 then b=1.
Q2.CAT 2022
How many female dancers are interested in attending a 2-day event?
- 2
- 1
- 0
- Cannot be determined
Show solution
No. of girls = 15

Let the no. of boys be x

No. of singers = 6

No of boys who are singers = 4

Therefore, no of girls who are singers = 2

No of dancers = 10

No of boys who are dancers = 6

Therefore, no. of girls who are dancers = 4

No. of boys who are neither singers nor dancers = x-10

No. of girls who are neither singers nor dancers = 9

Now we fill the above table,

using statements 1 and 2, we get the following table

Let the number of girls who are interested in attending a 2-day event be a and the number of girls who are dancers and are interested in 2-day event be b.

Now using statements 3 and 4, we get

$2\le0.18x-4\le6$

$6\le0.18x\le10$

0.18x should be integer for which x should be a multiple of 50, and 0.18x lies between 6 and 10; therefore, the only possible value of x is 50.

From statement 5, we can say that,

4+ 0.4a - b = 5+b +1

or, 0.4a = 2+2b

or, a = 5(1+b)

a should be a multiple of 5 as b is a whole number. So possible values of a can be 5, 10 or 15. Now, as the maximum value of b can be 4 and the maximum value of 0.4a-b can be 2, so the only possible value of a satisfying the conditions above is 5. If a= 5 then b=1.
Passage
A speciality supermarket sells 320 products. Each of these products was either a cosmetic product or a nutrition product. Each of these products was also either a foreign product or a domestic product. Each of these products had at least one of the two approvals - FDA or EU.

The following facts are also known:

1. There were equal numbers of domestic and foreign products.
2. Half of the domestic products were FDA approved cosmetic products.
3. None of the foreign products had both the approvals, while 60 domestic products had both the approvals.
4. There were 140 nutrition products, half of them were foreign products.
5. There were 200 FDA approved products. 70 of them were foreign products and 120 of them were cosmetic products.
Q3.CAT 2022
If 70 cosmetic products did not have EU approval, then how many nutrition products had both the approvals?
- 50
- 30
- 10
- 20
Show solution
It is given that the total number of products supermarket sells is 320.

cosmetic + nutrition = foreign + domestic = FDA + EU = 320 products

In statement 1, it is given that the number of foreign products is equal to the number of domestic products.

Foreign products = Domestic products = 320/2 = 160

In statement 2, it is given that half of the domestic products were FDA approved cosmetic products, i.e. domestic, cosmetic and FDA = 80

In statement 4, it is given that there were 140 nutrition products, half of them were foreign products. This implies remaining half are domestic.

In statement 5, it is given that there are 200 FDA approved products out of which 70 are foreign products and 120 are cosmetic products.
If 70 are foreign products, remaining 130 should be domestic products. In domestic products, FDA approved cosmetic products are 80. This implies FDA approved nutrition products are 130-80, i.e. 50.

There are 120 FDA approved cosmetic products.

Domestic, cosmetic and FDA approved = 80

This implies, Foreign, cosmetic and FDA approved is 120-80, i.e. 40.

There are 70 FDA approved foreign products.

This implies Foreign, nutrition and FDA approved is 70-40, i.e. 30.

Domestic and Cosmetic = 90

Domestic, comestic and FDA approved = 80

This implies, Domestic, cosmetic and FDA not approved is 90-80, i.e. 10.

Therefore, (domestic, cosmetic and only EU) = 10

Similarly, we get (domestic, nutrition and only EU) = 70-50 = 20

In the question, it is given that 70 cosmetic products did not have EU approval.

In foreign, 40 cosmetic products did not have EU approval. This implies 30 cosmetic products should have only FDA approval in domestic products.

According to the above statement, b = 30

a = 80 - 30 = 50

Given, a + c = 60

c = 60 - 50 = 10

Therefore, the number of nutrition products which had both the approvals is 10.

The answer is option C.
Passage
1000 patients currently suffering from a disease were selected to study the effectiveness of treatment of four types of medicines — A, B, C and D. These patients were first randomly assigned into two groups of equal size, called treatment group and control group. The patients in the control group were not treated with any of these medicines; instead they were given a dummy medicine, called placebo, containing only sugar and starch. The following information is known about the patients in the treatment group.

a. A total of 250 patients were treated with type A medicine and a total of 210 patients were treated with type C medicine.
b. 25 patients were treated with type A medicine only. 20 patients were treated with type C medicine only. 10 patients were treated with type D medicine only.
c. 35 patients were treated with type A and type D medicines only. 20 patients were treated with type A and type B medicines only. 30 patients were treated with type A and type C medicines only. 20 patients were treated with type C and type D medicines only.
d. 100 patients were treated with exactly three types of medicines.
e. 40 patients were treated with medicines of types A, B and C, but not with medicines of type D. 20 patients were treated with medicines of types A, C and D, but not with medicines of type B.
f. 50 patients were given all the four types of medicines. 75 patients were treated with exactly one type of medicine.
Q4.CAT 2020
How many patients were treated with medicine types B and D only?
Answer: 150
Show solution
Of the 1000 subjects, only 500 have been considered for the treatment. This constitutes our sample set. Thus the four drugs- A, B, C and D have been administered to this set of 500 individuals, while the rest 500 have been given the placebo. Based on the given information, we can then draw the following 4-set Venn diagram:

We can solve for the number of patients who were administered the drugs A, B and D excluding C by putting in the values for set A. The required value = 250 - (25+20+30+40+20+50+35) = 30. Based on condition (c), we know that 100 patients were treated with exactly three types of medicines. Thus, we can fill the slot for the number of patients who were administered only B, C and D excluding A by 100 - (40+20+30) = 10.

Similarly, based on condition (f), we know that the candidates who were administered only dug B are 75 - (25+20+10) = 20. Post this, we can easily calculate the number of people administered with only drugs B and C by 210 - (30+20+40+50+10+20+20) = 20. We can fill in the above values to obtain the following diagram:

The sum of all the values should add up to 500. On solving for 'x' [which represents the number of people who were administered drugs B and D only], we obtain x = 150. The final representation would appear as follows:

The number of people who were administered drugs B and D only were 150.
Passage
Ten musicians (A, B, C, D, E, F, G, H, I and J) are experts in at least one of the following three percussion instruments: tabla, mridangam, and ghatam. Among them, three are experts in tabla but not in mridangam or ghatam, another three are experts in mridangam but not in tabla or ghatam, and one is an expert in ghatam but not in tabla or mridangam. Further, two are experts in tabla and mridangam but not in ghatam, and one is an expert in tabla and ghatam but not in mridangam.

The following facts are known about these ten musicians.
1. Both A and B are experts in mridangam, but only one of them is also an expert in tabla.
2. D is an expert in both tabla and ghatam.
3. Both F and G are experts in tabla, but only one of them is also an expert in mridangam.
4. Neither I nor J is an expert in tabla.
5. Neither H nor I is an expert in mridangam, but only one of them is an expert in ghatam.
Q5.CAT 2020
Which of the following pairs CANNOT have any musician who is an expert in both tabla and mridangam but not in ghatam?
- F and G
- C and E
- A and B
- C and F
Show solution
Based on the given information, we can form the following Venn-diagram for ease of understanding:

The conditions help us to further bifurcate the individuals based on their expertise.

Mridangam: A and B (condition 1); one out of F and G (condition 3)

Tabla: one out of A and B (condition 1); F and G (condition 3); D (condition 2)

Ghatam: D (condition 2)

Based on condition 4, we infer that 'I' and 'J' are either experts in Ghatam or Mridangam. However, condition 5 adds that 'I' is not an expert in Mridangam. This helps us definitively zero-in on 'I' as an expert in Ghatam. Since 'I' is a Ghatam expert, J is an expert in Mridangam and 'H' is an expert in Tabla [based on conditions 4 and 5]. Thus, we can depict our understanding so far as follows:

Mridangam [total: 5] - A and B (condition 1); one out of F and G (condition 3); J (condition 4 and 5); one out of C and E {remaining experts}

Tabla [total: 6] - one out of A and B (condition 1); F and G (condition 3); D (condition 2); H (condition 4 and 5); one out of C and E {remaining experts}

Ghatam [total: 2] - D (condition 2); I (condition 4 and 5)

We observe that the pair C and E cannot have any musician who is an expert in both tabla and mridangam but not in ghatam. Hence, Option B is the correct answer.
Passage
A survey of 600 schools in India was conducted to gather information about their online teaching learning processes (OTLP). The following four facilities were studied.
F1: Own software for OTLP
F2: Trained teachers for OTLP
F3: Training materials for OTLP
F4: All students having Laptops
The following observations were summarized from the survey.
1. 80 schools did not have any of the four facilities - F1, F2, F3, F4.
2. 40 schools had all four facilities.
3. The number of schools with only F1, only F2, only F3, and only F4 was 25, 30, 26 and 20 respectively.
4. The number of schools with exactly three of the facilities was the same irrespective of which three were considered.
5. 313 schools had F2.
6. 26 schools had only F2 and F3 (but neither F1 nor F4).
7. Among the schools having F4, 24 had only F3, and 45 had only F2.
8. 162 schools had both F1 and F2.
9. The number of schools having F1 was the same as the number of schools having F4.
Q6.CAT 2020
What was the number of schools having only facilities F1 and F3?
Answer: 42
Show solution
Let the number of schools with exactly three of the facilities was the same irrespective of which three were considered be x.

Number of schools with none of the facilities be 'n' from 1, n=80.

Number of schools with only F1 and F2 be 'b'

Number of schools with only F1 and F3 be 'c'

Number of schools with only F1 and F4 be 'd'

From the information given in the question we will get the following Venn diagram.

From 5, b+141+3x=313 => b+3x=172....(i)

From 8, b+x+40+x=162 => b+2x=122....(ii)

(ii)-(i) gives x=50 => b=22

From 9, 237+3x+c+d=279+3x=d => c=42

Total number of schools =600 => 313+25+c+x+d+26+24+20+80=600 => d=20.

The final table looks like:

The total number of schools having only F1 and F3= c=42
Passage
Students in a college are discussing two proposals --
A: a proposal by the authorities to introduce dress code on campus, and
B: a proposal by the students to allow multinational food franchises to set up outlets on college campus.

A student does not necessarily support either of the two proposals.

In an upcoming election for student union president, there are two candidates in fray:
Sunita and Ragini. Every student prefers one of the two candidates.

A survey was conducted among the students by picking a sample of 500 students. The following information was noted from this survey.

1. 250 students supported proposal A and 250 students supported proposal B.
2. Among the 200 students who preferred Sunita as student union president, 80% supported proposal A.
3. Among those who preferred Ragini, 30% supported proposal A.
4. 20% of those who supported proposal B preferred Sunita.
5. 40% of those who did not support proposal B preferred Ragini.
6. Every student who preferred Sunita and supported proposal B also supported proposal A.
7. Among those who preferred Ragini, 20% did not support any of the proposals.
Q7.CAT 2019
What percentage of the students surveyed who supported both proposals A and B preferred Sunita as student union president?
- 40
- 25
- 20
- 50
Show solution
Total number of students surveyed= 500

Every student prefers one of the two candidates. Ragini(R) and Sunita(S).

Thus, R+S=500.

According to statement 2, "Among the 200 students who preferred Sunita as student union president, 80% supported proposal A."

The number of students who support Sunita(S)=200

The number of students who supported Ragini(R)=300

According to statements 2 and 3, 160 students who supported Sunita also supported the proposal A & 90 students who supported Ragini also supported proposal A.

According to statements 4 and 6, we can make the following Venn diagram for Sunita.

According to statement 5 and 7, we can make the following Venn diagram.

According to the Venn diagram, the students surveyed who supported both proposals A and B preferred Sunita as student union president $\frac{50}{50+50}$ % =50%
Passage
1600 satellites were sent up by a country for several purposes. The purposes are classified as broadcasting (B), communication (C), surveillance (S), and others (O). A satellite can serve multiple purposes; however a satellite serving either B, or C, or S does not serve O. The following facts are known about the satellites:

1. The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1.
2. The number of satellites serving all three of B, C, and S is 100.
3. The number of satellites exclusively serving C is the same as the number of satellites exclusively serving S. This number is 30% of the number of satellites exclusively serving B.
4. The number of satellites serving O is the same as the number of satellites serving both C and S but not B.
Q8.CAT 2018
If at least 100 of the 1600 satellites were serving O, what can be said about the number of satellites serving S?
- At most 475
- Exactly 475
- No conclusion is possible based on the given information
- At least 475
Show solution
It is given that a satellite serving either B, or C, or S does not serve O. So we can say that it's basically 3 satellites broadcasting (B), communication (C), surveillance (S) which can have intersections. Those satellites which are not part of any category are placed in others. We can draw the Venn diagram as follows.

1. The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1.
2. The number of satellites serving all three of B, C, and S is 100.
3. The number of satellites exclusively serving C is the same as the number of satellites exclusively serving S. This number is 30% of the number of satellites exclusively serving B.
4. The number of satellites serving O is the same as the number of satellites serving both C and S but not B.

Let '10x' be the number of satellites exclusively serving B. Then, the number of satellites exclusively serving C and S = 0.30*10x = 3x

Let 'y' be the number of satellites serving others(O).

Let 'z' be the number of satellites serving B, C but not S. Since the numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1. Therefore, we can can say that number of satellites serving B, S but not C = z.

It is given that

$\Rightarrow$ 10x+2z+2y+6x+100 = 1600

$\Rightarrow$ 8x+z+y = 750 ... (1)

The numbers of satellites serving B, C, and S (though maybe not exclusively) are in the ratio 2:1:1.

$\Rightarrow$ $\dfrac{10x+2z+100}{z+100+3x+y} = \dfrac{2}{1}$

$\Rightarrow$ $10x+2z+100=2(z+100+3x+y)$

$\Rightarrow$ $4x=100+2y$

$\Rightarrow$ $2x=50+y$

$\Rightarrow$ $y=2x-50$ ... (2)

We can substitute this in equation (1)

$\Rightarrow$ 8x+z+2x - 50 = 750

$\Rightarrow$ z = 800 - 10x ... (3)

Let us define boundary condition for x,

$\Rightarrow$ 2x - 50 $\geq$ 0

$\Rightarrow$ x $\geq$ 25

Also, 800 - 10x $\geq$ 0

$\Rightarrow$ x $\leq$ 80

Therefore, we can say that x $\epsilon$ [25, 80].

It is given that at least 100 of the 1600 satellites were serving O.

$\Rightarrow$ 2x - 50 $\geq$ 100

$\Rightarrow$ x $\geq$ 75

The number of satellites serving S = 100 + 800 - 10x + 2x - 50 + 3x = 850 - 5x

At x $_{min}$ = 75, the number of satellites serving S = 850 - 5*75 = 475

At x $_{max}$ = 80, the number of satellites serving S = 850 - 5*80 = 450

Hence, we can say that the number of satellites serving S must be from 425 to 475. Therefore, we can say that option A is the correct answer.
Passage
Fun Sports (FS) provides training in three sports - Gilli-danda (G), Kho-Kho (K), and Ludo (L). Currently it has an enrollment of 39 students each of whom is enrolled in at least one of the three sports. The following details are known:
1. The number of students enrolled only in L is double the number of students enrolled in all the three sports.
2. There are a total of 17 students enrolled in G.
3. The number of students enrolled only in G is one less than the number of students enrolled only in L.
4. The number of students enrolled only in K is equal to the number of students who are enrolled in both K and L.
5. The maximum student enrollment is in L.
6. Ten students enrolled in G are also enrolled in at least one more sport.
Q9.CAT 2018
Due to academic pressure, students who were enrolled in all three sports were asked to withdraw from one of the three sports. After the withdrawal, the number of students enrolled in G was six less than the number of students enrolled in L, while the number of students enrolled in K went down by one. After the withdrawal, how many students were enrolled in both G and K?
Answer: 2
Show solution
Let 'x' be the number of students enrolled in all three sports. Then the number of students enrolled only in L = 2x

It is given that there are a total of 17 students enrolled in G. Also, ten students enrolled in G are also enrolled in at least one more sport. Hence, the number of students enrolled in only G = 17 - 10 = 7

The number of students enrolled only in G is one less than the number of students enrolled only in L. Hence, the number of students enrolled only in L = 7+1

$\Rightarrow$ 2x = 8

$\Rightarrow$ x = 4

Let us assume that 'y' students are enrolled in K and L but not G. Then, the number of students enrolled only in K = y + 4

Let us assume that 'z' be the the number of students enrolled in G and K but not L. Then, the number of students enrolled G and L bot not K = 10 - 4 - z = 6 - z

It is given that a total of 39 students in the sports.

7 + z + 4 + 6 - z + 8 + y + y + 4 = 39

$\Rightarrow$ y = 5

Number of students enrolled in G = 17

Number of students enrolled in K = 9 + 4 + 5 + z = 18 + z

Number of students enrolled in L = 6 - z + 4 + 5 + 8 = 23 - z

It is given that the maximum student enrollment is in L.

$\Rightarrow$ 23 - z > 18 + z

$\Rightarrow$ 2z < 5

$\Rightarrow$ z < 2.5

Therefore, we can say that z can take three values = {0, 1, 2}

Hence, the number of students enrolled in K = 18 + z = {18, 19, 20}

It is given that after withdrawal the number of students enrolled in K went down by one. This one student must have left sports K. Hence we can say that the remaining 3 students must have left either G or L.

Before withdraw there were a total of 24 students were enrolled in exactly 1 sports, 11 students were enrolled in exactly 2 courses and 4 students were enrolled in all three courses.

The students which were enrolled in all three sports, withdrew from one of the sports. Hence, we can say that now the number of students who were enrolled in exactly 2 courses = 11 + 4 = 15.

It is given that the number of students enrolled in G was six less than the number of students enrolled in L. Let 'a' be the number of students who were enrolled in G and K but not L. Then, the number of students who were enrolled in L and K but not G = a + 5

Consequently, we can say that the number of students enrolled in G and L but not K = 15 - (2a + 5) = 10 - 2a

Number of students enrolled in this case = a + a+5 + 9 = 14 + 2a. We can see that '14+2a' is an even number. It is given that the number of students enrolled in K went down by one. Therefore, we can say that the number of students enrolled in K earlier was an odd number.

Hence, the number of students enrolled in K = 18 + z = {18, 19, 20}

We can see that only '19' is an odd number. Hence, we can say that the number of students enrolled in K after withdrawal = 18

$\Rightarrow$ 14 + 2a = 18

$\Rightarrow$ a = 2

From the diagram we can see that the number of students enrolled in both G and K = 2.
Passage
Applicants for the doctoral programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Engineering (BIE) have to appear for a Common Entrance Test (CET). The test has three sections: Physics (P), Chemistry (C), and Maths (M). Among those appearing for CET, those at or above the 80th percentile in at least two sections, and at or above the 90th percentile overall, are selected for Advanced Entrance Test (AET) conducted by AIE. AET is used by AIE for final selection.

For the 200 candidates who are at or above the 90th percentile overall based on CET, the following are known about their performance in CET:
1. No one is below the 80th percentile in all 3 sections.
2. 150 are at or above the 80th percentile in exactly two sections.
3. The number of candidates at or above the 80th percentile only in P is the same as the number of candidates at or above the 80th percentile only in C. The same is the number of candidates at or above the 80th percentile only in M.
4. Number of candidates below 80th percentile in P: Number of candidates below 80th percentile in C: Number of candidates below 80th percentile in M = 4:2:1.
BIE uses a different process for selection. If any candidate is appearing in the AET by AIE, BIE considers their AET score for final selection provided the candidate is at or above the 80th percentile in P. Any other candidate at or above the 80th percentile in P in CET, but who is not eligible for the AET, is required to appear in a separate test to be conducted by BIE for being considered for final selection. Altogether, there are 400 candidates this year who are at or above the 80th percentile in P.
Q10.CAT 2017
If the number of candidates who are at or above the 90th percentile overall and also at or above the 80th percentile in all three sections in CET is actually a multiple of 5, then how many candidates were shortlisted for the AET for AIE?
Answer: 170
Show solution
It is given that 200 candidates scored above 90th percentile overall in CET. Let the following Venn diagram represent the number of persons who scored above 80 percentile in CET in each of the three sections:

From (1), n = 0
From (2), d + e + f = 150
From (3), a = b = c
Since, there are a total of 200 candidates
a + b + c + g = 200 - 150 = 50
3a + g = 50 => a < 17
From (4), (b + f + c) : (a + d + b) : (a + e + c) = 4 : 2: 1
Or (2a + f) : (2a + d) : (2a + e) = 4 : 2 : 1
or, 6a + (d + e + f) = 7x
Or, 6a + 150 = 7x
So, a can be 3 or 10
x can be 24 or 30
2a + e can be 24 or 30 => e can be 18 or 10
2a + d can be 48 or 60 = > d can be 42 or 40
2a + f can be 96 or 120 => f can be 90 or 100
3a + g = 50 => g can be 41 or 20

Among the candidates who are at or above 90th percentile, the candidates who are at or above 80th percentile in at least two sections are selected for AET. Hence, the candidates represented by d, e, f and g are selected for AET.
BIE will consider the candidates who are appearing for AET and are at or above 80th percentile in P. Hence, BIE will consider the candidates represented by d, e and g, which can be 104 or 80.
BIE will conduct a separate test for the other students who are at or above 80th percentile in P. Given that there are a total of 400 candidates at or above 80th percentile in P, and since there are 104 or 80 candidates at or above 80th percentile in P and are at or above 90th percentile overall, there must be 296 or 320 candidates at or above 80th percentile in P who scored less than 90th percentile overall.

In this question, g = 20.
Number of candidates shortlisted for AET
= d + e + f + g
= 40 + 10 + 100 + 20
= 170

Hence, 170 is the correct answer.
Q11.CAT 1990
Choose the set in which the statements are most logically related.
A. All candid men are persons who acknowledge merit in a rival.
B. Some learned men are very candid.
C. Some learned men are not persons who acknowledge merit in a rival.
D. Some learned men are persons who are very candid.
E. Some learned men are not candid.
F. Some persons who recognize merit in a rival are learned.
- ABE
- ACF
- ADE
- BAF
Show solution
In option A, E is not a valid conclusion from A and B.

In option B, F cannot be inferred from A and C.

In option C, E cannot be concluded from A and D.

In option D, If some learned men are candid but all candid men recognize merit in a rival, it can be concluded that some men recognize merit in a rival are learned.

Hence, option D is the correct answer.
Q12.CAT 1990
Choose the set in which the statements are most logically related.
A. Some mammals are carnivores.
B. All whales are mammals.
C. All whales are aquatic animals.
D. All whales are carnivores.
E. Some aquatic animals are mammals.
F. Some mammals are whales.
- ADF
- ABC
- AEF
- BCE
Show solution
ADF is false as we cannot infer F(Some mammals are whales.) even if A(Some mammals are carnivores.) and D(All whales are carnivores.) are true.
ABC and AEF all statements are different, the set is not related.
All whales are mammals and all whales are aquatic animals. So, some aquatic animals are mammals. Therefore, BCE is valid combination.

Hence, option D is the correct answer.

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CAT DI Venn Diagrams Questions & Solutions

Passage

There are 15 girls and some boys among the graduating students in a class. They are planning a get-together, which can be either a 1-day event, or a 2-day event, or a 3-day event. There are 6 singers in the class, 4 of them are boys. There are 10 dancers in the class, 4 of them are girls. No dancer in the class is a singer.

Some students are not interested in attending the get-together. Those students who are interested in attending a 3-day event are also interested in attending a 2-day event; those who are interested in attending a 2-day event are also interested in attending a 1-day event.

The following facts are also known:
1. All the girls and 80% of the boys are interested in attending a 1-day event. 60% of the boys are interested in attending a 2-day event.
2. Some of the girls are interested in attending a 1-day event, but not a 2-day event; some of the other girls are interested in attending both.
3. 70% of the boys who are interested in attending a 2-day event are neither singers nor dancers. 60% of the girls who are interested in attending a 2-day event are neither singers nor dancers.
4. No girl is interested in attending a 3-day event. All male singers and 2 of the dancers are interested in attending a 3-day event.
5. The number of singers interested in attending a 2-day event is one more than the number of dancers interested in attending a 2-day event.

Q1.CAT 2022

How many boys are there in the class?

Answer: 50

Show solution

No. of girls = 15

Let the no. of boys be x

No. of singers = 6

No of boys who are singers = 4

Therefore, no of girls who are singers = 2

No of dancers = 10

No of boys who are dancers = 6

Therefore, no. of girls who are dancers = 4

No. of boys who are neither singers nor dancers = x-10

No. of girls who are neither singers nor dancers = 9

Now we fill the above table,

using statements 1 and 2, we get the following table

Let the number of girls who are interested in attending a 2-day event be a and the number of girls who are dancers and are interested in 2-day event be b.

Now using statements 3 and 4, we get

$2\le0.18x-4\le6$

$6\le0.18x\le10$

0.18x should be integer for which x should be a multiple of 50, and 0.18x lies between 6 and 10; therefore, the only possible value of x is 50.

From statement 5, we can say that,

4+ 0.4a - b = 5+b +1

or, 0.4a = 2+2b

or, a = 5(1+b)

a should be a multiple of 5 as b is a whole number. So possible values of a can be 5, 10 or 15. Now, as the maximum value of b can be 4 and the maximum value of 0.4a-b can be 2, so the only possible value of a satisfying the conditions above is 5. If a= 5 then b=1.

Q2.CAT 2022

How many female dancers are interested in attending a 2-day event?

2
1
0
Cannot be determined

Show solution

No. of girls = 15

Let the no. of boys be x

No. of singers = 6

No of boys who are singers = 4

Therefore, no of girls who are singers = 2

No of dancers = 10

No of boys who are dancers = 6

Therefore, no. of girls who are dancers = 4

No. of boys who are neither singers nor dancers = x-10

No. of girls who are neither singers nor dancers = 9

Now we fill the above table,

using statements 1 and 2, we get the following table

Let the number of girls who are interested in attending a 2-day event be a and the number of girls who are dancers and are interested in 2-day event be b.

Now using statements 3 and 4, we get

$2\le0.18x-4\le6$

$6\le0.18x\le10$

0.18x should be integer for which x should be a multiple of 50, and 0.18x lies between 6 and 10; therefore, the only possible value of x is 50.

From statement 5, we can say that,

4+ 0.4a - b = 5+b +1

or, 0.4a = 2+2b

or, a = 5(1+b)

Passage

A speciality supermarket sells 320 products. Each of these products was either a cosmetic product or a nutrition product. Each of these products was also either a foreign product or a domestic product. Each of these products had at least one of the two approvals - FDA or EU.

The following facts are also known:

1. There were equal numbers of domestic and foreign products.
2. Half of the domestic products were FDA approved cosmetic products.
3. None of the foreign products had both the approvals, while 60 domestic products had both the approvals.
4. There were 140 nutrition products, half of them were foreign products.
5. There were 200 FDA approved products. 70 of them were foreign products and 120 of them were cosmetic products.

Q3.CAT 2022

If 70 cosmetic products did not have EU approval, then how many nutrition products had both the approvals?

Show solution

It is given that the total number of products supermarket sells is 320.

cosmetic + nutrition = foreign + domestic = FDA + EU = 320 products

In statement 1, it is given that the number of foreign products is equal to the number of domestic products.

Foreign products = Domestic products = 320/2 = 160

In statement 2, it is given that half of the domestic products were FDA approved cosmetic products, i.e. domestic, cosmetic and FDA = 80

In statement 4, it is given that there were 140 nutrition products, half of them were foreign products. This implies remaining half are domestic.

In statement 5, it is given that there are 200 FDA approved products out of which 70 are foreign products and 120 are cosmetic products.
If 70 are foreign products, remaining 130 should be domestic products. In domestic products, FDA approved cosmetic products are 80. This implies FDA approved nutrition products are 130-80, i.e. 50.

There are 120 FDA approved cosmetic products.

Domestic, cosmetic and FDA approved = 80

This implies, Foreign, cosmetic and FDA approved is 120-80, i.e. 40.

There are 70 FDA approved foreign products.

This implies Foreign, nutrition and FDA approved is 70-40, i.e. 30.

Domestic and Cosmetic = 90

Domestic, comestic and FDA approved = 80

This implies, Domestic, cosmetic and FDA not approved is 90-80, i.e. 10.

Therefore, (domestic, cosmetic and only EU) = 10

Similarly, we get (domestic, nutrition and only EU) = 70-50 = 20

In the question, it is given that 70 cosmetic products did not have EU approval.

In foreign, 40 cosmetic products did not have EU approval. This implies 30 cosmetic products should have only FDA approval in domestic products.

According to the above statement, b = 30

a = 80 - 30 = 50

Given, a + c = 60

c = 60 - 50 = 10

Therefore, the number of nutrition products which had both the approvals is 10.

The answer is option C.

Passage

1000 patients currently suffering from a disease were selected to study the effectiveness of treatment of four types of medicines — A, B, C and D. These patients were first randomly assigned into two groups of equal size, called treatment group and control group. The patients in the control group were not treated with any of these medicines; instead they were given a dummy medicine, called placebo, containing only sugar and starch. The following information is known about the patients in the treatment group.

a. A total of 250 patients were treated with type A medicine and a total of 210 patients were treated with type C medicine.
b. 25 patients were treated with type A medicine only. 20 patients were treated with type C medicine only. 10 patients were treated with type D medicine only.
c. 35 patients were treated with type A and type D medicines only. 20 patients were treated with type A and type B medicines only. 30 patients were treated with type A and type C medicines only. 20 patients were treated with type C and type D medicines only.
d. 100 patients were treated with exactly three types of medicines.
e. 40 patients were treated with medicines of types A, B and C, but not with medicines of type D. 20 patients were treated with medicines of types A, C and D, but not with medicines of type B.
f. 50 patients were given all the four types of medicines. 75 patients were treated with exactly one type of medicine.

Q4.CAT 2020

How many patients were treated with medicine types B and D only?

Answer: 150

Show solution

Of the 1000 subjects, only 500 have been considered for the treatment. This constitutes our sample set. Thus the four drugs- A, B, C and D have been administered to this set of 500 individuals, while the rest 500 have been given the placebo. Based on the given information, we can then draw the following 4-set Venn diagram:

We can solve for the number of patients who were administered the drugs A, B and D excluding C by putting in the values for set A. The required value = 250 - (25+20+30+40+20+50+35) = 30. Based on condition (c), we know that 100 patients were treated with exactly three types of medicines. Thus, we can fill the slot for the number of patients who were administered only B, C and D excluding A by 100 - (40+20+30) = 10.

Similarly, based on condition (f), we know that the candidates who were administered only dug B are 75 - (25+20+10) = 20. Post this, we can easily calculate the number of people administered with only drugs B and C by 210 - (30+20+40+50+10+20+20) = 20. We can fill in the above values to obtain the following diagram:

The sum of all the values should add up to 500. On solving for 'x' [which represents the number of people who were administered drugs B and D only], we obtain x = 150. The final representation would appear as follows:

The number of people who were administered drugs B and D only were 150.

Passage

Ten musicians (A, B, C, D, E, F, G, H, I and J) are experts in at least one of the following three percussion instruments: tabla, mridangam, and ghatam. Among them, three are experts in tabla but not in mridangam or ghatam, another three are experts in mridangam but not in tabla or ghatam, and one is an expert in ghatam but not in tabla or mridangam. Further, two are experts in tabla and mridangam but not in ghatam, and one is an expert in tabla and ghatam but not in mridangam.

The following facts are known about these ten musicians.
1. Both A and B are experts in mridangam, but only one of them is also an expert in tabla.
2. D is an expert in both tabla and ghatam.
3. Both F and G are experts in tabla, but only one of them is also an expert in mridangam.
4. Neither I nor J is an expert in tabla.
5. Neither H nor I is an expert in mridangam, but only one of them is an expert in ghatam.

Q5.CAT 2020

Which of the following pairs CANNOT have any musician who is an expert in both tabla and mridangam but not in ghatam?

F and G
C and E
A and B
C and F

Show solution

Based on the given information, we can form the following Venn-diagram for ease of understanding:

The conditions help us to further bifurcate the individuals based on their expertise.

Mridangam: A and B (condition 1); one out of F and G (condition 3)

Tabla: one out of A and B (condition 1); F and G (condition 3); D (condition 2)

Ghatam: D (condition 2)

Based on condition 4, we infer that 'I' and 'J' are either experts in Ghatam or Mridangam. However, condition 5 adds that 'I' is not an expert in Mridangam. This helps us definitively zero-in on 'I' as an expert in Ghatam. Since 'I' is a Ghatam expert, J is an expert in Mridangam and 'H' is an expert in Tabla [based on conditions 4 and 5]. Thus, we can depict our understanding so far as follows:

Mridangam [total: 5] - A and B (condition 1); one out of F and G (condition 3); J (condition 4 and 5); one out of C and E {remaining experts}

Tabla [total: 6] - one out of A and B (condition 1); F and G (condition 3); D (condition 2); H (condition 4 and 5); one out of C and E {remaining experts}

Ghatam [total: 2] - D (condition 2); I (condition 4 and 5)

We observe that the pair C and E cannot have any musician who is an expert in both tabla and mridangam but not in ghatam. Hence, Option B is the correct answer.

Passage

A survey of 600 schools in India was conducted to gather information about their online teaching learning processes (OTLP). The following four facilities were studied.

F1: Own software for OTLP
F2: Trained teachers for OTLP
F3: Training materials for OTLP
F4: All students having Laptops

The following observations were summarized from the survey.

1. 80 schools did not have any of the four facilities - F1, F2, F3, F4.
2. 40 schools had all four facilities.
3. The number of schools with only F1, only F2, only F3, and only F4 was 25, 30, 26 and 20 respectively.
4. The number of schools with exactly three of the facilities was the same irrespective of which three were considered.
5. 313 schools had F2.
6. 26 schools had only F2 and F3 (but neither F1 nor F4).
7. Among the schools having F4, 24 had only F3, and 45 had only F2.
8. 162 schools had both F1 and F2.
9. The number of schools having F1 was the same as the number of schools having F4.

Q6.CAT 2020

What was the number of schools having only facilities F1 and F3?

Answer: 42

Show solution

Let the number of schools with exactly three of the facilities was the same irrespective of which three were considered be x.

Number of schools with none of the facilities be 'n' from 1, n=80.

Number of schools with only F1 and F2 be 'b'

Number of schools with only F1 and F3 be 'c'

Number of schools with only F1 and F4 be 'd'

From the information given in the question we will get the following Venn diagram.

From 5, b+141+3x=313 => b+3x=172....(i)

From 8, b+x+40+x=162 => b+2x=122....(ii)

(ii)-(i) gives x=50 => b=22

From 9, 237+3x+c+d=279+3x=d => c=42

Total number of schools =600 => 313+25+c+x+d+26+24+20+80=600 => d=20.

The final table looks like:

The total number of schools having only F1 and F3= c=42

Passage

Students in a college are discussing two proposals --
A: a proposal by the authorities to introduce dress code on campus, and
B: a proposal by the students to allow multinational food franchises to set up outlets on college campus.

A student does not necessarily support either of the two proposals.

In an upcoming election for student union president, there are two candidates in fray:
Sunita and Ragini. Every student prefers one of the two candidates.

A survey was conducted among the students by picking a sample of 500 students. The following information was noted from this survey.

1. 250 students supported proposal A and 250 students supported proposal B.
2. Among the 200 students who preferred Sunita as student union president, 80% supported proposal A.
3. Among those who preferred Ragini, 30% supported proposal A.
4. 20% of those who supported proposal B preferred Sunita.
5. 40% of those who did not support proposal B preferred Ragini.
6. Every student who preferred Sunita and supported proposal B also supported proposal A.
7. Among those who preferred Ragini, 20% did not support any of the proposals.

Q7.CAT 2019

What percentage of the students surveyed who supported both proposals A and B preferred Sunita as student union president?

Show solution

Total number of students surveyed= 500

Every student prefers one of the two candidates. Ragini(R) and Sunita(S).

Thus, R+S=500.

According to statement 2, "Among the 200 students who preferred Sunita as student union president, 80% supported proposal A."

The number of students who support Sunita(S)=200

The number of students who supported Ragini(R)=300

According to statements 2 and 3, 160 students who supported Sunita also supported the proposal A & 90 students who supported Ragini also supported proposal A.

According to statements 4 and 6, we can make the following Venn diagram for Sunita.

According to statement 5 and 7, we can make the following Venn diagram.

According to the Venn diagram, the students surveyed who supported both proposals A and B preferred Sunita as student union president $\frac{50}{50+50}$ % =50%

Passage

1600 satellites were sent up by a country for several purposes. The purposes are classified as broadcasting (B), communication (C), surveillance (S), and others (O). A satellite can serve multiple purposes; however a satellite serving either B, or C, or S does not serve O. The following facts are known about the satellites:

1. The numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1.
2. The number of satellites serving all three of B, C, and S is 100.
3. The number of satellites exclusively serving C is the same as the number of satellites exclusively serving S. This number is 30% of the number of satellites exclusively serving B.
4. The number of satellites serving O is the same as the number of satellites serving both C and S but not B.

Q8.CAT 2018

If at least 100 of the 1600 satellites were serving O, what can be said about the number of satellites serving S?

At most 475
Exactly 475
No conclusion is possible based on the given information
At least 475

Show solution

It is given that a satellite serving either B, or C, or S does not serve O. So we can say that it's basically 3 satellites broadcasting (B), communication (C), surveillance (S) which can have intersections. Those satellites which are not part of any category are placed in others. We can draw the Venn diagram as follows.

Let '10x' be the number of satellites exclusively serving B. Then, the number of satellites exclusively serving C and S = 0.30*10x = 3x

Let 'y' be the number of satellites serving others(O).

Let 'z' be the number of satellites serving B, C but not S. Since the numbers of satellites serving B, C, and S (though may be not exclusively) are in the ratio 2:1:1. Therefore, we can can say that number of satellites serving B, S but not C = z.

It is given that

$\Rightarrow$ 10x+2z+2y+6x+100 = 1600

$\Rightarrow$ 8x+z+y = 750 ... (1)

The numbers of satellites serving B, C, and S (though maybe not exclusively) are in the ratio 2:1:1.

$\Rightarrow$ $\dfrac{10x+2z+100}{z+100+3x+y} = \dfrac{2}{1}$

$\Rightarrow$ $10x+2z+100=2(z+100+3x+y)$

$\Rightarrow$ $4x=100+2y$

$\Rightarrow$ $2x=50+y$

$\Rightarrow$ $y=2x-50$ ... (2)

We can substitute this in equation (1)

$\Rightarrow$ 8x+z+2x - 50 = 750

$\Rightarrow$ z = 800 - 10x ... (3)

Let us define boundary condition for x,

$\Rightarrow$ 2x - 50 $\geq$ 0

$\Rightarrow$ x $\geq$ 25

Also, 800 - 10x $\geq$ 0

$\Rightarrow$ x $\leq$ 80

Therefore, we can say that x $\epsilon$ [25, 80].

It is given that at least 100 of the 1600 satellites were serving O.

$\Rightarrow$ 2x - 50 $\geq$ 100

$\Rightarrow$ x $\geq$ 75

The number of satellites serving S = 100 + 800 - 10x + 2x - 50 + 3x = 850 - 5x

At x $_{min}$ = 75, the number of satellites serving S = 850 - 5*75 = 475

At x $_{max}$ = 80, the number of satellites serving S = 850 - 5*80 = 450

Hence, we can say that the number of satellites serving S must be from 425 to 475. Therefore, we can say that option A is the correct answer.

Passage

Fun Sports (FS) provides training in three sports - Gilli-danda (G), Kho-Kho (K), and Ludo (L). Currently it has an enrollment of 39 students each of whom is enrolled in at least one of the three sports. The following details are known:
1. The number of students enrolled only in L is double the number of students enrolled in all the three sports.
2. There are a total of 17 students enrolled in G.
3. The number of students enrolled only in G is one less than the number of students enrolled only in L.
4. The number of students enrolled only in K is equal to the number of students who are enrolled in both K and L.
5. The maximum student enrollment is in L.
6. Ten students enrolled in G are also enrolled in at least one more sport.

Q9.CAT 2018

Due to academic pressure, students who were enrolled in all three sports were asked to withdraw from one of the three sports. After the withdrawal, the number of students enrolled in G was six less than the number of students enrolled in L, while the number of students enrolled in K went down by one. After the withdrawal, how many students were enrolled in both G and K?

Answer: 2

Show solution

Let 'x' be the number of students enrolled in all three sports. Then the number of students enrolled only in L = 2x

It is given that there are a total of 17 students enrolled in G. Also, ten students enrolled in G are also enrolled in at least one more sport. Hence, the number of students enrolled in only G = 17 - 10 = 7

The number of students enrolled only in G is one less than the number of students enrolled only in L. Hence, the number of students enrolled only in L = 7+1

$\Rightarrow$ 2x = 8

$\Rightarrow$ x = 4

Let us assume that 'y' students are enrolled in K and L but not G. Then, the number of students enrolled only in K = y + 4

Let us assume that 'z' be the the number of students enrolled in G and K but not L. Then, the number of students enrolled G and L bot not K = 10 - 4 - z = 6 - z

It is given that a total of 39 students in the sports.

7 + z + 4 + 6 - z + 8 + y + y + 4 = 39

$\Rightarrow$ y = 5

Number of students enrolled in G = 17

Number of students enrolled in K = 9 + 4 + 5 + z = 18 + z

Number of students enrolled in L = 6 - z + 4 + 5 + 8 = 23 - z

It is given that the maximum student enrollment is in L.

$\Rightarrow$ 23 - z > 18 + z

$\Rightarrow$ 2z < 5

$\Rightarrow$ z < 2.5

Therefore, we can say that z can take three values = {0, 1, 2}

Hence, the number of students enrolled in K = 18 + z = {18, 19, 20}

It is given that after withdrawal the number of students enrolled in K went down by one. This one student must have left sports K. Hence we can say that the remaining 3 students must have left either G or L.

Before withdraw there were a total of 24 students were enrolled in exactly 1 sports, 11 students were enrolled in exactly 2 courses and 4 students were enrolled in all three courses.

The students which were enrolled in all three sports, withdrew from one of the sports. Hence, we can say that now the number of students who were enrolled in exactly 2 courses = 11 + 4 = 15.

It is given that the number of students enrolled in G was six less than the number of students enrolled in L. Let 'a' be the number of students who were enrolled in G and K but not L. Then, the number of students who were enrolled in L and K but not G = a + 5

Consequently, we can say that the number of students enrolled in G and L but not K = 15 - (2a + 5) = 10 - 2a

Number of students enrolled in this case = a + a+5 + 9 = 14 + 2a. We can see that '14+2a' is an even number. It is given that the number of students enrolled in K went down by one. Therefore, we can say that the number of students enrolled in K earlier was an odd number.

Hence, the number of students enrolled in K = 18 + z = {18, 19, 20}

We can see that only '19' is an odd number. Hence, we can say that the number of students enrolled in K after withdrawal = 18

$\Rightarrow$ 14 + 2a = 18

$\Rightarrow$ a = 2

From the diagram we can see that the number of students enrolled in both G and K = 2.

Passage

Applicants for the doctoral programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Engineering (BIE) have to appear for a Common Entrance Test (CET). The test has three sections: Physics (P), Chemistry (C), and Maths (M). Among those appearing for CET, those at or above the 80th percentile in at least two sections, and at or above the 90th percentile overall, are selected for Advanced Entrance Test (AET) conducted by AIE. AET is used by AIE for final selection.

For the 200 candidates who are at or above the 90th percentile overall based on CET, the following are known about their performance in CET:
1. No one is below the 80th percentile in all 3 sections.
2. 150 are at or above the 80th percentile in exactly two sections.
3. The number of candidates at or above the 80th percentile only in P is the same as the number of candidates at or above the 80th percentile only in C. The same is the number of candidates at or above the 80th percentile only in M.
4. Number of candidates below 80th percentile in P: Number of candidates below 80th percentile in C: Number of candidates below 80th percentile in M = 4:2:1.

BIE uses a different process for selection. If any candidate is appearing in the AET by AIE, BIE considers their AET score for final selection provided the candidate is at or above the 80th percentile in P. Any other candidate at or above the 80th percentile in P in CET, but who is not eligible for the AET, is required to appear in a separate test to be conducted by BIE for being considered for final selection. Altogether, there are 400 candidates this year who are at or above the 80th percentile in P.

Q10.CAT 2017

If the number of candidates who are at or above the 90th percentile overall and also at or above the 80th percentile in all three sections in CET is actually a multiple of 5, then how many candidates were shortlisted for the AET for AIE?

Answer: 170

Show solution

It is given that 200 candidates scored above 90th percentile overall in CET. Let the following Venn diagram represent the number of persons who scored above 80 percentile in CET in each of the three sections:

From (1), n = 0
From (2), d + e + f = 150
From (3), a = b = c
Since, there are a total of 200 candidates
a + b + c + g = 200 - 150 = 50
3a + g = 50 => a < 17
From (4), (b + f + c) : (a + d + b) : (a + e + c) = 4 : 2: 1
Or (2a + f) : (2a + d) : (2a + e) = 4 : 2 : 1
or, 6a + (d + e + f) = 7x
Or, 6a + 150 = 7x
So, a can be 3 or 10
x can be 24 or 30
2a + e can be 24 or 30 => e can be 18 or 10
2a + d can be 48 or 60 = > d can be 42 or 40
2a + f can be 96 or 120 => f can be 90 or 100
3a + g = 50 => g can be 41 or 20

Among the candidates who are at or above 90th percentile, the candidates who are at or above 80th percentile in at least two sections are selected for AET. Hence, the candidates represented by d, e, f and g are selected for AET.
BIE will consider the candidates who are appearing for AET and are at or above 80th percentile in P. Hence, BIE will consider the candidates represented by d, e and g, which can be 104 or 80.
BIE will conduct a separate test for the other students who are at or above 80th percentile in P. Given that there are a total of 400 candidates at or above 80th percentile in P, and since there are 104 or 80 candidates at or above 80th percentile in P and are at or above 90th percentile overall, there must be 296 or 320 candidates at or above 80th percentile in P who scored less than 90th percentile overall.

In this question, g = 20.
Number of candidates shortlisted for AET
= d + e + f + g
= 40 + 10 + 100 + 20
= 170

Hence, 170 is the correct answer.

Q11.CAT 1990

Choose the set in which the statements are most logically related.
A. All candid men are persons who acknowledge merit in a rival.
B. Some learned men are very candid.
C. Some learned men are not persons who acknowledge merit in a rival.
D. Some learned men are persons who are very candid.
E. Some learned men are not candid.
F. Some persons who recognize merit in a rival are learned.

Show solution

In option A, E is not a valid conclusion from A and B.

In option B, F cannot be inferred from A and C.

In option C, E cannot be concluded from A and D.

In option D, If some learned men are candid but all candid men recognize merit in a rival, it can be concluded that some men recognize merit in a rival are learned.

Hence, option D is the correct answer.

Q12.CAT 1990

Choose the set in which the statements are most logically related.
A. Some mammals are carnivores.
B. All whales are mammals.
C. All whales are aquatic animals.
D. All whales are carnivores.
E. Some aquatic animals are mammals.
F. Some mammals are whales.

Show solution

ADF is false as we cannot infer F(Some mammals are whales.) even if A(Some mammals are carnivores.) and D(All whales are carnivores.) are true.

ABC and AEF all statements are different, the set is not related.

All whales are mammals and all whales are aquatic animals. So, some aquatic animals are mammals. Therefore, BCE is valid combination.

Hence, option D is the correct answer.

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