# LR - Venn Diagram - Previous Year CAT/MBA Questions

You can practice all previous year CAT questions from the topic LR - Venn Diagram. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

**Answer the next 4 questions based on the information given below.**

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 total of 250 patients were treated with type A medicine and a total of 210 patients were treated with type C medicine.
- 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.
- 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.
- 100 patients were treated with exactly three types of medicines.
- 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.
- 50 patients were given all the four types of medicines. 75 patients were treated with exactly one type of medicine.

**CAT 2020 LRDI Slot 1 | LR - Venn Diagram**

How many patients were treated with medicine type B?

Answer: 340

**Explanation** :

We can draw the following Venn diagram according to the information given in the question.

75 patients were treated with exactly one type of medicine.

∴ 25 + a + 20 + 10 = 75

⇒ a = 20

250 patients were treated with type A medicine

∴ 25 + 30 + 20 + 35 + 20 + 40 + 50 + e = 250.

⇒ e = 30

100 patients were treated with exactly three types of medicines.

∴ 20 + 40 + c + e = 100

⇒ c = 10

210 patients were treated with type C medicine

∴ 30 + 40 + b + 20 + 20 + 50 + c + 20 = 210

⇒ b = 20

Since patients were equally in treatment group and control group hence, there were total 500 patients who were in the treatment group and 500 in control group (who were not given any medicine).

Total patient in treatment group = 500 = 25 + 30 + 20 + 35 + 20 + 40 + 50 + e + a + b + c + d + 20 + 20 + 10.

⇒ d = 150

The completed Venn diagram:

∴ 340 patients were treated with Type B medicine.

Hence, 340.

Workspace:

**CAT 2020 LRDI Slot 1 | LR - Venn Diagram**

The number of patients who were treated with medicine types B, C and D, but not type A was:

Answer: 10

**Explanation** :

Consider the solution to the first question of this set.

The number of patients who were treated with medicine types B, C and D, but not type A is a = 10.

Hence, 10.

Workspace:

**CAT 2020 LRDI Slot 1 | LR - Venn Diagram**

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

Answer: 150

**Explanation** :

Consider the solution to the first question of this set.

The number of patients who were treated with medicine types B and D only = d = 150.

Hence, 150.

Workspace:

**CAT 2020 LRDI Slot 1 | LR - Venn Diagram**

The number of patients who were treated with medicine type D was:

Answer: 325

**Explanation** :

Consider the solution to the first question of this set.

The number of patients who were treated with medicine type D was 325.

Hence, 325.

Workspace:

**Answer the next 4 questions based on the information given below.**

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 table 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.

- Both A and B are experts in mridangam, but only one of them is also an expert in tabla.
- D is an expert in both tabla and ghatam.
- Both F and G are experts in tabla, but only one of them is also an expert in mridangam.
- Neither I nor J is an expert in tabla.
- Neither H nor I is an expert in mridangam, but only one of them is an expert in ghatam.

**CAT 2020 LRDI Slot 1 | LR - Venn Diagram**

Who among the following is DEFINITELY an expert in tabla but not in either mridangam or ghatam?

- A.
F

- B.
A

- C.
H

- D.
C

Answer: Option C

**Explanation** :

We can draw the following Venn diagram based on the information given in the main paragraph.

D is an expert in both tabla and ghatam.

Since I is neither an expert in table nor in mridangam hence, he is an expert in ghatam

∴ D and I are the only two experts in ghatam.

Now, J is not an expert in tabla hence, he can only be an expert in mridangam only.

Also, H is not an expert in mridangam hence, he can only be an expert in table only.

Both A and B are experts in mridangam, but only one of them is also an expert in tabla.

Both F and G are experts in tabla, but only one of them is also an expert in mridangam.

There are two experts in table and mridangam only one of them is either A or B and the other is either F or G.

∴ We can make the following Venn diagram

∴ H is an expert in table only.

Hence, option (c).

Workspace:

**CAT 2020 LRDI Slot 1 | LR - Venn Diagram**

Who among the following is DEFINITELY an expert in mridangam but not in either table or ghatam?

- A.
E

- B.
B

- C.
G

- D.
H

Answer: Option D

**Explanation** :

Consider the solution to the first question of this set.

J is an expert in mridangam only.

Hence, option (d).

Workspace:

**CAT 2020 LRDI Slot 1 | LR - Venn Diagram**

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

- A.
C and E

- B.
C and F

- C.
A and B

- D.
F and G

Answer: Option B

**Explanation** :

Consider the solution to the first question of this set.

Neither C nor E can be an expert in both table and mridangam.

Hence, option (b).

Workspace:

**CAT 2020 LRDI Slot 1 | LR - Venn Diagram**

If C is an expert in mridangam and F is not, then which are the three musicians who are experts in tabla but not in either mridangam or ghatam?

- A.
C, E and G

- B.
E, F and H

- C.
C, G and H

- D.
E, G and H

Answer: Option B

**Explanation** :

Consider the solution to the first question of this set.

If C is an expert in mridangam and F is not, we can make the following Venn diagram.

We can make the following Venn diagram

∴ E, F and H are experts in tabla but neither mridangam or ghatam.

Hence, option (b).

Workspace:

**Answer the next 4 questions based on the information given.**

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.

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

**CAT 2020 LRDI Slot 3 | LR - Venn Diagram**

What was the total number of schools having exactly three of the four facilities?

- A.
64

- B.
80

- C.
50

- D.
200

Answer: Option D

**Explanation** :

Given, the number of schools with exactly three of the facilities was the same irrespective of which three were considered.

Let us assume this number to be ‘a’ for every possible combination of three OTLPs.

The following diagram can be drawn from the given information.

It is also given that 162 schools had F1 and F2

∴ Number of students having only F1 and F2 = 162 – (a + 40 + a) = 122 – 2a.

Total schools having F2 = 313 = 162 + 30 + 26 + a + 45

⇒ a = 50

Total number of schools having F1 is equal to total number of schools having F2.

∴ (162 + 25 + y + 50 + x) = (50 + 40 + 50 + 24 + x + 50 + 45 + 20)

⇒ y = 42

Now there are a total of 600 schools

∴ 600 = 25 + 42 + 50 + x + 313 + 26 + 24 + 20 + 80

⇒ x = 20

Therefore, the complete Venn diagram is

Number of schools having exactly 3 of the 4 facilities = 50 + 50 + 50 + 50 = 200

Hence, option (d).

Workspace:

**CAT 2020 LRDI Slot 3 | LR - Venn Diagram**

What was the number of schools having facilities F2 and F4?

- A.
85

- B.
45

- C.
95

- D.
185

Answer: Option D

**Explanation** :

Consider the solution to the first question of this set.

Number of schools having facilities F2 and F4 = 40 + 50 + 50 + 45 = 185

Hence, option (d).

Workspace:

**CAT 2020 LRDI Slot 3 | LR - Venn Diagram**

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

Answer: 42

**Explanation** :

Consider the solution to the first question of this set.

Number of schools having only facilities F1 and F3 = 42

Hence, 42.

Workspace:

**CAT 2020 LRDI Slot 3 | LR - Venn Diagram**

What was the number of schools having only facilities F1 andF4?

Answer: 20

**Explanation** :

Consider the solution to the first question of this set.

Number of schools having only facilities F1 and F4 = 20.

Hence, 20.

Workspace:

**Answer the following question based on the information given below.**

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:

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

**CAT 2018 LRDI Slot 1 | LR - Venn Diagram**

What best can be said about the number of satellites serving C?

- A.
Cannot be more than 800

- B.
Must be at least 100

- C.
Must be between 450 and 725

- D.
Must be between 400 and 800

Answer: Option C

**Explanation** :

Let number of satellites exclusively serving B be ‘10x’ and number of satellites serving O be ‘z’.

∴ number of satellites exclusively serving C = number of satellites exclusively serving S = 3a

As the number of satellites serving C = number of satellites serving S, the number of satellites serving C and B but not S = number of satellites exclusively serving S and B but not C = y (assume).

∴ 10x + 2y + 100 + 6x + 2z

= 1600 ⇒ 8x + y + z

= 750 … (i)

From (1), 10x + 2y + 100

= 6x + 2y + 2z + 200

∴ z = 2x – 50 … (ii)

Thus, the minimum value that ‘x’ can takes is 25.

From (i) and (ii),

10x + y = 800 ⇒ y = 800 – 10x

Thus, the maximum value that ‘x’ can takes is 80.

The number of satellites serving

C = 3x + y + z + 100 = 3x + (800 – 10x) + (2x – 50) + 100 = 850 – 5x

As value of ‘x’ is between 25 and 80, value of (850 – 5x) must be between 450 and 725.

Hence, option 3.

Workspace:

**CAT 2018 LRDI Slot 1 | LR - Venn Diagram**

What is the minimum possible number of satellites serving B exclusively?

- A.
100

- B.
200

- C.
250

- D.
500

Answer: Option C

**Explanation** :

The number of satellites serving B exclusively = 10x.

As minimum value of ‘x’ is 25, minimum value of 10x is 250.

Hence, option 3.

Workspace:

**CAT 2018 LRDI Slot 1 | LR - Venn Diagram**

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

- A.
Exactly 475

- B.
At most 475

- C.
At least 475

- D.
No conclusion is possible based on the given information

Answer: Option B

**Explanation** :

Number of satellites serving S = 3x + y + z + 100 = 3x + (800 – 10x) + (2x – 50) + 100

= 850 – 5x

2x – 50 ≥ 100 ⇒ x ≥ 75 ⇒ (–5x) ≤ (–375)

∴ 850 – 5x ≤ 850 – 375 = 475

Thus, the number of satellites serving S can be at most 475.

Hence, option 2.

Workspace:

**CAT 2018 LRDI Slot 1 | LR - Venn Diagram**

If the number of satellites serving at least two among B, C, and S is 1200, which of the following MUST be FALSE?

- A.
All 1600 satellites serve B or C or S

- B.
The number of satellites serving B is more than 1000

- C.
The number of satellites serving C cannot be uniquely determined

- D.
The number of satellites serving B exclusively is exactly 250

Answer: Option C

**Explanation** :

The number of satellites serving at least two among B, C and S is 1200

i.e., 2y + z + 100 ≥ 1200

⇒ 1600 – 20x + 2x – 50 ≥ 1100

⇒ 450 ≥ 18x

⇒ 25 ≥ x

Earlier we have seen that the minimum value of ‘x’ is 25

∴ x = 25

[1]: z = 2x – 50 = 0 ⇒ All satellites serve B or C or S. Therefore, statement 1 is true.

[2]: The number of satellites serving B = 10x + 2y + 100 = 10(25) + 2(800 – 10x) + 100

= 1450

Therefore, statement 2 is true.

[3]: The number of satellites serving C

= 3x + y + z + 100 = 3x + (800 – 10x) + (2x – 50) + 100 = 850 – 5x = 725

Therefore, statement 3 is false.

[4]: The number of satellites serving B exclusively = 10x = 10 × 25 = 250

Therefore, statement 4 is true.

Hence, option 3.

Workspace:

**Answer the following question based on the information given below.**

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:

- The number of students enrolled only in L is double the number of students enrolled in all the three sports.
- There are a total of 17 students enrolled in G.
- The number of students enrolled only in G is one less than the number of students enrolled only in L.
- The number of students enrolled only in K is equal to the number of students who are enrolled in both K and L.
- The maximum student enrollment is in L.
- Ten students enrolled in G are also enrolled in at least one more sport.

**CAT 2018 LRDI Slot 2 | LR - Venn Diagram**

What is the minimum number of students enrolled in both G and L but not in K?

Answer: 4

**Explanation** :

Since 10 students who play G enrolled in at least one other sport, the number of students who enrolled only in G = 17 - 10 = 7. Therefore x = 7.

Also from statement 1, the number of students = who enrolled in

all the three sports $\frac{7+1}{2}$ = 4

As 17 students enrolled in G, the number of students who did not enrol in G = 39 - 17 = 22.

Therefore, x + 1 + z + y + z = 22. We know that x = 7 and y = 4. Therefore we have the following: 8 + 4 + 2z = 22 or z = 5.

From statement 5, 6-w>wor w=0 or 1 or 2.

Now first two questions can be answered.

The required answer is 6 - 2 = 4.

Answer: 4

Workspace:

**CAT 2018 LRDI Slot 2 | LR - Venn Diagram**

If the numbers of students enrolled in K and L are in the ratio 19:22, then what is the number of students enrolled in L?

- A.
17

- B.
19

- C.
22

- D.
18

Answer: Option C

**Explanation** :

If the ratio of the numbers of students enrolled in K and L are in the ratio 19:22,

$\frac{18+w}{23-w}=\frac{19}{22}$

Therefore w = 1.

Therefore total enrollment in L

= 23 − 1 = 22.

Hence, option 3.

Workspace:

**CAT 2018 LRDI Slot 2 | LR - Venn Diagram**

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

**Explanation** :

Out of 4 students who are enrolled in all the three, suppose ‘a’ students dropped out of L and ‘b’ students dropped out of K. Therefore the number of students who dropped out of G = 4 - a - b.

Therefore we have the following:

If the number of students enrolled in K reduced by 1 that means out of the 4 students who had enrolled in all the three, one student dropped out of K i.e. b = 1.

Now, if the number of students enrolled in G was 6 less than the number of students enrolled in L, we have the following:

(7 + w + a + 6 − w + b) + 6

= 6 − w + b + 9 − a − b + 8

∴19 + a + b = 23 − w − a

∴2a + b + w = 4

Since b = 1, the only solution for the equation 2a + b + w = 4 is a = 1, b = 1 and w = 1.

Now both the questions can be answered.

The required number of students = w + a = 2.

Answer: 2

Workspace:

**CAT 2018 LRDI Slot 2 | LR - Venn Diagram**

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 L?

- A.
7

- B.
5

- C.
8

- D.
6

Answer: Option D

**Explanation** :

The required number of students

= 6 − w + b = 6 − 1 + 1 = 6.

Hence, option 4.

Workspace:

**Answer the following question based on the information given below.**

Applicants for the doctoral programmes of Ambi Institute of Engineering (AIE) and Bambi Institute of Enginneering (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:

- No one is below the 80th percentile in all 3 sections.
- 150 are at or above the 80th percentile in exactly two sections.
- 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.
- 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 candidates is appearing in the AET by AIE, BIE consider their AET score for final selection provided the candidates 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 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.

**CAT 2017 LRDI Slot 1 | LR - Venn Diagram**

What best can be concluded about the number of candidates sitting for the separate test for BIE who were at or above the 90th percentile overall in CET?

- A.
3 or 10

- B.
10

- C.
5

- D.
7 or 10

Answer: Option A

**Explanation** :

Using the information given in the question let us represent it in the Venn diagram shown below. The diagram depicts the number of candidates getting 80 percentile and above in at least one or more of the subjects amongst students getting 90 percentile overall.

The number of candidates scoring 80 percentile and above in only Physics, only Chemistry and only Math is the same. Let this be ‘d’

Let ‘a’ – number of candidates scoring 80 percentile and above only in Physics and Math.

Let ‘b’ – number of candidates scoring 80 percentile above only in Physics and Chemistry.

Let ‘c’ – number of candidates scoring 80 percentile and above in Chemistry and Math.

Let ‘e’ – number of candidates scoring 80 percentile and above in all 3 subjects.

a + b + c = 150

Also a + b + c + 3d + e = 200

⇒ 3d + e = 50

Given that (2d + c) : (2d + a) : (2d + b) = 4 : 2 : 1

This implies 6d + a + b + c is a multiple of 7. We already know that a + b + c = 150. So 6d + 150 is a multiple of 7. This implies that 6d + 3 will also be a multiple of 7. So d will be 3, 10, 17. But as 3d + e = 50, it implies that d < 17. So d will be either 3 or 10.

The number of students who scored 90 percentile and above and scored at least 80 percentile in Physics (but not in Chemistry and Math) will be eligible for the BIE entrance test. This is equal to d which is either 3 or 10.

Hence, option 1.

Workspace:

**CAT 2017 LRDI Slot 1 | LR - Venn Diagram**

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, what is the number of candidates who are at or above the 90th percentile overall and at or above the 80th percentile in both P and M in CET?

Answer: 60

**Explanation** :

Using the information given in the question let us represent it in the Venn diagram shown below. The diagram depicts the number of candidates getting 80 percentile and above in at least one or more of the subjects amongst students getting 90 percentile overall.

The number of candidates scoring 80 percentile and above in exactly each of Physics, chemistry and Math is the same. Let this be ‘d’

Let ‘a’ – number of candidates scoring 80 percentile and above only in Physics and Math.

Let ‘b’ – number of candidates scoring 80 percentile above only in Physics and Chemistry.

Let ‘c’ – number of candidates scoring 80 percentile and above in Chemistry and Math.

Let ‘e’ – number of candidates scoring 80 percentile and above in all 3 subjects.

a + b + c = 150

Also a + b + c + 3d + e = 200

⇒3d + e=50

Given that (2d + c) : (2d + a) : (2d + b) = 4: 2: 1

This implies 6d + a + b + c is a multiple of 7. We already know that a + b + c= 150.

So 6d + 150 is a multiple of 7. This implies that 6d + 3 will also be a multiple of 7. So d will be 3, 10, 17. But as 3d + e = 50, it implies that d < 17. So d will be either 3 or 10.

Now 3d + e = 50

Also, d = 3 or 10

But it is given that e is a multiple of 5, so

e = 20

Now $\frac{20+c}{20+a}=\frac{2}{1},\frac{20+c}{20+b}=\frac{4}{1}$ and $\frac{20+a}{20+b}=\frac{2}{1}$

Solving the above expression we get the following equations:

c – 2a = 20 … (I)

c – 4b = 60 … (II)

a – 2b = 20 … (III)

Adding (I), (II) and (III) we get

–5b + 3c = 250 … (VI)

Solving (II) and (VI) we get

b = 10 and c = 100

∴ a =150 – 10 – 100 = 40

Now the number of candidates who scored 90 percentile overall and above and who score 80 percentile and above in P and M is a + e = 40 + 20 = 60

Answer : 60

Workspace:

**CAT 2017 LRDI Slot 1 | LR - Venn Diagram**

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: 70

**Explanation** :

Using the information given in the question let us represent it in the Venn diagram shown below. The diagram depicts the number of candidates getting 80 percentile and above in at least one or more of the subjects amongst students getting 90 percentile overall.

The number of candidates scoring 80 percentile and above in exactly each of Physics, chemistry and Math is the same. Let this be ‘d’

Let ‘a’ – number of candidates scoring 80 percentile and above only in Physics and Math.

Let ‘b’ – number of candidates scoring 80 percentile above only in Physics and Chemistry.

Let ‘c’ – number of candidates scoring 80 percentile and above in Chemistry and Math.

Let ‘e’ – number of candidates scoring 80 percentile and above in all 3 subjects.

a +b + c = 150

Also a + b + c + 3d + e = 200

⇒3d + e=50

Given that (2d + c) : (2d + a) : (2d + b)

= 4: 2: 1

This implies 6d + a + b + c is a multiple of 7. We already know that a + b + c= 150.

So 6d + 150 is a multiple of 7. This implies that 6d + 3 will also be a multiple of 7. So d will be 3, 10, 17. But as 3d + e = 50, it implies that d < 17. So d will be either 3 or 10.

Since the number of candidates who are at 90 percentile and above and also at or 80th percentile in all 3 sections is a multiple of 5 (which in the Venn Diagram is ‘e’), the Value of e(as explained in the previous questions) will have to be 20. Now the number of candidates who get 80 percentile in atleast 2 out of P, C and M which in the Venn Diagram will be the sum of a, b, c and e.

We already know a + b + c = 150 and e = 20

∴a + b + c + e = 150 + 20=70

Answer: 170

Workspace:

**CAT 2017 LRDI Slot 1 | LR - Venn Diagram**

If the number of candidates who are at or above the 90th percentile overall and also are at or above the 80th percentile in P in CET, is more than 100, how many candidates had to sit for the separate test for BIE?

- A.
299

- B.
310

- C.
321

- D.
330

Answer: Option A

**Explanation** :

The number of candidates scoring 80 percentile and above in exactly each of Physics, chemistry and Math is the same. Let this be ‘d’

Let ‘a’ – number of candidates scoring 80 percentile and above only in Physics and Math.

Let ‘b’ – number of candidates scoring 80 percentile above only in Physics and Chemistry.

Let ‘c’ – number of candidates scoring 80 percentile and above in Chemistry and Math.

Let ‘e’ – number of candidates scoring 80 percentile and above in all 3 subjects.

a +b + c = 150

Also a + b + c + 3d + e = 200

⇒3d+e=50

Given that (2d + c) : (2d + a) : (2d + b)

= 4: 2: 1

This implies 6d + a + b + c is a multiple of 7. We already know that a + b + c= 150.

So 6d + 150 is a multiple of 7. This implies that 6d + 3 will also be a multiple of 7. So d will be 3, 10, 17. But as 3d + e = 50, it implies that d < 17. So d will be either 3 or 10.

The number of candidates who are at or above 90th percentile overall and also at or above 80th percentile in P (as indicated in the Venn diagram) = a + b + d + e. As indicated in the answers to the previous question a = 3 or 10. Now we have already seen that if d = 10, a = 40, e As indicated in the answers to the previous question a = 3 or 10. Now we have So then a + b + d + e

= 42 + 18 + 3 + 41 = 104, which satisfies the condition given in the question.

Now the number of candidates who appear separately for the BIE test will be those who got 90 percentile and above overall and got 80 percentile and above only in P plus there who got 80 percentile only in P but got less than 90th percentile overall. Candidates who got 80th percentile only in P and got less than 90 in percentile overall = 400 – 104 =296

Number of candidates who get 80 percentile and above only in P and got 90 in percentile and above overall = d = 3

So, the number of candidates who sit for the separate test for BIE = 296 + 3 = 299

Hence, option 1.

Workspace:

**Answer the following question based on the information given below.**

Help Distress (HD) is an NGO involved in providing assistance to people suffering from natural disasters. Currently, it has 37 volunteers. They are involved in three projects: Tsunami Relief (TR) in Tamil Nadu, FloodRelief (FR) in Maharashtra, and Earthquake Relief (ER) in Gujarat. Each volunteer working with Help Distress has to be involved in at least one relief work project.

- A Maximum number of volunteers are involved in the FR project. Among them, the number of volunteers involved in FR project alone is equal to the volunteers having additional involvement in the ER project.
- The number of volunteers involved in the ER project alone is double the number of volunteers involved in all the three projects.
- 17 volunteers are involved in the TR project.
- The number of volunteers involved in the TR project alone is one less than the number of volunteers involved in ER Project alone.
- Ten volunteers involved in the TR project are also involved in at least one more project.

**CAT 2005 LRDI | LR - Venn Diagram**

Based on the information given above, the minimum number of volunteers involved in both FR and TR projects, but not in the ER project is:

- A.
1

- B.
3

- C.
4

- D.
5

Answer: Option C

**Explanation** :

17 volunteers are involved in the TR project and 10 in TR are also involved in other projects. Thus, 7 volunteers are involved only in TR.

∴ 8 volunteers are involved in ER alone.

∴ 4 volunteers are involved in all the three projects.

Let x people be involved in FR alone.

∴ Number of people involved in FR and ER but not TR = x – 4

Now, a + b + 4 = 10

∴ a + b = 6

Also, 7 + a + b + 4 + x + x – 4 + 8 = 37

∴ 2x = 16 or x = 8

Number of Volunteers involved in FR > Number of Volunteers involved in TR

And Number of Volunteers involved in FR > Number of Volunteers involved in ER

∴ 16 + a > 17 and 16 + a > 16 + b or a > b

∴ a and b can be (6, 0), (5, 1), (4, 2)

The minimum number of volunteers involved in both FR and TR projects, but not in the ER Project = minimum value of a = 4

Hence, option 3.

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