Venn Diagram - Previous Year CAT/MBA Questions
You can practice all previous year OMET questions from the topic Venn Diagram. This will help you understand the type of questions asked in OMET. It would be best if you clear your concepts before you practice previous year OMET questions.
400 students were admitted to the 2018-19 MBA batch. 200 of them did not choose “Business Statistics”. 100 of them did not choose “International Management’. There were 80 students who did not choose any of the two subjects. Find the number of students who chose both Business Statistics and International Management.
Answer: Option A
Number of students who chose Business statistics = 400 − 200 = 200
Number of students who chose International Management = 400 − 100 = 300
Number of students who chose at least one of the two subjects = 400 − 80 = 320
∴ Number of students who chose both the subjects = 200 + 300 − 320 = 500 − 320 = 180
Hence, option (a).
In a class of 60, along with English as a common subject, students can opt to major in Mathematics, Physics, Biology or a combination of any two. 6 students major in both Mathematics and Physics, 15 major in both Physics and Biology, but no one majors in both Mathematics and Biology. In an English test, the average mark scored by students majoring in Mathematics is 45 and that of students majoring in Biology is 60. However, the combined average mark in English, of students of these two majors, is 50. What is the maximum possible number of students who major ONLY in Physics?
None of the above
Answer: Option D
Let Tm and Tb be total scores of the students majoring in Mathematics and Biology respectively.
According to the given conditions,
Tm = (M + 6) × 45
Tb = (B + 15) × 60
Also, (Tm + Tb) = (B + P + 21) × 50
45(M + 6) + 60(B + 15) = 50M + 50B + 1050
10B – 5M + 270 + 900 = 1050
10B – 5M = –120
M = 2B + 24
Here, we need to determine the maximum value of P.
∴ We need to minimize the value of B. Minimum value of B can be 0.
∴ M = 24
Again, we know that,
M + B + P + 21 = 60
⇒ 24 + 0 + P + 21 = 60
⇒ P = 15
Hence, option (d).
A premier B-school, which is in process of getting an AACSB accreditation, has 360 second year students. To incorporate sustainability into their curriculum, it has offered 3 new elective subjects in the second year namely Green Supply Chain, Global Climate Change & Business and Corporate Governance. Twelve students have taken all the three electives, and 120 students study Green Supply Chain. There are twice as many students who study Green Supply Chain and Corporate Governance but not Global Climate Change and Business, as those who study both Green Supply Chain and Global Climate Change & Business but not Corporate Governance, and 4 times as many who study all the three. 124 students study Corporate Governance. There are 72 students who could not muster up the courage to take up any of these subjects. The group of students who study both Green Supply Chain and Corporate Governance but not global Climate Change & Business is exactly the same as the group made up to the students who study both Global Climate Change & Business and Corporate Governance. How many students study Global Climate Change & Business only?
Answer: Option B
The number of students who study each combination of subjects (based on the direct data) given is as shown below:
It is given that: (GSC and CG but not GCCB) = 4 times (all three electives)
∴ 2x = 4(12) i.e. x = 24
Also: (GSC and CG but not GCCB) = (all three electives) + (GCCB and CG but not GSC)
∴ (GCCB and CG but not GSC) = 2x − 12 = 2(24) − 12 = 36
So, the figure becomes:
Now, CG only = 124 − (48 + 12 + 36) = 28
∴ GCCB alone = 360 − 120 − 36 − 28 − 72 = 104
Hence, option 2.
In an amusement park along with the entry pass a visitor gets two of the three available rides (A, B and C) free. On a particular day 77 opted for ride A, 55 opted for B and 50 opted for C; 25 visitors opted for both A and C, 22 opted for both A and B, while no visitor opted for both B and C. 40 visitors did not opt for ride A and B, or both. How many visited with the entry pass on that day?
Answer: Option E
Let the Venn diagram be as shown in the figure,
No one can take all three rides, hence g = 0.
22 people take rides A and B,
∴ d = 22
25 people take rides A and C,
∴ f = 25
50 people take ride C,
∴ c = 50 – 25 = 25.
40 people don’t take A or B or both,
∴ 40 = c + h
⇒ h = 40 – 25 = 15
∴ Total number of people visiting the park = (77 + 55 + 50 – 25 – 22) + 15 = 150.
Hence, option (e).
290 students of MBA (International Business) in a reputed Business School have to study foreign language in Trimesters IV and V. Suppose the following information are given
i. 120 students study Spanish
ii. 100 students study Mandarin
iii. At least 80 students, who study a foreign language, study neither Spanish nor Mandarin
Then the number of students who study Spanish but not Mandarin could be any number from
80 to 170
80 to 100
50 to 80
20 to 110
Answer: Option D
Atleast 80 students study neither Spanish nor Mandarin.
Hence, maximum number of students who study atleast one language = 290 – 80 = 210
Minimum number of students who study both languages = 100 + 120 – 210 = 10
∴ Maximum number of students who study Spanish but not Mandarin = 120 – 10 = 110
Maximum number of students who study both languages = smaller value of 100 and 120 = 100
∴ Minimum number of students who study Spanish but not Mandarin = 120 – 100 = 20
Hence, the range could be any number from 20 to 110.
Hence, option 4.
In a certain village, 22% of the families own agricultural land, 18% own a mobile phone and 1600 families own both agricultural land and a mobile phone. If 68% of the families neither own agricultural land nor a mobile phone, then the total number of families living in the village is:
Answer: Option A
Let total number of families in the village be T
Number of families own agricultural land, n(A) = 0.22T
Number of families own mobile phone, n(M) = 0.18T
Number of families own both agricultural land and mobile phone, n(A ⋂ M) = 1600
Number of families own agricultural land or mobile phone, n(A ⋃ M) = T – 0.68T = 0.32T
∴ n(A ⋃ M) = n(A) + n(M) – n(A ⋂ M)
∴ n(A ⋂ M) = 0.08T
0.08T = 1600 ⇒ T = 20000
Hence, option 1.
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