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NIMCET Arithmetic Reasoning PYQ

NIMCET PYQ 2023
Thirty-six vehicles are parked in a parking lot in a single row. After the first car, there is one scooter. After the second car, there are two scooters. After the third car, there are three scooters and so on. Work out the number of scooters in the second half row.





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NIMCET Previous Year PYQNIMCET NIMCET 2023 PYQ

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NIMCET PYQ 2023
In a reality show, two judges independently provided marks based on the performance of the participants. If the marks provided by the second judge are given by y= 1+ x, where x is the marks provided by the first judge. Then for a participant





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NIMCET Previous Year PYQNIMCET NIMCET 2023 PYQ

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NIMCET PYQ 2023
On Monday, Akash ran 4 km less than the distance he ran on Tuesday. Sanjay, who ran the same distance on Monday and Tuesday, ran 5 km more on Tuesday than the distance Akash ran on Monday. Find the difference between the distances covered by Akash and Sanjay over the two days.





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NIMCET Previous Year PYQNIMCET NIMCET 2023 PYQ

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NIMCET PYQ 2023
Study the diagram and answer the question that follows.

How many players are there who are girls but not a coach.





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NIMCET Previous Year PYQNIMCET NIMCET 2023 PYQ

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NIMCET PYQ 2023
A university is offering elective courses in Mathematics, Economics and Sociology. Each of its 100 undergraduate students has to opt for at least one of these electives. Course enrollment data showed that 47 students enrolled for Mathematics, 47 students enrolled for Economics and 57 students enrolled for Sociology. If 7 students enrolled for all three courses, how many students enrolled for exactly one course?





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NIMCET Previous Year PYQNIMCET NIMCET 2023 PYQ

Solution

Venn Diagram Problem: Students in Elective Courses

Given:

  • Total students = 100
  • Math (M) = 47
  • Economics (E) = 47
  • Sociology (S) = 57
  • All three (M ∩ E ∩ S) = 7

Objective:

Find the number of students who enrolled in exactly one course.

Step-by-step:

Let’s use the formula for total union of 3 sets:

Total = M + E + S − (M∩E) − (E∩S) − (S∩M) + (M∩E∩S)

Let the number of students who enrolled in exactly 2 courses = x
So those who enrolled in all 3 = 7
Let the number of students who enrolled in exactly one course = y

Then, the total number of course enrollments =
1 × y + 2 × x + 3 × 7 = M + E + S = 47 + 47 + 57 = 151

⇒ y + 2x + 21 = 151
⇒ y + 2x = 130 — (1)

Also, total students = 100 = y + x + 7
⇒ y + x = 93 — (2)

Subtracting (2) from (1):

(y + 2x) − (y + x) = 130 − 93
⇒ x = 37

Now, from (2): y + 37 = 93 ⇒ y = 56

✅ Final Answer:

56 students enrolled in exactly one course.


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