NIMCET Arithmetic Reasoning Previous Year Questions (PYQs) – Page 1 of 3

Solution: Number of Scooters in Second Half of Row

Total vehicles = 36. Let number of cars = \(n\), scooters = \(\frac{n(n+1)}{2}\).
Solve \(n + \frac{n(n+1)}{2} = 36\) gives \(n=7\) cars and 28 scooters (total 35 vehicles).
Since total is 36, assume 1 extra scooter added somewhere.

The vehicle sequence (cars + scooters) goes as:
Car1, 1 scooter; Car2, 2 scooters; ... Car7, 7 scooters.

Half of 36 = 18 vehicles (second half is last 18).
Positions 19 to 36 in the sequence contain:
- Last 3 scooters of the 5th group (positions 18-20)
- All 6 scooters of the 6th group (positions 22-27)
- All 7 scooters of the 7th group (positions 29-35)

Counting scooters in these positions:
\(2 + 6 + 7 = 15\) scooters in the second half.

Final answer: 15 scooters.

Given: Total questions = 100, Part B = 23 questions (2 marks each).

Let Part A = \(a\) questions (1 mark), Part C = \(c\) questions (3 marks).

So, \(a + 23 + c = 100 \Rightarrow a + c = 77\).
Total marks = \(a + 46 + 3c\).
Part A marks ≥ 60% total marks:
\(\Rightarrow a \geq 0.6(a + 46 + 3c)\)
\(\Rightarrow a \geq 0.6a + 27.6 + 1.8c\)
\(\Rightarrow 0.4a \geq 27.6 + 1.8c\)
\(\Rightarrow a \geq 69 + 4.5c\).

Using \(a = 77 - c\):
\(77 - c \geq 69 + 4.5c \Rightarrow 8 \geq 5.5c \Rightarrow c \leq 1.45\).
So, \(c = 1\).

Answer: Part C has 1 question.

Distance Difference Between Akash and Sanjay

Let Akash’s Tuesday distance = \(x\) km.
Then Akash’s Monday distance = \(x - 4\) km.
Sanjay runs same distance on both days = \(y\) km.
Sanjay’s Tuesday distance is 5 km more than Akash’s Monday distance:
\[ y = (x - 4) + 5 = x + 1 \]

Total distance Akash ran in 2 days:
\[ (x - 4) + x = 2x - 4 \] Total distance Sanjay ran in 2 days:
\[ y + y = 2y = 2(x + 1) = 2x + 2 \]

Difference = Sanjay’s total - Akash’s total:
\[ (2x + 2) - (2x - 4) = 6 \text{ km} \]

Answer: The difference in total distances covered is 6 km.

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:

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):

Now, from (2): y + 37 = 93 ⇒ y = 56

✅ Final Answer:

56 students enrolled in exactly one course.