NIMCET 2027 Number System Previous Year Questions (PYQs) – Page 4 of 7

Custom Number System: Position of 10A

Given digits: 0, 1, A (base-3)

Convert 10A to base-10:

• 1 → 1
• 0 → 0
• A → 2
• 10A = 1×9 + 0×3 + 2 = 11

List of numbers in sequence:

1. 00
2. 01
3. 0A
4. 10
5. 11
6. 1A
7. A0
8. A1
9. AA
10. 100
11. 101
12. 10A

✅ Final Answer: 12

Solution (AP method):

Total children = 630. Let rows = n, first row = a, common difference d = -3.

Sum of n terms: \(S_n=\frac{n}{2}[2a+(n-1)d]=630\)

\(\Rightarrow 1260 = n\,[2a-3(n-1)] \;\Rightarrow\; 2a=\frac{1260}{n}+3(n-1)\)

\(\Rightarrow a=\frac{1}{2}\left(\frac{1260}{n}+3n-3\right)\). For a valid arrangement, a must be a positive integer and last row \(a-3(n-1)>0\).

• n = 3: \(a=\tfrac{1}{2}(420+9-3)=\tfrac{1}{2}\cdot 426=213\) ✔️ integer (valid)
• n = 4: \(a=\tfrac{1}{2}(315+12-3)=\tfrac{1}{2}\cdot 324=162\) ✔️ integer (valid)
• n = 5: \(a=\tfrac{1}{2}(252+15-3)=\tfrac{1}{2}\cdot 264=132\) ✔️ integer (valid)
• n = 6: \(a=\tfrac{1}{2}(210+18-3)=\tfrac{1}{2}\cdot 225=112.5\) ❌ not integer (invalid)

✅ Final Answer: 6 rows is not possible.

Let the number of children = n

Total sweets = 405

Each child gets = \( \frac{405}{n} \) sweets

Given: Each child’s sweets = \( \tfrac{1}{5} \) of the number of children

⇒ \( \frac{405}{n} = \frac{n}{5} \)

Cross multiply:

\( 405 \times 5 = n^2 \)
⇒ \( n^2 = 2025 \)
⇒ \( n = \sqrt{2025} = 45 \)

✅ Final Answer: The number of children = 45