NIMCET 2025 Previous Year Questions (PYQs) – Page 1 of 12

Let the first row have $n$ children and total rows be $r$. Then the numbers per row form an AP with difference $-3$ and sum $630$:

$\displaystyle \frac{r}{2}\big(2n-3(r-1)\big)=630$ $ \;\Rightarrow\; 1260=r\big(2n-3(r-1)\big)$.

Hence $r\mid1260$. Also the last row must be positive: $n-3(r-1)>0$.

Testing the options (and ensuring $n$ is integer and last term positive):

• $r=4$: $1260/4=315$, $2n=315+3\cdot3=324 \Rightarrow n=162$, last term $=162-9=153>0$ ✓
• $r=5$: $1260/5=252$, $2n=252+3\cdot4=264 \Rightarrow n=132$, last term $=132-12=120>0$ ✓
• $r=6$: $1260/6=210$, $2n=210+3\cdot5=225 \Rightarrow n=112.5$ (not integer) ✗

Therefore, the impossible number of rows is $\boxed{6}$.

Increased by 15% ⇒ New value = 1.15x

Given: 1.15x = 161

x = 161 ÷ 1.15

x = 140

✅ Final Answer: The original number is 140.

We separate the integer and fractional parts.

Integer part: $10101_2 = 1\times2^4 + 0\times2^3 + 1\times2^2 + 0\times2^1 + 1\times2^0 $$= 16 + 0 + 4 + 0 + 1 = 21$

Fractional part: $.10101_2 = 1\times2^{-1} + 0\times2^{-2} + 1\times2^{-3} + 0\times2^{-4} + 1\times2^{-5}$

= 0.5 + 0 + 0.125 + 0 + 0.03125 

= 0.65625

Therefore, $10101.10101_2 = 21 + 0.65625 = {21.65625}$

A debugger is a tool used to test and debug programs. It allows programmers to pause execution at specific points (breakpoints), run code step-by-step, and inspect variables and registers to find logic or runtime errors.

Hence, the correct option is:

✅ Option 3: allows to set breakpoints, execute a segment of program, and display contents of registers.