NIMCET Number System Previous Year Questions (PYQs) – Page 1 of 7

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 $\boxed{6}$.

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

First convert both binary numbers into decimal.

$1100_2=12$

$1011_2=11$

Now multiply:

$12\times 11=132$

Convert $132$ into binary:

$132=128+4$

$128=2^7$ and $4=2^2$

So,

$132=10000100_2$

Therefore,

$1100_2\times 1011_2=10000100_2$

Let’s first convert both binary numbers into decimal (two’s complement form):

• $A = 10011001_2$ → negative because MSB = 1 → invert bits: $01100110_2 = 102_{10}$ → add 1 → $103_{10}$ So, $A = -103$.
• $B = 11010111_2$ → MSB = 1 (negative) → invert bits: $00101000_2 = 40_{10}$ → add 1 → $41_{10}$ So, $B = -41$.

Add them:

$A + B = -103 + (-41) = -144$

Now, in 8-bit two’s complement, the range is $-128$ to $+127$. Since $-144$ is out of range, overflow occurs.

But let’s compute the 8-bit result (ignoring overflow):

$10011001 + 11010111 =$ 10011001
+ 11010111
= 101110000 (9 bits)

Drop the carry beyond 8 bits → 01110000.

Therefore, the resulting 8-bit binary value is:

✅ Result = 01110000₂

(Overflow occurred, actual signed result would have been −144, but the 8-bit stored value is +112.)