NIMCET 2027 2015 Previous Year Questions (PYQs) – Page 7 of 9

Case 1: Exactly one block of five consecutive 0’s.
The block $00000$ can start at positions 1 to 6, so 6 choices.
The remaining 5 positions can be filled freely with 0 or 1, except they should not create another block of five 0’s.
Valid fillings = $2^5 - 1 = 31$.
So number of strings with exactly one block of five 0’s is
$6 \times 31 = 186$

Case 2: Exactly one block of five consecutive 1’s.
By symmetry, the count is the same.
$186$

Case 3: One block of five 0’s and one block of five 1’s.
This is possible only when the blocks do not overlap.
The only such strings are
$0000011111$ and $1111100000$

So total such strings = $2$.
Using inclusion–exclusion principle:
$186 + 186 + 2 = 222$
Final Answer: $222$

(1) $a^2 + 2b^2 - 2bc = 100$ 

(2) $2ab - c^2 = 100$ 

Equating equation (1) and equation (2): $a^2 + 2b^2 - 2bc = 2ab - c^2$ 

Rearranging all terms to one side: $a^2 - 2ab + 2b^2 - 2bc + c^2 = 0$ 

Split $2b^2$ into $b^2 + b^2$: 

$(a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) = 0$ 

$(a - b)^2 + (b - c)^2 = 0$ 

Since $a, b, c$ are real numbers (positive integers), the sum of two non-negative squares can only be zero if each term is zero: 

$a - b = 0 \implies a = b$ 

$b - c = 0 \implies b = c$ 

Thus: $a = b = c$ Substitute $a = b = c$ into equation (2): 

$2(c)(c) - c^2 = 100$ 

$2c^2 - c^2 = 100$ 

$c^2 = 100$ 

Since $c$ is a positive integer: $c = 10 \implies a = b = c = 10$ 

 Now, compute the required value: $\frac{a + b}{c} = \frac{10 + 10}{10} = \frac{20}{10} = 2$