f(x+2)=f(x+1)+f(x)
For f(15), value of x = 13
So, f(15) = f(14)+f(13)
Now, for f(14), x = 12 and for f(13), x = 11
Therefore, f(15) = f(13)+f(12)+f(12)+f(11)
=> f(15) = 3f(12)+2f(11)
=> f(15) = 3(f(11)+f(10))+2f(11)
=> f(15) = 5f(11)+3f(10)
=> 617-5(91) = 3f(10)
=> f(10) = 162/3 = 54

Given that $f(x + 2) = f(x) + f(x + 1)$ for all positive integers x , and $f(11) = 91$, $f(15) = 617$ and we need to find the value of $f(10)$

$f(x + 2) = f(x) + f(x + 1)$ - this means that f(any positive integer) = sum of f of previous two numbers
=> f(15) = f(13) + f(14) ...(1)
f(14) = f(12) + f(13) ...(2)
f(13) = f(11) + f(12) ....(3)
f(12) = f(10) + f(11)

Putting value of f(12) in (3) we get
f(13) = f(11) + f(10) + f(11) = f(10) + 2*f(11)

Putting value of f(13) and f(12) in (2) we get
f(14) = f(10) + f(11) + f(10) + 2*f(11) = 2*f(10) + 3*f(11)

Putting value of f(14) and f(13) in (1) we get
f(15) = f(10) + 2*f(11) + 2*f(10) + 3*f(11) = 3*f(10) + 5*f(11)
=> f(10) = $\frac{f(15) - 5*f(11) }{3}$ = $\frac{617 - 5*91}{3}$ = $\frac{617-455}{3}$ = $\frac{162}{3}$ = 54

So, Answer will be B
Hope it helps!

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