If Zeba were younger by 5 yr than what she really is, then the square of, her age (in years) would have been 11, more than five times her actual age,
what is her age now?
Let the actual age of Zeba = x yr,
Herage when she was 5 yr younger = (x – 5)yr.
Now, by given condition,
Square of her age = 11 more than five times her actual age
$(x-5)^{2}=5 \times$ actual age $+11$
$\Rightarrow \quad(x-5)^{2}=5 x+11$
$\Rightarrow \quad x^{2}+25-10 x=5 x+11$
$\Rightarrow \quad x^{2}-15 x+14=0$
$\Rightarrow \quad x^{2}-14 x-x+14=0$ [by splitting the middle term]
$\Rightarrow \quad x(x-14)-1(x-14)=0$
$\Rightarrow \quad(x-1)(x-14)=0$
$\Rightarrow \quad x=14$
[here, x ≠ 1 because her age is x – 5. So, x – 5 = 1 – 5 = – 4 i.e., age cannot be negative] Hence, required Zeba’s age now is 14 yr.