**Question:**

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?

**Solution:**

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.