Question

If zeba were younger by 5 years than what she really is than the square of her age ( in year) would have been 11 more than five time her actual age . what is her age now ?

Answer 1

55 YEARS 11 MORE THAN 5 TIME HER AGE MEANT LET HER AGE BE 5x =11 BRING 5 THERE 11X5=55 WELCOME MENTION NOT

Answer 2

$\bf\huge\underline{Solution:-}$

Let the actual age of Zeba = x years

Her age when she was 5 years younger = (x - 5) years

Now, by given condition,

Square of her age = 11 more than five times her actual age

${x - 5}^{2} = 5 \times actual \: age + 11 \\ => {x - 5}^{2} = 5x + 11 \\ => {x}^{2} + 25 - 10x = 5x + 11 \\ => {x}^{2} - 15x + 14 = 0 \\ => {x}^{2} - 14x - x + 14$

[by splitting the middle term]

$=> x(x - 14) - 1(x - 14) = 0 \\ =>(x - 1(x - 14) </strong><strong>\</strong><strong>\</strong><strong> </strong><strong>=> x = 14$

[here x is not equal to 1 because her age is x - 5. So, x - 5 = 1 - 5 = 4 = -4 i.e. age cannot be negetive]

Hence, present age of Zeba is 14 years.