Question

Amita was given test in Mathematics by a tutor the square of the actual marks scored by was nine times the marks reported by the parents if the actual marks scored by 10 less than the score reported by her to the parents then find the actual score of Amita

Answer 1

Step-by-step explanation:

Given  

Amita was given test in Mathematics by a tutor the square of the actual marks scored by was nine times the marks reported by the parents

Let the actual score of Amita be a  

Score reported by Amita to  her parents = a + 10

Now according to question we have  

        a^2 = 9 (a + 10)

 So a^2 = 9a + 90

    a ^2 – 9a – 90 = 0  

   a^2 – 15a + 6a – 90 = 0

 a (a – 15) + 6 (a – 15)

(a – 15) (a + 6) = 0

So a = 15 since a ≥ 0

Therefore actual score of Amita will be 15

Answer 2

The actual score of Amita is 15 marks.

Step-by-step explanation:

We are given that the square of the actual marks scored by Amita was nine times the marks reported by her to the parents.

Also, the actual marks scored by her was 10 less than the score reported by her to the parents.

Let the actual score of Amita be $x$ marks.

So, according to the question ;

                      $x^{2} = 9(x+10)$

                      $x^{2} = 9x+90$  

                      $x^{2} - 9x-90=0$

Now, we will use the method of middle term splitting here, i.e.;

                      $x^{2} - 15x+6x-90 =0$

                      $x(x - 15)+6(x-15) =0$

                       $(x+6)(x-15) =0$

This means, either (x + 6) = 0  or  (x - 15) = 0.

(x + 6) = 0 is not possible as from this we will get negative value of x and (x > 0) .

Therefore, (x - 15) = 0 which means  x = 15 marks.