Question

Two students appeared at an examination. One of them secured 9 marks more than the other and his marks were 56% of the sum of their marks. What are the marks obtained by both of them?
$
  {\text{A}}{\text{. 25 and 78}} \\
  {\text{B}}{\text{. 58 and 77}} \\
  {\text{C}}{\text{. 33 and 42}} \\
  {\text{D}}{\text{. 56 and 63}} \\
$

Answer 1

Hint: In this question, we would assume one of the numbers as x and then proceed by applying the given constraints. After applying the constraints we would get a linear equation in x. This will help us simplify the question and reach the answer.

Complete step-by-step answer:

Let the marks of a student be x. So, the marks of another number is (x+9).
We have been also given that (x+9) is 56% of the sum of their both marks.
So $x + 9 = \dfrac{{56}}{{100}}\left( {x + 9 + x} \right)$
$ \Rightarrow x + 9 = \dfrac{{56}}{{100}}\left( {2x + 9} \right)$
$ \Rightarrow 100\left( {x + 9} \right) = 56\left( {2x + 9} \right)$
$ \Rightarrow 100x + 900 = 112x + 504$
$ \Rightarrow 12x = 396$
$ \Rightarrow x = 33$
So, the marks are 33 and 42
Hence, option C is correct.

Note: Whenever we face such types of problems the value point to remember is that we need to have a good grasp over linear equations in one variable and number theory. In these types of questions, we should always assume one of the numbers and then applying constraints move ahead. This helps in getting us the required expressions and gets us on the right track to reach the answer.