Question

In an examination first division marks are $ 60\% $ . A student secures $ 538{\text{ marks}} $ and misses the first division by $ 2{\text{ marks}} $ . Find the total marks of the examination.

Answer 1

Hint: Here we can suppose that the maximum marks of the examination be $ x $
Then we can say that $ 60\% $ of the total marks $ = \dfrac{{60x}}{{100}} $ and this is the marks for the first division and student secures $ 538{\text{ marks}} $ and this is $ 2{\text{ marks}} $ less than required for first division so we can say that $ \dfrac{{60x}}{{100}} = 538 + 22 = 560 $ and we can calculate $ x $ from here.

Complete step-by-step answer:
Here we are given that in an examination first division marks are $ 60\% $ . A student secures $ 538{\text{ marks}} $ and misses the first division by $ 2{\text{ marks}} $ .
So if we suppose that the maximum marks be $ x $
Then $ 60\% $ of the marks is the marks for the first division. Hence we can say that:
First division marks $ = 60\% {\text{ of }}x = \dfrac{{60x}}{{100}} $ $ - - - \left( 1 \right) $
Now we are also given in the problem that student secures $ 538{\text{ marks}} $ and misses the first division by $ 2{\text{ marks}} $
Hence we can say that first division marks are $ 2 $ more than $ 538 $
Hence we can say that first division marks are $ 538 + 2 = 540 $ $ - - - - (2) $
Now we can equate equations (1) and (2) we will get:
\[
  \dfrac{{60x}}{{100}} = 540 \\
  x = \dfrac{{\left( {540} \right)\left( {100} \right)}}{{60}} = 900 \\
 \]
Hence we can say that the maximum marks that can be obtained by a student is $ 900{\text{ marks}} $
Therefore in these types of problems we must know what the statement is saying and the terms we need to equate to get the required value.

Note: Here the student needs to formulate the equation in a proper way. If we are given that $ x $ score $ n $ marks more than $ y $ then this can be written in the form of equation as:
 $ {\text{score of }}x = {\text{score of }}y + n $
So the equation must be formed in a proper way.