Question

At an examination in which maximum marks are 500. K got 10 percent less than L, L got 25 percent more than M and M got 20 percent less than N. If K got 360 marks what percentage of the marks was obtained by N?

Answer 1

Hint: Here, according to the problem statement we will be obtaining the individual marks obtained by each L, M and N. Then, we will be using the formula i.e., Percentage of marks$ = \left( {\dfrac{{{\text{Marks obtained}}}}{{{\text{Maximum marks}}}}} \right) \times 100$ in order to find the percentage of the marks obtained by N.

Complete step-by-step answer:
Given, Maximum marks = 500
Marks obtained by K = 360
Let us suppose that the marks obtained by L, M and N are x, y and z respectively
It is given that the marks obtained by K is 10 percent less than that obtained by L
i.e., Marks obtained by K $ = \left( {1 - \dfrac{{10}}{{100}}} \right) \times $Marks obtained by L
$
   \Rightarrow 360 = \left( {\dfrac{{100 - 10}}{{100}}} \right) \times x \\
   \Rightarrow 360 = \left( {\dfrac{{90}}{{100}}} \right) \times x \\
   \Rightarrow x = \dfrac{{360 \times 100}}{{90}} = 400 \\
 $
Therefore, the marks obtained by L is 400 marks.
It is also given that the marks obtained by L is 25 percent more than that obtained by M
i.e., Marks obtained by L $ = \left( {1 + \dfrac{{25}}{{100}}} \right) \times $Marks obtained by M
$
   \Rightarrow 400 = \left( {\dfrac{{100 + 25}}{{100}}} \right) \times y \\
   \Rightarrow 400 = \left( {\dfrac{{125}}{{100}}} \right) \times y \\
   \Rightarrow y = \dfrac{{400 \times 100}}{{125}} = 320 \\
 $
Therefore, the marks obtained by M is 320 marks.
It is also given that the marks obtained by M is 20 percent less than that obtained by N
i.e., Marks obtained by M $ = \left( {1 - \dfrac{{20}}{{100}}} \right) \times $Marks obtained by N
$
   \Rightarrow 320 = \left( {\dfrac{{100 - 20}}{{100}}} \right) \times z \\
   \Rightarrow 320 = \left( {\dfrac{{80}}{{100}}} \right) \times z \\
   \Rightarrow z = \dfrac{{320 \times 100}}{{80}} = 400 \\
 $
Therefore, the marks obtained by N is 400 marks.
As we know that
Percentage of the marks obtained by N$ = \left( {\dfrac{{{\text{Marks obtained by N}}}}{{{\text{Maximum marks}}}}} \right) \times 100$
$ \Rightarrow $ Percentage of the marks obtained by N$ = \left( {\dfrac{{{\text{400}}}}{{{\text{500}}}}} \right) \times 100 = 80$ percent.
Therefore, 80 percent is the required percentage of the marks obtained by N.

Note: In this particular problem, the percentage of the marks obtained by K, L and M are given by $\left( {\dfrac{{{\text{360}}}}{{{\text{500}}}}} \right) \times 100 = 72$ percent , $\left( {\dfrac{{400}}{{{\text{500}}}}} \right) \times 100 = 80$ percent and $\left( {\dfrac{{{\text{320}}}}{{{\text{500}}}}} \right) \times 100 = 64$ percent respectively. Also, it is important to note that none of the marks obtained should ever exceed the maximum marks.