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# A got $30\%$marks but failed by 120 marks. B got $60\%$ marks which is 240 marks more than the required marks to pass. Then what is the pass percentage?A.35B.40C. 45D.50  Verified
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Hint: In this question A got $30\%$ but he failed by 120 marks. B got $60\%$ which is 240 marks more than the required marks to pass. So, we have to find the passing percentage. Let the total marks of the exam be $x$. Now we try to find out the $30\%$ of the total marks obtained by A. Similarly we try to find out $60\%$ marks of total marks that are obtained by B. Now we also assume that the passing marks is P. From the above assumption and marks obtained by candidate two equations will be formed. So, with the help of two equations we can find out the two unknown variables.

Complete step by step solution:
According to the given information, we will proceed to the question from the starting statement. As ‘A’ got$30\%$ , ‘B’ got $60\%$ and followed by other given information in the question.
Step1: ‘A’ got $30\%$ marks but failed by 120 marks. So, let the total number of marks be$x$. So, we find out the $30\%$ of the total marks $x$. And one more assumption is let the passing marks be $P$. So, the marks obtained by ‘A’
$\begin{array}{l} 30\% ofx\\ = x \times \dfrac{{30}}{{100}} = \dfrac{{3x}}{{10}} \end{array}$

‘A’ failed by 120 marks, so the marks obtained by ‘A’ will be 120 less than from the passing marks, so,
$\begin{array}{l} \dfrac{{3x}}{{10}} = (P - 120)\\ \Rightarrow 3x = 10P - 1200\\ \Rightarrow 3x - 10P = - 1200\\ 10P - 3x = 1200 \to equation(1) \end{array}$
Step2: The marks obtained by student ‘B’ is $60\%$ that is 240 marks more than required passing marks. So, ‘B’ got $60\%$ out of the total marks.
$\begin{array}{l} 60\% ofx\\ = x \times \dfrac{{60}}{{100}}\\ = \dfrac{{6x}}{{10}} \end{array}$

Since passing marks be$P$ and marks obtained by ‘B’ is 240 more than it so
$\begin{array}{l} \dfrac{{6x}}{{10}} = P + 240\\ \Rightarrow \dfrac{{6x}}{{10}} - \dfrac{P}{1} = 240\\ \Rightarrow 6x - 10P = 2400\\ 6x - 10P = 2400 \to equation(2) \end{array}$
Step3: We have got two equations {equation (1) and equation (2)}.
Now we will try to solve it. So, we add equation (1) and equation (2).
$\begin{array}{l} 10P - 3x = 1200\{ equation(1)\} \\ \underline { + 6x - 10P = 2400\{ equation(2)\} } \\ 3x = 3600\\ \Rightarrow x = \dfrac{{3600}}{3}\\ x = 1200 \end{array}$
Step4: Now, total marks of the exam are 1200 and we have to find passing marks$P$ . So, we put the value of $x$in . So,
$\begin{array}{l} 10P - 3x = 1200\{from equation(1)\} \\ \Rightarrow 10P - 3 \times 1200 = 1200(x = 1200)\\ \Rightarrow 10P = 1200 + 3600\\ \Rightarrow P = \dfrac{{4800}}{{10}}\\ P = 480 \end{array}$
Hence, the passing marks of the exam is $480$ .
So, passing percentage
$\begin{array}{l} = \dfrac{{480}}{{1200}} \times 100\\ = 40\% \end{array}$

Note: It is based on linear equation conversion from the statement of the question. When we are converting it into a linear equation, follow all the conditions carefully because misinterpretation will lead to a wrong answer.