
In an examination of n questions, a student replied $15$ out of the first $20$ questions correctly. Of the remaining questions, he answered one-third correctly. All the questions have the same credit. If the student gets $50\% $ marks, the value of n is:
A)$20$
B)$40$
C)$50$
D)100
Answer
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Hint: According to the question, the sum of the number of correct answers should be equal to $50\% $ of the total number of questions because all the questions have the same credit.
Complete step-by-step answer:
Since all the questions have the same credit,
Let the marks obtained for each correct answer be $x$.
Total number of questions $ = n$
Maximum marks that could be obtained by a student = Total number of questions $ \times $ marks awarded for each question $ = nx$ (3)
Out of $20$questions, $15$ were answered correctly by the student.
Marks awarded $ = 15x$ (1)
Remaining number of questions $ = n - 20$
No. of questions answered correctly out of the remaining questions $ = \dfrac{1}{3}\left( {n - 20} \right)$
Marks awarded in this case would be $ = \dfrac{1}{3}\left( {n - 20} \right)x$ (2)
Hence, from equation (1) and (2) we can conclude that the marks obtained by the student must be
$ = 15x + \dfrac{1}{3}\left( {n - 20} \right)x$ (4)
Also, we are given that student gets $50\% $ marks, i.e., $50\% $ of maximum marks
$
= 50\% \times nx \\
= \dfrac{{50}}{{100}} \times nx = \dfrac{{nx}}{2} \\
$ (5)
So, equation (4) must be equal to equation (5)
$
15x + \dfrac{1}{3}\left( {n - 20} \right)x = \dfrac{{nx}}{2} \\
15 + \dfrac{n}{3} - \dfrac{{20}}{3} = \dfrac{n}{2} \\
15 - \dfrac{{20}}{3} = \dfrac{n}{2} - \dfrac{n}{3} \\
\dfrac{{15 \times 3 - 20}}{3} = \dfrac{{3n - 2n}}{6} \\
\dfrac{n}{6} = \dfrac{{45 - 20}}{3} \\
\dfrac{n}{6} = \dfrac{{25}}{3} \\
n = \dfrac{{25 \times 6}}{3} \\
n = 50 \\
$
C is the correct answer.
Note: In these types of questions, students should try to obtain an equation using the information given in the question and then put it equal to the value that we were given in the question. Students must write all the steps for calculation and not skip them to avoid mistakes.
Complete step-by-step answer:
Since all the questions have the same credit,
Let the marks obtained for each correct answer be $x$.
Total number of questions $ = n$
Maximum marks that could be obtained by a student = Total number of questions $ \times $ marks awarded for each question $ = nx$ (3)
Out of $20$questions, $15$ were answered correctly by the student.
Marks awarded $ = 15x$ (1)
Remaining number of questions $ = n - 20$
No. of questions answered correctly out of the remaining questions $ = \dfrac{1}{3}\left( {n - 20} \right)$
Marks awarded in this case would be $ = \dfrac{1}{3}\left( {n - 20} \right)x$ (2)
Hence, from equation (1) and (2) we can conclude that the marks obtained by the student must be
$ = 15x + \dfrac{1}{3}\left( {n - 20} \right)x$ (4)
Also, we are given that student gets $50\% $ marks, i.e., $50\% $ of maximum marks
$
= 50\% \times nx \\
= \dfrac{{50}}{{100}} \times nx = \dfrac{{nx}}{2} \\
$ (5)
So, equation (4) must be equal to equation (5)
$
15x + \dfrac{1}{3}\left( {n - 20} \right)x = \dfrac{{nx}}{2} \\
15 + \dfrac{n}{3} - \dfrac{{20}}{3} = \dfrac{n}{2} \\
15 - \dfrac{{20}}{3} = \dfrac{n}{2} - \dfrac{n}{3} \\
\dfrac{{15 \times 3 - 20}}{3} = \dfrac{{3n - 2n}}{6} \\
\dfrac{n}{6} = \dfrac{{45 - 20}}{3} \\
\dfrac{n}{6} = \dfrac{{25}}{3} \\
n = \dfrac{{25 \times 6}}{3} \\
n = 50 \\
$
C is the correct answer.
Note: In these types of questions, students should try to obtain an equation using the information given in the question and then put it equal to the value that we were given in the question. Students must write all the steps for calculation and not skip them to avoid mistakes.
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