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Hint: Take ‘y’ as the number of unattempted questions and try to find inequality or relation between them. Then find the conditions for which ‘x’ and ‘y’ should satisfy. Then tdo hit and trial method and find out all the values of x and y and count them to get the desired results.

“Complete step-by-step answer:”

We are given a total of 30 questions with certain conditions which are:

(i) If a question is left unattempted by the student, then the student gets 1 mark.

(ii) If a question is correctly marked by the student, then the student gets 4 marks.

(iii) If a question marked by the student is wrong, then the student gets 0 marks.

Now the information is given that the students has marked ‘x’ number of questions correctly and got 60 marks.

Now let us consider ‘y’ be the number of unattempted questions by the student, so the number of questions answered wrongly are $\left( 30-x-y \right)$ as total number of question were 30

So, total marks scored by a student is in form of x, y is $\left\{ \left( 4\times x \right)+\left( 1\times y \right)+0\times \left( 30-x-y \right) \right\}$which is equal to $\left( 4x+y \right).$

In the question total marks given is 60

So, we can say that,

$4x+y=60.........\left( 1 \right)$

By further modification of equation (i) we can say,

$\begin{align}

& y=60-4x \\

& \therefore y=4\left( 15-x \right)........\left( ii \right) \\

\end{align}$

Other conditions or relations about ‘x’ and ‘y’ we can say that $x\le 30,y\le 30$and $x+y\le 30$ as the total number of correctly answered questions and the number of unattended questions will be less than or equal to 30.

From equation (ii) we can say that ‘x’ will always be less than or equal to 15 because ‘y’ cannot be a negative.

So, now we will check values of ‘x’ for all the numbers less than 15 or equal to 15, until the condition $x+y\le 30$ is satisfied for equation (ii).

For x = 15,

$\begin{align}

& y=4\left( 15-x \right) \\

& \Rightarrow y=4\left( 15-15 \right)=0 \\

\end{align}$

And,

$\begin{align}

& x+y\le 30 \\

& \Rightarrow 15+0=15\le 30 \\

\end{align}$

This is true, hence it satisfies.

For $x=14$ ,

$\begin{align}

& y=4\left( 15-x \right) \\

& \Rightarrow y=4\left( 15-14 \right)=4 \\

\end{align}$

And $x+y\le 30$

$14+14=18\le 30$ which is true, hence it satisfies.

For $x=13$ ,

$\begin{align}

& y=4\left( 15-x \right) \\

& \Rightarrow y=4\left( 15-13 \right)=8 \\

\end{align}$

And $x+y\le 30$

$13+8=21\le 30$ which is true, hence it satisfies.

For $x=12$ ,

$\begin{align}

& y=4\left( 15-x \right) \\

& \Rightarrow y=4\left( 15-13 \right)=12 \\

\end{align}$

And $x+y\le 30$

$12+12=24\le 30$ which is true, hence it satisfies.

For $x=11$ ,

$\begin{align}

& y=4\left( 15-x \right) \\

& \Rightarrow y=4\left( 15-11 \right)=16 \\

\end{align}$

And $x+y\le 30$

$11+16=27\le 30$ which is true, hence it satisfies.

For $x=10$ ,

$\begin{align}

& y=4\left( 15-x \right) \\

& \Rightarrow y=4\left( 15-10 \right)=20 \\

\end{align}$

And $x+y\le 30$

$10+20=30\le 30$ which is true, hence it satisfies.

For $x=9$ ,

$\begin{align}

& y=4\left( 15-x \right) \\

& \Rightarrow y=4\left( 15-9 \right)=24 \\

\end{align}$

And $x+y\le 30$

$9+24\le 30$ which is not true, hence it does not satisfy.

For all the numbers $x\le 9$ the condition $x+y\le 30$ does not satisfy.

So only $x$ = 15, 14, 13, 12, 11, 10 satisfies. So, the number of possible values for $x$ is 6

Hence the correct answer is option (c).

Note: Students must be careful while finding the relation or forming the equation between ‘x’ and ‘y’ . They should also be cautious about the conditions for which $x$ and $y$ should satisfy. The inequality must be formed correctly for the problem to be correct. They should also check cases one by one to avoid any problems. For these types of questions students must not go for three variables simultaneously, rather they should stick for two variables only. While finding the number of values, they should check whether the condition satisfies all the conditions or not.

“Complete step-by-step answer:”

We are given a total of 30 questions with certain conditions which are:

(i) If a question is left unattempted by the student, then the student gets 1 mark.

(ii) If a question is correctly marked by the student, then the student gets 4 marks.

(iii) If a question marked by the student is wrong, then the student gets 0 marks.

Now the information is given that the students has marked ‘x’ number of questions correctly and got 60 marks.

Now let us consider ‘y’ be the number of unattempted questions by the student, so the number of questions answered wrongly are $\left( 30-x-y \right)$ as total number of question were 30

So, total marks scored by a student is in form of x, y is $\left\{ \left( 4\times x \right)+\left( 1\times y \right)+0\times \left( 30-x-y \right) \right\}$which is equal to $\left( 4x+y \right).$

In the question total marks given is 60

So, we can say that,

$4x+y=60.........\left( 1 \right)$

By further modification of equation (i) we can say,

$\begin{align}

& y=60-4x \\

& \therefore y=4\left( 15-x \right)........\left( ii \right) \\

\end{align}$

Other conditions or relations about ‘x’ and ‘y’ we can say that $x\le 30,y\le 30$and $x+y\le 30$ as the total number of correctly answered questions and the number of unattended questions will be less than or equal to 30.

From equation (ii) we can say that ‘x’ will always be less than or equal to 15 because ‘y’ cannot be a negative.

So, now we will check values of ‘x’ for all the numbers less than 15 or equal to 15, until the condition $x+y\le 30$ is satisfied for equation (ii).

For x = 15,

$\begin{align}

& y=4\left( 15-x \right) \\

& \Rightarrow y=4\left( 15-15 \right)=0 \\

\end{align}$

And,

$\begin{align}

& x+y\le 30 \\

& \Rightarrow 15+0=15\le 30 \\

\end{align}$

This is true, hence it satisfies.

For $x=14$ ,

$\begin{align}

& y=4\left( 15-x \right) \\

& \Rightarrow y=4\left( 15-14 \right)=4 \\

\end{align}$

And $x+y\le 30$

$14+14=18\le 30$ which is true, hence it satisfies.

For $x=13$ ,

$\begin{align}

& y=4\left( 15-x \right) \\

& \Rightarrow y=4\left( 15-13 \right)=8 \\

\end{align}$

And $x+y\le 30$

$13+8=21\le 30$ which is true, hence it satisfies.

For $x=12$ ,

$\begin{align}

& y=4\left( 15-x \right) \\

& \Rightarrow y=4\left( 15-13 \right)=12 \\

\end{align}$

And $x+y\le 30$

$12+12=24\le 30$ which is true, hence it satisfies.

For $x=11$ ,

$\begin{align}

& y=4\left( 15-x \right) \\

& \Rightarrow y=4\left( 15-11 \right)=16 \\

\end{align}$

And $x+y\le 30$

$11+16=27\le 30$ which is true, hence it satisfies.

For $x=10$ ,

$\begin{align}

& y=4\left( 15-x \right) \\

& \Rightarrow y=4\left( 15-10 \right)=20 \\

\end{align}$

And $x+y\le 30$

$10+20=30\le 30$ which is true, hence it satisfies.

For $x=9$ ,

$\begin{align}

& y=4\left( 15-x \right) \\

& \Rightarrow y=4\left( 15-9 \right)=24 \\

\end{align}$

And $x+y\le 30$

$9+24\le 30$ which is not true, hence it does not satisfy.

For all the numbers $x\le 9$ the condition $x+y\le 30$ does not satisfy.

So only $x$ = 15, 14, 13, 12, 11, 10 satisfies. So, the number of possible values for $x$ is 6

Hence the correct answer is option (c).

Note: Students must be careful while finding the relation or forming the equation between ‘x’ and ‘y’ . They should also be cautious about the conditions for which $x$ and $y$ should satisfy. The inequality must be formed correctly for the problem to be correct. They should also check cases one by one to avoid any problems. For these types of questions students must not go for three variables simultaneously, rather they should stick for two variables only. While finding the number of values, they should check whether the condition satisfies all the conditions or not.

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