Answer

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Hint: The question is related to the linear equation in two variables. Try to make two equations using the information given in the problem statement and solve them simultaneously.

In the question, it is given that Yash scored \[40\] marks in a test, getting \[3\] marks for each right answer and losing \[1\] mark for each wrong answer. So, we will consider $x$ as the total number of questions that were answered correctly by Yash and $y$ as the total number of answers that were answered incorrectly.

Now, in the first case, it is given that for each correct answer $3$ marks are awarded. So, for $x$ correct answers, total marks awarded are equal to $3\times x=3x$ . Also, for each wrong answer, \[1\] mark is deducted. So, for $y$ wrong answers, total marks deducted is equal to $1\times y=y$ . So, the total marks obtained will be $3x-y$ . But it is given that the total marks obtained are \[40\]. So. $3x-y=40..........(i)$

Now, in the second case, it is given that for each correct answer $4$ marks are awarded. So, for $x$ correct answers, total marks awarded are equal to $4\times x=4x$ . Also, for each wrong answer, \[2\] mark is deducted. So, for $y$ wrong answers, total marks deducted is equal to $2\times y=2y$ . So, the total marks obtained will be $4x-2y$ . But it is given that the total marks obtained are \[50\]. So. $4x-2y=50..........(ii)$

Now, we will solve the linear equations to find the values of $x$ and $y$ .

From equation$(i)$ , we have $3x-y=40$

$\Rightarrow y=3x-40$

On substituting $y=3x-40$ in equation$(ii)$ , we get

$4x-2(3x-40)=50$

$\Rightarrow 4x-6x+80=50$

$\Rightarrow 80-2x=50$

$\Rightarrow 2x=80-50$

$\Rightarrow 2x=30$

$\Rightarrow x=15$

Now, on substituting $x=15$ in the equation$(i)$ , we get

$(3\times 15)-y=40$

$\Rightarrow 45-y=40$

$\Rightarrow y=5$

So, the total number of questions in the test is equal to$15+5=20$.

Note: While solving the equations, make sure that the substitutions are done correctly and no sign mistakes are present. Sign mistakes can cause the final answer to be wrong.

In the question, it is given that Yash scored \[40\] marks in a test, getting \[3\] marks for each right answer and losing \[1\] mark for each wrong answer. So, we will consider $x$ as the total number of questions that were answered correctly by Yash and $y$ as the total number of answers that were answered incorrectly.

Now, in the first case, it is given that for each correct answer $3$ marks are awarded. So, for $x$ correct answers, total marks awarded are equal to $3\times x=3x$ . Also, for each wrong answer, \[1\] mark is deducted. So, for $y$ wrong answers, total marks deducted is equal to $1\times y=y$ . So, the total marks obtained will be $3x-y$ . But it is given that the total marks obtained are \[40\]. So. $3x-y=40..........(i)$

Now, in the second case, it is given that for each correct answer $4$ marks are awarded. So, for $x$ correct answers, total marks awarded are equal to $4\times x=4x$ . Also, for each wrong answer, \[2\] mark is deducted. So, for $y$ wrong answers, total marks deducted is equal to $2\times y=2y$ . So, the total marks obtained will be $4x-2y$ . But it is given that the total marks obtained are \[50\]. So. $4x-2y=50..........(ii)$

Now, we will solve the linear equations to find the values of $x$ and $y$ .

From equation$(i)$ , we have $3x-y=40$

$\Rightarrow y=3x-40$

On substituting $y=3x-40$ in equation$(ii)$ , we get

$4x-2(3x-40)=50$

$\Rightarrow 4x-6x+80=50$

$\Rightarrow 80-2x=50$

$\Rightarrow 2x=80-50$

$\Rightarrow 2x=30$

$\Rightarrow x=15$

Now, on substituting $x=15$ in the equation$(i)$ , we get

$(3\times 15)-y=40$

$\Rightarrow 45-y=40$

$\Rightarrow y=5$

So, the total number of questions in the test is equal to$15+5=20$.

Note: While solving the equations, make sure that the substitutions are done correctly and no sign mistakes are present. Sign mistakes can cause the final answer to be wrong.

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