In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. If she/he attempts all 60 questions and secures 130 marks, the number of questions she/he attempts wrongly, are?
Understanding the Examination Scoring Problem
This problem involves calculating the number of wrong answers a student attempted in an examination. We are given the total number of questions, the marks awarded for each correct answer, the marks deducted for each wrong answer, and the total marks secured by the student.
Setting Up the Equations
Let's define the variables:
- Let $C$ represent the number of questions answered correctly.
- Let $W$ represent the number of questions answered wrongly.
From the problem statement, we know the following:
- The total number of questions attempted is 60. So, the sum of correct and wrong answers is 60: $C + W = 60 \quad (1)$
- The student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. The total marks secured are 130. This can be written as: $(4 \times C) - (1 \times W) = 130$ $4C - W = 130 \quad (2)$
Solving the System of Equations
We now have a system of two linear equations with two variables:
- $C + W = 60$
- $4C - W = 130$
We need to find the value of $W$ (the number of wrong answers).
Step 1: Express C in terms of W
From equation (1), we can express the number of correct answers ($C$) in terms of the number of wrong answers ($W$):
$C = 60 - W$Step 2: Substitute C into the second equation
Now, substitute this expression for $C$ into equation (2):
$4(60 - W) - W = 130$Step 3: Simplify and solve for W
Distribute the 4:
$240 - 4W - W = 130$Combine the $W$ terms:
$240 - 5W = 130$To isolate the term with $W$, subtract 130 from both sides and add $5W$ to both sides:
$240 - 130 = 5W$ $110 = 5W$Now, divide by 5 to find the value of $W$:
$W = \frac{110}{5}$ $W = 22$Conclusion
The number of questions the student attempted wrongly is 22.
Let's verify this: If $W = 22$, then $C = 60 - 22 = 38$. The total score would be $(4 \times 38) - (1 \times 22) = 152 - 22 = 130$. This matches the given total score.
Therefore, the number of questions attempted wrongly is 22.
Important Questions from Algebra (Notes)
- What is the remainder when 2023²⁰²⁴ + 2025²⁰²⁴ is divided by 2024?
Match List-I with List-II
List-1 List-II (A) If $\begin{bmatrix}\lambda-1 & 0 \\ 0 & \lambda-1 \end{bmatrix} $, then $\lambda$ is (I) 0 (B) If A=$ \begin{bmatrix}1 & 2 \\2 & 4 \end{bmatrix} $, then $\Delta$ is (II) 1 (C) If A = $ \begin{bmatrix}1 & 0 \\0 & \frac{1}{2} \end{bmatrix} $, then $|A^{-1}|$ is (III) -2 (D) If $ \begin{bmatrix}a+1 & 1 \\1 & 2 \end{bmatrix} = \begin{bmatrix}-1 & 1 \\1 & 2 \end{bmatrix} $, then a is (IV) 2 Choose the correct answer from the options given below:
- If (x - 1) is a factor of $2x^2 - 5x + k = 0$, then the value of k is:
- If $x = (2+\sqrt{3})^{\frac{1}{3}} + (2+\sqrt{3})^{-\frac{1}{3}}$ and $x^3-3x + k = 0$, then the value of k is:
If \[ \begin{vmatrix} \dfrac{1}{a(b+c)} & \dfrac{1}{b(c+a)} & \dfrac{1}{c(a+b)} \\ \dfrac{bc}{a(b+c)} & \dfrac{ca}{b(c+a)} & \dfrac{ab}{c(a+b)} \\ 1 & 1 & 1 \end{vmatrix} = k, \text{ then the value of } k \text{ is:} \]
