Question

In an examination,there were 75 questions.3 marks were awarded for each correct answer,1 mark was deducted for each wrong answer and 1 mark was awarded for each unattempted question.Rayan scored a total of 97 marks in the examination.If the number of unattempted questions was higher than the number of attempted questions,then the maximum number of correct answers that Rayan could have given in the examination is

Answer 1

The correct answer is :24
Let's break down the information given in the problem:
Correct answers:+3 marks
Wrong answers: -1 mark
Unattempted questions:+1 mark
Let's denote the following variables:
C: Number of correct answers
W: Number of wrong answers
U: Number of unattempted questions
We are given the following information:
C+W+U=75(total questions)
3C-W+U=97(total marks)
We also know that U(unattempted questions) is greater than the number of attempted questions(C+W),so U>C+W.
We need to find the maximum value of C(number of correct answers).
From the equations above, we can solve for U in terms of C:
U=97-3C+W
Since U>C+W,we can substitute the value of U:
97-3C+W>C+W
Simplify:
97-3C>C
Subtract C from both sides:
97>4C
Divide by 4:
C<24.25
Since the number of correct answers (C) must be a whole number,the maximum possible value for C is 24.
Therefore,the maximum number of correct answers Rayan could have given in the examination is 24.

Answer 2

Let's assume the following :
x be the number of correct questions.
y be the number of incorrect questions.
z be the number of unattempted questions.
Given :
x + y + z = 75 …… (i)
3x - y + z = 97 …… (ii)
(2) - (1) ⇒ x - y = 11
(1) + (2) ⇒ 2x + z = 86
z > x + y
z > 75 - z
z > 37.5
From the above equation, we can assume that minimum possible value of z is 38.
2x + 38 = 86
2x = 48
x = 24
Therefore , the maximum number of correct questions solved is 24.