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In a competitive exam, 3 marks are to be awarded for every correct answer and for every wrong answer, 1 mark will be deducted. Stephen scored 40 marks in this exam. Had 4 marks been awarded for each correct answer and 2 marks deducted for each incorrect answer,Stephen would have scored 50 marks. How many questions were there in the test ?
Solve by -
Substitution Method
Quick !
Answers
Step-by-step explanation:
Let the number of correct answers be “x”
And the number of wrong answers Be “y”
When 3 marks are given for each correct answer and 1 mark deducted for each wrong answer ,his score Is 40 marks
So 3x-y = 40 —(1)
His score would have been 50 marks if 4 marks were given for each correct answer and 2 marks deducted for each wrong answer .
So., 4x-2y = 50–(2)
From equation (1), y = 3x-40
Substituting Eq(1) in Eq(2)
= 4x -2(3x-40)= 50
=4x-6x+80=50
=-2x = 50-80
=-2x = -30
=x = -30/-2
=>
Substitute the value of d in Eq(1)
3(15) -y = 40
45-y = 40
y = 45-40
y = 5
Therefore ,Total number of question = 15+5 =
• Let 'M' be the number for every correct answer.
And
• 'N' be the number for every wrong answer.
• Total number of questions = M + N
» In a competitive exam, 3 marks are to be awarded for every correct answer and for every wrong answer, 1 mark will be deducted. Stephen scored 40 marks in this exam.
A.T.Q.
→ 3M - N = 40
→ - N = 40 - 3M
→ N = 3M - 40 _______ (eq 1)
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» 4 marks been awarded for each correct answer and 2 marks deducted for each incorrect answer,Stephen would have scored 50 marks.
A.T.Q.
→ 4M - 2N = 50
On taking two common from both sides, we get;
→ 2M - N = 25
→ 2M - (3M - 40) = 25 [From (eq 1)]
→ 2M - 3M + 40 = 25
→ - M + 40 = 25
→ - M = 25 - 40
→ - M = - 15
→ M = 15
Stephen gave 15; correct answers.
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Put value of M in (eq 1)
→ N = 3(15) - 40
→ N = 45 - 40
→ N = 5
He/she gave 5 wrong answers.
_____________________________
So,
Total number of questions = Correct answers + wrong answers
=> 15 + 5
=> 20
_____________________________
20 questions were there in the test.
_________ [ ANSWER ]
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