In a competitive exam, 3 marks are to be awarded for every correct answer and for every wrong answer, 1 mark will be deducted. Madhu scored 40 marks in this exam. Had 4 marks been awarded for each correct answer and 2 marks deducted for each incorrect answer,Madhu would have scored 50 marks. How many questions were there in the test? (Madhu attempted all the questions)
Solve by a) Substitution Method b) Elimination Method
Answers
Let the number of correct answers = x ,
the number of wrong answers = y
When 3 marks are given for each
correct answer and 1 mark deducted
for each wrong answer , his score is 40
marks .
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.
4x - 2y = 50 ----( 2 )
i ) Substitution method :
from equation ( 1 ) , y = 3x - 40
substitute it in equation ( 2 ) ,
4x - 2( 3x - 40 ) = 50
4x - 6x + 80 = 50
=> -2x = 50 - 80
=> -2x = -30
x = ( -30 )/( -2 )
x = 15
substitute the value of x in equation ( 1 ),
we get
3( 15 ) - y = 40
45 - y = 40
- y = 40 - 45
- y = -5
y = 5
Total number of questions = x + y
= 15 + 5
= 20
ii ) Elimination method :
Do [ 2 × ( 1 ) - ( 2 ) ] we get ,
6x - 2y = 80
4x - 2y = 50
_____________
2x = 30
=> x = 30/2 = 15
substitute x = 15 in equation ( 1 ) ,
3 ( 15 ) - y = 40
45 - y = 40
- y = 40 - 45
- y = -5
y = 5
Therefore ,
total number of questions = x + y
= 15 + 5
= 20
I hope this helps you.
: )
Answer:
Step-by-step explanation:
Let the number of correct answers = x ,
the number of wrong answers = y
When 3 marks are given for each
correct answer and 1 mark deducted
for each wrong answer , his score is 40
marks .
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.
4x - 2y = 50 ----( 2 )
i ) Substitution method :
from equation ( 1 ) , y = 3x - 40
substitute it in equation ( 2 ) ,
4x - 2( 3x - 40 ) = 50
4x - 6x + 80 = 50
=> -2x = 50 - 80
=> -2x = -30
x = ( -30 )/( -2 )
x = 15
substitute the value of x in equation ( 1 ),
we get
3( 15 ) - y = 40
45 - y = 40
- y = 40 - 45
- y = -5
y = 5
Total number of questions = x + y
= 15 + 5
= 20
ii ) Elimination method :
Do [ 2 × ( 1 ) - ( 2 ) ] we get ,
6x - 2y = 80
4x - 2y = 50
_____________
2x = 30
=> x = 30/2 = 15
substitute x = 15 in equation ( 1 ) ,
3 ( 15 ) - y = 40
45 - y = 40
- y = 40 - 45
- y = -5
y = 5
Therefore ,
total number of questions = x + y
= 15 + 5
= 20
I hope this helps you.
: )