SOLUTION: in competiative examination, there were 60 questions, the correct answer would carry 2 marks and for incorrect 1 mark would be subtracted. yash had attempted all the questions and

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Question 1095782: in competiative examination, there were 60 questions, the correct answer would carry 2 marks and for incorrect 1 mark would be subtracted. yash had attempted all the questions and he got total 90 marks. then how many questions he got wrong?
Found 2 solutions by KMST, MathTherapy:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
THE FIFTH-GRADER SOLUTION:
With all the answers right, a student would get
2%2A60=120 marks.
Each wrong answer makes a student lose 2%2B1=3 marks,
2 because he did not choose a right answer,
and 1 more because he chose a wrong answer.
A student answered every question and lost 120-90=30 of the possible marks
must have %2230+%F7+3%22=10 wrong answers.
He answered the other 60-10=50 questions correctly.

THE ALGEBRA CLASS SOLUTION:
r= number of right answers
w= number of wrong answersin
As all questions were answered,
r%2Bw=60 .
The marks for tthe test are
2r-w=90

Then, you have to solve the system of linear equations system%28r%2Bw=60%2C2r-w=90%29 .
There are many ways to do that.

COMBINATIONS:
You could do
2r-w=90
%22%2B%22
%22+%22r%2Bw=60
%22---------%22
3r %22=%22150 ---> %22r+=+150+%F7+3%22 --->r=50
Then you could just substitute that value for r in r%2Bw=60
to get 50%2Bw=60 ---> w=60-50 ---> w=10 .
Otherwise, you could keep doing combinations of equations,
and multiply
2r-w=90 times (-1) to get -2r%2Bw=-90 ,
r%2Be=60 times 2 to get 2r%2B2w=1200 ,
and add them up to get
-2r%2Bw=-90
%22%2B%22
2r%2B2w=120
%22___________%22
%22+%22 %22+%22 %22+%223w=30 ---> %22w+=+30+%F7+3%22 ---> w=10
Teachers call that adding up of equations a "method" to solve systems of linear equations by making making combinations of the equations, often calling it the "combinations method." Your teacher may use another name.
Why it works and is reasonable:
If r%2Bw and 60 are the same number,
and adding the same number to both sides of 2r-w=90 we get an equivalent equation,
we can add r%2Bw to the left side and 60 to the right side,
to get the "combination" equation
3r=90%2B60 .
We could also multiply both sides of r%2Bw=60 times (-2) and add them to 2r-w=90

SUBSTITUTION:
What teachers call "substitution method"
requires solving one of the equations for one of the variables,
and then substituting the expression found into the other equation.
You choose which equation to solve and which variable to solve for.
There are 4 possible choices.
One of them is shown below.

2r-w=90 2r=90%2Bw 2r-90=w

system%28r%2Bw=60%2Cw=2r-90%29 r%2B2r-90=60 3r-90=603r=60%2B903r=150r=150%2F3r=50

Answer by MathTherapy(10557) About Me  (Show Source):
You can put this solution on YOUR website!

in competiative examination, there were 60 questions, the correct answer would carry 2 marks and for incorrect 1 mark would be subtracted. yash had attempted all the questions and he got total 90 marks. then how many questions he got wrong? 
Let number of wrong answers be W

The number of correct answers = 60 - W

We then get the following equation: - 1(W) + 2(60 - W) = 90

Solve this for W, or the number of wrong answers. You should get 10.

That's all. Nothing COMPLEX and/or CONFUSING.