Algebra
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Q.

In an objective examination of 90 questions, 5 marks are allotted for every correct answer and 2 marks are deducted for every wrong answer. After attempting all the 90 questions a students got a total of 387 marks. Find the number of questions that he attempted wrong.

 A.

9

 B.

10

 C.

11

 D.

12

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Solution:
Option(A) is correct

Let the wrong questions be $x$. 

We get the equations:

$(90−x)×5−x×2=387$

$⇒ x=9$

Edit: For an alternative solution to the question, check comment by Sravan Reddy.


(3) Comment(s)


Sravan Reddy
 ()

I generally go other way in calculating marks. Almost similar but slightly different...

=> Total marks $= 90 * 5 = 450$

=> Marks gone per wrong question $= 5+2 = 7$

=> total marks gone $= 450-387 = 63$

So, 9 questions wrong.

Took 30 sec with mind calculation :)



Pooja
 ()

this ques is wrong. and anything multiplied by 2 gives even answer and 387 is not even. correct the question


Shireen
 ()

So are you implying with the given marking scheme its not possible to score 387 marks?

Well lets calculate total marks scored as per the given answer.

Total Questions = 90

So Maximum marks $= 90*5=450$

Now as per the solution, total WRONG answers $= 9$

so total -ve marks = $9*2=18$ (2 marks are deducted for every wrong answer)

Since, student has attempted ALL the 90 questions, total number of Correct answers $= 90-9=81$

Total marks for correct answers $=81*5=405$ (5 marks are allotted for every correct answer)

Now based on the marking scheme,

Total marks scored = Marks for correct ans - Marks for wrong ans

So, Total marks $=405-18=\bo 387$

Apparently THE QUESTION IS RIGHT.

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