Question

A test has 120 questions. A student gets 4 marks for every correct answer, loses 2 marks for every wrong answer and loses 1 mark for every un-attempted question. If the net score of a student who took that test is 228 marks, then the number of questions he answered correctly could be at the most how much?​

Answer 1

Answer:

What is the number of maximum wrong answers?

Let’s assume, X= no of questions attempted correctly

Y = no of questions attempted incorrectly

Z = no of unattempted questions

As per given conditions, total no of questions is 120. Hence, X+Y+Z=120

Total marks obtained = 228. Hence, 4X-2Y-Z=228

Adding equation 1 and 2,

5X-Y=348

Using above equation we need to find maximum value of Y such that X+Y should not exceed 120.

So, if we put value of X as 70 then we get y =2

If we put X as 80, we get Y as 52, but this will violate condition that X + Y less than or equal to 120

So put X as 79, we get Y as 47, again this will violate condition that X + Y less than or equal to 120

Put X as 78, we will get Y as 42 which satisfies our constraint also( X+Y=120).

Hence Max value of X that is= attempted questions is 78 and max wrong questions = y= 42.

Answer 2

Answer: The maximum number of correct questions can be 78.

Step-by-step explanation:

Let the number of un-attempted questions be x.

Therefore, number of attempted questions = 120 - x

Let the number of correct questions be y.

Therefore, number of incorrect questions = 120 - x - y

Total marks given for correct questions = 4 × y

Total marks given for correct questions =  4y

Total marks given for incorrect questions =  -2 × (120 - x - y)

Total marks given for incorrect questions =  2x + 2y - 240

Total marks given for un-attempted questions  =  -1 × x

Total marks given for un-attempted questions  =  -x

Total marks given to student = 4y + (2x + 2y - 240) + (-x)

Total marks given to student =  6y + x - 240

228 = 6y + x - 240

6y + x  =  468

Constraints :

Number of questions attempted, un-attempted, correct and incorrect are integers .∴  x, y ∈ I.

Number of attempted question > 0 ⇒ x > 0

Number of correct questions > 0 ⇒ y > 0

Number of incorrect questions > 0 ⇒ 120 - x - y > 0 ⇒ x + y < 120

$6y + x = 468$

⇒ $6y = 468 - x$

⇒ $y = \frac{468-x}{6}$

⇒ $y = 78 - \frac{x}{6}$

As x and y are integers, we can choose only multiples of 6 for x.

As x increases, y decreases.

We have to maximize the number of correct questions which is equal to y. Therefore we need to minimize x.

Minimum value of x = 0

$y = 78 - \frac{x}{6}$

y = 78 - 0

y = 78

Checking the constrain x + y < 120 at this solution.

x + y < 120

0 + 78 < 120

78 < 120

The constrain is satisfied. Hence, the maximum number of correct questions can be 78.

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