IPMAT/Algebra•12mos
solulu

Minima Maxima

There are 80 questions in a test. Each correct answer fetches 1 mark, each wrong answer & unanswered question attract a penalty of 1/4 mark & 1/8 mark respectively each. Frodo scored 23 marks in the test. What is the minimum possible number of questions wrongly answered by him?

I have no idea how to tackle these type of "common sense/ no formula" type questions.

Is there any advice on these types of questions? theyre kind of like minima maxima.

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Comments (2)

M
meow_woof
•12mos
Correct

Let number of correct, wrong and unanswered question be x, y, z respectively.

Then marks is given by x-y/4-z/8

That is 1/8[8x- 2y - z]

This is equal to 23

1/8[8x-2y-z] = 23

8x-2y-z = 184 ....(i)

Also as per the question, x + y + z = 80

This is 2x + 2y + 2z = 160 ...(ii)

Now (ii) + (i)

= 10 x + z = 344

=> z = 344 - 10x

Again putting it in (i)

8x - 2y -344 + 10x = 184

18 x - 2y =528

9 x - y = 264

=> 9 x = 264+ y

We know that x and y are whole numbers (number of question cannot be negative nor can they be in fractions)

We can just put all the minimum value of y to find the solution as this is prettu simple equation.

at y = 0, x = 264/9 that is 29.3 Not accepted

at y = 1, x= 265/9 that is 29.4 not accepted.

Looking at the pattern put y = 6 and we get x = 270/9 = 30

So minimum value of y is 6.

To check if this is the solution put the values of x, y and z in original equation. x = 30 y = 6 z = 344 -10 x =44

x - y/4 - z/8

30 - 6/4 - 44/8

= 23.

This question is mostly about constructing the equations.

2
S
solulu
OP•12mos

Okay thank you

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