In an examination, the Data Insights section has 20 questions. Each correct answer awards 3 points, while each incorrect answer results in a 1-point deduction, and there is no penalty for unanswered questions. During his only attempt, Paul scored 35 points on this section. Based on the above information, select for Unattempted the maximum number of questions that Paul could have left unanswered and for Incorrect the maximum number of questions that Paul could have answered incorrectly and still scored 35 points. Make only two selections, one in each column.

I solved it as follows => 3x -y + 0*z = 35 & x+y+z = 20 Now x = 20-y-z 3x= 35+y We eliminate x since X is not asked 35-y = 20-y-z => 15 = 2y-z Since 2y is even Z can only be odd Look for Z values as Odd from option and see corresponding y values we get values satisfying as (y,z) is (4,3) and (1,7) => Max value of Z possible is 7 and max value of y is 4.

I like the solution - it’s helpful. how would one figure out if the cases being asked for are separate in the exam ?

I like the solution - it’s helpful.

I made a structure out of this question - and I am posting so that I don't forget and others can also benefit. see 3c - incorrect + 0* unattempt = 35 & C + Inc + Unat = 20 No 3C + 3Inc + 3Unat = 60 => 4In + 3Un = 25 ------ Now this is the guiding eqn and you have to find pairs of inc and unattempted which will make it right eg -> 1in * 7un satisfies solution - 7 unattempt is max 4 incorrect and 3 unattempt is also the max solution for incorrect

I don’t quite agree with the solution. 11 questions correctly would be 33 points which does not amount to. 35

That’s exactly why the solution says the minimum correct must be 12 , not 11. If Paul got 11 correct, then 11 * 3 = 33. Even if he left the other 9 blank (no penalty), his score would still be 33, not 35. To reach 35, he must cross 35 and then come back down if needed. With 12 correct we get 12 * 3 = 36. Then, if 1 is wrong, he loses 1 point, hence 35. So 12 correct and 1 incorrect is the only way to hit 35 with the minimum correct count, which then leaves the maximum of 7 unattempted. Review again!

This is a great question that’s helpful for learning.

I like the solution - it’s helpful.

I don’t quite agree with the solution. If paul answered 13 questions correctly and 4 incorrectly that would mean he unattempted 3 questions but in the answer it is mentioned 7 is the correct.

That’s because those are two different parts of the question. The case with 13 correct and 4 incorrect (3 unattempted) gives the maximum number of incorrect questions Paul could have while still scoring 35. The case with 12 correct and 1 incorrect (7 unattempted) gives the maximum number of unattempted questions possible while still scoring 35. Each part is optimized separately, not both at once.