# Calculating a “killer question” for a total test score

all -

I'm trying to figure out if this can be done at all, and if so, how; unfortunately, my skill set isn't up to the task, and I'm hoping to find an answer here - or at least guidance on how to approach it.

I need to score a 10-question test that has weighted questions on a pass/fail basis, with 70% as the cut score. So far, so good - sum the total difficulties, multiply by 0.7, then check to see if the candidate's score is above or below (the framework we are using does all this automatically.) However, I now want to add another condition: if the candidate does a certain thing (something I can test for), I need the score to always be a fail (below 70%.)

Changing the scoring framework to do this would be problematic in lots of ways - but treating the condition test as a hidden 11th question would be trivial. The challenge, then, is this:

Given 10 questions with associated difficulty scores (as below), is it possible to calculate a difficulty score for question #11 such that failing it will produce a score below 70% in all cases, but passing it will result in the same pass/fail rate as if q#11 wasn't present? The weight can't be negative.

Hopefully, I've managed to phrase this in a coherent manner; please feel free to ask any questions you need to clarify.

Example of question weights:

1. 2
2. 3
3. 2
4. 2
5. 3
6. 2
7. 1
8. 3
9. 2
10. 3
11. ???
• Are there partial marks for a question or do you always get either 0 or the full mark. – FullofDill Sep 14 '17 at 20:10
• @FullofDill It's pass/fail, no partial scores - so 69.999% is a fail, while 70% is a pass. – bluewater_sailor Sep 14 '17 at 20:13
• The whole test is pass/fail, but what about individual questions? For instance, can I score a '1' on question 1 or only 0/2? – FullofDill Sep 14 '17 at 20:15
• Only 0 or 2 - no partials for individual questions either. – bluewater_sailor Sep 14 '17 at 20:18
• @bluewater_sailor Can you change the threshold? Could you make it 80% instead of 70%? – Théophile Sep 14 '17 at 20:22

No it can't be done. For simplicity we scale all the scores such that the original ones sum to 1. Now we define x to be the weight of question 11 so the new total score is $1 + x$. Then for a student who gets all questions but 11 right to still fail we need $$1 < 0.7(1+x) \iff x > 3/7 \approx 0.43$$ but for a student with 60% on the first 10 questions and question 11 right to also fail we need $$\frac{0.6+x}{1+x} < 0.7 \iff x < 1/3 \approx 0.33$$ a contradiction.

Remark: This assumes it to be possible to score somewhere between 58% and 70% on the original test, which is certainly true for the scores you posted: Answering only the first 6 questions correctly will get you 60%

• THAT is just perfect. I couldn't even figure out how to reason about the numbers here, and you've just demonstrated it. Awesome! – bluewater_sailor Sep 14 '17 at 20:46
• \iff $\iff$ and \implies $\implies$ are usually preferred to those little stub arrows ;) – gen-z ready to perish Sep 14 '17 at 22:04

Note: in what follows I am treating $70$ as an absolute score. I believe the conditions of the post can not be satisfied as written, but if you let $70$ be an absolute score instead of a proportion of the total you can do it.

Let question $11$ be worth $31$ and let the other ten questions sum to $55.7$ (so $5.57$ each).

Then if you miss $11$ your maximum score is $69$ so you are sure to fail.

In order to pass you need question $11$ plus $70\%$ of the rest. Thus the rest has to have value $$100\times \left(1-\frac {.31}{.7}\right)\approx 55.714$$ It follows that, having gotten $11$ right you need to pass the test in the ordinary way.

Note: with these weights, your maximum score is $86.7$. Not sure if that's a problem.

• Is there a way to turn this around? In other words, given an arbitrary set of weights for the different questions, how would I calculate the weight for #11? – bluewater_sailor Sep 14 '17 at 20:29
• Sure. If the total is of the rest is $X$, so $X=55.7$ in my scheme, then the score for $\#11$ is $70\times \left(1-\frac X{100}\right)$ – lulu Sep 14 '17 at 20:32
• Is this the sort of thing you were looking for? – lulu Sep 14 '17 at 20:32
• I'm going to try it out with some score scenarios, but - yes, thank you! That is exactly what I was looking for! – bluewater_sailor Sep 14 '17 at 20:33
• Well, I'm not sure it's what you want. I was working with $70$ as an absolute score, not a ratio. I don't think my approach works with the ratio. – lulu Sep 14 '17 at 20:34