Guess or leave it blank: negative-marking calculator for Mathematik (CBK) at WU

In the Mathematik exam of the CBK at WU Vienna, part of the bachelor's programme in Business, Economics and Social Sciences, you decide for every question whether to tick an answer or leave it blank. Wrong answers cost points, but only beyond a certain number, and a well-prepared student can guess their way below the pass mark.

The scoring rule is in the course syllabus, which is in German like the exam itself. You can practise with WU's exercises on Canvas, WU's learning platform, in the forum there, and at the walk-in tutorial WU runs for exam preparation. This calculator shows, for your own situation, what guessing earns on average and how often it costs you the pass.¹

The negative-marking rule of the Mathematik (CBK) exam

The exam has 20 questions with 5 options each, exactly one of them right. A right answer earns 1 point, a wrong one costs 0.25 points, a blank counts 0. WU counts your first 3 wrong answers like blanks, so they cost nothing. You need 11 points to pass.¹

In points: right answers minus 0.25 times the wrong answers from the fourth on. With 11 right and 4 wrong you have 11 − 0.25 = 10.75 and fail. With 12 right and 7 wrong you have 12 − 4 · 0.25 = 11 and pass. The syllabus does not say how points above 11 translate into grades.

If you sat WU's entrance test, forget its rule here. There, questions with several right options give partial credit and a question never goes below zero, as our guide to partial credit shows.² The Mathematik exam has no partial credit, and wrong answers are counted across the whole exam, not per question.

What a guess is worth in the Mathematik (CBK) exam

From here on this is our own calculation from the syllabus rule, not advice from WU. A blind guess among all five options is right one time in five. While your three free wrong answers last, it costs nothing and earns 0.2 points on average. After that it earns 0.2 · 1 − 0.8 · 0.25 = 0 points on average: from the fourth wrong answer on, blind guessing breaks even.

Our calculation

Our calculation from the rule in the syllabus: if you rule out k of the 5 options for certain, a guess is right with probability p = 1/(5 − k). While the free wrong answers last, a miss costs nothing and the guess is worth p · 1 on average. From the fourth wrong answer on a miss costs 0.25: p · 1 − (1 − p) · 0.25.

ruled out	chance of a hit	wrong 1–3 free	from 4th wrong
0	1/5	0.2	0
1	1/4	0.25	0.0625
2	1/3	≈ 0.33	≈ 0.17
3	1/2	0.5	0.375

Blind guess, from the fourth wrong answer on: 1/5 · 1 − 4/5 · 0.25 = 0.2 − 0.2 = 0.

Source of the rule: WU Vienna, syllabus of the course Mathematik (CBK), course catalogue (in German): „Pro richtiger Antwort gibt es 1 Punkt, für eine falsche Antwort werden 0,25 Punkte abgezogen, eine Enthaltung wird mit 0 Punkten bewertet. Hierbei werden die ersten 3 falschen Antworten wie Enthaltungen gezählt und führen damit zu keinem Punktabzug.“

Every option you can rule out for certain tips the calculation in your favour. With four options left, a guess after the free wrong answers is worth 1/16 of a point on average, with three 1/6, with two 3/8. On average, then, guessing never loses points. Close to the pass mark, though, what counts is how often you reach 11, and that is a different calculation.

Negative-marking calculator for Mathematik (CBK)

Enter how many questions you are sure of, how many you guess on, and how many options you can rule out on each guessed question. The calculator shows your expected points, your chance of at least 11 points and every possible result. The preset is 11 sure answers and 9 blind guesses.

Negative-marking calculator

Mathematik exam (CBK): 20 questions, pass mark 11 points

Negative-marking calculator

Answers you’re sure of01234567891011121314151617181920Guesses01234567891011121314151617181920Ruled out per guessnone1 of 52 of 53 of 5

Of the 9: guess 3, leave 6 blank. With 3 guesses you pass for certain; with all 9 there is a 43.6% chance you drop below 11 points.

With 9 guesses:on average 11.75 points (without guessing 11)Chance of passing 56.4% (without guessing 100%)

How the calculator works

Per guess: no option ruled out, chance of a hit 1/5. While the 3 free wrong answers last, a guess earns 1/5 point on average and costs nothing. After that: 1/5 · 1 − 4/5 · 0.25 = 0 points.

Worst case (every guess wrong): 9.5 points. Best case (every guess right): 20 points.

hits	points	probability
6	17	0.3%
5	15.75	1.7%
4	14.5	6.6%
3	13.25	17.6%
2	12	30.2%
1	10.75	30.2%
0	9.5	13.4%

Chance of passing by number of guesses:

guesses	0	1	2	3	4	5	6	7	8	9
chance	100%	100%	100%	100%	59.0%	67.2%	73.8%	79.0%	83.2%	56.4%

The calculator counts sure answers as right. If one of them is wrong, it uses up one of the three free wrong answers. Every number is our calculation from the rule in the WU syllabus: +1 per right answer, −0.25 per wrong answer from the fourth on, pass from 11 points.

The calculator counts your sure answers as right. If one of them is wrong, it uses up one of the three free wrong answers, and every guess after that gets more expensive.

When guessing costs you the pass in Mathematik (CBK)

With 11 sure answers you pass as long as you do not guess. If you guess blind on the other 9 questions, your expected points rise to 11.75, but there is a 43.6% chance that you fail: hit none or only one of the 9 and you are left with 9.5 or 10.75 points.

With s sure answers, s at least 11, you keep 11 points even if every guess is wrong as long as you guess on no more than 3 + 4 · (s − 11) questions. That is 3 guesses with 11 sure answers, 7 with 12, and every remaining question from 13 on.

Below 11 sure answers it turns around: without guessing you fail. But more guesses are not automatically better. With s sure answers and g guesses you need enough hits for s plus hits minus 0.25 times the wrong answers from the fourth on to reach 11. Every five extra guesses therefore demand one more hit, and that is exactly where the chance drops.

An example with 10 sure answers and blind guesses, by our calculation: with 4 guesses one hit is enough and your chance of passing is 59.0%. With 5 guesses you need two hits and the chance falls to 26.3%. With 9 guesses it climbs back to 56.4%, with 10 it falls to 32.2%. If you can rule out two options on every question, 4 guesses give you 80.2%, 6 only 64.9% and 9 give 85.7%. The calculator tells you for every input whether fewer guesses would be better.

You have no calculator in the exam room, but the rule is short. First count your sure answers. From 11 on, guess at most as often as the formula above allows. Below 11, guess on every remaining question except one: you then need exactly 12 minus s hits, and guessing the last question as well always demands one more. By our calculation that is the best number of guesses for 6 to 9 sure answers, and for 10 as soon as you can rule out at least one option per question. Only with 10 sure answers and blind guesses are 4 guesses slightly better, 59.0% against 56.4%. Only you know how sure your sure answers really are, which is why it pays to measure it while you practise.

Three practice questions in the Mathematik (CBK) format

The three questions are our own, in the format of the exam: five options, exactly one right. Each wrong option comes from a typical mistake. The solution shows what each choice does to your score, and how to rule options out before you finish calculating. Unlike the exam, they are in English here.

Practice question 1Elementary financial mathematics

A student pays 1,000 MU (monetary units) into a savings account at the end of each year for 8 years; the account pays 3% p.a. What is the balance right after the last payment (rounded)?

1. a)1,266.77 MU
2. b)8,000.00 MU
3. c)8,892.34 MU
4. d)9,159.11 MU
5. e)10,159.11 MU

Exactly one option is right.

Solution and points for every choice+

a)

Wrong. Only one payment compounded for 8 years: 1,000 · 1.03⁸.

0 or −0.25

b)

Wrong. The payments added up without interest: 8 · 1,000.

0 or −0.25

c)

Right. Future value of an ordinary annuity (payments at the end of each period): 1,000 · (1.03⁸ − 1) / 0.03 ≈ 8,892.34. The last payment has earned no interest yet.

+1

d)

Wrong. Calculated as an annuity due, as if each payment had been in the account one year longer: 8,892.34 · 1.03.

0 or −0.25

e)

Wrong. Calculated with 9 payments instead of 8: 1,000 · (1.03⁹ − 1) / 0.03.

0 or −0.25

–

Left blank.

0

A wrong answer counts 0 while it is one of your first three wrong answers in the exam, −0.25 after that.

Ruling out before you finish. With interest the balance must exceed the sum of the payments, 8,000. That rules out a and b and leaves three.

A guess among the 3 left has a 1-in-3 chance. It is worth ≈ 0.33 points on average while your three free wrong answers last, and ≈ 0.17 after that.

Practice question 2Differential calculus

Demand for a product is q(p) = 200 − 4p, with the price p in MU and the quantity q in units. What is the price elasticity of demand at the price p = 30?

1. a)−4
2. b)−1.5
3. c)−0.67
4. d)−0.05
5. e)−120

Exactly one option is right.

Solution and points for every choice+

a)

Wrong. Only the derivative q′(p) = −4, without the factor p/q.

0 or −0.25

b)

Right. ε = q′(p) · p / q(p) = −4 · 30 / 80 = −1.5. At p = 30, demand is 200 − 120 = 80 units.

+1

c)

Wrong. The fraction upside down: q / (q′ · p) = 80 / (−120).

0 or −0.25

d)

Wrong. Without the price: q′ / q = −4 / 80.

0 or −0.25

e)

Wrong. Not divided by the quantity: q′ · p = −4 · 30.

0 or −0.25

–

Left blank.

0

A wrong answer counts 0 while it is one of your first three wrong answers in the exam, −0.25 after that.

Ruling out before you finish. With linear demand, demand is elastic above half the maximum price, so the absolute value of the elasticity is greater than 1. The maximum price here is 50 (where q = 0), half of it 25, and 30 is above. That rules out c and d.

A guess among the 3 left has a 1-in-3 chance. It is worth ≈ 0.33 points on average while your three free wrong answers last, and ≈ 0.17 after that.

Practice question 3Probability

In a course, 60% of the participants handed in every exercise sheet. Those who did pass the exam with probability 70%, the others with probability 40%. A randomly chosen person has passed. What is the probability that they handed in every exercise sheet (rounded)?

1. a)0.42
2. b)0.58
3. c)0.60
4. d)0.72
5. e)0.70

Exactly one option is right.

Solution and points for every choice+

a)

Wrong. The probability of “handed in and passed”: 0.6 · 0.7.

0 or −0.25

b)

Wrong. The probability of passing: 0.6 · 0.7 + 0.4 · 0.4.

0 or −0.25

c)

Wrong. The share before knowing “passed”, the condition ignored.

0 or −0.25

d)

Right. Bayes’ theorem: 0.42 / 0.58 ≈ 0.724. Of everyone who passes (0.58), 0.42 handed in.

+1

e)

Wrong. The condition reversed: P(passed | handed in) instead of P(handed in | passed).

0 or −0.25

–

Left blank.

0

A wrong answer counts 0 while it is one of your first three wrong answers in the exam, −0.25 after that.

Ruling out before you finish. Those who hand in pass more often, so knowing “passed” makes “handed in” more likely than the 0.60 from before. That rules out a, b and c and leaves two.

A guess among the 2 left has a 1-in-2 chance. It is worth 0.5 points on average while your three free wrong answers last, and 0.375 after that.

In the third question two options are left after ruling out. A guess between them is worth 3/8 of a point on average after the free wrong answers, against 0 for a blind guess.

How to practise ruling out options for Mathematik (CBK)

1. Sit a mock exam of 20 questions against the clock. For each answer, note whether you were sure, guessed, or ruled options out.
2. Score it with the exam's rule and count separately: how many sure answers were wrong, how many guesses hit. If your sure answers are right less often than you thought, plan fewer guesses for the exam.
3. Practise ruling out on questions you cannot finish: sign, order of magnitude, bounds, as in the three questions above.
4. Put your numbers into the calculator and decide how many guesses you will go up to in the exam.

What else the exam asks, which topics it covers and when it takes place in 2026/27 is in our overview of the Mathematik exam in the CBK. The unipass course for Mathematik (CBK) opens before the January 2027 exam week, with mock exams planned that are scored exactly this way.

Sources

1. 1Syllabus 0003 Mathematik (LVP), Wintersemester 2026/27WU Wien, Vorlesungsverzeichnis · undated page · checked on 27 September 2026
2. 2Informationen zum gemischten Teilpunktesystem, Aufnahmeprüfung WISO (PDF)WU Wien · document dated 8 May 2026 · checked on 26 September 2026