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# Wrong = % Right (Posted on 2007-06-16) Difficulty: 2 of 5
I finished grading tests for two classes when I noticed something interesting about the worst scoring student for each.

In the first class a student who got 46 problems wrong got 46 percent right (to the nearest whole percent.)
How many problems were there on this test?

In the second class there were 35 problems but again the lowest scoring student got the same number wrong as percent right (again rounded.)
How many questions did this student get wrong?

What is the smallest number of questions for which this kind of thing can happen?
(Aside from the trivial one question test.)

What if the percent must be exact instead of rounded?

[Note: This actually happened.]

See The Solution Submitted by Jer    
Rating: 3.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution To The Third Part : Approximate Case | Comment 5 of 6 |
(In reply to Solution To The First Two Parts by K Sengupta)

Let the total number of questions be B and the number of
questions answered incorrectly be A.

Then, by the problem:

(i) A< B
(ii) (B-A) = Round(100*(A/B))

If possible, let A=1. Then, from  (ii), we obtain:
B - 1.5 < = 100/B < B-0.5
Or, 2*B^2 - 3B < =200, and 2*B^2 - B > 200. No positive integer B can satisfy both the above relationships imultaneously. Accordingly, no solution is possible for A = 1.

If possible, let A=2.

Then, from (ii), we obtain:

B - 2.5 < = 100/B < B-1.5
Or,  2*B^2 - 5B < =400, and 2*B^2 - 3B > 400.

The only positive integer value of B satisfying both the above relationships occur at B=15.

Consequently, the smallest total number of questions for which this can happen is 15 (with the number of questions answered incorrectly being 2).

Edited on July 12, 2007, 12:25 pm
  Posted by K Sengupta on 2007-06-17 05:24:40

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