A student took some exams. The arithmetical average of his grades is 25. He takes another exam today and he gets a 30, so the arithmetical average of his grades becomes 26. How many exams did he take, including today’s one?

Edit: I know how to solve this, it’s from a book, I found it interesting and I posted it for you!


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  • 1
    $\begingroup$ sorry to close your question! Unfortunately we're quite strict around here about puzzles being puzzles rather than routine mathematics problems; this one is basically a matter of writing down an equation and solving it. Take a look at xnor's answer to the meta question linked in the close message, which lays out the key distinction very nicely. $\endgroup$ – Gareth McCaughan May 21 at 13:06
  • $\begingroup$ @GarethMcCaughan don’t worry, I didn’t know. Thanks for telling me! $\endgroup$ – Marybnq May 21 at 14:40




Let he gave n exams earlier Then total initial total=25n So new average=(25n+30)/(n+1)=26 Solving we get n=4, so he gave n+1=5 exams including today's

  • $\begingroup$ Almost right... $\endgroup$ – Marybnq May 21 at 11:33
  • $\begingroup$ This is correct; the question asks how many exams were taken, including today's. $\endgroup$ – Jeff Zeitlin May 21 at 11:49
  • $\begingroup$ rot13: Lbh pbhyq nyfb tb nobhg guvf ol fnlvat gur qvssrerapr bs 5 rknz cbvagf tvirf n 1 cbvag nirentr vapernfr, fb 5/1 = ahzore bs rknzf. Unys n cbvag jbhyq or 5/0.5 = 10 rknzf, rgp. $\endgroup$ – mkinson May 21 at 12:03
  • $\begingroup$ @JeffZeitlin exactly, be careful. $\endgroup$ – Marybnq May 21 at 12:39
  • $\begingroup$ Just to let you know, it was 4 $\endgroup$ – Marybnq May 21 at 21:47

This question might be better fit on the Math Stack Exchange, but I will answer this anyway :)

We get

$$\frac{x+30}{y}=26$$ $$\frac{x}{y}=25$$ where $x$ is the total of his exams' results excluding today's exam, and $y$ is the number of exams.

We know that






That's how many exams he took, including today's one.

  • $\begingroup$ I know the solution, I found it challenging and I wanted to post it😅 $\endgroup$ – Marybnq May 21 at 11:31
  • $\begingroup$ By the way it’s wrong, try again! $\endgroup$ – Marybnq May 21 at 11:31
  • $\begingroup$ @Marybnq it is? Hmm.... $\endgroup$ – Mr Pie May 21 at 11:47
  • $\begingroup$ Yes, try solving it keeping in mind the arithmetical average rule $\endgroup$ – Marybnq May 21 at 12:43
  • $\begingroup$ That second fraction should be x / (y-1) as the previous average would not include today's exam $\endgroup$ – PunPun1000 May 21 at 19:51

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