Easy math, but not sure what I did wrong.
Alex took 31 exams during 5 years studying at the MaPhAs University. Each year, he took more exams than the previous year. During his 5th year, he took three times as many exams as in the 1st year. How many exams did Alex take during his 4th year at the university?

8

Step-by-step explanation:

In his 5th year, he took 3 times as many exams as the first year.  So the number of exams taken in the 5th year must be a multiple of 3.

If a₁ = 1, then a₅ = 3.  However, this isn't possible because we need 4 integers between them, and a sum of 31.

If a₁ = 2, then a₅ = 6.  Same problem as before.

If a₁ = 3, then a₅ = 9.  This is a possible solution.

If a₁ = 4, then a₅ = 12.  If we assume a₂ = 5, a₃ = 6, and a₄ = 7, then the sum is 34, so this is not a possible solution.

Therefore, Alex took 3 exams in his first year and 9 exams in his fifth year.  So he took 19 exams total in his second, third, and fourth years.

3 < a₂ < a₃ < a₄ < 9

If a₂ = 4, then a₃ = 7 and a₄ = 8.

If a₂ = 5, then a₃ = 6 and a₄ = 8.

If a₂ = 6, then there's no solution.

So Alex must have taken 8 exams in his fourth year.

94

step-by-step explanation:

average the numbers and use a variable to figure out the minimum to get a 90.

(79+94+91+92+x)/5=90

x = 94

if im not mistaking its

step-by-step explanation: