Pytagoriade

Two fifth-graders teams compete in math competitions - Mathematical Olympiad and Pytagoriade. Of the 33 students who competed in at least one of the contests, 22 students. Students who competed only in Pytagoriade were twice as many who just competed in the Mathematical Olympiad. Students who competed in both competitions were four times more than those who only competed in Pytagoriade.

a) How many students competed in both competitions?
b) How many students competed in just one competition?
c) How many students competed in no more than one competition?

Correct answer:

a =  16
b =  6
c =  17

Step-by-step explanation:


p+a+o = 22
p = 2 o
a = 4p
b = p+o
c = 33-a


p+a+o = 22
p = 2·o
a = 4·p
b = p+o
c = 33-a

a+o+p = 22
2o-p = 0
a-4p = 0
b-o-p = 0
a+c = 33

Row 3 - Row 1 → Row 3
a+o+p = 22
2o-p = 0
-o-5p = -22
b-o-p = 0
a+c = 33

Row 5 - Row 1 → Row 5
a+o+p = 22
2o-p = 0
-o-5p = -22
b-o-p = 0
c-o-p = 11

Pivot: Row 2 ↔ Row 4
a+o+p = 22
b-o-p = 0
-o-5p = -22
2o-p = 0
c-o-p = 11

Pivot: Row 3 ↔ Row 5
a+o+p = 22
b-o-p = 0
c-o-p = 11
2o-p = 0
-o-5p = -22

Row 5 - -1/2 · Row 4 → Row 5
a+o+p = 22
b-o-p = 0
c-o-p = 11
2o-p = 0
-5.5p = -22


p = -22/-5.5 = 4
o = 0+p/2 = 0+4/2 = 2
c = 11+o+p = 11+2+4 = 17
b = 0+o+p = 0+2+4 = 6
a = 22-o-p = 22-2-4 = 16

a = 16
b = 6
c = 17
o = 2
p = 4

Our linear equations calculator calculates it.



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