Problem K. 505. (September 2016)

K. 505. A certain job was completed by three workers. They manufactured machine parts (each machine part was made by one particular worker). The whole task took 15 hours, and the total wages paid to the three of them were \(\displaystyle 142\;000\) forints (HUF, Hungarian currency). The wages were proportional to the quantity of machine pars manufactured. It took the first worker 12 minutes to make one machine component. The second worker got twice as much money as the first worker, and the third one was paid \(\displaystyle 8\;000\) forints less than the second one. How many machine components did each of the three workers make?

(6 pont)

Deadline expired on October 10, 2016.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. Ha az első ember \(\displaystyle x\) Ft-ot kapott, akkor a második \(\displaystyle 2x\)-et, a harmadik pedig \(\displaystyle 2x–8000\)-et. Ezeket összegezve \(\displaystyle 5x–8000=142\,000\), azaz \(\displaystyle x=30\,000\). Az első ember \(\displaystyle 15\cdot60:12=75\) db alkatrészt készített. Egy alkatrész munkadíja tehát \(\displaystyle 30\,000 : 75 = 400\) Ft. A második ember ehhez képest kétszer annyit, tehát \(\displaystyle 150\) db alkatrészt készített. A harmadik ember ennél \(\displaystyle 20\)-szal kevesebbet, vagyis \(\displaystyle 130\)-at.

Tehát az első ember 75 darabot készített 900 perc alatt \(\displaystyle 30\,000\) Ft-ért, a második 150 darabot szintén 900 perc alatt 60000 Ft-ért, míg a harmadik 130 darabot készített ugyanennyi idő alatt, melyért \(\displaystyle 52\,000\) Ft-ot kapott.

Statistics:

180 students sent a solution.
6 points:	154 students.
5 points:	13 students.
4 points:	5 students.
3 points:	5 students.
0 point:	2 students.
Unfair, not evaluated:	1 solutions.

Problems in Mathematics of KöMaL, September 2016