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Question Number 180648 by lazyboy last updated on 15/Nov/22

there are 40 people   (teachers students and workers)   in a school.and there are 40 apples   on the table.one teacher can eat   4 apples.and one student can eat    2 apples. 4 workers can eat one     apple.find the numbers of teacher     students  and workers       who eat apples.  please help me.

$${there}\:{are}\:\mathrm{40}\:{people} \\ $$$$\:\left({teachers}\:{students}\:{and}\:{workers}\right) \\ $$$$\:{in}\:{a}\:{school}.{and}\:{there}\:{are}\:\mathrm{40}\:{apples} \\ $$$$\:{on}\:{the}\:{table}.{one}\:{teacher}\:{can}\:{eat} \\ $$$$\:\mathrm{4}\:{apples}.{and}\:{one}\:{student}\:{can}\:{eat} \\ $$$$\:\:\mathrm{2}\:{apples}.\:\mathrm{4}\:{workers}\:{can}\:{eat}\:{one}\: \\ $$$$\:\:{apple}.{find}\:{the}\:{numbers}\:{of}\:{teacher} \\ $$$$\:\:\:{students}\:\:{and}\:{workers} \\ $$$$\:\:\:\:\:{who}\:{eat}\:{apples}. \\ $$$${please}\:{help}\:{me}. \\ $$

Commented by Acem last updated on 14/Nov/22

 To avoid solving by experiment, it′s better to   find equations  4a +2b + (c/4)= 40 ... (1)   a+b+c= 40           ... (2)            ?                         ...(3)

$$\:{To}\:{avoid}\:{solving}\:{by}\:{experiment},\:{it}'{s}\:{better}\:{to} \\ $$$$\:{find}\:{equations} \\ $$$$\mathrm{4}{a}\:+\mathrm{2}{b}\:+\:\frac{{c}}{\mathrm{4}}=\:\mathrm{40}\:...\:\left(\mathrm{1}\right) \\ $$$$\:{a}+{b}+{c}=\:\mathrm{40}\:\:\:\:\:\:\:\:\:\:\:...\:\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\left(\mathrm{3}\right) \\ $$$$ \\ $$

Commented by Acem last updated on 15/Nov/22

Either there′s a way to solve such this problem   otherwise it shoudn′t ask such questions...

$${Either}\:{there}'{s}\:{a}\:{way}\:{to}\:{solve}\:{such}\:{this}\:{problem} \\ $$$$\:{otherwise}\:{it}\:{shoudn}'{t}\:{ask}\:{such}\:{questions}... \\ $$

Commented by Acem last updated on 15/Nov/22

At distribution of the estate, we calculate   the shares and it ratio, but we never count how   many sons, daughters, sisters, wife...etc he had

$${At}\:{distribution}\:{of}\:{the}\:{estate},\:{we}\:{calculate} \\ $$$$\:{the}\:{shares}\:{and}\:{it}\:{ratio},\:{but}\:{we}\:{never}\:{count}\:{how} \\ $$$$\:{many}\:{sons},\:{daughters},\:{sisters},\:{wife}...{etc}\:{he}\:{had} \\ $$

Commented by Acem last updated on 15/Nov/22

Anyway i hated this question so much   because it gives to the children an impression   that the workers are less value than the others

$${Anyway}\:{i}\:{hated}\:{this}\:{question}\:{so}\:{much} \\ $$$$\:{because}\:{it}\:{gives}\:{to}\:{the}\:{children}\:{an}\:{impression} \\ $$$$\:{that}\:{the}\:{workers}\:{are}\:{less}\:{value}\:{than}\:{the}\:{others} \\ $$

Commented by a.lgnaoui last updated on 15/Nov/22

  32a+24b+17c=400   (3)

$$ \\ $$$$\mathrm{32}{a}+\mathrm{24}{b}+\mathrm{17}{c}=\mathrm{400}\:\:\:\left(\mathrm{3}\right) \\ $$$$ \\ $$

Commented by Frix last updated on 15/Nov/22

Acem, there is a way to solve this problem.  Not everything is solveable with formulas,  i.e. Diophantine Equations.    Another example, I once played the game  Tetris. You got points for each filled line,  depending on how many lines you filled at  once:  1 line = 1 point  2 lines = 3 points  3 lines = 6 points  4 lines = 10 points  If you made 100 points, which possible  combinations of lines did you fill when  you filled 31 lines?

$$\mathrm{Acem},\:\mathrm{there}\:{is}\:\mathrm{a}\:\mathrm{way}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{problem}. \\ $$$$\mathrm{Not}\:\mathrm{everything}\:\mathrm{is}\:\mathrm{solveable}\:\mathrm{with}\:\mathrm{formulas}, \\ $$$$\mathrm{i}.\mathrm{e}.\:\mathrm{Diophantine}\:\mathrm{Equations}. \\ $$$$ \\ $$$$\mathrm{Another}\:\mathrm{example},\:\mathrm{I}\:\mathrm{once}\:\mathrm{played}\:\mathrm{the}\:\mathrm{game} \\ $$$$\mathrm{Tetris}.\:\mathrm{You}\:\mathrm{got}\:\mathrm{points}\:\mathrm{for}\:\mathrm{each}\:\mathrm{filled}\:\mathrm{line}, \\ $$$$\mathrm{depending}\:\mathrm{on}\:\mathrm{how}\:\mathrm{many}\:\mathrm{lines}\:\mathrm{you}\:\mathrm{filled}\:\mathrm{at} \\ $$$$\mathrm{once}: \\ $$$$\mathrm{1}\:\mathrm{line}\:=\:\mathrm{1}\:\mathrm{point} \\ $$$$\mathrm{2}\:\mathrm{lines}\:=\:\mathrm{3}\:\mathrm{points} \\ $$$$\mathrm{3}\:\mathrm{lines}\:=\:\mathrm{6}\:\mathrm{points} \\ $$$$\mathrm{4}\:\mathrm{lines}\:=\:\mathrm{10}\:\mathrm{points} \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{made}\:\mathrm{100}\:\mathrm{points},\:\mathrm{which}\:\mathrm{possible} \\ $$$$\mathrm{combinations}\:\mathrm{of}\:\mathrm{lines}\:\mathrm{did}\:\mathrm{you}\:\mathrm{fill}\:\mathrm{when} \\ $$$$\mathrm{you}\:\mathrm{filled}\:\mathrm{31}\:\mathrm{lines}? \\ $$

Answered by Acem last updated on 15/Nov/22

@LazyBoy, as long as you′re lazy, so relax and   go to your school without solving it,  telling your   teacher that you refuse the injustice against   the workers. I think that your school administration   would give you a badge of honor.

$$@{LazyBoy},\:{as}\:{long}\:{as}\:{you}'{re}\:{lazy},\:{so}\:{relax}\:{and} \\ $$$$\:{go}\:{to}\:{your}\:{school}\:{without}\:{solving}\:{it},\:\:{telling}\:{your} \\ $$$$\:{teacher}\:{that}\:{you}\:{refuse}\:{the}\:{injustice}\:{against} \\ $$$$\:{the}\:{workers}.\:{I}\:{think}\:{that}\:{your}\:{school}\:{administration} \\ $$$$\:{would}\:{give}\:{you}\:{a}\:{badge}\:{of}\:{honor}. \\ $$

Answered by mr W last updated on 15/Nov/22

x teachers  y students  z workers with z=4k  x+y+4k=40  ⇒x=40−y−4k    4x+2y+k=40  4(40−y−4k)+2y+k=40  2y+15k=120  ⇒y=15n+45 ≥0 ⇒n≥−3  ⇒k=−2n+2  ⇒x=−7n−13 ≥0 ⇒n≤−2  ⇒z=4(−2n+2)=8−8n ≥0 ⇒n≤1  general solution is  x=−7n−13  y=15n+45  z=8−8n  with n=−3 or −2  i.e. (x,y,z)=(8,0,32) or (1,15,24)

$${x}\:{teachers} \\ $$$${y}\:{students} \\ $$$${z}\:{workers}\:{with}\:{z}=\mathrm{4}{k} \\ $$$${x}+{y}+\mathrm{4}{k}=\mathrm{40} \\ $$$$\Rightarrow{x}=\mathrm{40}−{y}−\mathrm{4}{k} \\ $$$$ \\ $$$$\mathrm{4}{x}+\mathrm{2}{y}+{k}=\mathrm{40} \\ $$$$\mathrm{4}\left(\mathrm{40}−{y}−\mathrm{4}{k}\right)+\mathrm{2}{y}+{k}=\mathrm{40} \\ $$$$\mathrm{2}{y}+\mathrm{15}{k}=\mathrm{120} \\ $$$$\Rightarrow{y}=\mathrm{15}{n}+\mathrm{45}\:\geqslant\mathrm{0}\:\Rightarrow{n}\geqslant−\mathrm{3} \\ $$$$\Rightarrow{k}=−\mathrm{2}{n}+\mathrm{2} \\ $$$$\Rightarrow{x}=−\mathrm{7}{n}−\mathrm{13}\:\geqslant\mathrm{0}\:\Rightarrow{n}\leqslant−\mathrm{2} \\ $$$$\Rightarrow{z}=\mathrm{4}\left(−\mathrm{2}{n}+\mathrm{2}\right)=\mathrm{8}−\mathrm{8}{n}\:\geqslant\mathrm{0}\:\Rightarrow{n}\leqslant\mathrm{1} \\ $$$${general}\:{solution}\:{is} \\ $$$${x}=−\mathrm{7}{n}−\mathrm{13} \\ $$$${y}=\mathrm{15}{n}+\mathrm{45} \\ $$$${z}=\mathrm{8}−\mathrm{8}{n} \\ $$$${with}\:{n}=−\mathrm{3}\:{or}\:−\mathrm{2} \\ $$$${i}.{e}.\:\left({x},{y},{z}\right)=\left(\mathrm{8},\mathrm{0},\mathrm{32}\right)\:{or}\:\left(\mathrm{1},\mathrm{15},\mathrm{24}\right) \\ $$

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