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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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