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Question-212209




Question Number 212209 by efronzo1 last updated on 06/Oct/24
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Answered by som(math1967) last updated on 06/Oct/24
 let no of solders in A,B,C are  x,y,z  ∴37x+23y=29x+29y  ⇒ 8x=6y⇒ x:y=3:4   23y+41z=33y+33z  ⇒10y=8z⇒y:z=4:5  ∴x:y:z=3:4:5   say x=3k,y=4k,z=5k   average all solders  =((37×3k+23×4k+41×5k)/(3k+4k+5k))  =((111k+92k+205k)/(12k))  =34
$$\:{let}\:{no}\:{of}\:{solders}\:{in}\:{A},{B},{C}\:{are} \\ $$$${x},{y},{z} \\ $$$$\therefore\mathrm{37}{x}+\mathrm{23}{y}=\mathrm{29}{x}+\mathrm{29}{y} \\ $$$$\Rightarrow\:\mathrm{8}{x}=\mathrm{6}{y}\Rightarrow\:{x}:{y}=\mathrm{3}:\mathrm{4} \\ $$$$\:\mathrm{23}{y}+\mathrm{41}{z}=\mathrm{33}{y}+\mathrm{33}{z} \\ $$$$\Rightarrow\mathrm{10}{y}=\mathrm{8}{z}\Rightarrow{y}:{z}=\mathrm{4}:\mathrm{5} \\ $$$$\therefore{x}:{y}:{z}=\mathrm{3}:\mathrm{4}:\mathrm{5} \\ $$$$\:{say}\:{x}=\mathrm{3}{k},{y}=\mathrm{4}{k},{z}=\mathrm{5}{k} \\ $$$$\:{average}\:{all}\:{solders} \\ $$$$=\frac{\mathrm{37}×\mathrm{3}{k}+\mathrm{23}×\mathrm{4}{k}+\mathrm{41}×\mathrm{5}{k}}{\mathrm{3}{k}+\mathrm{4}{k}+\mathrm{5}{k}} \\ $$$$=\frac{\mathrm{111}{k}+\mathrm{92}{k}+\mathrm{205}{k}}{\mathrm{12}{k}} \\ $$$$=\mathrm{34} \\ $$

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