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Between-12-p-m-today-and-12-p-m-tomorrow-how-many-times-do-the-hour-hand-and-the-minute-hand-on-a-clock-form-an-angle-of-120-

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Question Number 144025 by ZiYangLee last updated on 20/Jun/21
Between 12 p.m. today and 12 p.m. tomorrow, how many times do the hour hand and the minute hand on a clock form an angle of 120°?
$$\mathrm{Between}\:\mathrm{12}\:\mathrm{p}.\mathrm{m}.\:\mathrm{today}\:\mathrm{and}\:\mathrm{12}\:\mathrm{p}.\mathrm{m}. \\ $$$$\mathrm{tomorrow},\:\mathrm{how}\:\mathrm{many}\:\mathrm{times}\:\mathrm{do}\:\mathrm{the} \\ $$$$\mathrm{hour}\:\mathrm{hand}\:\mathrm{and}\:\mathrm{the}\:\mathrm{minute}\:\mathrm{hand}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{clock}\:\mathrm{form}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{120}°? \\ $$
Answered by mr W last updated on 21/Jun/21
at time h:m the position of minute hand is m×6°, the position of hour hand is (h+(m/(60)))×30°. (0≤h≤11, 0≤m<60) m×6=(h+(m/(60)))×30+120 ⇒11m=60(h+4) 0≤m=((60(h+4))/(11))<60 ⇒0≤h≤6 ⇒7 times m×6=(h+(m/(60)))×30+240 ⇒11m=60(h+8) 0≤((60(h+8))/(11))<60 ⇒0≤h≤2 ⇒ 3 times m×6=(h+(m/(60)))×30−120 ⇒11m=60(h−4): 0≤((60(h−4))/(11))<60 ⇒4≤h≤11 ⇒ 8 times m×6=(h+(m/(60)))×30−240 ⇒11m=60(h−8): 0≤((60(h−8))/(11))<60 ⇒8≤h≤11 ⇒ 4 times totally 7+3+8+4=22 times in 12 hours or 44 times in a day.
$${at}\:{time}\:{h}:{m}\:{the}\:{position}\:{of}\:{minute} \\ $$$${hand}\:{is}\:{m}×\mathrm{6}°,\:{the}\:{position}\:{of}\:{hour} \\ $$$${hand}\:{is}\:\left({h}+\frac{{m}}{\mathrm{60}}\right)×\mathrm{30}°. \\ $$$$\left(\mathrm{0}\leqslant{h}\leqslant\mathrm{11},\:\mathrm{0}\leqslant{m}<\mathrm{60}\right) \\ $$$$ \\ $$$${m}×\mathrm{6}=\left({h}+\frac{{m}}{\mathrm{60}}\right)×\mathrm{30}+\mathrm{120} \\ $$$$\Rightarrow\mathrm{11}{m}=\mathrm{60}\left({h}+\mathrm{4}\right) \\ $$$$\mathrm{0}\leqslant{m}=\frac{\mathrm{60}\left({h}+\mathrm{4}\right)}{\mathrm{11}}<\mathrm{60} \\ $$$$\Rightarrow\mathrm{0}\leqslant{h}\leqslant\mathrm{6}\:\Rightarrow\mathrm{7}\:{times} \\ $$$$ \\ $$$${m}×\mathrm{6}=\left({h}+\frac{{m}}{\mathrm{60}}\right)×\mathrm{30}+\mathrm{240} \\ $$$$\Rightarrow\mathrm{11}{m}=\mathrm{60}\left({h}+\mathrm{8}\right) \\ $$$$\mathrm{0}\leqslant\frac{\mathrm{60}\left({h}+\mathrm{8}\right)}{\mathrm{11}}<\mathrm{60} \\ $$$$\Rightarrow\mathrm{0}\leqslant{h}\leqslant\mathrm{2}\:\Rightarrow\:\mathrm{3}\:{times} \\ $$$$ \\ $$$${m}×\mathrm{6}=\left({h}+\frac{{m}}{\mathrm{60}}\right)×\mathrm{30}−\mathrm{120} \\ $$$$\Rightarrow\mathrm{11}{m}=\mathrm{60}\left({h}−\mathrm{4}\right): \\ $$$$\mathrm{0}\leqslant\frac{\mathrm{60}\left({h}−\mathrm{4}\right)}{\mathrm{11}}<\mathrm{60} \\ $$$$\Rightarrow\mathrm{4}\leqslant{h}\leqslant\mathrm{11}\:\Rightarrow\:\mathrm{8}\:{times} \\ $$$$ \\ $$$${m}×\mathrm{6}=\left({h}+\frac{{m}}{\mathrm{60}}\right)×\mathrm{30}−\mathrm{240} \\ $$$$\Rightarrow\mathrm{11}{m}=\mathrm{60}\left({h}−\mathrm{8}\right): \\ $$$$\mathrm{0}\leqslant\frac{\mathrm{60}\left({h}−\mathrm{8}\right)}{\mathrm{11}}<\mathrm{60} \\ $$$$\Rightarrow\mathrm{8}\leqslant{h}\leqslant\mathrm{11}\:\Rightarrow\:\mathrm{4}\:{times} \\ $$$$ \\ $$$${totally}\:\mathrm{7}+\mathrm{3}+\mathrm{8}+\mathrm{4}=\mathrm{22}\:{times}\:{in}\:\mathrm{12}\:{hours} \\ $$$${or}\:\mathrm{44}\:{times}\:{in}\:{a}\:{day}. \\ $$

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