Question

A person who left home between 4 p.m. and 5 p.m. returned between 5 p.m. and 6 p.m. and found that the hands of his watch had exactly exchanged place, when did he go out ?

Answer 1

@nathan917

Answer 2

5:30 and 6:20?

Answer 3

How ?

Answer 4

do you have a grandfather clock at home?

Answer 5

No My grand father is dead

Answer 6

https://www.google.com/search?q=grandfather+clock&source=lnms&tbm=isch&sa=X&ei=oh-MUdaGC42MigKunYDQDg&sqi=2&ved=0CAcQ_AUoAQ&biw=1366&bih=667#imgrc=5gG_oWcUiRa2oM%3A%3Bn3Wb8flcHE_glM%3Bhttp%253A%252F%252F4.bp.blogspot.com%252F-o6peqAk6-gU%252FUC1esSBtqtI%252FAAAAAAAABC0%252FZASN2SgZ3fc%252Fs1600%252FGrandfather-Clock.jpg%3Bhttp%253A%252F%252Fclocksgalleries.blogspot.com%252F2013%252F04%252Fgrandfather-clock-faces.html%3B1600%3B1600

Answer 7

that is a grandfather clock

Answer 8

ok

Answer 9

You should be able to find what you need here:

https://sites.google.com/site/mymathclassroom/trigonometry/clock-angle-problems/clock-angle-problem-formula

Answer 10

Ok Thank you Sir

Answer 11

Yes. Thank you i was lost.

Answer 12

left at 4:25 and came back at 5:20

Answer 13

oops i was an hour off

Answer 14

0.5˚ per minute and long hand moves 6˚ per minute

let t₁ = time he left home in minutes.
let t₂ = time he returned home in minutes.

comparing the hour hand when he left and the minute hand when he returned

0.5*t₁ = 6*(t₂ - 300)
t₁ = 12*(t₂ - 300)
t₁ = 12t₂ - 3600 <---equation 1

comparing the hour hand when he returned and the minute hand when he left

0.5*t₂ = 6*(t₁ - 240)
t₂ = 12*(t₁ - 240)
t₂ = 12t₁ - 2880 <---equation 2

equating 1 and 2 will give us the ff. equation

t₁ = 12(12t₁ - 2880) - 3600
t₁ = 144t₁ - 34560 - 3600
t₁ = 144t₁ - 38160
144t₁ - t₁ = 38160
143t₁ = 38160
t₁ = 266.8531469 or 4:26.85 pm

t₂ = 12t₁ - 2880
t₂ = 12(266.8531469) - 2880
t₂ = 322.2377628 or 5:22.24 pm

therefore he left the house at 4:26.85 pm and returned at 5:22.24 pm

Answer 15

Got it.

Answer 16

Thanks again