All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 3 grandfather clock problems (Posted on 2006-03-27)
I Draw the face of a clock numbered with roman numerals in the usual way. Explain how to draw 4 rays radiating from the center such that the sum of the numerals in each sector is 20.

II At what time are the two hands of a clock situated so that, reckoning in minutes from XII, one is exactly the square of the distance of the other?

III At what time between three and four o’clock is the minute hand the same distance from VIII as the hour hand is from XII?

 No Solution Yet Submitted by Jer Rating: 3.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Part III solution | Comment 8 of 9 |

If the hour and minute hands moved steadily and fluidically through the passage of time, the answer is:
3:23 & 1/13 minutes.

At 3 O'Clock the hour hand is 15 minute marks from the XII, and the minute hand is 40 minute marks from the VIII.  Putting the rate in minutes of the movement of each hand where these hands are equidistant from the respective hour marks, the equations are:
H = M/12
3 O'Clock [III] hour mark = 15 minute mark
[XIII] hour mark = 40 minute mark
15 + M(1/12) = 40 - M
Solving for M, gives:
M = 300/13 = 23 1/13

Edited on March 30, 2006, 1:43 am
 Posted by Dej Mar on 2006-03-29 18:43:05

 Search: Search body:
Forums (0)