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A Fun Question...

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For what fraction of a day wil a 12 hour digital clock that shows only the hours and minutes display at least one 5 or 6

Nov 24, 2018
edited by Guest  Nov 24, 2018

#1
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from 5:00 - 6:59 there will be a 5 or 6 visible at all times

from H:50 - H:59 there will be a 5 visible at all times, 1<= H <= 4 and 7 <= H <= 12

from H:M5 - H:M7 there will be a 5 or 6 visible at all times, 1 <= H <= 4 and 7<= H <= 12, 0 <= M <= 4

Summing all this up

2 hrs = 120 minutes for the first category

10 minutes apiece of 10 hrs = 100 min for the second category

2 minutes 5 times an hr for 10 hrs = 100 minutes

So a total of 320 minutes out of a total of 12 x 60 = 720 minutes

p = 320/720 = 4/9

Nov 24, 2018

#1
+1

from 5:00 - 6:59 there will be a 5 or 6 visible at all times

from H:50 - H:59 there will be a 5 visible at all times, 1<= H <= 4 and 7 <= H <= 12

from H:M5 - H:M7 there will be a 5 or 6 visible at all times, 1 <= H <= 4 and 7<= H <= 12, 0 <= M <= 4

Summing all this up

2 hrs = 120 minutes for the first category

10 minutes apiece of 10 hrs = 100 min for the second category

2 minutes 5 times an hr for 10 hrs = 100 minutes

So a total of 320 minutes out of a total of 12 x 60 = 720 minutes

p = 320/720 = 4/9

Rom Nov 24, 2018