I am preparing for the KVPY exam which will be held on 27th of October,2013.I was going through the previous years question paper that I found this problem.

A 12 hour digital clock displays the hour and the minute of a day.Due to defect in the clock whenever the digit 1 is supposed to be displayed it displays 7.What fraction of the day will the clock show the correct time?

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TopNewest\[ \frac{1}{4} \]

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Sorry, I mean \[ \frac{1}{2} \]

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There are 45 minutes which doesn't show one in an hour, and there are 16 hours which doesn't show one in a day. There are 60 minutes in an hour, and 24 hours in a day. Therefore, the fraction is \[ \frac{45 \times 16}{60 \times 24} \] which can be simplified into \[ \frac{1}{2} \]

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It is a 12 hour digital clock. So, how can there be 16 hours when clock doesn't show 1?

The hour display shows 1 only when it is 1 o' clock, be it a.m. or p.m. And the minute display shows 1 when it displays 01,11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 31, 41, 51.

So, I think the answer should be \(\frac{11 \times 14 +60}{60 \times 12}\)

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Well, Takeda did it over \(24\) hours, which is basically multiplying by \(\dfrac{2}{2}\).

Your problem is that the hour display also shows 1 at 10, 11, and 12. The clock shows the correct time if there are no ones. For each hour without a 1 (there are 8), there are 45 minutes without a 1.

Then, the answer is \(\dfrac{8}{12}\cdot\dfrac{45}{60}=\dfrac{1}{2}\).

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Takeda is correct Maharnab. There are really 16 hours when the clock does not show 1. In 12 hours the clock does show 1 at 1, 10, 11, and 12 o clock and in 24 hour system it will show these all two times which sums up to 8 hours and 24 - 8 is 16 hours.

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173/180

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