LR - Clocks - Previous Year CAT/MBA Questions

Answer: Option B

Video Explanation

Text Explanation :

The angle between the two hands at h hours and m minutes = $\left|30\mathrm{h}-\frac{11}{2}\mathrm{m}\right|$

∴ Angle at 8:48 am = $\left|30\times 8-\frac{11}{2}\times 48\right|$ = |240 - 264| = 24°.

Now, the angle between the two hands should increase by 50% i.e., 12°.

Relative speed of the two hands = 6 – ½ = 11/2°/min.

∴ Time taken for angle to increase 12° = $\frac{12}{11/2}$ = $\frac{24}{11}$ minutes.

Hence, option (b).
 

Concept:

Angle between hands of a Clock.

Answer: Option B

Text Explanation :

In a watch that is running correct, the minute hand should cross the hour hand once in every 65 + $\frac{5}{11}$ min. So they should ideally cross three times once in $3\times \left(\frac{720}{11}\right)=\frac{2060}{11}$ min = 196.36 min. But in the watch under consideration they meet after every 3 hr, 18 min and 15 s, i.e. (3 × 60 + 18 + $\frac{15}{60}$) = $\frac{793}{4}$ min = 198.25 min. In other words, our watch is actually losing time (as it is slower than the normal watch). Hence, when our watch elapsed 198.25 min, it actually should have elapsed 196.36 min. So in a day, when our watch will elapse (60 × 24) = 1440, it should actually elapse $\left(1440\times \frac{196.36}{198.25}\right)$ = 1426. 27. Hence, the amount of time lost by our watch in one day = (1440 – 1426.27) = 13.73, i.e. 13 min and 50 s (approximately).