# LR - Clocks - Previous Year CAT/MBA Questions

The best way to prepare for LR - Clocks is by going through the previous year **LR - Clocks CAT questions**.
Here we bring you all previous year LR - Clocks CAT questions along with detailed solutions.

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**CAT 2023 QA Slot 1 | LR - Clocks CAT Question**

The minor angle between the hour hand and minute hand of a clock was observed at 8:48 am. The minimum deviation (in min) after 8:48 am when the angle increases by 50% is?

- (a)
36/11

- (b)
24/11

- (c)
2

- (d)
4

Answer: Option B

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**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**:

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**CAT 1996 LRDI | LR - Clocks CAT Question**

In a watch, the minute hand crosses the hour hand for the third time exactly after every 3 hr 18 min and 15 s of watch time. What is the time gained or lost by this watch in one day?

- (a)
14 min 10 s lost

- (b)
13 min 50 s lost

- (c)
13 min 20 s gained

- (d)
14 min 40 s gained

Answer: Option B

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**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).

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