# Exercise : 1

(b) 18°

(c) 24°

(d) 60°

(e) None of these

**Ans.b**

In a clock, each minute makes 6°

∴ 3 minutes will make 6 × 3 = 18°

(b) 4 : 38 am

(c) 4 : 35 am

(d) 4 : 39 am

(e) None of these

**Ans.b**

Here H × 30 = 4 × 30 = 120°.

(Since initially the hour hand is at 4. ∴ H = 4).

Required angle A = 90° and since, H × 30 > A° so, there will be two timings.

Required time T = ^{2}⁄_{11}(H × 30 ± A) minutes past H.

∴ One timing = ^{2}⁄_{11}(4 × 30 + 90) minutes past 4

= 38^{2}⁄_{11} minutes past 4.

Or 4 : 38 approx.

(b) 115°

(c) 95°

(d) 135°

(e) None of these

**Ans.a**

At 2'O Clock, Minute Hand will be 10 × 6 = 60°

behind the Hour Hand.

In 30 minutes, Minute Hand will gain (5^{1}⁄_{2})° × 30

= 150 + 15 = 165°

∴ Angle between Hour Hand and Minute Hand

= 165 – 60 = 105°

(b) 80°

(c) 88°

(d) 96°

(e) None of these

**Ans.c**

In 1 hour, the minute hand gains 330° over the hour hand.

i.e. in 60 minute, the minute hand gains 330° over the hour hand.

∴ In 16 minutes, the minute hand gains over the hour hand by ^{330°}⁄_{60} × 16° = 88°

^{5}⁄

_{61}minutes past 5

(b) 5 minutes past 6

(c) 5 minutes to 6

(d) 59

^{1}⁄

_{64}minutes past 5

(e) None of these

**Ans.a**

Time interval indicated by incorrect clock

= 6 p.m – 1 p.m = 5hrs.

Time gained by incorrect clock in one hour

= + 1 min. = +^{1}⁄_{60} hr.

Using the formula, \(\frac{True \; time\; interval}{Time \; interval \; in \; incorrect \; clock}\)

= \(\frac{1}{1 + hour \; gained \; in \; 1 \; hour \; by \; incorrect \; clock}\)

⇒ \(\frac{True \; time \; interval}{5} = \frac{1}{1 + \frac{1}{60}}\)

⇒ True time interval = \(\frac{5 \times 60}{61} = 4\tfrac{56}{61}\)

∴ True time = 1 p.m. + \(4\tfrac{56}{61}\) hrs.

= 5 p.m. + ^{56}⁄_{61} hrs. = 5 p.m. + ^{56}⁄_{61} × 60 min.

= 55^{5}⁄_{61} minutes past 5.

(b) 2 : 54 pm

(c) 2 : 23 pm

(d) 2 : 48 pm

(e) None of these

**Ans.b**

Time from noon on Sunday to 3 pm on Wednesday = 75 hours.

24 hours 2 minutes of the first clock = 24 hours of the correct one.

⇒ 1 hour of the first clock = 24 × (30/721) hours of correct one.

⇒ 75 ours of the first clock

= 24 × 30 × (75/721) hours of correct one

= 54000/721 hours = 74 hours 53.7 min.

Hence the answer is 2 : 54 pm.

^{4}⁄

_{11}minutes past 9

(b) 16

^{4}⁄

_{11}minutes past 8

(c) 55

^{5}⁄

_{61}minutes past 7

(d) 55

^{5}⁄

_{61}minutes past 8

(e) None of these

**Ans.a**

At 9’o clock, the Minute Hand is ahead of Hour Hand by 45 minutes. The hands will be opposite to each other when there is a space of 30 minutes between them.

This will happen when the Minute Hand gains 15 minutes space over Hour Hand.

Time taken by Minutes Hand to gain 15 minutes

= 15 × (1 + ^{1}⁄_{11}) = 15 + ^{15}⁄_{11} = 15 + 1^{4}⁄_{11} + 16^{4}⁄_{11} minutes.

Hence the Hands are opposite to each other at 16^{4}⁄_{11} minutes past 9.

(b) 4 : 45 am

(c) 4 : 20 am

(d) 5 : 00 am

(e) None of these

**Ans.a**

The clock gains 15 min in 24 hours.

Therefore, in 16 hours, it will gain 10 minutes.

Hence, the time shown by the clock will be 4.10 am.

(b) 107°

(c) 106°

(d) 108°

(e) None of these

**Ans.d**

Required angle = 240 – 24 × (11/2)

= 240 – 132 = 108°

(b) 13 min. 50 seconds lost

(c) 13 min. 20 seconds gained

(d) 14 min. 40 seconds gained

(e) None of these

**Ans.b**

In a watch than is running correct the minute hand should cross the hour hand once in every 65 + ^{5}⁄_{11} min.

So they should ideally cross 3 times once in

\(3 \times \left ( \frac{720}{11} \right )\frac{-2060}{11} \; min = 196.36 \; minutes.\)But in the watch under consideration, they meet after every 3 hr,18 min and 15 seconds,

i.e, (3 × 60 + 18 + ^{15}⁄_{60}) = ^{793}⁄_{4} min.

Thus, our watch is actually losing time (as it is slower than the normal watch). Hence when our watch elapsed

\(\left ( 1440 \times \frac{196.36}{198.25} \right ) = 1426.27.\)Hence the amount of time lost by our watch in one day = (1440 - 1426.27) = 13.73 i.e. 13 min and 50s (approx).