Quantitative Aptitude · Reasoning

Time and Clocks

491 Questions

Time and clocks problems evaluate quantitative aptitude by testing concepts on clock gains, losses, and relative speeds of hands. Questions also cover global time zones and standard time calculations. These logical reasoning topics frequently appear in SSC and banking exams.

Clock gain or lossTime zonesAngle between handsStandard time calculationPrecise time measurement

Time and Clocks Questions

Multiple choice physics our solar system planets of the solar system solar system and sun introduction to solar system

The difference in the length of a mean solar day and a sidereal day is about

  1. $1$ minute
  2. $4$ minute
  3. $15$ minute
  4. $56$ minute
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solar day is the time taken by earth to complete one rotation about its axis with respect to sun. Sidereal day is the time taken by earth to complete one rotation about its axis with respect to distant star. There is a difference of $4$ minutes between solar day and sidereal day. Solar day is longer as earth has to rotate greater angle to attain same position with respect to sun due to its own motion (revolution) round the sun. Its position with respect to distant star remains almost fixed.

Multiple choice maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

The minute hand of a clock is $\displaystyle \sqrt{21}$ cm long. The area described by the minute hand on the face of the clock between $7$ am and $7.05$ am is

  1. $5.5$ $\displaystyle cm^{2}$
  2. $22$ $\displaystyle cm^{2}$
  3. $11$ $\displaystyle cm^{2}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given $r=\sqrt{21}cm $

Angle made by minute hand in $1$ minute $=$ $\dfrac { { 360 }^{ 0 } }{ { 60 }} ={ 6 }^{ 0 }$

Therefore, angle made in $5$ minutes $=$ ${ 6 }^{ 0 }\times 5={ 30 }^{ 0 }$

Area of the sector$=\dfrac{\theta}{360^\circ}\times \pi \times r^2$

Here $\theta=30^\circ$

Hence, area swept in $5$ minutes $=$ $\dfrac { { 30 }^{ 0 } }{ { 360 }^{ 0 } } \times \dfrac { 22 }{ 7 } \times { \left( \sqrt { 21 }  \right)  }^{ 2 }$

                                                      $=$ $\dfrac { 1 }{ 12 } \times \dfrac { 22 }{ 7 } \times 21$

                                                      $=$ $5.5$ ${ cm }^{ 2 }$
Multiple choice maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

The length of a minute hand of a wall clock is $8.4\ cm$. Find the area swept by it in half an hour.

  1. $100\ cm^{2}$
  2. $110.88\ cm^{2}$
  3. $120\ cm^{2}$
  4. $130\ cm^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that minute hand covers $180^{o}$ in half an hour, which is a semicircle, hence area is

 $\Rightarrow \dfrac{1}{2}(\pi)(r^{2})=0.5\times3.1428\times(8.4)^{2}=110.88 \,cm^{2}$

Multiple choice maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

The minute hand of a clock is $8: cm$ long. Find the area swept by the minute hand between $8.30: a.m.$ and $9.05: a.m.$

  1. $\displaystyle 117\frac{1}{3}\:cm^{2}$
  2. $\displaystyle 107\frac{1}{3}\:cm^{2}$
  3. $\displaystyle 217\frac{1}{3}\:cm^{2}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Angle made the centre by each $5$ minutes =$\dfrac{360}{12}$
                                                              =$30^o$


Angle covered between $8.30$am to $9.5$a.m is $210^o$

Therefore,
$Area=\pi(8)^2\dfrac{210}{360}$
        $=\dfrac{22}{7}\times 8\times 8\times \dfrac{210}{360}$
        $=117\dfrac{1}{3} cm^2$

Multiple choice maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

The minute hand of a clock is $10$ cm long. Find the area of the face of the clock described by the minute hand between $9$A.M and $9.35$A.M.

  1. $90.165cm^2$
  2. $112.6cm^2$
  3. $156.4cm^2$
  4. $183.3cm^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have,
Angle described by the minute hand in one minute $=6^o$


$\therefore$ Angle described by the minute hand in $35$ minutes $=(6\times 35)^o=210^o$

$\therefore$ Area swept by the minute hand in $35$ minutes.


$=$ Area of a sector of angle $210^o$ in a circle of radius $10$ cm

 $=\dfrac {\theta}{360} \pi {r _1}^2$

$= \dfrac{210}{360}\times \dfrac{22}{7}\times (10)^2\ cm^2$.....

$=183.3\ cm^2$

Multiple choice maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

The minute hand of a clock is 7 cm long Find the area  traced out by the minute hand of the clock between 6 pm to 6:30 pm

  1. $\displaystyle 14.4cm^{2}$
  2. $\displaystyle 15.4cm^{2}$
  3. $\displaystyle 7.2cm^{2}$
  4. $\displaystyle 6.42cm^{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The total angle of $ 12 $ hours in a clock is 
$ { 360 }^{ 0 } $.
$ => 24 $ half hours $ = { 360 }^{ 0 } $.
This means for one half an hour, angle $ = \frac {{ 360 }^{ 0 }}{24} = { 15 }^{ 0 } $

Area of a sector of a circle of radius 'r' and angle 
$ \theta = \frac { \theta  }{ 360 } \pi {r}^{2}$
Hence, area of the sector of the circle of  radius $ 7 $ cm and angle $

{ 15 }^{ 0 } = \frac { 15 }{ 360 } \times \frac { 22 }{ 7 } \times 7 \times

7\quad = 6.42  {cm}^{2} $

Multiple choice maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

The minute hand of a clock is $7\ cm$ long. Find the area traced by it on the clock face between $4{:}15$ p.m. and $4{:}35$ p.m.

  1. $59\ cm^{2}$
  2. $65\ cm^{2}$
  3. $51.3\ cm^{2}$
  4. $45\ cm^{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Time $= 20$ min

Angle made by minute hand in 1 minute  $=\dfrac { { 360 }^{ 0 } }{ { 60 }^{ 0 } } ={ 6 }^{ 0 }$

$\therefore $  In $20$ min  $={ 6 }^{ 0 }\times 20={ 120 }^{ 0 }$

$\therefore $  Area swept  $=\dfrac { \theta  }{ { 360 }^{ 0 } } \times \pi { r }^{ 2 }=\dfrac { { 120 }^{ 0 } }{ { 360 }^{ 0 } } \times \dfrac { 22 }{ 7 } \times 7\times 7=\dfrac { 154 }{ 3 } =51.3{ cm }^{ 2 }$

Multiple choice civics why do we need a parliament? parliamentary procedures parliamentary system parliamentary system and procedures

The time gap between the question hour and the agenda is known as zero hour.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Zero hour is neither mentioned in the Constitution nor the Rules of Procedure of the Houses. It is an Indian innovation in the field of parliamentary procedures and has been in existence since 1962. It is an informal device available to the members of the Parliament to raise matters without any prior notice. It starts immediately after the question hour and lasts until the agenda for the day is taken up.

Multiple choice chemistry the s-block elements (alkali and alkaline earth metals) some important compounds of calcium some important compounds of magnesium and calcium compounds of s block elements

Final setting time of cement should not be more than:

  1. 1 hour

  2. 2 hours

  3. 5 hours

  4. 10 hours
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Initial setting time of cement should be more than 45 mins. This is to allow enough time for transportation and placing of concrete before setting starts. Final setting time should be less than 375 minutes. 375 minutes is equivalent to 10.25 hours. Hence, the final setting time of cement should not be more than 10 hours.

Multiple choice commercial applications generally accepted accounting principles (gaap) acccounting cycle meaning, need and objectives of accounting accounting process

Standard hour is a _____________.

  1. Unit of work

  2. Unit of time

  3. Both (A) and (B)

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The standard hours per unit is derived from the labor routing, which is a compilation of the normal amount of time expected to be required to manufacture a unit. Therefore, the standard hours allowed is 750 hours, which is calculated as 500 units multiplied by 1.5 hours per unit.