Clock Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to Clock. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.

Q 1 - At what angle the hands of a clock are inclined at 15 minute past 5?

A - 72$\frac{1}{2}^{\circ}$

B - 67$\frac{1}{2}^{\circ}$

C - 58$\frac{1}{2}^{\circ}$

D - 64^o

Answer : B

Explanation

Angle 15 minute past 5 = 60 + 7$\frac{1}{2}$ = 67$\frac{1}{2}^{\circ}$

Q 2 - The reflex angle between the hands of a clock at 10:25 is?

A - 180^o

B - 192$\frac{1}{2}^{\circ}$

C - 195^o

D - 197$\frac{1}{2}^{\circ}$

Answer : D

Explanation

By hour hand $\frac{360}{12} \times \frac{125}{12} = 312 \frac{1}{2}^{\circ}$

By min hand $\frac{360}{12} \times 25^{\circ}$ = 150^o

Reflex angle = 360^o - (312$\frac{1}{2}^{\circ}$ - 150^o) = 360^o - 162$\frac{1}{2}^{\circ}$ = 197$\frac{1}{2}^{\circ}$

Q 3 - At what time between 8 and 9 O`clock will the hands of a clock be in straight line?

A - 11$\frac{11}{11}$ min past 8

B - 11$\frac{10}{11}$ min past 8

C - 11$\frac{9}{11}$ min past 8

D - 11$\frac{8}{11}$ min past 8

Answer : B

Explanation

$\left ( \frac{5n - 30}{11} \right )$ × 12 min past (n>6)

$\frac{120}{11}$ = 11 $\frac{10}{11}$ past 8

Q 4 - At what time between 8 and 9 O`clock in the morning, will the hands of a clock be at right angle?

A - $\frac{574}{11}$ min past 8, $\frac{384}{11}$ min past 8

B - $\frac{575}{11}$ min past 8, $\frac{377}{11}$ min past 8

C - $\frac{576}{11}$ min past 8, $\frac{376}{11}$ min past 8

D - $\frac{577}{11}$ min past 8, $\frac{374}{11}$ min past 8

Answer : A

Explanation

(5n±8) × $\frac{12}{11}$; n = 8

$\frac{574}{11}$ min. past 8, $\frac{384}{11}$ min past 8

Q 5 - Angle covered by hour hand in 1 min is?

A - $\frac{1}{2}^{\circ}$

B - $\frac{1}{3}^{\circ}$

C - $\frac{1}{4}^{\circ}$

D - $\frac{1}{5}^{\circ}$

Answer : A

Explanation

12 hrs = 360^o

1 hrs = $\frac{360}{12}$ = 30^o, $\frac{30}{60}$ = $\frac{1}{2}^{\circ}$ in 1 min.

Q 6 - In the morning 6 and 7 O`clock, what should be a particular time to make right angle?

A - 48$\frac{2}{11}$ min past 6 and 16$\frac{3}{11}$ min past 6

B - 49$\frac{1}{11}$ min past 6 and 16$\frac{4}{11}$ min past 6

C - 49$\frac{2}{11}$ min past 6 and 16$\frac{3}{11}$ min past 6

D - 49$\frac{3}{11}$ min past 6 and 16$\frac{4}{11}$ min past 6

Answer : B

Explanation

(5n ± 15) × $\frac{12}{11}$ min past 6, here n = 6

Therefore 49$\frac{1}{11}$ min past 6 and 16$\frac{4}{11}$ min past 6

Q 7 - How much time does a clock gain if it coincide per 62 min in a day?

A - 80$\frac{2}{341}$

B - 81$\frac{2}{341}$

C - 82$\frac{2}{341}$

D - 83$\frac{2}{341}$

Answer : A

Explanation

($\frac{720}{11}$ - x) × ($\frac{60 × 24}{x}$); given x = 62 then 80$\frac{2}{341}$

Q 8 - We decided to go a place at 30 min before 8.40 a.m, Ram found himself 40 min earlier than Hari who is 20 min late. Then what was the perfect time?

Answer : B

Explanation

Ram`s reaching time = 8.10 am

Hari found Ram at 45 min earlier = 8.50 am

The perfect time = 8.30 am

Q 9 - A clock was late by 5 minutes at noon on Sunday, but on Friday at 3P.M it is found that clock was fast by 10 minutes. Find the time when time was correct.

A - On Monday at 6P.M

B - On Tuesday at 5A.M

C - On Tuesday at 5P.M

D - On Thursday at 7A.M

Answer : B

Explanation

It is 123 hours from Sunday noon to Friday 3P.M. Therefore clock gained 15 minutes in 123 hrs, thus 5 minute can be gained in $\frac{120}{15}$ × 5 hrs = 40 hrs = on Tuesday at 5 A.M

Q 10 - At what time between 7 O`clock and 8 O`clock in the morning will the hands of a clock be at right angle?

A - $\frac{500}{11}$ min past 7 and $\frac{340}{11}$ min past 7

B - $\frac{440}{11}$ min past 7 and $\frac{280}{11}$ min past 7

C - $\frac{600}{11}$ min past 7 and $\frac{240}{11}$ min past 7

D - $\frac{340}{11}$ min past 7 and $\frac{120}{11}$ min past 7

Answer : C

Explanation

(5n ± 15) × $\frac{12}{11}$; here n = 7

Then, $\frac{600}{11}$ min past 7 and $\frac{240}{11}$ min past 7

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