TIFR CSE 2010 | Part A | Question: 2

Question

The hour hand and the minute hands of a clock meet at noon and again at mid-night. In between they meet $N$ times, where $N$ is.:

• A. $6$
• B. $11$
• C. $12$
• D. $13$
• E. None of the above

Answer 1

Clearly , Both hands meet at 12:00.

Apart from this During the day they meet at 1:05 , 2:10, 3:15 ...so on upto 12:00.

Total = 12

We have to exclude 12:00 .. so 11 such combinations

Comments

The meeting times are not perfect :)

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@Arjun sir, can u check the answer by pramod. 

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The question is for the number of meets in between noon and mid-night, which says to exclude the last meet.
Hence, the answer is 10.

Answer 2

Answer:- E

Hr hand & Minute hand will meet at12PM(Noon) should be excluded

1st meet :- Between 1 and 2

2nd meet :- Between 2 and 3

3rd meet :- Between 3 and 4

4th meet :- Between 4 and 5

5th meet :- Between 5 and 6

6th meet :- Between 6 and 7

7th meet :- Between 7 and 8

8th meet :- Between 8 and 9

9thmeet :- Between 9 and 10

10th meet :- Between 10 and 11

now after that, Hr hand & Minute hand will meet at 12AM(midnight)

so, only 10 meets between noon & midnight

Answer 3

Its 11 times,  option B is correct

Answer 4

$$
\begin{align}
&60 \text{ min} \rightarrow \dfrac{360}{12} = 30^{\circ} \text{ hour hand moves}\\
&1 \text{ min} \rightarrow 0.5^{\circ} \text{ hour hand moves}  \\
&1^{\circ} \rightarrow \dfrac{1}{6} \text{ min} \ (360^{\circ }\rightarrow 60 \text{ min})\\
&\text{Speed of minute hand} (m_v) = 6^{\circ}\;\text{min}^{-1} = \Big(\dfrac{1}{10}\Big)^{\circ} \; s^{-1}\\
&\text{Speed of hour hand} (h_v) = 0.5^{\circ}\;\text{min}^{-1} = \Big(\dfrac{1}{120}\Big)^{\circ} \; s^{-1}\\
&\text{Current position of hour hand} (h_c) \\
&\text{time} = \dfrac{\text{Angular Distance}}{\text{Angular Speed}} \\
\\
\hline
\\
&t_h = \dfrac{x^{\circ}}{h_v}\\
&t_m = \dfrac{h_c \times 30+x}{m_v} \\
&\text{Now }, t_m = t_h\\
&\dfrac{(h_c \times 30)^{\circ}+x^{\circ}}{m_v} = \dfrac{x^{\circ}}{h_v} \\
&x^{\circ} = \dfrac{(30 \times h_c)^{\circ} \times h_v}{m_v-h_v} = \Big(\dfrac{30}{11}h_c\Big)^{\circ} =
\Big(\dfrac{30}{11}h_c\Big) \times \dfrac{1}{6} \text{ min} = \Big(\dfrac{5}{11}h_c\Big) \text{ min} =
\Big(\dfrac{300}{11}h_c\Big) \text{ s}\\
&\text{time when hour and minute hand meet} = \underbrace{h_c}_{\text{Hour}}: \underbrace{h_c\times
5}_{\text{Minute}}:\underbrace{\dfrac{300}{11}h_c}_{\text{Seconds}} \\
\\
\hline
\\
&\begin{array}{|c|c|}
\hline
\textbf{h}_c & \textbf{meet at} \\
\hline
1 & 1:05:27 \\
2 & 2:10:54 \\
3 & 3:16:21 \\
\vdots & \vdots \\
10 & 10: 54: 32 \\
11 & 12:00:00 \\
\hline
\end{array}
\end{align}
$$