Clock

1.An accurate clock shows 8 o'clock in the morning. Through how may degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon?

1. 144º
2. 150º
3. 168º
4. 180º

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Angle traced be the hour hand in 6 hours \\ = {(\frac{360}{12} \times 6)}^{\circ} = 180^{\circ} \end{aligned}

$\eqalign{ & {\text{Angle}}\,{\text{traced}}\,{\text{be}}\,{\text{the}}\,{\text{hour}}\,{\text{hand}}\,{\text{in}}\,{\text{6}}\,{\text{hours}} \cr & = {\left( {\frac{{360}}{{12}} \times 6} \right)^ \circ } = {180^ \circ } \cr}$

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2.The reflex angle between the hands of a clock at 10.25 is:

1. 180º
2. 192^1º/₂
3. 195º
4. 197^1º/₂

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Angle traced by hour hand in \frac{125}{12} h r s \\ = {(\frac{360}{12} \times \frac{125}{12})}^{\circ} = 312 \frac{1^{\circ}}{2} \\ Angle traced by minute hand in 25 min \\ = {(\frac{360}{60} \times 25)}^{\circ} = 150^{\circ} \\ ∴ Reflex angle \\ = 360^{\circ} - {(312 \frac{1}{2} - 150)}^{\circ} \\ = 360^{\circ} - 162 \frac{1^{\circ}}{2} \\ = 197 \frac{1}{2} \end{aligned}

$\eqalign{ & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{hour}}\,{\text{hand}}\,{\text{in}}\,\frac{{125}}{{12}}hrs \cr & = {\left( {\frac{{360}}{{12}} \times \frac{{125}}{{12}}} \right)^ \circ } = 312\frac{{{1^ \circ }}}{2} \cr & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{minute}}\,{\text{hand}}\,{\text{in}}\,{\text{25}}\,\min \cr & = {\left( {\frac{{360}}{{60}} \times 25} \right)^ \circ } = {150^ \circ } \cr & \therefore {\text{Reflex}}\,{\text{angle}} \cr & = {360^ \circ } - {\left( {312\frac{1}{2} - 150} \right)^ \circ } \cr & = {360^ \circ } - 162\frac{{{1^ \circ }}}{2} \cr & = 197\frac{1}{2} \cr}$

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3.A clock is started at noon. By 10 minutes past 5, the hour hand has turned through:

1. 145º
2. 150º
3. 155º
4. 160º

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Angle traced by hour hand in 12 h r s = 360^{\circ} \\ Angle traced by hour hand in 5 h r s 10 min . i . e ., \\ \frac{31}{6} h r s = {(\frac{360}{12} \times \frac{31}{6})}^{\circ} = 155^{\circ} \end{aligned}

$\eqalign{ & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{hour}}\,{\text{hand}}\,{\text{in}}\,12\,hrs = {360^ \circ } \cr & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{hour}}\,{\text{hand}}\,{\text{in}}\,5\,hrs\,10\,\min .\,i.e., \cr & \frac{{31}}{6}hrs = {\left( {\frac{{360}}{{12}} \times \frac{{31}}{6}} \right)^ \circ } = {155^ \circ } \cr}$

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4.How much does a watch lose per day, if its hands coincide every 64 minutes?

1. 32 ⁸/₁₁ min.
2. 36 ⁵/₁₁ min.
3. 90 min.
4. 96 min.

Answer:Option 1

Explanation:

Solution:

\begin{aligned} 55 min . spaces are covered in 60 min \\ 60 min . spaces are covered in \\ = (\frac{60}{55} \times 60) min . \\ = 65 \frac{5}{11} min . \\ Loss in 64 min . \\ = (65 \frac{5}{11} - 64) = \frac{16}{11} min . \\ Loss in 24 h r s \\ = (\frac{16}{11} \times \frac{1}{64} \times 24 \times 60) min \\ = 32 \frac{8}{11} min \end{aligned}

$\eqalign{ & 55\,\min .\,{\text{spaces}}\,{\text{are}}\,{\text{covered}}\,{\text{in}}\,60\,\min \cr & 60\,\min .\,{\text{spaces}}\,{\text{are}}\,{\text{covered}}\,{\text{in}} \cr & = \left( {\frac{{60}}{{55}} \times 60} \right)\,\min . \cr & = 65\frac{5}{{11}}\,\min . \cr & {\text{Loss}}\,{\text{in}}\,64\,\min . \cr & = \left( {65\frac{5}{{11}} - 64} \right) = \frac{{16}}{{11}}\,\min . \cr & {\text{Loss}}\,{\text{in}}\,24\,hrs \cr & = \left( {\frac{{16}}{{11}} \times \frac{1}{{64}} \times 24 \times 60} \right)\,\min \cr & = 32\frac{8}{{11}}\,\min \cr}$

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5.At what time between 7 and 8 o'clock will the hands of a clock be in the same straight line but, not together?

1. 5 min. past 7
2. 5 ²/₁₁ min. past 7
3. 5 ³/₁₁ min. past 7
4. 5 ⁵/₁₁ min. past 7

Answer:Option 1

Explanation:

Solution:

When the hands of the clock are in the same straight line but not together, they are 30 minute spaces apart.
At 7 o'clock, they are 25 min. spaces apart.
∴ Minute hand will have to gain only 5 min. spaces.
55 min. spaces are gained in 60 min.
5 min. spaces are gained in

\begin{aligned} (\frac{60}{55} \times 5) min = 5 \frac{5}{11} min \\ ∴ Required time = 5 \frac{5}{11} min . p a s t 7 \end{aligned}

$\eqalign{ & \left( {\frac{{60}}{{55}} \times 5} \right)\,\min = 5\frac{5}{{11}}\,\min \cr & \therefore {\text{Required}}\,{\text{time}} = 5\frac{5}{{11}}\,\min .\,past\,7 \cr}$

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6.At what time between 5.30 and 6 will the hands of a clock be at right angles?

1. 43 ⁵/₁₁ min. past 5
2. 43 ⁷/₁₁ min. past 5
3. 40 min. past 5
4. 45 min. past 5

Answer:Option 1

Explanation:

Solution:

At 5 o'clock, the hands are 25 min. spaces apart.
To be at right angles and that too between 5.30 and 6, the minute hand has to gain (25 + 15) = 40 min. spaces.
55 min. spaces are gained in 60 min.
40 min. spaces are gained in

\begin{aligned} (\frac{60}{55} \times 40) min = 43 \frac{7}{11} min \\ ∴ Required time = 43 \frac{7}{11} min . p a s t 5 \end{aligned}

$\eqalign{ & \left( {\frac{{60}}{{55}} \times 40} \right)\,\min = 43\frac{7}{{11}}\,\min \cr & \therefore {\text{Required}}\,{\text{time}} = 43\frac{7}{{11}}\,\min .\,past\,5 \cr}$

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7.The angle between the minute hand and the hour hand of a clock when the time is 4.20, is:

1. 0º
2. 10º
3. 5º
4. 20º

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Angle traced by hour hand in \frac{13}{3} hrs \\ = {(\frac{360}{12} \times \frac{13}{3})}^{\circ} = 130^{\circ} \\ Angle traced by min . hand in 20 min \\ = {(\frac{360}{60} \times 20)}^{\circ} = 120^{\circ} \\ ∴ Required angle \\ = {(130 - 120)}^{\circ} \\ = 10^{\circ} \end{aligned}

$\eqalign{ & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{hour}}\,{\text{hand}}\,{\text{in}}\,\frac{{13}}{3}\,{\text{hrs}} \cr & = {\left( {\frac{{360}}{{12}} \times \frac{{13}}{3}} \right)^ \circ } = {130^ \circ } \cr & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{min}}{\text{.}}\,{\text{hand}}\,{\text{in}}\,{\text{20}}\,{\text{min}} \cr & = {\left( {\frac{{360}}{{60}} \times 20} \right)^ \circ } = {120^ \circ } \cr & \therefore {\text{Required}}\,{\text{angle}} \cr & = {\left( {130 - 120} \right)^ \circ } \cr & = {10^ \circ } \cr}$

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8.At 3:40, the hour hand and the minute hand of a clock form an angle of:

1. 120°
2. 125°
3. 130°
4. 135°

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Angle traced by hour hand in 12 hrs = 36 0^{\circ} \\ Angle traced by it in \frac{11}{3} hrs \\ = {(\frac{360}{12} \times \frac{11}{3})}^{\circ} = 110^{\circ} \\ Angle traced by min . hand in 60 min = 36 0^{\circ} \\ Angle traced by it in 40 min . \\ = {(\frac{360}{60} \times 40)}^{\circ} = 240^{\circ} \\ ∴ Required angle \\ = {(240 - 110)}^{\circ} = 130^{\circ} \end{aligned}

$\eqalign{ & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{hour}}\,{\text{hand}}\,{\text{in}}\,12\,{\text{hrs}}\,{\text{ = }}\,{\text{36}}{{\text{0}}^ \circ } \cr & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{it}}\,{\text{in}}\,\frac{{11}}{3}\,{\text{hrs}} \cr & = {\left( {\frac{{360}}{{12}} \times \frac{{11}}{3}} \right)^ \circ } = {110^ \circ } \cr & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{min}}{\text{.}}\,{\text{hand}}\,{\text{in}}\,60\,{\text{min}}\,{\text{ = }}\,{\text{36}}{{\text{0}}^ \circ } \cr & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{it}}\,{\text{in}}\,{\text{40}}\,{\text{min}}. \cr & = {\left( {\frac{{360}}{{60}} \times 40} \right)^ \circ } = {240^ \circ } \cr & \therefore {\text{Required}}\,{\text{angle}} \cr & = {\left( {240 - 110} \right)^ \circ } = {130^ \circ } \cr}$

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9.How many times are the hands of a clock at right angle in a day?

1. 22
2. 24
3. 44
4. 48

Answer:Option 1

Explanation:

Solution:

In 12 hours, they are at right angles 22 times.
∴ In 24 hours, they are at right angles 44 times.

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10.The angle between the minute hand and the hour hand of a clock when the time is 8.30, is:

1. 80°
2. 75°
3. 60°
4. 105°

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Angle traced by hour hand in \frac{17}{2} hrs \\ = {(\frac{360}{12} \times \frac{17}{2})}^{\circ} = 255 \\ Angle traced by min . hand in 30 min \\ = {(\frac{360}{60} \times 30)}^{\circ} = 180 \\ ∴ Required angle \\ = {(255 - 180)}^{\circ} = 75^{\circ} \end{aligned}

$\eqalign{ & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{hour}}\,{\text{hand}}\,{\text{in}}\,\frac{{17}}{2}\,{\text{hrs}} \cr & \,{\text{ = }}\,{\left( {\frac{{360}}{{12}} \times \frac{{17}}{2}} \right)^ \circ } = 255 \cr & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{min}}{\text{.}}\,{\text{hand}}\,{\text{in}}\,{\text{30}}\,{\text{min}} \cr & = {\left( {\frac{{360}}{{60}} \times 30} \right)^ \circ } = 180 \cr & \therefore {\text{Required}}\,{\text{angle}} \cr & = {\left( {255 - 180} \right)^ \circ } = {75^ \circ } \cr}$

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