Clock - Aptitude MCQ Questions and Solutions with Explanations

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

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

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

Time from 7 a.m. to 4:15 p.m. = 9 hrs 15 min. = $$\frac{{37}}{4}$$ hrs
3 min. 5 sec. of this c;ocl = 3 min. of the correct clock
⇒ $$\frac{{37}}{{720}}$$ hrs. of this clock = $$\frac{1}{{20}}$$ hrs of the correct clock
⇒ $$\frac{{37}}{4}$$ hrs. of this clock = $$\left( {\frac{1}{{20}} \times \frac{{720}}{{37}} \times \frac{{37}}{4}} \right)$$     hrs. of the correct clock
= 9 hrs. of the correct clock
∴ The correct time is 9 hrs. after 7 a.m. i.e., 4 p.m.

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

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
$$\eqalign{ & = \left( {\frac{{60}}{{55}} \times 5} \right){\kern 1pt} \min . \cr & = 5\frac{5}{{11}}{\kern 1pt} \min . \cr & \therefore {\text{Required time}} = 5\frac{5}{{11}}{\kern 1pt} \min .{\kern 1pt} \,{\text{past}}{\kern 1pt} 7 \cr} $$

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
$$\eqalign{ & = \left( {\frac{{60}}{{55}} \times 40} \right){\kern 1pt} {\kern 1pt} \min . \cr & = 43\frac{7}{{11}}{\kern 1pt} {\kern 1pt} \min . \cr & \therefore {\text{Required time}} = 43\frac{7}{{11}}{\kern 1pt} {\kern 1pt} \min .\,{\text{past}}\,5 \cr} $$

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

$$\eqalign{ & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{hour}}\,{\text{hand}}\,{\text{in}}\,\frac{{21}}{4}\,{\text{hrs}} \cr & = {\left( {\frac{{360}}{{12}} \times \frac{{21}}{4}} \right)^ \circ } = 157{\frac{1}{2}^ \circ } \cr & {\text{Angle}}\,{\text{traced}}\,{\text{by}}\,{\text{min}}{\text{.}}\,{\text{hand}}\,{\text{in}}\,15\,{\text{min}} \cr & = {\left( {\frac{{360}}{{60}} \times 15} \right)^ \circ } = {90^ \circ } \cr & \therefore {\text{Required}}\,{\text{angle}} \cr & = {\left( {157\frac{1}{2}} \right)^ \circ } - {90^ \circ } \cr & = 67{\frac{1}{2}^ \circ } \cr} $$

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