Speed Time And Distance - Aptitude MCQ Questions and Solutions with Explanations

Time taken to cover 275 km = $$\frac{{275}}{{50}}$$ h = 5.5 h
Time taken to cover 315 km = $$\frac{{315}}{{70}}$$ h = 4.5 h
Average speed of entire journey
$$\eqalign{ & = \frac{{{\text{Total distance}}}}{{{\text{Total time taken}}}} \cr & = \frac{{590}}{{10}} \cr & = 59{\text{ Answer}} \cr} $$

Walking speed = 5 kmph
Jogging speed = 10 kmph
Running speed = 20 kmph
$$\eqalign{ & {\text{Average speed}} = \frac{{20}}{{\frac{5}{5} + \frac{5}{{10}} + \frac{5}{{10}} + \frac{5}{{20}}}} \cr & = \frac{{20}}{{2 + \frac{5}{{20}}}} \cr & = \frac{{20 \times 20}}{{45}} \cr & = \frac{{80}}{9}{\text{ kmph}} \cr} $$

Let speed of the bus = x
Relative speed in same direction = 75 - x
Total distance = 80 + 40 = 120 m
$$\eqalign{ & {\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & 75 - x = \frac{{120}}{{15}} \times \frac{{18}}{5} \cr & 75 - x = \frac{{144}}{5} \cr & 75 - x = 28.8 \cr & x = 75 - 28.8 \cr & x = 46.2{\text{ km/hr}} \cr} $$

$$\eqalign{ & \left( {\frac{{{\text{Faster speed}} - {\text{Slower speed}}}}{{{\text{Faster speed}}}} \times 60} \right){\text{min/hr}} \cr & = \frac{{80 - 64}}{{80}} \times 60 \cr & = \frac{{16}}{{80}} \times 60 \cr & = 12{\text{ min}} \cr} $$

\[\begin{array}{*{20}{c}} {}&{\text{A}}&:&{\text{B}} \\ {{\text{Time}}}&8&{}&6 \end{array}\]
And 7 × A = B × 5
Distance A : B = 5 : 7
\[\begin{array}{*{20}{c}} {{\text{Ratio}}}&{\text{A}}&:&{\text{B}} \\ {}&8&:&6 \\ {}&5&:&7 \\ {}&{\overline {\,40\,} }&{}&{\overline {\,42\,} } \\ {}&{20}&:&{21} \end{array}\]

Let the speed of the A be a and B be b
According to the question,
$$\frac{{\text{d}}}{{\text{a}}} - \frac{{\text{d}}}{{\text{b}}} = 3 - - - - \left( {\text{i}} \right)$$
Again,
$$\frac{{\text{d}}}{{\text{b}}} - \frac{{\text{d}}}{{2{\text{a}}}} = 1 - - - - \left( {{\text{ii}}} \right)$$
By adding equation (i) and (ii) we get,
$$\eqalign{ & \frac{{\text{d}}}{{2{\text{a}}}} = 4 \cr & \Rightarrow \frac{{\text{d}}}{{\text{a}}} = 4 \times 2 \cr & \Rightarrow \frac{{\text{d}}}{{\text{a}}} = 8 \cr} $$
So, it will take 8 hours by A to cover the distance d
∴ A requires 8 hours to walk 'd' km.

$$\eqalign{ & \frac{{{{\text{S}}_{\text{C}}}}}{{{{\text{S}}_{\text{B}}}}} = \frac{{\frac{{60}}{{\frac{3}{4}}}}}{{\frac{{60}}{{\frac{3}{2}}}}} \cr & {{\text{S}}_{\text{C}}}:{{\text{S}}_{\text{B}}} = 2:1 \cr} $$

Let the distance travelled on foot be $$d$$
Time taken on foot + Time taken on bicycle = Total time taken
$$\eqalign{ & \Rightarrow \frac{d}{6} + \frac{{42 - d}}{{10}} = 5 \cr & \Rightarrow \frac{{d \times 5 + 3 \times \left( {42 - d} \right)}}{{30}} = 5 \cr & \Rightarrow 5d + 126 - 3d = 5 \times 30 \cr & \Rightarrow 2d + 126 = 150 \cr & \Rightarrow 2d = 150 - 126 \cr & \Rightarrow 2d = 24 \cr & \Rightarrow d = 12 \cr} $$
∴ The distance travelled on foot is 12 km

$$\eqalign{ & {\text{Total distance}} \cr & = 6 \times 2 + 2 \times 1 + \frac{{1 \times 18}}{5} \times 1 + 4 \times 3 \cr & = 12 + 2 + 12 + 3.6 \cr & = 29.6\,{\text{km}} \cr & {\text{Total time}} \cr & = 2 + 1 + 1 + \frac{1}{2} + 1 + 3 \cr & = 8\frac{1}{2} \cr & = \frac{{17}}{2}{\text{hr}}{\text{.}} \cr & {\text{Average speed}} \cr & = \frac{{29.6}}{{17}} \times 2 \cr & = \frac{{59.2}}{{17}} \cr & = 3.4823 \cr} $$

$$\eqalign{ & \Rightarrow \frac{{1125}}{{\left( {11 - x} \right)\frac{5}{{18}}}} = 810 \cr & \Rightarrow \frac{{1125 \times 18}}{{\left( {11 - x} \right) \times 5}} = 810 \cr & \Rightarrow 5 = 11 - x \cr & \Rightarrow x = 6 \cr} $$
Distance of thief = 810 × 6 × $$\frac{5}{{18}}$$ = 1350 m Ans.