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

Let tistance between A and B be $$x$$ km, then
$$\eqalign{ & \frac{x}{9} - \frac{x}{{10}} = \frac{{36}}{{60}} \cr & \Rightarrow \frac{x}{{90}} = \frac{3}{5} \cr & \Rightarrow x = \frac{3}{5} \times 90 \cr & \Rightarrow x = 54\,{\text{km}} \cr & {\text{Hence option B is correct}} \cr} $$

---

$$\eqalign{ & \text{Alternate Solution:} \cr & {\text{Here,}}\,{s_1} = 9,\,{t_1} = x \cr & {s_2} = 10,\,{t_2} = x - \frac{{36}}{{60}} \cr & {s_1}{t_1} = {s_2}{t_2} \cr & \Rightarrow 9 \times x = 10\times\left( {x - \frac{{36}}{{60}}} \right) \cr & \Rightarrow 9 9x = 10x - 6 \cr & \Rightarrow 9 x = 6 \cr & \therefore {\text{ Distance }}\,{\text{travelled}} \cr & = s_1 \times t_1 \cr & = 9 \times 6 \cr & = 54\,\,{\text{km}} \cr} $$

Let the speed of Ravi = x km/hr
Then Ajay's speed will be = (x + 4) km/hr
Total distance, covered by Ajay :
= (60 + 12) km
= 72 km
Total distance, covered by Ravi :
= (60 - 12) km
= 48 km
According to the question,
They run at same time
$$\eqalign{ & \Rightarrow \frac{{72}}{{(x + 4)}} = \frac{{48}}{x} \cr & \Rightarrow 72x = 48x + 192 \cr & \Rightarrow 24x = 192 \cr & \Rightarrow x = 8{\text{ km/hr}} \cr} $$
Therefore, Ravi's speed = 8 km/hr

Total distance
= $$4 \times 3\frac{3}{4}$$
= 15 km
Time taken on cycle :
= $$\frac{{15}}{{16.5}} \times 60$$
= 54.55 minutes

As we know :
⇒ 1 m/s = $$\frac{{18}}{5}$$ km/hr
⇒ 20 m/s = $$\frac{{18}}{5}$$ × 20 km/hr
                = 72 km/hr

Let the speed of train = $$x$$ km/hr
Length of train = 300 mitres
Their relative speed in same direction :
= ($$x$$ - 3) km/hr
According to the question,
$$\frac{{\left( {300 + 0} \right){\text{m}}}}{{\left( {x - 3} \right) \times \frac{5}{{18}}{\text{ m/s}}}} = 33$$
[Here man's length is 0 metre]
$$\eqalign{ & \Rightarrow \frac{{300 \times 18}}{{\left( {x - 3} \right) \times 5}} = 33 \cr & \Rightarrow \frac{{100 \times 18}}{{5x \times 15}} = 11 \cr & \Rightarrow 1800 = 55x - 165 \cr & \Rightarrow 55x = 1965 \cr} $$
∴ Speed of the train :
$$\eqalign{ & \Rightarrow x = \frac{{1965}}{{55}} \cr & \Rightarrow x = 35\frac{8}{{11}}{\text{ km/hr}} \cr} $$

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ A }}:{\text{ B}} \cr & {\text{Ratio of speed = }}\,4\,\,:\,\,5 \cr & {\text{Ratio of time = }}\,\,\,\,5{\text{ }}:{\text{ }}4{\text{ }} \cr & (5 - 4){\text{ unit = 15 unit}} \cr & {\text{1 unit = 15 min}} \cr & {\text{So, time taken by B :}} \cr & {\text{ = 4}} \times {\text{15}} \cr & {\text{ = 60 min}} \cr & {\text{ = 1 hr}} \cr & {\text{Distance = S }} \times {\text{ T }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = 50}} \times {\text{1 }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = 50 km}} \cr} $$

Time taken to ride one way :
= $$\frac{3}{2}$$
= 1.5 hrs
Time taken to walk one way :
= 4.5 - 1.5
= 3 hrs
Time taken to walk both way :
= 3 × 2
= 6 hours

Distance covered in 1^st 2 hr :
= 2 × 20
= 40 km
Time taken to cover 60 km distance
= $$\frac{60}{10}$$
= 6 hours
$$\eqalign{ & {\text{Average speed}} \cr & = \frac{{{\text{Total distance}}}}{{{\text{Total time}}}} \cr & = \frac{{100}}{{2 + 6}} \cr & = \frac{{100}}{8} \cr & = 12\frac{1}{2}{\text{ km/hr}} \cr} $$

Circumference of wheel = 2πr
⇒ 2πr
⇒ 2 × $$\frac{22}{7}$$ × $$\frac{70}{2}$$
⇒ 220 cm
Speed per hours :
= $$\frac{{220 \times 400 \times 60}}{{1000 \times 100}}$$
= 52.8 km/hr

If the required distance be x km, then
$$\eqalign{ & \frac{x}{3} - \frac{x}{4} = \frac{1}{2} \cr & \Rightarrow \frac{{4x - 3x}}{{12}} = \frac{1}{2} \cr & \Rightarrow \frac{x}{{12}} = \frac{1}{2} \cr & \Rightarrow x = 6\,\,{\text{km}} \cr} $$