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

Let the usual speed be S.
So, increased speed = $$\frac{{3{\text{S}}}}{2}$$
Let usual time taken be T
Thus, S × T = D, and also $$\frac{{3{\text{S}}}}{2}$$ × (T - 40) = D. (T - 40, Because he started 30 minutes later and reached 10 minutes earlier, means he saved 40 minutes)
$$\frac{{3{\text{S}}}}{2}$$ × (T - 40) = ST
$$\frac{{3 \times {\text{ST}}}}{2}$$  - 60S = ST
$$\frac{1}{2}$$ × ST = 60S
T = 120 minutes = 2 hours

Alternatively,
As, ST = D [S ∝ $$\frac{1}{{\text{T}}}$$ , speed is inversely proportional to time] Speed increase to $$\frac{{3}}{{2}}$$ times so time decreases to $$\frac{{2}}{{3}}$$ times .
Net decrease = $$\frac{{1}}{{3}}$$ = 40 minutes.
Hence actual time taken
= 3 × 40 = 120 mnts.
= 2 hours

The ratio of speeds is 2 : 5. So when slower one completes 2 one way journeys (and reaches its starting point), faster one travels 5 one way journeys (and reaches the other end). So that means after the faster one has traveled 5 one way journeys, both of them have reached same end and in next 5 one-way journey of faster runner, both reach their starting position simultaneously. Now most important to observe is that FASTER one will always meet the SLOWER one EXACTLY ONCE in each of its one-way journey, except when both of them have started with the same starting point.
Once you reduce that for the 5^th time, they'll meet at A, and the next 5 rounds will have 4 meetings.
Then, just add 5 + 4 + 5 + 4 + 5 = 23 or just go by the position of Plant after every 5 rounds : A, B, A, B, A. The cycle repeats.
So, total distance will be 0 meters.

Suppose both moves with the same speed 60 km/h then they will meet max 10 times. So answer will be less than 10. Thus correct answer will be 8.

P_____60m___R__Xm____40m____Q
Total distance = 100 + X
Initially, A traveled 60m and B traveled (40 + X)m. Since time is same. So, Speed ∝ Distance.
$$\frac{{{{\text{S}}_{\text{a}}}}}{{{{\text{S}}_{\text{b}}}}} = \frac{{60}}{{40 + {\text{X}}}}$$
Second time,
So, now total distance traveled by A = 100 + X + 40
Total distance traveled by B = 100 + X + 60 + X
These will be in the ratio of Speed of A and B
Thus,
$$\frac{{{{\text{S}}_{\text{a}}}}}{{{{\text{S}}_{\text{b}}}}} = \frac{{60}}{{40 + {\text{X}}}} = \frac{{100 + {\text{X}} + 40}}{{100 + {\text{X}} + 60 + {\text{X}}}}$$
X = 40
Thus Total distance = 100 + X = 100 + 40 = 140 m

Let the speed of the rickshaw = S km/h. Let after time T they meet.
So, 60t + ST = $$\frac{{4 \times 60}}{5}$$ $$\left[ {{\text{Since,}}\,\,48\,{\text{min}} = \frac{4}{5}\,\,{\text{hour}}} \right]$$
Rohan saves 16 min. by making Shikha move as well. So, she saves 16 min by moving for distance X (Let).
2X = $$\frac{{60 \times 16}}{5}$$
X = 8 km
Speed of Auto rickshaw = 12 km. $$\left( {{\text{ST}} = 8\,{\text{km,}}\,\,{\text{T}} = \frac{2}{3}} \right)$$

Let usual speed be X kmph, then new speed will be (x + 400) kmph.
Time taken to cover 1600 km with speed X kmph,
= $$\frac{{1600}}{{\text{x}}}$$
Time taken to cover 1600 km with Speed (x + 400) kmph,
= $$\frac{{1600}}{{{\text{x}} + 400}}$$
Now,
Time difference = 40 minutes.
$$\frac{{1600}}{{\text{x}}} - \frac{{1600}}{{{\text{x}} + 400}} = \frac{{40}}{{60}}$$     hours
x² + 400x - 960000 = 0 On solving,
x = -1200, 800
Speed cannot be negative, So usual speed will be 800 km/hour

P__150m___T______x m (Let)_______Q.
Let Policeman caught thief at a distance (x + 150)m. And Thief has traveled x m.
Speed of Policeman
$$\eqalign{ & = 12\,\,{\text{kmph}} \cr & = \frac{{12 \times 5}}{{18}} \cr & = \frac{{60}}{{18}}\,\,{\text{m/sec}} \cr} $$
Speed of thief
$$\eqalign{ & = 10\,\,{\text{km}} \cr & = \frac{{10 \times 5}}{{18}} \cr & = \frac{{50}}{{18}}\,\,{\text{m/sec}} \cr} $$
In this case time is constant means Policeman covered (x + 150)m in same time thief covered x m.
Thus,
$$\eqalign{ & \frac{{{\text{Speed of the thief}}}}{{{\text{Speed of Policeman}}}} = \frac{x}{{150 + x}} \cr & \Rightarrow \frac{{50}}{{60}} = \frac{x}{{150 + x}} \cr & \Rightarrow 7500 + 50x = 60x \cr & \Rightarrow 10x = 7500 \cr & \Rightarrow x = 750\,{\text{m}} \cr} $$
So, Thief has traveled 750 m before the caught.

Ratio of speed of Bus and Train = 15 : 27
Let speed of the bus is 15X and Speed of the Train is 27X
Car Covered 720 km in 9 hours.
So, Speed of the Car = $$\frac{{720}}{9}$$  = 80 kmph
Given, Speed of the bus is $$\frac{3}{4}$$ of Car, So speed of the Bus,
= $$\frac{{80 \times 3}}{4}$$   = 60 kmph
Thus,
15X = 60
X = 4
So, Speed of the train = 27X = 27 × 4 = 108 kmph.
Hence, Train will cover distance in 7 hours,
= 108 × 7
= 756 km

P__________3.5Km__________M__1km____Q
Speed of A = 1 km in 18 min = $$\frac{{1000}}{{18}}$$  = 55.55 m/min.
When A travels 3.5 km. B already has been traveled (4.5 + 1) = 5.5 km.
A will take time to travel 3.5 km = $$\frac{{3500}}{{55.55}}$$  = 63 min.
So, B will take 66 min to travel 5.5 km (As B has started 3 min before of A)
Thus, speed of B = $$\frac{{5500}}{{66}}$$  = 83.33 m/min = 5 km/h

P __________100km__________Q
Let car takes time T hours to reach destination.
So, Bike will take $$\left( {{\text{T}} + \frac{{25}}{{60}}} \right)$$
Let speed of the bike = S kmph
Speed of Car = S + 50% of S = $$\frac{{3{\text{S}}}}{2}$$ kmph
For the both the case distance is constant. And when distance remain constant then time is inversely proportional to speed (As ST + D)
So,
$$\eqalign{ & \frac{{ {\frac{{3S}}{2}} }}{{\left( S \right)}} = \frac{{ {T + {\frac{5}{{12}}} } }}{T} \cr & 3T = 2T + \frac{{10}}{{12}} \cr & T = \frac{{10}}{{12}}{\text{hours}} \cr & {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{car}} \cr & \frac{{3S}}{2} = \frac{{100}}{{ {\frac{{10}}{{12}}} }} \cr & \frac{{3S}}{2} = 120 \cr & S = 80\,kmph \cr & {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{bike}} = 80\,kmph \cr} $$