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

Let Speed of the man is x kmph.
Distance covered in 10 minutes at 20 kmph = distance covered in 8 minutes at (20 + x) kmph.
Or, $$20 \times \frac{{10}}{{60}} = \frac{8}{60}\times\left(20+x\right) $$
Or, 200 = 160 + 8x
Or, 8x = 40
Hence, x = 5kmph.

Detailed Explanation:
A _____________M_______________B
A = Bus Terminal.

B = Let meeting point of first bus and the man and this distance is covered by Bus in 10 minutes. I.e. Distance A to be is covered first bus in 10 min. As AB distance can be covered by second bus in 10 minutes as well.

Distance Covered by Bus in 10 min = AB = $$\frac{{20}}{{60}} \times 10$$   = $$\frac{{10}}{3}$$ km.

Now, M is the Meeting Point of Second Bus with Man. Man covered distance B to M in 8 minutes.
Now, Relative distance of both Man and Bus will be same as both are traveling in opposite direction of each other. Let Speed of the man = x kmph.

Relative speed = 20 + x

To meet at Point M, bus and Man has covered the distance (AB) in 8 minutes with relative speed. And Same AB distance is covered by bus in 10 minutes. Thus, Distance covered in 8 minutes with relative speed (20 + x) kmph = distance covered by bus in 10 minuted with speed 20 kmph.

1^st method:
$$\frac{4}{3}$$ of usual time = Usual time + 16 minutes;
Hence, $$\frac{1}{3}{\text{rd}}$$  of usual time = 16 minutes;
Thus, Usual time = 16 × 3 = 48 minutes.

2^nd method:
When speed goes down to
$$\frac{3}{4}{\text{th}}$$  (i.e. 75%) time will go up to $$\frac{4}{3}{\text{rd}}$$  (or 133.33%) of the original time.
Since, the extra time required is 16 minutes; it should be equated to $$\frac{1}{3}{\text{rd}}$$  of the normal time.
Hence, the usual time required will be 48 minutes.

Difference in time of departure between two trains = 45 min. = $$\frac{{45}}{{60}}$$ hour = $$\frac{{3}}{{4}}$$ hour.
Let the distance be x km from Delhi where the two trains will be together.
Time taken to cover x km with speed 136 kmph be t hour
and time taken to cover x km with speed 100 kmph (As the train take 45 mins. more) be
$$ {{\text{t}} + \frac{3}{4}} $$
= $$ {\frac{{4{\text{t}} + 3}}{4}} $$

Now,
100 × $$ {\frac{{4{\text{t}} + 3}}{4}} $$  = 136t
Or, 25(4t + 3) = 136t
Or, 100t + 75 = 136t
Or, 36t = 75
Or, t = $$\frac{{75}}{{36}}$$ = 2.083 hours
Then, distance x km = 136 × 2.083 ≈ 283.33 km.

Time taken in walking both the ways = 7 hours 45 minutes -------- (i)
Time taken in walking one way and riding back = 6 hours 15 minutes ----------- (ii)
By the equation (ii) × 2 - (i), we have,
Time taken by the man in riding both ways,
= 12 hours 30 minutes - 7 hours 45 minutes
= 4 hours 45 minutes.

$$\eqalign{ & {\text{Let the total distance be 100 km}}. \cr & {\text{Average speed}} \cr & = \frac{{{\text{total}}\,{\text{distance}}\,{\text{covered}}}}{{{\text{time}}\,{\text{taken}}}} \cr & = \frac{{100}}{{ { {\frac{{30}}{{20}}} + {\frac{{60}}{{40}}} + {\frac{{10}}{{10}}} } }} \cr & = \frac{{100}}{{ { {\frac{3}{2}} + {\frac{3}{2}} + 1 } }} \cr & = \frac{{100}}{{ {\frac{{ {3 + 3 + 2} }}{2}} }} \cr & = \frac{{ {100 \times 2} }}{8} \cr & = 25\,\text{kmph} \cr} $$

Alternate

10% of journey's = 40 km
Then, total journey = 400 kms
$$\eqalign{ & {\text{And,}}\,{\text{Average speed}} \cr & = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr & 30\% {\text{ of journey}} \cr & = 400 \times \frac{{30}}{{100}} \cr & = 120{\text{ km}} \cr & \cr & 60\% {\text{ of journey}} \cr & = 400 \times \frac{{60}}{{100}} \cr & = 240{\text{ km}} \cr & \cr & 10\% {\text{ of journey}} \cr & = 400 \times \frac{{10}}{{100}} \cr & = 40{\text{ km}} \cr & {\text{Average speed}} \cr & = \frac{{400}}{{\frac{{120}}{{20}} + \frac{{240}}{{40}} + \frac{{40}}{{10}}}} \cr & = \frac{{400}}{{ {6 + 6 + 4} }} \cr & = \frac{{400}}{{16}} \cr & \therefore {\text{Average speed}} = 25{\text{ km/hr}} \cr} $$

A →_______60Km_________← B
Let the speed of A = x kmph and that of B = y kmph

According to the question;
x × 6 + y × 6 = 60
Or, x + y = 10 --------- (i)
And,
$$\left( {\frac{{2{\text{x}}}}{3} \times 5} \right) + \left( {2{\text{y}} \times 5} \right) = 60$$
Or, 10x + 30y = 180
Or, x + 3y = 18 ---------- (ii)
From equation (i) × 3 - (ii)
3x + 3y - x - 3y = 30 - 18
Or, 2x = 12
Hence, x = 6 kmph

Alternate

$$\because $$ They meet after 6 hours if they walk towards each other i.e., their speed will be added.
So, their relative speed in opposite direction
$$ = \frac{{{\text{Distance }}}}{{{\text{Time }}}} = \frac{{60}}{6}$$
Relative speed in opposite direction :
$$\left( \rightleftharpoons \right) = 10{\text{ km/h}}.....{\text{(i)}}$$
According to the question,
$$\eqalign{ & \Rightarrow \frac{2}{3}A + 2B = \frac{{60}}{5} \cr & \Rightarrow \frac{2}{3}A + 2B = 12 \cr & \Rightarrow A + 3B = 18 \cr & \Rightarrow B's{\text{ Speed = }}\frac{{18 - A}}{3} \cr & \Rightarrow A + B = 10 \cr & \Rightarrow A + \frac{{18 - A}}{3} = 10 \cr & \Rightarrow 3A + 18 - A = 30 \cr & \Rightarrow 2A = 12 \cr & \Rightarrow A{\text{'s speed = 6 km/h}} \cr} $$

Time taken to complete the revolution:
A → $$\frac{{12}}{{4}}$$ = 3 hours
B → $$\frac{{12}}{{3}}$$ = 4 hours
C → 12 × $$\frac{{2}}{{3}}$$ = 8 hours
Required time,
= LCM of 3, 4, 8.
= 24 hours.

Ajay → (x + 4) kmph.
A ________ 60 km _________ B
Ravi → x kmph.

Let the speed of Ravi be x kmph;
Hence, Ajay's speed = (x + 4) kmph;
Distance covered by Ajay = 60 + 12 = 72 km;
Distance covered by Ravi = 60 - 12 = 48 km.

According to question,
$$\eqalign{ & \frac{{72}}{{x + 4}} = \frac{{48}}{x} \cr & {\text{or,}}\,\frac{3}{{x + 4}} = \frac{2}{x} \cr & {\text{or,}}3x = 2x + 8 \cr & {\text{or,}}x = 8\,{\text{kmph}} \cr} $$

Ratio of speed = 3 : 4
Ratio of time taken = 4 : 3 (As Speed ∝ $$\frac{1}{{{\text{Time}}}},$$   When distance remains constant.)
Let time taken by A and B be 4x and 3x hour respectively.

Then,
4x - 3x = $$\frac{{20}}{{60}}$$
Or, x = $$\frac{{1}}{{3}}$$
Hence, time taken by A = 4x
hours = 4 × $$\frac{1}{3}$$ = $$1\frac{1}{3}$$ hours.

$$\eqalign{ & {\text{Average}}\,{\text{speed}} \cr & = \frac{{2xy}}{{ {x + y} }} \cr & = \frac{{2 \times 6 \times 3}}{{ {6 + 3} }} \cr & = \frac{{36}}{9} \cr & = 4\,{\text{kmph}} \cr} $$