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

In a 6 × 6 grid of a chessboard, each row and each column contains 3 white and 3 black squares placed alternatively
There are a total of 18 black and 18 white squares
For every black square chosen to put one coin, we cannot choose any white square present in its row or column.
There are 3 white squares in its row and 3 white square in its column for every black square.
Hence for every black square chosen, we can choose (18 −6) = 12 white squares.
Total number of possibilities where a black square and a white square can be chosen so that they do not fall in the same row or in the same column,
= 18 × 12
= 216
So, there are 216 ways of placing the coins that are identical.

Average speed for this case is given by,
$$ {\frac{{2xy}}{{x + y}}} ,$$   where X and Y are two different speeds.
Average speed,
$$\eqalign{ & = {\frac{{2 \times 36 \times 45}}{{36 + 45}}} \cr & = \frac{{36 \times 90}}{{81}} \cr & = 40\,\,{\text{kmph}} \cr} $$

Let speed of C = X
Then Speed of B = 3X
Then Speed of A = 6X
Ratio of the speed of A and C = 1 : 6
So, Greater the speed less time taken in journey.
C’s speed is 6 times less than A So A will take $$\frac{1}{6}$$ of the total time taken C to covered same distance.
So, Time taken by A
$$\eqalign{ & = \frac{3}{{2 \times 6}} \cr & = \frac{1}{4}\,\,{\text{hours}} \cr & = 15\,\,{\text{minutes}} \cr} $$

Area of square = side × side= 25 sq. km.
Side = 5 km.
perimeter = 4 × 5 = 20 km.
Time taken to complete one round with speed 3 kmph,
= $$\frac{{20}}{3}$$
= 6.66 hours
= 6 hour 40 minutes

Train was halted for half an hour
So, total time taken in Journey = 23 - $$\frac{1}{2}$$ = 22.5 hours
Average speed in Whole Journey = $$\frac{{1000}}{{22.5}}$$  = 44.5 km/hr
The average speed on return journey is 20% less than onward journey.
Therefore, ratio of average speed of onward and return journey,
$$\eqalign{ & = \frac{{100}}{{80}} \cr & = \frac{5}{4} \cr} $$
Let average speed of onward journey = 5x
Average speed on return journey = 4x
Average speed on whole journey = $$\frac{{5x + 4x}}{2}$$
44.5 = $$\frac{{5x + 4x}}{2}$$
89 = 9x
x = 9.88
Average speed on return
= 9.88 × 4
= 39.52
= 40 km/hr (Approx.)

Let the speeds of boat and stream was $$s$$ and $$v$$ km/hr respectively
Then, Actual Speed Downstream = $$\left(s + v\right)$$  km/hr
Actual Speed upstream = $$\left(s - v\right)$$  km/hr
According to question,
$$\eqalign{ & \frac{d}{{s + v}} + \frac{d}{{s - v}} = 5\,{\text{hr}}{\text{.}}\,15\,{\text{min}}{\text{.}} \cr & \Rightarrow \frac{d}{{s + v}} + \frac{d}{{s - v}} = \frac{{21}}{4}\,.\,.....\left( 1 \right) \cr & {\text{and}} \cr & \frac{{2d}}{{s - v}} = 7 \cr & \Rightarrow \frac{d}{{s - v}} = \frac{7}{2}\,......\left( 2 \right) \cr & {\text{By equation }}\left( 1 \right) - \left( 2 \right), \cr & \frac{d}{{s + v}} = \frac{{21}}{4} - \frac{7}{2} \cr & \Rightarrow \frac{d}{{s + v}} = \frac{{21 - 14}}{4} \cr & \Rightarrow \frac{d}{{s + v}} = \frac{7}{4} \cr & \Rightarrow \frac{{2d}}{{s + v}} = \frac{7}{2} \cr & \cr} $$
Hence, he takes $$\frac{{7}}{{2}}$$ hours to row 2d km distance downstream

Let the distance of the station from the house of the person = x km
$$\eqalign{ & {\text{Difference}}\,{\text{of}}\,{\text{time}} \cr & = 8 + 7 \cr & = 15\,{\text{minutes}} \cr & = \frac{1}{4}\,hr \cr & {\text{Since}},\, \cr & {\text{Time}} = \frac{{{\text{Distance}}}}{{{\text{Speed}}}} \cr & \therefore \frac{x}{6} - \frac{x}{{10}} = \frac{1}{4} \cr & \Rightarrow \frac{{10x - 6x}}{{60}} = \frac{1}{4} \cr & \Rightarrow \frac{{2x}}{{30}} = \frac{1}{4} \cr & \Rightarrow x = \frac{{15}}{4} = 3\frac{3}{4}km \cr} $$

Let length of the Platform is X m and Train is Y m.
Speed of the train = 54 kmph = $$\frac{{54 \times 5}}{{18}}$$ = 15 m/sec.
To cross the platform, train needs to travel (X + Y) m in 30 sec.
$$\eqalign{ & {\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & 15 = \frac{{{\text{X}} + {\text{Y}}}}{{30}} \cr & {\text{X}} + {\text{Y}} = 450\,.\,.\,.\,.\,.\,.\,.\,.\,.\,.\,.\left( 1 \right) \cr} $$
Now Platform is renovated and its length is doubled. So, Now, train need to travel (2X + Y) m to cross the platform.
Thus,
$$\eqalign{ & {\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & 15 = \frac{{{\text{2X}} + {\text{Y}}}}{{46}} \cr & {\text{2X}} + {\text{Y}} = 690\,.\,.\,.\,.\,.\,.\,.\,.\,.\,.\,.\left( 2 \right) \cr} $$
Multiplying equation (1) by (2)
2X + 2Y = 900 ------------------ (3)
Now, equation (2) - (3)
2X + Y - 2X - 2Y = 690 - 900
- Y = - 210
Y = 210
Length of the train = 210 metres

$$\eqalign{ & {\text{Speed}} = {\frac{{600}}{{5 \times 60}}} {\text{ m/sec}} = 2{\text{m/sec}} \cr & {\text{Converting}}\,{\text{m/sec}}\,{\text{to}}\,{\text{km/hr}} \cr & = {2 \times \frac{{18}}{5}} {\text{ km/hr}} \cr & = 7.2\,{\text{ km/hr}} \cr} $$

$$\eqalign{ & {\text{Distance}} = {240 \times 5} = 1200\,{\text{km}} \cr & {\text{Speed}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}} \cr & {\text{Speed}} = \frac{{1200}}{{ {\frac{5}{3}} }}\,{\text{ km/hr}}\, \cr & \left[ {{\text{We}}\,{\text{can}}\,{\text{write}}\,1\frac{2}{3}\,{\text{hours}}\,{\text{as}}\,\frac{5}{3}\,{\text{hours}}} \right] \cr & \therefore {\text{Required}}\,{\text{speed}} \cr & = {1200 \times \frac{3}{5}} \,{\text{ km/hr}} \cr & = 720\,{\text{ km/hr}} \cr} $$