Quantitative Aptitude
SPEED TIME AND DISTANCE MCQs
Time And Distance, Time & Distance, Speed Time & Distance
Let Abhay's speed be x km/hr.
Then, \(\frac{30}{x}-\frac{30}{2x}=3\)
6x = 30
x = 5 km/hr.
Let the distance travelled by x km.
Then,\(\frac{x}{10}-\frac{x}{15}=2\)
3x - 2x = 60
x = 60 km.
Time taken to travel 60 km at 10 km/hr = \(\left(\frac{60}{10}\right)hrs = 6hrs.\)
So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.
Therefore Required speed = \(\left(\frac{60}{5}\right)kmph.=12kmph.\)
Let the speed of the train be x km/hr and that of the car be y km/hr.
Then,\(\frac{120}{x}+\frac{480}{y}=8 \Rightarrow\frac{1}{x}+\frac{4}{y}=\frac{1}{15}.....(i)\)
And,\(\frac{200}{x}+\frac{400}{y}=\frac{25}{3} \Rightarrow\frac{1}{x}+\frac{2}{y}=\frac{1}{24}.....(ii)\)
Solving (i) and (ii), we get: x = 60 and y = 80.
Therefore Ratio of speeds = 60 : 80 = 3 : 4.
Let the distance travelled on foot be x km.
Then, distance travelled on bicycle = (61 -x) km.
So, \(\frac{x}{4}+\frac{(61-x)}{9}=9\)
9x + 4(61 -x) = 9 x 36
5x = 80
x = 16 km.
Let distance = x km and usual rate = y kmph.
Then, \(\frac{x}{y}-\frac{x}{y+3}=\frac{40}{60} \Rightarrow2y\left(y+3\right)=9x .......(i) \)
And, \(\frac{x}{y-2}-\frac{x}{y}=\frac{40}{60} \Rightarrow y\left(y-2\right)=3x .......(ii)\)
On dividing (i) by (ii), we get: x = 40.
 - Let the length of the train be x metres.
Then, the train covers x metres in 8 seconds and (x + 180) metres in 20 seconds.
∴ x/8 =( x + 180) / 20 ⇔ 20x = 8(x + 180) ⇔ x = 120.
∴ Length of the train = 120 m.
Speed of the train = [120/8]m/sec =[15 x 18/5]kmph = 54 kmph.
 - Relative speed = (40-20) km/hr = [20 x 5/18] m/sec = [50/9] m/sec.
Length of faster train = [50/9 x 5] m = 250/9 m = 27 7/9 m.
 - Relative speed = (36 + 45) km/hr
= [81 x 5/18] m/sec = [45/2] m/sec.
Length of train = [45/2 x 8] m = 180 m.
 - Speed of the first train = [120 / 10] m/sec = 12 m/sec.
Speed of the second train = [120 / 15] m/sec = 8 m/sec.
Relative speed = (12 + 8) = m/sec = 20 m/sec.
∴ Required time = (120 + 120) / 20 secc = 12 sec
 - Let the speed of each train be x m/sec.
Then, relative speed of the two trains = 2x m/sec.
So, 2x = (120 + 120)/12 ⇔ 2x = 20 ⇔ x = 10.
∴ Speed of each train = 10 m/sec = [10 x 18/5] km/hr = 36 km/hr