Practice MCQ Questions and Answer on Speed Time and Distance
441.
R jogs at twice the speed of walking and runs at twice the speed of jogging. From his home to office, he covers half of the distance by walking and the rest by jogging. From his office to home, he covers half the distance jogging and the rest by running. What is his average speed (in km/h) in a complete round from his home to office and back home if the distance between his office and home is 10 km and he walks at the speed of 5 km/h?
In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is:
Two persons A and B start simultaneously from P and Q respectively. A meets B at a distance of 60m from P. After A reaches Q and B reaches P, they turn around and start walking in opposite direction, now B meets A at a distance of 40m from Q. Find distance between P and Q?
(A) 160 m
(B) 100 m
(C) 140 m
(D) 105 m
Solution:
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
444.
The speed of a train is 220% of the speed of a car. The car covers a distance of 950 km in 19 hours. How much distance will the train cover in $$3\frac{1}{2}$$ÃÂÃÂ hours?
A motor car starts with the speed of 70 km/hr with its speed increasing every two hours by 10 kmph. In how many hours will it cover 345 kms ?
(A) $$2\frac{1}{2}$$ hrs
(B) 4 hrs 5 min
(C) $$4\frac{1}{2}$$ hrs
(D) Cannot be determined
Solution:
Distance covered in first 2 hours : = (70 × 2) km = 140 km Distance covered in next 2 hours : = (80 × 2) km = 160 km Remaining distance : = 345 - (140 + 160) = 45 km Speed in the fifth hour = 90 km/hr Time taken to cover 45 km : = $$\frac{45}{90}$$ hr = $$\frac{1}{2}$$ hr ∴ Total time taken : = 2 + 2 + $$\frac{1}{2}$$ = $$4\frac{1}{2}$$ hrs
446.
A bus moving at a speed of 45 km/hr catches a truck 150 metres ahead going in the same direction in 30 seconds. The speed of the truck is
(A) 27 km/hr
(B) 24 km/hr
(C) 25 km/hr
(D) 28 km/hr
Solution:
Let the speed of truck is = $$x$$ km/h Their relative speed in same direction = (45 - $$x$$) km/h (Here (45 - $$x$$) has been written because bus crosses the truck which is running 150 meters ahead of it. i.e. Truck speed will be lower than that of bus) According to the question $$\eqalign{ & {\text{Time}} = \frac{{{\text{Total distance}}}}{{{\text{Total speed}}}} \cr & \frac{{150}}{{\left( {45 - x} \right) \times \frac{5}{{18}}}} = 30 \cr & \frac{{150 \times 18}}{{\left( {45 - x} \right) \times 5}} = 30 \cr & x = 27{\text{ km/h}} \cr} $$ So, speed of the truck is 27 km/h
447.
When do the two hands of a clock of just after 3 pm make 30ÃÂÃÂÃÂðangel between them?
A person travels 5x distance at a speed of 5 km/h, x distance at a speed of 5 km/h, and 4x distance at a speed of 6 km/h, and takes a total of 112 minutes. What is the total distance (in km) travelled by the person?
A, B and C start together from the same place to walk round a circular path of length 12km. A walks at the rate of 4 km/h, B 3 km/h and C $$\frac{3}{2}$$ km/h. They will meet together at the starting place at the end of:
(A) 10 hours
(B) 12 hours
(C) 15 hours
(D) 24 hours
Solution:
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.
450.
A train running at $$\frac{7}{11}$$ of its own speed reached a place in 22 hours. How much time could be saved if the train would have run at its own speed ?
(A) 7 hr
(B) 8 hr
(C) 14 hr
(D) 16 hr
Solution:
New speed = $$\frac{7}{11}$$ of usual speed ∴ New time = $$\frac{11}{7}$$ of usual time So, $$\frac{11}{7}$$ of usual time = 22 hrs ⇒ Usual time : = $$\frac{22 × 7}{11}$$ = 14 hrs Hence, time saved = (22 - 14) = 8 hrs