Practice MCQ Questions and Answer on Speed Time and Distance
421.
Two rifles are fired from the same place at a difference of 11 minutes 45 seconds. But a man who is coming towards the place in a train hears the second sound after 11 minutes. Find the speed of train. (speed of sound = 330 m/s)
(A) 72 kmph
(B) 36 kmph
(C) 81 kmph
(D) 108 kmph
Solution:
Distance travelled by man-train in 11 minute = distance traveled by sound in (11 min 45 sec - 11 min) = 45 sec. Let the speed of train be x kmph. Now, Or, $$x \times \frac{{11}}{{60}} = \frac{{45}}{{3600}} \times 330 \times \frac{{18}}{5}$$ Or, x = 81 kmph.
422.
Two persons A and B started from two different places towards each other. If the ratio of their speed be 3 : 5, then what is the ratio of distance covered by A and B respectively till the point of meeting?
(A) 1 : 2
(B) 3 : 4
(C) 3 : 5
(D) 5 : 3
Solution:
When time is constant the distance covered by A and B will be in the ratio of their speeds. i.e. ratio of distance is 3 : 5 as time is constant.
423.
Two guns were fired from the same place at an interval of 8 minutes. A person approaching the place observes that 5 minutes 52 seconds have elapsed between the hearing of the sound of the two guns. If the velocity of the sound is 330 m/sec, the man was approaching the place at what speed (in km/hr) ?
(A) 24 km/hr
(B) 27 km/hr
(C) 30 km/hr
(D) 36 km/hr
Solution:
Let the speed of the man be x m/sec Then, distance travelled by the man in 5 min 52 sec = Distance travelled by sound in 8 sec $$\eqalign{ & \Leftrightarrow x \times 352 = 330 \times 8 \cr & \Leftrightarrow x = \left( {\frac{{330 \times 8}}{{352}}} \right){\text{m/sec}} \cr & \Leftrightarrow x = \left( {\frac{{330 \times 8}}{{352}} \times \frac{{18}}{5}} \right){\text{km/hr}} \cr & \Leftrightarrow x = 27{\text{ km/hr}} \cr} $$
424.
A cyclist moving on a circular track of radius 100 meters completes one revolution in 2 minutes. What is the average speed of cyclist (approx.)?
A man can cross a downstream river by steamer in 40 minutes and same by boat in 1 hour. If the time of crossing Upstream by streamer is 50% more than downstream time by steamer and the time required by boat to cross same river by boat in upstream is 50% more than time required by in downstream. What is the time taken for the man to cross the river downstream by steamer and then return to same place by boat half the way and by steamer the rest of the way?
(A) 85 minutes
(B) 115 minutes
(C) 120 minutes
(D) 125 minutes
Solution:
Downstream (steamer) = 40 min Downstream (boat) = 60 min Upstream (steamer) = 60 min Upstream (boat) = 90 min Required time = 40 + 30 + 45 = 115 minutes
426.
A train is 250 m long. If the train takes 50 second to cross a tree by the railway line, then the speed of the train on Km /hr is :
Two men start together to walk to a certain destination, one at 3 kmph and another at 3.75 kmph. The latter arrives half an hour before the former. The distance is :
(A) 6 km
(B) 7.5 km
(C) 8 km
(D) 9.5 km
Solution:
Let the distance be x km Then, $$\eqalign{ & \Leftrightarrow \frac{x}{3} - \frac{x}{{3.75}} = \frac{1}{2} \cr & \Leftrightarrow 2.5x - 2x = 3.75 \cr & \Leftrightarrow x = \frac{{3.75}}{{0.50}} \cr & \Leftrightarrow x = \frac{{15}}{2} \cr & \Leftrightarrow x = 7.5{\text{ km}} \cr} $$
428.
A and B start from the same point and in the same direction at 7 am to walk around a rectangular field 400 m ÃÂÃÂÃÂÃÂÃÂÃÂ 300 m. A and B walk at the rate of 3 km/hr and 2.5 km/hr respectively. How many times shall they cross each other if they continue to walk till 12.30 pm ?
(A) Not even once
(B) Once
(C) Twice
(D) Thrice
Solution:
Perimeter of the field = 2(400 + 300) m = 1400 m = 1.4 km Since A and B move in the same direction, so they will first meet each other when there is a difference of one round i.e., 1.4 km between the two. Relative speed of A and B = (3 - 2.5) km = 0.5 km/hr Time take to cover 1.4 km at this speed : $$\eqalign{ & = \left( {\frac{{1.4}}{{0.5}}} \right){\text{ hr}} \cr & = 2\frac{4}{5}{\text{ hr}} \cr & = 2{\text{ hr 48 min}} \cr} $$ So, they shall first cross each other at 9.48 am And again 2 hr 48 min after 9.48 am i,e., 12.36 pm Thus, till 12.30 pm they will cross each other once.
429.
Jane travelled $$\frac{4}{7}$$ as many miles on foot as by water and $$\frac{2}{5}$$ as many miles on horseback as by water. If she covered total of 3036 miles, how many miles did she travel on foot ?
(A) 1540 miles
(B) 880 miles
(C) 756 miles
(D) 616 miles
Solution:
Suppose Jane travelled x miles by water, $$\frac{4x}{7}$$ miles on foot and $$\frac{2x}{5}$$ miles on horseback. Then, $$\eqalign{ & \Leftrightarrow x + \frac{{4x}}{7} + \frac{{2x}}{5} = 3036 \cr & \Leftrightarrow \frac{{69x}}{{35}} = 3036 \cr & \Leftrightarrow x = \left( {\frac{{3036 \times 35}}{{69}}} \right) \cr & \Leftrightarrow x = 1540 \cr} $$ ∴ Distance travelled on foot : $$\eqalign{ & = \left( {\frac{4}{7} \times 1540} \right){\text{ miles}} \cr & {\text{ = 880 miles}} \cr} $$
430.
A person can row $$7\frac{1}{2}$$ km an hour in still water. Finds that it takes twice the time to row upstream than the time to row downstream. The speed of the stream is:
(A) 2 kmph
(B) 2.5 kmph
(C) 3 kmph
(D) 4 kmph
Solution:
Let the distance covered be x km and speed of stream = y kmph. Speed downstream = $$\frac{{15}}{2} + y$$ kmph Speed upstream = $$\frac{{15}}{2} - y$$ kmph $$\eqalign{ & {\text{According}}\,{\text{to}}\,{\text{question,}} \cr & {\frac{{2x}}{{ { {\frac{{15}}{2}} + y} }}} = {\frac{x}{{ { {\frac{{15}}{2}} - y} }}} \cr & or,\,15 - 2y = {\frac{{15}}{2}} + y \cr & or,3y = 15 - {\frac{{15}}{2}} = \frac{{15}}{2} \cr & or,y = \frac{{15}}{6} = 2.5\,{\text{kmph}} \cr} $$