Practice MCQ Questions and Answer on Speed Time and Distance

Solution: $$\eqalign{ & 15\min = \frac{1}{4}{\text{hrs}} \cr & {\text{1 hr }} \to {\text{5 kms}} \cr & \frac{1}{4}{\text{hr}} \to \frac{5}{4}{\text{ kms}} \cr & {\text{So, length of the bridge}} \cr & = \frac{5}{4}{\text{ km}} \cr & {\text{ = 1250 metres}} \cr} $$

Solution: He proceeds at $$\frac{3}{4}$$ S where S is his usual speed means $$\frac{1}{4}$$ decrease in the speed which will lead to $$\frac{1}{3}$$ increase in the time. Now the main difference comes in those 24km and the change in difference of time = 35 - 25 m = 10 m. ⇒ $$\frac{1}{3}$$ × T = 10 where T is the time required to cover the distance of (74 - 50) = 24 km. T = 30 min = 0.5 hours. Speed of the dog = $$\frac{{24}}{{0.5}}$$  = 48 kmph.

Solution: 1 hour 45 minutes = $$1 + \frac{{45}}{{60}}$$  = $$\frac{7}{4}$$ hours. Speed of the motorboat up-stream, $$\eqalign{ & = \frac{{{\text{Distance}}}}{{{\text{Time}}\,\,{\text{Taken}}}} \cr & = \frac{{56\,{\text{km}}}}{{\frac{7}{4}{\text{hours}}}} \cr & = \frac{{56 \times 4}}{7} \cr & = 32\,{\text{kmph}} \cr} $$ Let the speed of the current = x kmph Hence, 36 - x = 32 Or, x = 36 - 32 = 4 kmph Speed of boat down the stream = 36 + 4 = 40 kmph. ∴ Time taken to cover 56 km at 40 kmph = $$\frac{{56}}{{40}}$$ = $$\frac{7}{5}$$ hours or 1 hours 24 minutes.

Solution: $$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{A}}\,{\text{:}}\,{\text{B}}\,\,\,\,\,\,\,{\text{length}} \cr & {\text{Ratio of A's and }} \to 5:3\,\,\,\left( {5x:3x} \right) \cr & {\text{ B's length}} \cr & {\text{Ratio of A's and }} \to 6:5\,\,\,\left( {6y:5y} \right) \cr & {\text{ B's speed}} \cr} $$ We know that, When a train crosses a pole, i.e., it covers the distance equal to its length Time taken by train A to cross the pole : $$ = \frac{{{\text{Total distance}}}}{{{\text{Speed}}}} = \frac{{5x}}{{6y}}$$ Time taken by train B to cross the pole : $$ = \frac{{{\text{Total distance}}}}{{{\text{Speed}}}} = \frac{{3x}}{{5y}}$$ Ratio of the their time : $$\eqalign{ & = {\text{A}}:{\text{B}} \cr & = \frac{{5x}}{{6y}}:\frac{{3x}}{{5y}} \cr & = 25:18 \cr} $$

Solution: Let A's speed = x km/hr Then, B's speed = (x + 4) km/hr Clearly, time taken by B to cover (60 + 12) i.e., 72 km = Time taken by A to cover (60 - 12) i.e., 48 km $$\eqalign{ & \therefore \frac{{72}}{{x + 4}} = \frac{{48}}{x} \cr & \Rightarrow 72x = 48x + 192 \cr & \Rightarrow 24x = 192 \cr & \Rightarrow x = 8 \cr} $$ Hence, A's speed = 8 km/hr

Solution: Let the cyclist's speed without wind be x km/hr And the speed of the wind be y km/hr Then, $$\eqalign{ & \Rightarrow \frac{1}{{x + y}} = \frac{3}{{60}} \cr & \Rightarrow x + y = 20.....(i) \cr} $$ And $$\eqalign{ & \Rightarrow \frac{1}{{x - y}} = \frac{4}{{60}} \cr & \Rightarrow x - y = 15.....(ii) \cr} $$ Adding (i) and (ii), we get: 2x = 35 or x = 17.5 Putting x = 17.5 in (i), we get : y = 2.5 Time taken to drive 17.5 km without wind = 1 hr Time taken to drive 1 km without wind $$\eqalign{ & = \left( {\frac{1}{{17.5}}} \right){\text{ hr}} \cr & {\text{ = }}\left( {\frac{1}{{17.5}} \times 60} \right){\text{ min}} \cr & {\text{ = 3}}\frac{3}{7}{\text{ min}} \cr} $$

Solution: Average speed : $$\eqalign{ & = \frac{{\left( {2 \times \frac{5}{2} \times \frac{{13}}{4}} \right)}}{{\left( {\frac{5}{2} + \frac{{13}}{4}} \right)}}{\text{ km/hr}} \cr & {\text{ = }}\left( {\frac{{65}}{4} \times \frac{4}{{23}}} \right){\text{ km/hr}} \cr & = \left( {\frac{{65}}{{23}}} \right){\text{ km/hr}} \cr} $$ Total time taken : = 4 hr 36 min $$= 4\frac{36}{60}\,\, hr$$ $$= 4\frac{3}{5}\,\, hr$$ $$= \frac{23}{5}\,\, hr$$ Total distance covered uphill and downhill : $$\eqalign{ & = \left( {\frac{{65}}{{23}} \times \frac{{23}}{5}} \right){\text{ km}} \cr & = 13{\text{ km}} \cr} $$ ∴ Distance walked uphill : $$\eqalign{ & = \left( {\frac{{13}}{2}} \right){\text{ km}} \cr & = 6\frac{1}{2}{\text{ km}} \cr} $$

Solution: $$\eqalign{ & {\text{Total distance}} = \frac{{{S_1} \times {S_2}}}{{{S_1} + {S_2}}} \times {\text{Total time}} \cr & = \frac{{5\frac{1}{2} \times 4}}{{5\frac{1}{2} + 4}} \times 4\frac{3}{4} \cr & = \frac{{\frac{{11}}{2} \times 4}}{{\frac{{11}}{2} + 4}} \times \frac{{19}}{4} \cr & = \frac{{22}}{{\frac{{19}}{2}}} \times \frac{{19}}{4} \cr & = \frac{{44}}{{19}} \times \frac{{19}}{4} \cr & = 11{\text{ km}} \cr & {\text{One side distance}} = 11{\text{ km}} \cr & {\text{Both side distance}} = 2 \times 11 = 22{\text{ km}} \cr} $$

Solution: The sum of speeds would be 0.8 m/s (relative speed in opposite direction). Also if we go by option (B) the speeds will be 0.03 and 0.05 m/s respectively. At this speed the overlapping would occur in every 60 second. Alternate : Let their speeds be x m/sec and y m/sec respectively. Then, $$\eqalign{ & \frac{{1200}}{{x + y}} = 15 \cr & \Rightarrow x + y = 80.....(i) \cr} $$ And, $$\eqalign{ & \frac{{1200}}{{x - y}} = 60 \cr & \Rightarrow x - y = 20.....(ii) \cr} $$ Adding (i) and (ii), we get : 2x = 100 or x = 50 Putting x = 50 in (i), we get : y = 30 Hence, speed of slower plane : = 30 m/sec = 0.03 km/sec

Solution: When the athlete walks at $$\frac{5}{4}$$ of his usual speed then he takes $$\frac{4}{5}$$ of his usual time $$\left( {{\text{As}}\,\,{\text{Speed}}\, \propto \frac{1}{{{\text{Time}}}}} \right)$$    and he saves 5 minutes. → Usual time - $$\frac{4}{5}$$ × usual time = 5 minutes → $$\frac{1}{5}$$ × Usual time = 5 minutes Usual time = 25 minutes.