Practice MCQ Questions and Answer on Speed Time and Distance

411.

Two trains 150 m and 120 m long respectively moving from opposite direction cross each other in 10 sec. If the speed of the second train is 43.2 km/hr, then the speed of the first train is :

(A) 54 km/hr
(B) 50 km/hr
(C) 52 km/hr
(D) 51 km/hr

Solution: Let the speed of second train = x km/hr Their relative speed in opposite direction : = (43.2 + x) km/hr According to the question, $\begin{aligned} Time = \frac{l_{1} + l_{2}}{Speed} \\ \Rightarrow 10 \sec = \frac{(150 + 120) m}{(43.2 + x) \times \frac{5}{18} m/s} \\ \Rightarrow 10 \sec = \frac{270 \times 18}{(43.2 + x) \times 5} \\ \Rightarrow 43.2 \times 5 + 5 x = 486 \\ \Rightarrow x = \frac{486 - 216}{5} \\ \Rightarrow x = 54 \end{aligned}$ ∴ Speed of the second train = 54 km/hr

412.

2 km 5 metres is equal to ?

(A) 2.5 km
(B) 2.005 km
(C) 2.0005 km
(D) 2.05 km

Solution: We know, 1 km = 1000 metres ⇒ 2 km 5 metres = 2 km + $\frac{5}{1000}$ km = 2 km + 0.005 km = 2.005 km

413.

A man travels some distance at a speed of 12 km/hr and returns at a speed of 9 km/hr. If the total time taken by him is 2 hr 20 min, the distance is :

(A) 35 km
(B) 21 km
(C) 9 km
(D) 12 km

Solution: Let the distance be x km According to the question, $\begin{aligned} \frac{x}{12} + \frac{x}{9} = 2 + \frac{20}{60} \\ \frac{3 x + 4 x}{36} = \frac{7}{3} \\ \frac{7 x}{36} = \frac{7}{3} \\ x = 12 km \end{aligned}$

414.

A man completes a certain journey by a car. If he covered 30% of the distance at the speed of 20kmph. 60% of the distance at 40km/h and the remaining of the distance at 10 kmph, his average speed is:

(A) 25 km/h
(B) 28 km/h
(C) 30 km/h
(D) 33 km/h

Solution: $\begin{aligned} Let the total distance be 100 km . \\ Average speed \\ = \frac{total distance covered}{time taken} \\ = \frac{100}{\frac{30}{20} + \frac{60}{40} + \frac{10}{10}} \\ = \frac{100}{\frac{3}{2} + \frac{3}{2} + 1} \\ = \frac{100}{\frac{3 + 3 + 2}{2}} \\ = \frac{100 \times 2}{8} \\ = 25 kmph \end{aligned}$ Alternate 10% of journey's = 40 km Then, total journey = 400 kms $\begin{aligned} And, Average speed \\ = \frac{Total distance}{Total time} \\ 30 % of journey \\ = 400 \times \frac{30}{100} \\ = 120 km \\ 60 % of journey \\ = 400 \times \frac{60}{100} \\ = 240 km \\ 10 % of journey \\ = 400 \times \frac{10}{100} \\ = 40 km \\ Average speed \\ = \frac{400}{\frac{120}{20} + \frac{240}{40} + \frac{40}{10}} \\ = \frac{400}{6 + 6 + 4} \\ = \frac{400}{16} \\ ∴ Average speed = 25 km/hr \end{aligned}$

415.

A man can walk uphill at the rate of $2 \frac{1}{2}$ km/hr and downhill at the rate of $3 \frac{1}{4}$ km/hr. If the total time required to walk a certain distance up the hill and return to the starting point was 4 hr 36 min, then what was the distance walked up the hill by the man ?

(A) 4 km
(B) $4 \frac{1}{2}$ km
(C) $5 \frac{1}{2}$ km
(D) $6 \frac{1}{2}$ km

Solution: Average speed : $\begin{aligned} = \frac{(2 \times \frac{5}{2} \times \frac{13}{4})}{(\frac{5}{2} + \frac{13}{4})} km/hr \\ = (\frac{65}{4} \times \frac{4}{23}) km/hr \\ = (\frac{65}{23}) km/hr \end{aligned}$ Total time taken : = 4 hr 36 min $= 4 \frac{36}{60} h r$ $= 4 \frac{3}{5} h r$ $= \frac{23}{5} h r$ Total distance covered uphill and downhill : $\begin{aligned} = (\frac{65}{23} \times \frac{23}{5}) km \\ = 13 km \end{aligned}$ ∴ Distance walked uphill : $\begin{aligned} = (\frac{13}{2}) km \\ = 6 \frac{1}{2} km \end{aligned}$

416.

One third of a certain journey is covered at the speed of 80 km/hr one fourth of the journey at the speed of 50 km/hr and the rest at the speed of 100 km/hr what will be the average speed (in km/hr) for the whole journey?

(A) 75
(B) 67
(C) 66.66
(D) 76.66

Solution: $\begin{aligned} Average speed = \frac{Total Journey}{Time Taken} \\ \Rightarrow Remaining Distance \\ = 1 - (\frac{1}{3} + \frac{1}{4}) \\ = \frac{5}{12} km \\ ∴ Average speed \\ = \frac{1}{\frac{1}{3 \times 80} + \frac{1}{50 \times 4} + \frac{1 \times 5}{100 \times 12}} \\ = \frac{1}{\frac{5 + 6 + 5}{1200}} \\ = \frac{1200}{6} \\ = 75 km/hr \end{aligned}$

417.

A car after traveling 18 km from a point A developed some problem in the engine and the speed became $\frac{4}{5}$ th of its original speed. As a result, the car reached point B 45 minutes late. If the engine had developed the same problem after traveling 30 km from A, then it would have reached 36 minutes late. The original speed of the car (in km/h) is:

(A) 25 kmph
(B) 30 kmph
(C) 20 kmph
(D) 35 kmph

Solution: He proceeds at $\frac{4}{5}$ S where S is his usual speed means $\frac{1}{5}$ decrease in speed which will lead to $\frac{1}{4}$ increase in time. Now the main difference comes in those 12km (30 - 18) and the change in difference of time = (45 - 36) min = 9 min Thus, $\frac{1}{4}$ × T = 9 min where T is the time required to cover the distance of (30 - 18) = 12 km T = 36 min = $\frac{36}{60}$ hours = 0.6 hours. Speed of the car = $\frac{12}{0.6}$ = 20 kmph.

418.

A train takes 45 minutes to cover a certain distance at a speed of 80 km/h. If the speed is increased by 125%, then how long will it take the train to cover $\frac{8}{5}$ of the same distance?

(A) 25 minutes
(B) 30 minutes
(C) 32 minutes
(D) 28 minutes

Solution: Distance = Speed × Time Distance = 45 min × 80 After incrassation the speed $= \frac{80 \times 225}{100} = 180 km/hr$ $\begin{aligned} 45 \times 80 \times \frac{8}{5} = 180 \times t {s_{1} \times t_{1} = s_{2} \times t_{2}} \\ t = 32 minutes \end{aligned}$

419.

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, $\frac{4 x}{7}$ miles on foot and $\frac{2 x}{5}$ miles on horseback. Then, $\begin{aligned} \Leftrightarrow x + \frac{4 x}{7} + \frac{2 x}{5} = 3036 \\ \Leftrightarrow \frac{69 x}{35} = 3036 \\ \Leftrightarrow x = (\frac{3036 \times 35}{69}) \\ \Leftrightarrow x = 1540 \end{aligned}$ ∴ Distance travelled on foot : $\begin{aligned} = (\frac{4}{7} \times 1540) miles \\ = 880 miles \end{aligned}$

420.

A tiger is 50 of its own leaps behinds a deer. The tiger takes 5 leaps and per minutes to the deer's 4. If the tiger and the deer cover 8 m and 5 m per leap respectively, what distance will the tiger have to run before it caches the deer?

(A) 600 m
(B) 700 m
(C) 800 m
(D) 1000 m

Solution: Speed of tiger = 40 m/min Speed of deer = 20 m/min. Relative speed = 40 - 20 = 20 m/min. Initial difference in distance = 50 × 8 = 400 m Time take to catch = $\frac{400}{20}$ = 20 min. Distance traveled in 20 min,= 20 × 40 = 800 m

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Practice MCQ Questions and Answer on Speed Time and Distance

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