Practice MCQ Questions and Answer on Speed Time and Distance

441.

A and B start moving towords each other from places X and Y, respectively, at the same time on the day. The speed of A is 20% more than of B. After meeting on the way, A and B take p hours and $7 \frac{1}{5}$ ÃÂÃÂ hours, respectively. What is the value of p?

(A) 4.5
(B) 5
(C) 5.5
(D) 6

Solution: $\begin{matrix} A & B \\ S \to & 6 & : & 5 \end{matrix}$ $\begin{aligned} \frac{V 1}{V 2} = \sqrt{\frac{T 2}{T 1}} \\ \frac{6}{5} = \sqrt{\frac{\frac{36}{5}}{p}} \\ \frac{36}{25} = \frac{36}{5 p} \\ p = 5 \end{aligned}$

442.

Two planes move along a circle of circumference 1.2 km with constant speeds. When they move in different directions, they meet every 15 seconds and when they move in the same direction, one plane overtakes the other every 60 seconds. Find the speed of the slower plane.

(A) 0.04 km/s
(B) 0.03 km/s
(C) 0.05 km/s
(D) 0.02 km/s

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, $\begin{aligned} \frac{1200}{x + y} = 15 \\ \Rightarrow x + y = 80. . . . . (i) \end{aligned}$ And, $\begin{aligned} \frac{1200}{x - y} = 60 \\ \Rightarrow x - y = 20. . . . . (i i) \end{aligned}$ 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

443.

In a race of 1000m, A can beat B by 100m. In a 400m, B beats C by 40m. In a race of 500m. A will beat C by

(A) 95m
(B) 50m
(C) 45m
(D) 60m

Solution: When A runs 1000m, B runs 900m. Hence, when A runs 500m, B runs 450m. Again, when B runs 400m, C runs 360m. And, when B runs 450m, C runs = 360 × $\frac{450}{400}$ = 405m. Required distance = 500 - 405 = 95 meter. That means when A runs 500 meter then B can run 450m then C runs 405m.

444.

With an average speed of 50 km/hr, a train reaches its destination in time. If it goes with an average speed of 40 km/hr, it is late by 24 min. The total journey is :

(A) 30 km
(B) 40 km
(C) 70 km
(D) 80 km

Solution: Difference between timings = 24 min = $\frac{24}{60}$ hr = $\frac{2}{5}$ hr Let the length of the journey be x km Then, $\begin{aligned} \Leftrightarrow \frac{x}{40} - \frac{x}{50} = \frac{2}{5} \\ \Leftrightarrow \frac{x}{200} = \frac{2}{5} \\ \Leftrightarrow x = (\frac{2}{5} \times 200) \\ \Leftrightarrow x = 80 km \end{aligned}$

445.

Two cyclists start on a circular track from a given point but in opposite directions with speeds of 7 m/sec and 8 m/sec respectively. If the circumference of the circle is 300 metres, after what time will they meet at the starting point ?

(A) 20 sec
(B) 100 sec
(C) 200 sec
(D) 300 sec

Solution: Time taken by the two cyclists to cover one round of the track is $\frac{300}{7}$ sec and $\frac{300}{8}$ sec respectively. ∴ Required time : = L.C.M. of $\frac{300}{7}$ and $\frac{300}{8}$ = 300 sec

446.

Places A and B are 396 km apart. Train X leaves from A for B and train Y leaves from B for A at the same time on the same day on parallel tracks. Both trains meet after $5 \frac{1}{2}$ ÃÂÃÂ hours. The speed of Y is 10 km/h more than that of X. What is the speed (in km/h) of Y?

(A) 41
(B) 54
(C) 31
(D) 56

Solution: $A \overset{396}{\to} B$ $\begin{aligned} \frac{396}{A + B} = \frac{11}{2} \\ X + Y = \frac{396 \times 2}{11} \\ X + Y = 36 \times 2 \\ X + Y = 72 \\ Let, \\ X = a \\ Y = a + 10 \\ a + a + 10 = 72 \\ 2 a = 62 \\ a = 31 km/h \\ Y = 31 + 10 = 41 \end{aligned}$

447.

A man travels for 5 hours 15 minutes. If he covers the first half of the journey at 60 km/hr and rest at 45 km/hr. Find the total distance travelled by him ?

(A) $1028 \frac{6}{7}$ km
(B) $189$ km
(C) $378$ km
(D) $270$ km

Solution: Let ther distance covered be 2x km $Time = \frac{Distance}{Speed}$ Time taken to covers the first half and second half of the journey in t1 hours and t2 hours $\begin{aligned} \Rightarrow \frac{a}{60} = t_{1} . . . . . (i) \\ \Rightarrow \frac{a}{45} = t_{2} . . . . . (i i) \end{aligned}$ Adding (i) and (ii) we get $\begin{aligned} \Leftrightarrow \frac{a}{60} + \frac{a}{45} = t_{1} + t_{2} \\ \Leftrightarrow \frac{a}{60} + \frac{a}{45} = 5 \frac{15}{60} \\ \Leftrightarrow \frac{a}{60} + \frac{a}{45} = 5 \frac{1}{4} \\ \Leftrightarrow \frac{3 a + 4 a}{180} = \frac{21}{4} \\ \Leftrightarrow 7 a = \frac{21}{4} \times 180 \\ \Rightarrow a = \frac{21 \times 180}{4 \times 7} \\ \Rightarrow a = 135 km \end{aligned}$ ∴ Length of total journey : = (2 × 135) km = 270 km

448.

A is twice as fast as B and B is thrice as fast as C. The journey covered by C in $1 \frac{1}{2}$ hours will be covered by A in :

(A) 15 min
(B) 2 min
(C) 30 min
(D) 1 hour

Solution: Distance cover by C in 90 mintue will cover by B $\frac{90}{3}$ = 30 minute Distance cover by B in 30 mintue will cover by A $\frac{30}{2}$ = 15 minute

449.

A person covers 40% of the distance from A to B at 8 km/h, 40% of the remaining distance at 9 km/h and the rest at 12 km/h. His average speed (in km/h) for the journey is:

(A) $9 \frac{5}{8}$
(B) $9 \frac{1}{3}$
(C) $9 \frac{3}{8}$
(D) $9 \frac{2}{3}$

Solution: $\begin{aligned} Let total distance = 100 \\ Average speed = \frac{Total Distance}{Total Time} \\ = \frac{100}{\frac{40}{8} + \frac{24}{9} + \frac{36}{12}} \\ = \frac{100}{5 + \frac{8}{3} + 3} \\ = \frac{100 \times 3}{32} \\ = \frac{75}{8} \\ = 9 \frac{3}{8} \end{aligned}$

450.

A train without stoppage travels with an average speed of 80 km/h and with stoppage, it travels with an average speed of 64 km/h. For how many minutes does the train stop on an average per hour?

(A) 12
(B) 8
(C) 10
(D) 14

Solution: $\begin{aligned} (\frac{Faster speed - Slower speed}{Faster speed} \times 60) min/hr \\ = \frac{80 - 64}{80} \times 60 \\ = \frac{16}{80} \times 60 \\ = 12 min \end{aligned}$

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Practice MCQ Questions and Answer on Speed Time and Distance

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