Practice MCQ Questions and Answer on Speed Time and Distance

451.

A train passes two bridges of length 800 m and 400 m in 100 seconds and 60 seconds respectively. The length of the train is:

(A) 80m
(B) 90m
(C) 200m
(D) 150m

Solution: 1st Method: Let length of the train be x m and speed of the train is s kmph. Speed, s = $\frac{x + 800}{100}$ . . . . . (i) Speed, s = $\frac{x + 400}{60}$ . . . . . (ii) Equating equation (i) and (ii), we get, Or, $\frac{x + 800}{100}$ = $\frac{x + 400}{60}$ Or, 5x + 200 = 3x + 2400 Or, 2x = 400 Or, x = 200m 2nd Method: As in both cases, the speed of the train is constant, and then we have; Time α distance $\begin{aligned} \frac{100}{60} = \frac{x + 800}{x + 400} \\ or, x = 200 m \end{aligned}$

452.

Points 'A' and 'B' are 70 km apart on a highway and two cars start at the same time. If they travel in the same direction, they meet in 7 hours, but if they travel towards each other they meet in one hour. Find the speed of the two cars (in km/hr).

(A) 20, 30
(B) 40, 30
(C) 30, 50
(D) 20, 40

Solution: Let the speed of the cars be S1 and S2 And S1 - S2 = $\frac{70}{7}$ = 10 . . . . . (i) Also S1 + S2 = $\frac{70}{1}$ = 70 . . . . . (ii) From equation (i) and (ii) S1 = $\frac{10 + 70}{2}$ = 40 km /hr And S2 = $\frac{70 - 10}{2}$ = 30 km / hr ∴ Required speeds are 40 km/hr and 30 km/hr

453.

A person travels 285 km in 6 hours in two stages. In the first part of the journey, he travels by bus at the speed of 40 km per hour. In the second part of the journey, he travels by train at the speed of 55 km per hour. How much distance did he travel by train ?

(A) 145 km
(B) 165 km
(C) 185 km
(D) 205 km

Solution: Let the distance travelled by train be x km Then, distance travelled by bus = (285 - x) km $\begin{aligned} ∴ (\frac{285 - x}{40}) + \frac{x}{55} = 6 \\ \Leftrightarrow \frac{(285 - x)}{8} + \frac{x}{11} = 30 \\ \Leftrightarrow \frac{11 (285 - x) + 8 x}{88} = 30 \\ \Leftrightarrow 3135 - 11 x + 8 x = 2640 \\ \Leftrightarrow 3 x = 495 \\ \Leftrightarrow x = 165 \end{aligned}$ Hence, distance travelled by train = 165 km

454.

A student goes to school at the rate of $2 \frac{1}{2}$ km/hr and reaches 6 min late. If he travels at the speed of 3 km/hr he is 10 min early. What is the distance to the school ?

(A) 4 km
(B) $3 \frac{1}{4}$ km
(C) 1 km
(D) $3 \frac{1}{2}$ km

Solution: Let the distance between school and home be = D According to the given information : $\frac{D}{2 \frac{1}{2}} - \frac{D}{3} = 16 minutes$ $\begin{aligned} ∴ \frac{2 D}{5} - \frac{D}{3} = \frac{16}{60} \\ \Rightarrow \frac{6 D - 5 D}{15} = \frac{16}{60} \\ \Rightarrow D = \frac{16}{60} \times 15 \\ \Rightarrow D = 4 km \end{aligned}$

455.

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

456.

A car travels the first one-third of a certain distance with a speed of 10 km/hr, the next one-third distance with a speed of 20 km/hr, and the last one-third distance with a speed of 60 km/hr. The average speed of the car for the whole journey is :

(A) 18 km/hr
(B) 24 km/hr
(C) 30 km/hr
(D) 36 km/hr

Solution: Let the whole distance travelled be x km and the average speed of the car for the whole journey be y km/hr Then, $\begin{aligned} \Leftrightarrow \frac{(\frac{x}{3})}{10} + \frac{(\frac{x}{3})}{20} + \frac{(\frac{x}{3})}{60} = \frac{x}{y} \\ \Leftrightarrow \frac{x}{30} + \frac{x}{60} + \frac{x}{180} = \frac{x}{y} \\ \Leftrightarrow \frac{1}{18} y = 1 \\ \Leftrightarrow y = 18 km/hr \end{aligned}$

457.

A is chasing B in the same interval of time. A jumps 8 times, while B jumps 6 times. But the distance covered by A in 7 jumps is the same as that of B in 5 jumps. The ratio between the speeds of A and B is . . . . . . . .

(A) 32 : 24
(B) 30 : 56
(C) 20 : 21
(D) 28 : 36

Solution: $\begin{matrix} A & : & B \\ Time & 8 & 6 \end{matrix}$ And 7 × A = B × 5 Distance A : B = 5 : 7 $\begin{matrix} Ratio & A & : & B \\ 8 & : & 6 \\ 5 & : & 7 \\ \overset{―}{40} & \overset{―}{42} \\ 20 & : & 21 \end{matrix}$

458.

A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is :

(A) 100 kmph
(B) 110 kmph
(C) 120 kmph
(D) 130 kmph

Solution: Let speed of the car be x kmph Then, speed of the train : $= \frac{150}{100} x = (\frac{3}{2} x) kmph$ $\begin{aligned} ∴ \frac{75}{x} - \frac{75}{\frac{3}{2} x} = \frac{125}{10 \times 60} \\ \Leftrightarrow \frac{75}{x} - \frac{50}{x} = \frac{5}{24} \\ \Leftrightarrow x = (\frac{25 \times 24}{5}) \\ \Leftrightarrow x = 120 kmph \end{aligned}$

459.

The speed of A and B are in the ratio 3 : 4. A takes 20 minutes more than B to reach a destination. Time in which A reach the destination?

(A) $1 \frac{1}{3}$ hours
(B) 2 hours
(C) $2 \frac{2}{3}$ hours
(D) $1 \frac{2}{3}$ hours

Solution: Ratio of speed = 3 : 4 Ratio of time taken = 4 : 3 (As Speed ∝ $\frac{1}{Time},$ When distance remains constant.) Let time taken by A and B be 4x and 3x hour respectively. Then, 4x - 3x = $\frac{20}{60}$ Or, x = $\frac{1}{3}$ Hence, time taken by A = 4x hours = 4 × $\frac{1}{3}$ = $1 \frac{1}{3}$ hours.

460.

A runs twice as fast as B and B runs thrice as fast as C. The distance covered by C in 72 minutes, will be covered by A in :

(A) 12 min
(B) 16 min
(C) 18 min
(D) 24 min

Solution: Let C's speed = x km/hr Then, B's speed = 3x km/hr And A's speed = 6x km/hr ∴ Ratio of speeds of A, B, C = 6x : 3x : x = 6 : 3 : 1 Ratio of times taken = $\frac{1}{6}$ : $\frac{1}{3}$ : 1 = 1 : 2 : 6 If C takes 6 minutes, then A takes 1 minute If C takes 72 minutes, then A takes $(\frac{1}{6} \times 72)$ min = 12 min

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Practice MCQ Questions and Answer on Speed Time and Distance

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