Practice MCQ Questions and Answer on Speed Time and Distance

291.

An ant climbing up a vertical pole ascends 12 meters and slips down 5 meters in every alternate hour. If the pole is 63 meters high how long will it take it to reach the top?

(A) 18 hours
(B) 17 hours
(C) 16 hours 35 min
(D) 16 hours 40 min

Solution: Since it climbs 7 meter in every 2 hr. At the end of 14 hours it would have climbed up to 49 mt. In 15th hour it will reach 61 mt. Then drops back to 56 mt in the 16th hour. It will take another 35 min to travel remaining 7 mt since it will be ascending in the 17th hour. Thus, 16 hr and 35 min. Detail Explanation: Given, Ant ascends 12 meters in first hours and slips down 5 meters in second hour. This is repeated in every alternate hour. It means in every two hours Ant ascends (12 - 5 = 7 meters) Therefore, in ends of 14 hours, Ant ascends = $\frac{14 \times 7}{2}$ = 49 meters. So, In next hour i.e 15 hour, Ant will ascends another 12 meter. Till end of 15th hour, Ant climbed = 49 + 12 = 61 meters. But in 16th hour, Ant will slips down 5 meters (61 - 5 = 56 meters). Remaining pole to climb, = 63 - 56 = 7

292.

Two men start together to walk a certain distance, one at 4 km/hr and another at 3 km/hr. The former arrives half an hour before the latter. Find the distance :

(A) 8 km
(B) 7 km
(C) 6 km
(D) 9 km

Solution: If the required distance be x km, then $\begin{aligned} \frac{x}{3} - \frac{x}{4} = \frac{1}{2} \\ \Rightarrow \frac{4 x - 3 x}{12} = \frac{1}{2} \\ \Rightarrow \frac{x}{12} = \frac{1}{2} \\ \Rightarrow x = 6 km \end{aligned}$

293.

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 : $\frac{D}{2 \frac{1}{2}} - \frac{D}{3} = 16 minutes$ $\begin{aligned} ∴ \frac{2 D}{5} - \frac{D}{3} = \frac{16}{60} \\ \Rightarrow \frac{6 D - 5 D}{15} = \frac{16}{60} \\ \Rightarrow D = \frac{16}{60} \times 15 \\ \Rightarrow D = 4 km \end{aligned}$

294.

A car completed a journey of 400 km in $12 \frac{1}{2}$ hrs. The first $\frac{3}{4}$ of the journey was done at 30 km/hr. Calculate the speed for the rest of the journey.

(A) 45 km/hr
(B) 40 km/hr
(C) 25 km/hr
(D) 30 km/hr

Solution: $\begin{aligned} Total journey = 400 \\ \frac{3}{4} Journey = 400 \times \frac{3}{4} \\ = 300 km \end{aligned}$ Remaining journey = 100 km Let the speed of car for the rest of journey = x km/hr According to the question, $\begin{aligned} \frac{300}{30} + \frac{100}{x} = 12 \frac{1}{2} \\ 10 + \frac{100}{x} = \frac{25}{2} \\ \frac{100}{x} = \frac{25 - 20}{2} \\ x = \frac{100 \times 2}{5} \\ x = 40 km/hr \end{aligned}$

295.

A man goes to a place on bicycle at speed of 16 km/hr and comes back at lower speed. If the average speed is 6.4 km/hr in total, then the return speed (in km/hr) is :

(A) 10 km/hr
(B) 8 km/hr
(C) 6 km/hr
(D) 4 km/hr

Solution: $\begin{aligned} Average speed : \\ = \frac{2 x y}{x + y} \\ = \frac{2 \times 16 \times y}{16 + y} \\ \Rightarrow 6.4 = \frac{32 y}{16 + y} \\ \Rightarrow 102.4 + 6.4 = 32 y \\ \Rightarrow 102.4 = 25.6 y \\ \Rightarrow y = \frac{102.4}{25.6} \\ \Rightarrow y = 4 km/hr \end{aligned}$

296.

The speed $3 \frac{1}{3}$ m/sec when expressed in km/hr becomes :

(A) 8 km/hr
(B) 9 km/hr
(C) 10 km/hr
(D) 12 km/hr

Solution: 1 m/sec = $\frac{18}{5}$ km/hr $\frac{10}{3}$ m/sec = $\frac{10}{3}$ × $\frac{18}{5}$ = 12 km/hr

297.

A takes 2 hours more than B to walk d km, but if A doubles his speed, then he can make it in 1 hour less than B. How much times does B required for walking d km ?

(A) $\frac{d}{2}$ hours
(B) 3 hours
(C) 4 hours
(D) $\frac{2 d}{3}$ hours

Solution: Suppose B takes x hours to walk d km Then, A takes (x + 2) hours to walk d km A's speed = $(\frac{d}{x + 2})$ km/hr and B's speed = $(\frac{d}{x})$ km/hr A's new speed = $(\frac{2 d}{x + 2})$ km/hr $\begin{aligned} ∴ \frac{d}{(\frac{d}{x})} - \frac{d}{(\frac{2 d}{x + 2})} = 1 \\ \Leftrightarrow x - (\frac{x + 2}{2}) = 1 \\ \Leftrightarrow x - 2 = 2 \\ \Leftrightarrow x = 4 hours \end{aligned}$

298.

If A travels to his school from his house at the speed of 3 km/h, then he reaches the school 5 minutes late. If he travels at the speed of 4 km/h, he reaches the school 5 minutes earlier than school time. The distance of his school from his house is:

(A) 1 km
(B) 2 km
(C) 3 km
(D) 4 km

Solution: Let the distance between school and home be x km. The difference of time when A goes school to school with these two different speed is 10 min $\begin{aligned} = \frac{10}{60} hour \\ Now, \\ \frac{x}{3} - \frac{x}{4} = \frac{10}{60} \\ Or, \frac{x}{12} = \frac{1}{6} \\ Or, x = \frac{12}{6} \\ = 2 km \end{aligned}$

299.

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

300.

A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.

(A) 220 km
(B) 224 km
(C) 230 km
(D) 234 km

Solution: $\begin{aligned} \frac{(1 / 2) x}{21} + \frac{(1 / 2) x}{24} = 10 \\ \Rightarrow \frac{x}{21} + \frac{x}{24} = 20 \\ \Rightarrow 15 x = 168 \times 20 \\ \Rightarrow x = \frac{168 \times 20}{15} \\ \Rightarrow x = 224 k m \end{aligned}$

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Practice MCQ Questions and Answer on Speed Time and Distance

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