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Q:

A motor boat takes 12 hours to go downstream and it takes 24 hours to return the same distance. what is the time taken by boat in still water?

A) 15 h	B) 16 h
C) 8 h	D) 20 h

Answer & Explanation Answer: B) 16 h

Explanation:

If t1 and t2 are the upstream and down stream times. Then time taken in still water is given by

$\frac{2 \times t 1 \times t 2}{t 1 + t 2} = \frac{2 \times 12 \times 24}{36} = 16 h$

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Filed Under: Boats and Streams

Q:

Two horses start trotting towards each other, one from A to B and another from B to A. They cross each other after one hour and the first horse reaches B, 5/6 hour before the second horse reaches A. If the distance between A and B is 50 km. what is the speed of the slower horse?

A) 30 km/h	B) 15 km/h
C) 25 km/h	D) 20 km/h

Answer & Explanation Answer: D) 20 km/h

Explanation:

If the speed of the faster horse be $f_{s}$ and that of slower horse be $s_{s}$ then

$f_{s} + s_{s} = \frac{50}{1} = 50$

and $\frac{50}{s_{s}} - \frac{50}{f_{s}} = \frac{5}{6}$

Now, you can go through options.

The speed of slower horse is 20km/h

Since, 20+30=50

and $\frac{50}{20} - \frac{50}{30} = \frac{5}{6}$

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Q:

A train overtakes two girls who are walking in the opposite direction in which the train is going at the rate of 3 km/h and 6km/h and passes them completely in 36 seconds and 30 seconds respectively. The length of the train is:

A) 120m	B) 150m
C) 125m	D) None of these

Answer & Explanation Answer: B) 150m

Explanation:

Let the length of the train e x meter, and let the speed of train be y km/h, then

$x = (y + 3) \times \frac{5}{18} \times 36$ --------(1) and

$x = (y + 6) \times \frac{5}{18} \times 30$ ---------(2)

From eq (1) and (2), we get

y = 12 km/h

Therefore, $x = (y + 3) \times \frac{5}{18} \times 36$ => x= 150 m

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Q:

If Deepesh had walked 20 km/h faster he woulh have saved 1 hour in the distance of 60km. what is the usual speed of Deepesh?

A) 100	B) 120
C) 150	D) None of these

Answer & Explanation Answer: A) 100

Explanation:

Let the original speed be x km/h, then

$\frac{600}{x} - \frac{600}{(x + 20)} = 1$

$\Rightarrow 600 (\frac{x + 20 - x}{x (x + 20)}) = 1$

$\Rightarrow x^{2} + 120 x - 12000 = 0$

$\Rightarrow (x - 120) (x - 100) = 0$

Therefore, original speed = 100km/h

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Q:

A Person X started at 3 hours earlier at 40km/h from a place P, then another person Y followed him at 60km/h. Started his journey at 3 O'clock, afternoon. What is the diference in time when X was30 km ahead of Y and when Y was 30 km ahead of X?

A) 2h	B) 3h
C) 3.5h	D) 4.25h

Answer & Explanation Answer: B) 3h

Explanation:

Time ( when X was 30 km ahead of Y) = (120-30)/20 =4.5h

Time ( when Y was 30 km ahead of X) = (120+30)/20 = 7.5 h

Thus, required difference in time = 3h

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Q:

A person goes to his office at 1/3rd of the speed at which he returns from his office. If the average speed during the whole trip is 12 m /h . what is the speed of the person while he was going to his office?

A) 10	B) 6
C) 8	D) can't be determined

Answer & Explanation Answer: C) 8

Explanation:

u = k , v= 3k

$∴ \frac{2 u v}{u + v} \Rightarrow \frac{2 \times k \times 3 k}{(k + 3 k)} = 12$ => k = 8km/h

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Q:

Arjun and Bhishma are running on a circular track of length 600m. Speed of Arjun is 30 m/s and that of Bhishma is 20 m/s .They start fro the same point at the same time in the same direction. When will they meet again for the first time?

Answer

Sol : $Time = \frac{Circumference}{Relative Speed} = \frac{600}{10} = 60 \sec .$

Actually Arjun has to make a lead of 600 m, because when Arjun will be 600m ahead of Bhishma, they will be together again as a person when completes the total length it starts retracing he same path and thus Arjun and Bhishma can be together again.

Since , they make a difference of 10 m in 1 second. so, he will create 600m difference in 60 second.

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Subject: Time and Distance Exam Prep: AIEEE , Bank Exams , CAT
Job Role: Bank Clerk , Bank PO

Q:

In a one km race A gives B a start of 100m and in a one km race B gives a start of 80 m to C. In a 1 km race who will win and by how much distance from the worst performer between two losers?

Answer

Ratio of speeds of A:B = 1000:900 = 100:90

Ratio of speeds of B:C = 1000:920 = 100:92

Therefore, when A moves 1000m , B moves 900 m and when B moves 900m, C moves 828m

Thus, $\begin{matrix} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; 828m\; \; \;\; \; 900m\;\; \; \; \; 1000m\\ A...............|........|..........|\\ \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; C\; \; \; \; \; \; \; \; B\; \; \; \; \; \; \; \; \; A \end{matrix}$

Since , C moves 8 % less than B in the same time . Thus, C is the worst performer and A will win by him by 172 m

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Subject: Time and Distance

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