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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:

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

Q:

A contractor undertook a project to complete it in 20 days which needed 5 workers to work continuously for all the days estimated. But before the start of the work the client wanted to complete it earlier than the scheduled time, so the contractor calculated that he needed to increase 5 additional men every 2 days to complete the work in the time the client wanted it:

If the work was further increased by 50% but the contractor continues to increase the 5 workers o every 2 days then how many more days are required over the initial time specified by the client.

A) 1 day	B) 2 days
C) 5 days	D) None of these

Answer & Explanation Answer: B) 2 days

Explanation:

Total work = 100+50 = 150man-days

In 8 days 100 man-days work has been completed. Now on 9th and 10th day there will be 25 workers. So in 2 days they wll complete additional 50 man- days work. Thus the work requires 2 more days.

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Filed Under: Time and Work

Q:

In a public bathroom there are n taps 1,2,3...n. Tap1 and Tap2 take equal time to fill the tank while Tap3 takes half the time taken by Tap2 and Tap4 takes half the time taken by Tap3. Similarly each next number of tap takes half the time taken by previous number of Tap i.e, K-th Tap takes half the time taken by (K-1)th Tap. If the 10th tap takes 2 hours to fill the tank alone then what is the ratio of efficiency of 8th tap and 12th tap, respectively?

A) 4:1	B) 5:3
C) 16:1	D) 1:16

Answer & Explanation Answer: D) 1:16

Explanation:

Time taken by 8th tap = 2 x 2 x 2= 8 hours

Time taken by 12th tap = 2x (1/2) x (1/2) = 1/2 hour

Ratio of time taken by 8th tap and 12th tap = 8 : 1/2 =16:1

Therefore, Ratio of efficiencies of 8th tap and 12th tap =1:16

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Filed Under: Time and Work

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