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#### Solving Time-Speed-Distance problems without using equations

(Photo credit: Michael Gallacher)

I guess my first fascination with problems of Time, Speed and Distance began when I first watch a film called Henna. An important portion of the plot, if you can call it that, had Rishi Kapoor floating from India to Pakistan in a river without drowning. I remember arguing with my friends that if he could float that long he could swim back to India as well. My friends nullified the argument by saying,

Speed River > Speed Rishi Kapoor

I know that the reference is a little dated for most readers, but Zeba Bhaktiyar made me look beyond reason. In this post we will discuss some of the ideas that have helped me solve TSD problems without forming too many equations.

Funda 1: Average Speed

We know that the average speed during a journey is given by (Total Distance Covered) / (Total Time Taken); but there are a few special cases which might help in solving questions,

- If the distance covered is constant (d1 = d2 = d3 = dn) in each part of the journey, then the average speed is the Harmonic Mean of the values.

SpeedAvg = n / (1/s1 + 1/s2 + 1/s3 ... + 1/sn)

- If the time taken is constant (t1 = t2 = t3 = tn) in each part of the journey then the average speed is Arithmetic Mean of the values.

SpeedAvg = (s1 + s2 + s3 ... +sn)/n

Funda 2: Using Progressions (Arithmetic & Harmonic)

In many questions, you will come across a situation when a person is going from point A to point B at various speeds and taking various times. We know that if distance is constant, speed and time are inversely proportional to each other. But this information can also be used to deduce the following two facts,

- If the various speeds which are mentioned are in AP, then the corresponding times taken will be in HP.

- If the various speeds which are mentioned are in HP, then the corresponding times taken will be in AP.

Let us use these ideas to solve couple of quant questions.

Example 1

Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30, 40 and 60 km per hour respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start?

(Some useless information: Arun Barun Kiranmala is a 1968 Bangladeshi film. Now you can guess what inspires the CAT question setters. Here is a song from the film.)

Solution As you can see that the speeds are in HP, so we can say that the times taken will be in AP. Time difference between Arun and Barun is 2 hours, so the time difference between Barun and Kiranbala will also be 2 hours.

Hence, Kiranbala started 4 hours after Arun.

Example 2

Rishi Kapoor can swim a certain course against the river flow in 84 minutes; he can swim the same course with the river flow in 9 minutes less than he can swim in still water. How long would he take to swim the course with the river flow?

Solution

Let us say Speed of Rishi Kapoor in still water is RK and Speed of the river is R. Hence, Rishi Kapoors speeds against the river flow, in still water and with the river flow are,

RK R, RK and RK + R.

As you can see, they are in AP.

Hence, the corresponding times taken will be in HP.

Let us say that the time taken to row down with the stream is t, then 84, t+9 and t are in HP. So,

t + 9 = (2 * 84 * t) / (84 + t)

? t2 + 93t + 756 = 168t

? t2 - 75t + 756 = 0

? t = 63 or 12

Funda 3: Special Case

Let us say that two bodies a & b start at the same time from two points P & Q towards each other and meet at a point R in between. After meeting at R, a takes ta time to reach its destination (Q) and b takes tb time to reach its destination (P). Then,

Sa / Sb = ?(tb / ta)

Also, the time taken by a & b to meet (i.e. to reach point R from P & Q respectively) is given by,

t = ?(ta * tb)

Note: The same formulae will be valid if two bodies a & b start at different times from two points P & Q towards each other. They meet at a point R in between after travelling for ta and tb time respectively. After meeting, they take the same amount of time (t) to reach their respective destinations (Q & P).

I hope that these ideas will help you reduce the number of equations that you form while solving TSD problems if not completely eliminate them.

Author Ravi Handa has taught Quantitative Aptitude at IMS for 4 years. An alumnus of IIT Kharagpur where he studied a dual-degree in computer science, he has also written a book on business awareness.

• @ravihanda : please explain how to calculate the squarero.... 02 Sep '13.
• sir thank you !. 11 Sep '13.
shiwami sharma @SHIWAMI 12
please explain me HP.i didnt get exmple no 2.
shiwami sharma @SHIWAMI 12
hi everybody.. ijust start preparng for cAT exam .please guide me ..from where i start or what to do??? am in muddle stage.please help me out
Siddharth pote @Sid56
RAvi sir thank you very much for this article.........This is very helpful......But im not getting pranav_maheshwa answer on @ankit87sharma question......can u pls explain it
raj man @able1
sir is there any text book that could provide me all methods you teach
raj man @able1
ranging from proportionality to AP GP in time speed and distance
Avishek Mukherjee @avidexstr 5
Love the Rishi Kapoor problem.
praveen kumar @shalloit 1
the above question cannot be solved until and unless you have dont have the speed of anyone. either the speed of the first person or the second person should be given
Ashima Dewan @earth21
@ravihanda : please explain how to calculate the squareroot of [(16*60+20)*(21*60+20)] in regard to ques by @ ksagarwa ? It gets reduced to sqrt(1254400). how can such tedious calculations be done during exams ?

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