**PSA_LEVEL-UP_QNR48.Amit starts on his bike from his hometown Hazipur for Vizag at a high but constant speed. Just after 100 km he feels tired and from this point he rides at a lower speed by reducing the speed by 25%. He finds he got Vizag late by 45 minutes. He does some calculation and finds that if he had slowed down after 148 km (and not after 100 km), he would have reached late anyway but this time by 35 mins. What is Amit’s original speed?**

**a. 48 km/h**

**b. 24 km/h**

**c. 56 km/h**

**d. 96 km/h**

Ans:

Let us tackle this question using conventional method and later using a shortcut.

Conventional method:

——————————

Assume that the distance between Hazipur and Vizag is ‘D’ km.

Amit starts at a certain speed say ‘s’ and travels for 100 km (say till R); afterward he rides at a speed of ‘3s/4’ (speed reduces by 25% or one fourth).

|H|————————-|R|————————————-|V|

<————100km—><——(D-100 )km————–>

If he usally takes ‘t’ mins of time, he has taken ‘t+45’ min this time. Thus, equation that can be formed here is:

t+45 = ^{100}/_{s} + ^{(D-100)}/_{(3s/4)} ————(i)

|H|——————————-|R|——————————-|V|

<————148km———><——(D-148)km———>

Similarly, when he finds that if he had slowed down after 148 km (and not after 100 km), the time taken would have been:

t+35 = ^{148}/_{s} + ^{(D-148)}/_{(3s/4)} ————(ii)

Subtracting (ii) from (i),

10 = ^{-48}/_{s} + ^{48}/_{(3s/4)}

=>10 = ^{-48}/_{s} + ^{64}/_{(3s/4)}

=>10= ^{16}/_{s}

=>s= 1.6 km/min = 1.6 *60 km/hr = 96 km/hr

**Shortcut**

** ————-**

If you refer to the above two journey figures, you would find that in both cases speed for the first 100 km is ‘s’. Similarly, speed for the stretch (D-148 distance from Vizag) is ‘3s/4’. The only difference is the stretch from 100 km mark till 148 km i.e a distance of 48 km where Amit travels at speed ‘s’ in the first case and ‘3s/4’ in the second. This distance is marked (by vertical red dotted lines) in the below figure.

SxT=D.

If speed reduces by ¼th, speed becomes ¾th. Since, distance (=48 km) is constant, time would become 4/3 times the original value.Therefore, reduction in time is ^{1}/_{3}rd.

Also from the question, reduction in time is given as 10 minutes. (He was 45 minutes late and 35 minutes late in the second case).

Now, the simple question is-> ^{1}/_{3}rd of ‘*WHAT*‘ is 10?

As you guess it right, ‘*WHAT*‘ = 30 minutes. Thus, Amit takes 30 minutes to travel this 48 km stretch at his original speed ‘s’.

Speed, s= ^{48}/_{(30/60)} = 96 km/hr.

This entire calculation can be done in less than a minute.

——

Hi,

I am following your website for quite sometime. Are you going to publish videos on speed, time and distance.

Thanks.

Hi Sekhar,

We missed out your (and lot of other visitors’) comment due to site issues. You may have already noticed some videos on Time, Speed and Distance with some really good questions and “how-to-solve” approaches! Hope, those (& videos of other topics) help you in your preparation for any aptitude examinations or enriching your general knowledge about the topic.

We appreciate your comments! Please do not hesitate to drop a note on any further queries that you may have in future.

Thanks,

Admin (onlinepsa.in)