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Time Speed and Distance - Quick Approach

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Time Speed and Distance - Quick Approach [#permalink]

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15 Sep 2010, 07:48
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Time Speed and Distances

I expect after reading this, you guys won't fear from time speed and distance problems.

# Remember the basic formula : Distance = Time x Speed

This formula involves 3 variable and large number of permutations and combination are possible to twist the problem. Lets analyze three main cases :

1. Distance is constant, speed and time are varied.

=> $$Time$$ α $$\frac{1}{speed}$$ => $$\frac{T_{1}}{T_{2}} = \frac{V_{2}}{V_{1}}$$

Result: If the speed is doubled, the time is halved and if the speed is $$\frac{m}{n}$$ 'th of the original speed, the time required is $$\frac{n}{m}$$ 'th.

Lets practice this.

Eg1. A person leaves his home everyday at 11:00 am and reaches his office at 12:00 pm. One day he left his house at normal time but traveled the first half of the distance at speed of $$\frac{2}{3}$$ of the normal speed. What should be the speed of second half so that he reaches at the same time?

Solution:1
Time taken to reach the office normally = 60 minutes.
=> at half the distance time taken = 30 minutes. If the traveling speed is $$\frac{2}{3}$$ of the normal speed for the first half, the time taken is $$\frac{3}{2}$$ of the time taken to reach the first half.

=> $$\frac{3}{2}* 30$$ = 45 minutes.
To reach the office after 1 hour he needs to travel the second half in 60-45 = 15 minutes.

With the normal speed he travels the second half in 30 minutes, now using the above result if he needs to cover the second half in 15 minutes, he should double his speed.

Eg2. If in the eg1, if he travels the second half at $$\frac{3}{2}$$ of the original speed, at what time will he reach the office?

Solution 2:
Using eg1, he will reach the first half in 45 minutes i.e. 11:45.

If he travels the second half at speed = $$\frac{3}{2}$$ of normal speed, time required will be

=$$\frac{2}{3}$$ of the normal time = $$\frac{2}{3}*30$$ = 20 minutes.

Thus he will reach the office at 12:05 pm i.e. 5 minutes late.

Eg3. If 'GG' travels at the $$\frac{3}{4}$$ normal speed, he is late by 15 minutes. How much time usually he takes to reach the office?

Solution 3:
New speed = $$\frac{3}{4}$$ of original
New time taken = $$\frac{4}{3}$$ of original

Difference in time = new time - old time = $$\frac{4t}{3} - t = \frac{t}{3}= 15$$ minutes
=> t = 45 minutes

Lets crack a tough problem.

Eg4. A, B, and C starts from the same place and travel in the same direction at speeds of 30,40,60 respectively.
B starts 2 hours after A, but B and C overtakes A at the same instant. How many hours after A did C start?

Looks daunting? Don't worry. Lets crack it.

Solution 4:
Since the distance traveled by each of them is same, we can use the concept we had already discussed.

Time taken by A = T , speed of A = 30

Time taken by B = T -2 , speed of A = 40

Time taken by C = T - c , speed of A = 60

Now we have , $$\frac{T}{(T-2)} = \frac{40}{30}$$
=> T = 8

Also we have, $$\frac{T}{(T-c)} = \frac{60}{30}$$ , put T=8
we get c=4

=> C started 4 hours after A.

2. Time is constant, speed and Distance are varied.

$$Distance$$ α $$speed$$ => $$\frac{D_{1}}{D_{2}} = \frac{V_{1}}{V_{2}}$$

Result : If the speed of travel is doubled, the distance traveled in the same time is doubled.
If the speed is $$\frac{m}{n}$$ 'th of the original speed, the Distance traveled in the same time is $$\frac{m}{n}$$ 'th. of the original distance.

Lets practice this.

Eg5. X and Y run a race between A and B stations, 5 Kms apart. X starts at 9 AM from A at speed of 5 km/h, reaches B and returns back to A at same speed. Y starts at 9:45 AM from A at speed 10 km/h, reaches B and comes back to A at same speed.
At what time do X and Y first meet each other?

Solution 5: Since distance is 5 Kms, X reaches B at 10:00 AM. In 15 minutes (at 10:00 AM) Y has traveled 2.5 kms i.e. half the distance between stations.

Now X is traveling towards A and Y towards B. From 10:00 to the time till they reach they travel for the same time.
=> Ratio of their speed = Ratio of distance traveled by them

=> $$10/5 = \frac{D_{Y}}{D_{X}}$$

=> $$2 * D_{X} = D_{Y}$$ => Distance traveled by Y = twice that of X.
=> Distance traveled by Y = $$\frac{2}{3} *$$ half of AB = $$\frac{2}{3} * \frac{5}{2}$$ = 5/3

Time taken = $$\frac{5}{3} * \frac{60}{10}= 10$$ minutes => they will meet at 10:10 Am.

Will be updated soon with more Time speed and distance related fundamentals, keep visiting the thread

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Re: Time Speed and Distance - Quick Approach [#permalink]

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15 Sep 2010, 11:26
good work keep going............
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Re: Time Speed and Distance - Quick Approach [#permalink]

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15 Sep 2010, 11:43
Hey, Thanks gurpreet !
I would also really appreciate if you could share something similar for Work Rate..I am having a rough time with it

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Re: Time Speed and Distance - Quick Approach [#permalink]

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15 Sep 2010, 12:11
Thank you very much for this post. One question. Where are you getting 2/3 in Solution 5?
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Kudos [?]: 1885 [0], given: 235

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GMAT 2: 710 Q50 V35
Re: Time Speed and Distance - Quick Approach [#permalink]

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15 Sep 2010, 16:14
dokiyoki wrote:
Hey, Thanks gurpreet !
I would also really appreciate if you could share something similar for Work Rate..I am having a rough time with it

I will update the thread with work-Rate as well. Give me some days.

cheetarah1980 wrote:
Thank you very much for this post. One question. Where are you getting 2/3 in Solution 5?

Since Distance traveled by Y = twice that of X.

=> distances traveled are in the ratio ( Ratio of Y and X) = 2:1
=> Distance traveled by Y = 2a , distance traveled by X = a => total distance = 3a
=> distance traveled by y = 2/3 * total distance
here total distance is half of AB as at 10:00 both have a gap equal to AB.

I hope it helps.
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Re: Time Speed and Distance - Quick Approach [#permalink]

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25 Jan 2012, 12:43
Hi, can you tell me how in example 5 you got the following?

=> $$2 * D_{X} = D_{Y}$$ => Distance traveled by Y = twice that of X.
=> Distance traveled by Y = $$\frac{2}{3} *$$ half of AB = $$\frac{2}{3} * \frac{5}{2}$$ = 5/3

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Re: Time Speed and Distance - Quick Approach [#permalink]

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26 Jan 2012, 03:38
Thanu1083 wrote:
Hi, can you tell me how in example 5 you got the following?

=> $$2 * D_{X} = D_{Y}$$ => Distance traveled by Y = twice that of X.
=> Distance traveled by Y = $$\frac{2}{3} *$$ half of AB = $$\frac{2}{3} * \frac{5}{2}$$ = 5/3

At 10:00, X is at B and Y is at mid point of AB. Now, together they need to cover the distance between them which is half the distance between A and B. When they meet, they would have traveled for the same time (starting at 10:00). Hence, D1/D2 = V1/V2

The ratio of speeds of X: Y is 1:2 (Since speed of X is 5 kmph and speed of Y is 10 kmph)
From the theory given in the original post, distance traveled will also be in the ratio 1:2. This means X will travel 1/3rd of the total distance between them (half of AB) and Y will travel 2/3rd of the total distance between them (half of AB)

I am a fan of ratios too and have discussed these methods on my blog. You can check them out at:
http://www.veritasprep.com/blog/2011/03 ... of-ratios/
http://www.veritasprep.com/blog/2011/03 ... os-in-tsd/
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Re: Time Speed and Distance - Quick Approach [#permalink]

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26 Jan 2012, 05:06
Thanu1083 wrote:
Hi, can you tell me how in example 5 you got the following?

=> $$2 * D_{X} = D_{Y}$$ => Distance traveled by Y = twice that of X.
=> Distance traveled by Y = $$\frac{2}{3} *$$ half of AB = $$\frac{2}{3} * \frac{5}{2}$$ = 5/3

I'd approach this question in a different manner:
X and Y run a race between A and B stations, 5km apart. X starts at 9am from A at speed of 5km/h, reaches B and returns back to A at same speed. Y starts at 9:45am from A at speed 10km/h, reaches B and comes back to A at same speed. At what time do X and Y first meet each other?

X needs an hour to reach station B (time=distance/rate=5/5=1 hour), so X reaches B at 10:00am;

At 10:00am Y has traveled for 15 minutes (1/4th of an hour) hence covered 1/4*10=2.5km, so half of the distance;

Now, the distance left to cover for both of them is another 2.5km and as combined rate of X and Y is (5+10)=15km/h, then they'll cover it in 2.5/15=5/30 hours=10min;

Thus the will meet at 10:00am+10min=10:10am.

Hope it helps.
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Re: Time Speed and Distance - Quick Approach [#permalink]

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19 Dec 2013, 13:30
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Re: Time Speed and Distance - Quick Approach [#permalink]

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18 Aug 2017, 23:54
Bunuel wrote:
Thanu1083 wrote:
Hi, can you tell me how in example 5 you got the following?

=> $$2 * D_{X} = D_{Y}$$ => Distance traveled by Y = twice that of X.
=> Distance traveled by Y = $$\frac{2}{3} *$$ half of AB = $$\frac{2}{3} * \frac{5}{2}$$ = 5/3

I'd approach this question in a different manner:
X and Y run a race between A and B stations, 5km apart. X starts at 9am from A at speed of 5km/h, reaches B and returns back to A at same speed. Y starts at 9:45am from A at speed 10km/h, reaches B and comes back to A at same speed. At what time do X and Y first meet each other?

X needs an hour to reach station B (time=distance/rate=5/5=1 hour), so X reaches B at 10:00am;

At 10:00am Y has traveled for 15 minutes (1/4th of an hour) hence covered 1/4*10=2.5km, so half of the distance;

Now, the distance left to cover for both of them is another 2.5km and as combined rate of X and Y is (5+10)=15km/h, then they'll cover it in 2.5/15=5/30 hours=10min;

Thus the will meet at 10:00am+10min=10:10am.

Hope it helps.

I have a query in the solution provided. So, X reached B at 10:00 am. However, by that time (15 mins before that ) Y has already started ,so before reaching B it will meet Y. Hence, I was adding the 10 mins in 9:45am instead 10:00 am

Am I missing something here?

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Re: Time Speed and Distance - Quick Approach [#permalink]

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19 Aug 2017, 03:12
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2201neha wrote:
Bunuel wrote:
Thanu1083 wrote:
Hi, can you tell me how in example 5 you got the following?

=> $$2 * D_{X} = D_{Y}$$ => Distance traveled by Y = twice that of X.
=> Distance traveled by Y = $$\frac{2}{3} *$$ half of AB = $$\frac{2}{3} * \frac{5}{2}$$ = 5/3

I'd approach this question in a different manner:
X and Y run a race between A and B stations, 5km apart. X starts at 9am from A at speed of 5km/h, reaches B and returns back to A at same speed. Y starts at 9:45am from A at speed 10km/h, reaches B and comes back to A at same speed. At what time do X and Y first meet each other?

X needs an hour to reach station B (time=distance/rate=5/5=1 hour), so X reaches B at 10:00am;

At 10:00am Y has traveled for 15 minutes (1/4th of an hour) hence covered 1/4*10=2.5km, so half of the distance;

Now, the distance left to cover for both of them is another 2.5km and as combined rate of X and Y is (5+10)=15km/h, then they'll cover it in 2.5/15=5/30 hours=10min;

Thus the will meet at 10:00am+10min=10:10am.

Hope it helps.

I have a query in the solution provided. So, X reached B at 10:00 am. However, by that time (15 mins before that ) Y has already started ,so before reaching B it will meet Y. Hence, I was adding the 10 mins in 9:45am instead 10:00 am

Am I missing something here?

Notice that both start from A. Next, the fact that by 10am Y has travelled for 15 minutes (1/4th of an hour) and covered 1/4*10=2.5km, is already accounted in the solution. Pay attention to the highlighted part.

Hope it helps.
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Re: Time Speed and Distance - Quick Approach [#permalink]

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19 Aug 2017, 19:24
Thanks Bunuel . Got it what I was missing. :)

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Re: Time Speed and Distance - Quick Approach   [#permalink] 19 Aug 2017, 19:24
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