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One hour after Matthew started waking from x to y, a distanc

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One hour after Matthew started waking from x to y, a distanc  [#permalink]

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New post Updated on: 09 Jan 2013, 03:06
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One hour after Matthew started waking from x to y, a distance of 45km, Johnny started walking along the same road from y to x. Matthew's walking rate was 3 km per hour and Johnny's was 4km per hour, how many km had Johnny walked when they met?

A. 24
B. 23
C. 22
D. 21
E. 19.5

Originally posted by chiccufrazer1 on 08 Jan 2013, 14:45.
Last edited by Bunuel on 09 Jan 2013, 03:06, edited 1 time in total.
Renamed the topic, edited the question and added OA.
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Re: one hour after matthew started walking from x to y,a distanc  [#permalink]

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New post 08 Jan 2013, 23:05
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1
chiccufrazer1 wrote:
One hour after matthew started waking from x to y,a distance of 45km,johnny started walking along the same road from y to x..matthew's walking rate was 3 km per hour and johnny's was 4km per hour,how many km had johnny walked when they met?

A.24
B.23
C.22
D.21
E.19.5


Just an alternative method...

After the first hour the distance is 42 km (45-3). Now the problem can be treated as if bothof them started at the same time. Since the speeds are in the ratio 3 : 4, the distances will also be in the same ratio. Splitting 42 in that ratio we get 18 : 24. So answer is 24.

Hence A.
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Re: one hour after matthew started walking from x to y,a distanc  [#permalink]

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New post 08 Jan 2013, 19:39
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chiccufrazer1 wrote:
One hour after matthew started waking from x to y,a distance of 45km,johnny started walking along the same road from y to x..matthew's walking rate was 3 km per hour and johnny's was 4km per hour,how many km had johnny walked when they met?

A.24
B.23
C.22
D.21
E.19.5


The best method to solve such questions is the summation of distance method.
Please note that when Matthew and Johny would meet, the sum of the distance covered will be the lenth of the journey i.e. 45 km.
Distance covered by Matthew=\(3(t+1)\)
Distance covered by Johny=\(4t\)
where t=time taken by Johny.
Therefore \(3(t+1) + 4t = 45\)
t will come as 6.
Distance covered by Johny is \(4t\) or \(4*6\)=\(24km\) Ans.
+1A
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Re: one hour after matthew started walking from x to y,a distanc  [#permalink]

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New post 08 Jan 2013, 22:09
wow..u have done in a simple way indeed..i tried using the train's formula which states that (v2+v1)*t=D,but it did not work..would you please explain to me why i had problems using that formula?

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Re: one hour after matthew started walking from x to y,a distanc  [#permalink]

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New post 08 Jan 2013, 22:16
I assume that v1 and v2 are respective speeds.
Here the formula won't work because the two start at different times. Had they started at same time, the formula would have been justified.
I hope you getting my point.
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Re: one hour after matthew started walking from x to y,a distanc  [#permalink]

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New post 08 Jan 2013, 23:33
MacFauz wrote:
chiccufrazer1 wrote:
One hour after matthew started waking from x to y,a distance of 45km,johnny started walking along the same road from y to x..matthew's walking rate was 3 km per hour and johnny's was 4km per hour,how many km had johnny walked when they met?

A.24
B.23
C.22
D.21
E.19.5


Just an alternative method...

After the first hour the distance is 42 km (45-3). Now the problem can be treated as if bothof them started at the same time. Since the speeds are in the ratio 3 : 4, the distances will also be in the same ratio. Splitting 42 in that ratio we get 18 : 24. So answer is 24.

Hence A.


what would happen if the were walking in the same direction?

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Re: one hour after matthew started walking from x to y,a distanc  [#permalink]

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New post 09 Jan 2013, 02:12
1
chiccufrazer1 wrote:
MacFauz wrote:
chiccufrazer1 wrote:
One hour after matthew started waking from x to y,a distance of 45km,johnny started walking along the same road from y to x..matthew's walking rate was 3 km per hour and johnny's was 4km per hour,how many km had johnny walked when they met?

A.24
B.23
C.22
D.21
E.19.5


Just an alternative method...

After the first hour the distance is 42 km (45-3). Now the problem can be treated as if bothof them started at the same time. Since the speeds are in the ratio 3 : 4, the distances will also be in the same ratio. Splitting 42 in that ratio we get 18 : 24. So answer is 24.

Hence A.


what would happen if the were walking in the same direction?

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In such a case, the total distance would be irrelevant.

In the 1st hour Mathew would have travelled 3 kms.
Let us say that after this first hour it takes Matthew and Johnny "t" hours to meet.
So, distance travelled by Matthew and distance travelled by Johnny should be the same.

Distance travelled by Matthew = 3 + 3*t
Distance travelled by Johnny = 4*t

3 + 3*t = 4*t. So, t = 3 hrs. So distance travelled would be = 4*3 = 12 kms
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Re: one hour after matthew started walking from x to y,a distanc  [#permalink]

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New post 09 Jan 2013, 02:55
I used the train formula but modified the distance travelled.

After one hour, mathew has travelled 3 kms so 42 kms left when Johnny starts.

so distance to travel is 42 kms for both, in bwtween which they will meet,

No d = 42 kms
t = hrs
speed = 3+4 (since they moving in opposite direction to each other) = 7

s = d/t
7 = 42/t
t = 6 hrs

So distance travelled by John = 4*6 = 24 kms

Option A
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Re: One hour after Matthew started waking from x to y, a distanc  [#permalink]

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New post 09 Jan 2013, 05:07
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Quote:
One hour after Matthew started waking from x to y, a distance of 45km, Johnny started walking along the same road from y to x. Matthew's walking rate was 3 km per hour and Johnny's was 4km per hour, how many km had Johnny walked when they met?


Since M started walking one hour before J the distance they have to travel together has to be 45km-3km/h*1h=42km

The combinded speed of M and J is 3km/h+4km/h=7km/h. They thus need 42km/7(km/h)=6h to cover the 42km between them. Therefore J has to walk 4km/h*6h=24km
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Re: One hour after Matthew started waking from x to y, a distanc  [#permalink]

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New post 09 Jan 2013, 09:15
If Matthew has an hour heads start then that means the distance between them when Johnny starts walking is 42kms. Each hr their combined distance travelled is 7 kms (3km/hr + 4km/hr). That must mean that it will take 6 hours for them to meet each other.

In 6 hrs Johnny walked 6x4 = 24 kms.
Answer: A
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Re: one hour after matthew started walking from x to y,a distanc  [#permalink]

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New post 10 Jan 2013, 02:21
bhavinshah5685 wrote:
I used the train formula but modified the distance travelled.

After one hour, mathew has travelled 3 kms so 42 kms left when Johnny starts.

so distance to travel is 42 kms for both, in bwtween which they will meet,

No d = 42 kms
t = hrs
speed = 3+4 (since they moving in opposite direction to each other) = 7

s = d/t
7 = 42/t
t = 6 hrs

So distance travelled by John = 4*6 = 24 kms

Option A


can we approach this problem in the same way(willard types at an average speed of 40 words per minute and george types at 60 words per minute.if willard begins typing a manuscript at 4:20 and george begins typing an identical copy of the manuscript at 5:00,at what time will they be typing the same word?
A.5:10 B.5:30 C.5:40 D.6:00 E.6:20''

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Re: One hour after Matthew started waking from x to y, a distanc  [#permalink]

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New post 19 Mar 2013, 06:43
speed of mathew = 3 kph
let time be 't'
tot dist travelled by mathew = 3t

johny starts an hour later ...so john's time would be = t-1
speed of johny = 4 kph
tot dist travelled bby johny = 4(t-1)

since tot dist given is 45 kms so...
3t + 4(t-1) = 45
7t - 4 = 45
7t = 41
t= 41/7 hrs

so tot dist travelled by johny is = 4(t-1)
= 4 ( 41/7 - 1)
= 4 ( 34/7)
= 136/7
= 19.42 = 19.5 (approx0

so answer is E.

if its wrong...please explain me the reason...
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Re: One hour after Matthew started waking from x to y, a distanc  [#permalink]

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New post 20 Mar 2013, 04:47
Perhaps wrote:
One hour after Matthew started waking from x to y, a distance of 45km, Johnny started walking along the same road from y to x. Matthew's walking rate was 3 km per hour and Johnny's was 4km per hour, how many km had Johnny walked when they met?

A. 24
B. 23
C. 22
D. 21
E. 19.5

speed of mathew = 3 kph
let time be 't'
tot dist travelled by mathew = 3t

johny starts an hour later ...so john's time would be = t-1
speed of johny = 4 kph
tot dist travelled bby johny = 4(t-1)

since tot dist given is 45 kms so...
3t + 4(t-1) = 45
7t - 4 = 45
7t = 41
t= 41/7 hrs


so tot dist travelled by johny is = 4(t-1)
= 4 ( 41/7 - 1)
= 4 ( 34/7)
= 136/7
= 19.42 = 19.5 (approx0

so answer is E.

if its wrong...please explain me the reason...


The red part above is not correct: 7t - 4 = 45 --> 7t = 49 --> t = 7 --> 4(t - 1) = 24.

Hope it helps.
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Re: One hour after Matthew started waking from x to y, a distanc  [#permalink]

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New post 20 Mar 2013, 06:08
oops... :roll:
thanks bunuel :)
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Re: One hour after Matthew started waking from x to y, a distanc  [#permalink]

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New post 18 Mar 2017, 19:52
By the time Jhonny Started, Mathew had already traveled 3 KM.
So the distance left between jho & Mat is 42 km.
Relative Speed = 3 + 4 = 7 km as both are moving towards each other.
So Both will meet after 42/7 = 6 hours.
So Math travels = 4 * 6 =24 Km

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Re: One hour after Matthew started waking from x to y, a distanc  [#permalink]

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New post 13 Aug 2017, 21:01
chiccufrazer1 wrote:
One hour after Matthew started waking from x to y, a distance of 45km, Johnny started walking along the same road from y to x. Matthew's walking rate was 3 km per hour and Johnny's was 4km per hour, how many km had Johnny walked when they met?

A. 24
B. 23
C. 22
D. 21
E. 19.5


3(x+1) + 4(x) =45
3x + 3 +4x =45
7x =42
x= 6

4(6)= 24

A
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Re: One hour after Matthew started waking from x to y, a distanc   [#permalink] 13 Aug 2017, 21:01
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