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At the same time Fredrick started walking toward Bernard's house, a

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Bunuel
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At the same time Fredrick started walking toward Bernard's house, a [#permalink] New post   28 Mar 2017, 02:49
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A
B
C
D
E

Difficulty:

25% (medium)

Question Stats:

74% (01:38) correct 26% (01:29) wrong based on 130 sessions
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At the same time Fredrick started walking toward Bernard's house, a distance of 70 blocks, Barnard left his house along the same route to meet him. If Fredrick was traveling at 6 blocks every ten minutes and Bernard was traveling at 8 blocks every ten minutes, how long did it take them to meet?

A. 5 minutes
B. 10 minutes
C. 35 minutes
D. 50 minutes
E. 70 minutes
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D
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At the same time Fredrick started walking toward Bernard's house, a [#permalink] New post   28 Mar 2017, 03:43
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Bunuel wrote:
At the same time Fredrick started walking toward Bernard's house, a distance of 70 blocks, Barnard left his house along the same route to meet him. If Fredrick was traveling at 6 blocks every ten minutes and Bernard was traveling at 8 blocks every ten minutes, how long did it take them to meet?

A. 5 minutes
B. 10 minutes
C. 35 minutes
D. 50 minutes
E. 70 minutes


So both of them will cover 70 blocks in 50 mins walking at their respective rates.

Hence option D is correct
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Re: At the same time Fredrick started walking toward Bernard's house, a [#permalink] New post   28 Mar 2017, 04:00
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add the rates i.e 6/10 + 8/10 = 14/10 blocks per minute.

total blocks to be covered = 70

hence time = 70 / (14/10) = 50 mins

ans D
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Re: At the same time Fredrick started walking toward Bernard's house, a [#permalink] New post   30 Mar 2017, 16:39
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Bunuel wrote:
At the same time Fredrick started walking toward Bernard's house, a distance of 70 blocks, Barnard left his house along the same route to meet him. If Fredrick was traveling at 6 blocks every ten minutes and Bernard was traveling at 8 blocks every ten minutes, how long did it take them to meet?

A. 5 minutes
B. 10 minutes
C. 35 minutes
D. 50 minutes
E. 70 minutes


This is a converging rate problem in which we can use the following formula:

distance of Fredrick + distance of Bernard = total distance

Since Fredrick walks 6 blocks every 10 minutes, his rate is 6/10 = 3/5 blocks per minute. Since Bernard walks 8 blocks every 10 minutes, his rate is 8/10 = 4/5 blocks per minute. Since distance = rate x time, and if we let t = the time they travel before they meet, Frederick’s distance is (3/5)t and Bernard’s distance is (4/5)t. Thus:

(3/5)t + (4/5)t = 70

7t/5 = 70

7t = 350

t = 50

Answer: D
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Re: At the same time Fredrick started walking toward Bernard's house, a [#permalink] New post   21 Apr 2017, 23:15
Bunuel wrote:
At the same time Fredrick started walking toward Bernard's house, a distance of 70 blocks, Barnard left his house along the same route to meet him. If Fredrick was traveling at 6 blocks every ten minutes and Bernard was traveling at 8 blocks every ten minutes, how long did it take them to meet?

A. 5 minutes
B. 10 minutes
C. 35 minutes
D. 50 minutes
E. 70 minutes


Option:D
Time taken: 1:06
Here we go
add the two speeds 8+6=14
14 blocks/10 minutes
divide it by number of blocks
70/14/10
70*10/14
700/14
=50 minutes
Experts Please need your critical analysis on my approach will it work always?
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Re: At the same time Fredrick started walking toward Bernard's house, a [#permalink]
21 Apr 2017, 23:15
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