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Bunuel
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A train meets with an accident and travels at 4/7th of its regular spe 10 Jan 2020, 05:29
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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76% (02:33) correct 24% (02:48) wrong based on 321 sessions

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A train meets with an accident and travels at 4/7th of its regular speed hereafter and hence it reaches its destination 36 minutes late. How much time does the train take to reach its destination from the site of the accident had it traveled at its regular speed?

A. 40 minutes.
B. 42 minutes.
C. 46 minutes.
D. 48 minutes.
E. 50 minutes.
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D
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A train meets with an accident and travels at 4/7th of its regular spe 10 Jan 2020, 08:09
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Bunuel
A train meets with an accident and travels at 4/7th of its regular speed hereafter and hence it reaches its destination 36 minutes late. How much time does the train take to reach its destination from the site of the accident had it traveled at its regular speed?

A. 40 minutes.
B. 42 minutes.
C. 46 minutes.
D. 48 minutes.
E. 50 minutes.
Show more

D = s1 * t1 = s2 * t2
s1/s2 = t2/t1
Let t be normal travel time in minutes
7/4 = (t+36)/t
t = 48 minutes

IMO D

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A train meets with an accident and travels at 4/7th of its regular spe 10 Jan 2020, 05:30
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Bunuel
A train meets with an accident and travels at 4/7th of its regular speed hereafter and hence it reaches its destination 36 minutes late. How much time does the train take to reach its destination from the site of the accident had it traveled at its regular speed?

A. 40 minutes.
B. 42 minutes.
C. 46 minutes.
D. 48 minutes.
E. 50 minutes.
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TWIN QUESTION: https://gmatclub.com/forum/a-train-meet ... 14109.html
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A train meets with an accident and travels at 4/7th of its regular spe 10 Jan 2020, 08:01
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We know that if the distance is constant, then speed is inversely proportional to time.

Let's assume the original speed of \(7x\), so new speed is \(4x\).
Therefore, the time ratio will be \(\frac{4y}{7y}\)

Now, if the train was travelling at a speed of 7x then the time taken would be 4y.
But, now the speed is 4x and hence the time taken is 7y.

This difference of speeds creates a delay of 36 minutes = \(7y-4y = 3y\)

\(y = 12\)

The original time taken by the train to reach its destination from the site of accident = \(4y = 4 \times 12 = 48\)

OA, D

Bunuel
A train meets with an accident and travels at 4/7th of its regular speed hereafter and hence it reaches its destination 36 minutes late. How much time does the train take to reach its destination from the site of the accident had it traveled at its regular speed?

A. 40 minutes.
B. 42 minutes.
C. 46 minutes.
D. 48 minutes.
E. 50 minutes.
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A train meets with an accident and travels at 4/7th of its regular spe 04 Nov 2020, 06:04
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you could have used t and s/r to denote time and speed, why confuse so much.

uchihaitachi
We know that if the distance is constant, then speed is inversely proportional to time.

Let's assume the original speed of \(7x\), so new speed is \(4x\).
Therefore, the time ratio will be \(\frac{4y}{7y}\)

Now, if the train was travelling at a speed of 7x then the time taken would be 4y.
But, now the speed is 4x and hence the time taken is 7y.

This difference of speeds creates a delay of 36 minutes = \(7y-4y = 3y\)

\(y = 12\)

The original time taken by the train to reach its destination from the site of accident = \(4y = 4 \times 12 = 48\)

OA, D

Bunuel
A train meets with an accident and travels at 4/7th of its regular speed hereafter and hence it reaches its destination 36 minutes late. How much time does the train take to reach its destination from the site of the accident had it traveled at its regular speed?

A. 40 minutes.
B. 42 minutes.
C. 46 minutes.
D. 48 minutes.
E. 50 minutes.
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A train meets with an accident and travels at 4/7th of its regular spe 04 Nov 2020, 08:32
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Over the 2 scenarios, the Distance remains the same (from the point of the accident to the Destination)

Over this Constant Distance, Speed is Inversely Proportional to the Time spent traveling


From on-time, Usual Speed ———-> to the Reduced Speed after the accident:


The Speed has Decreased by (3/7) = (n)/(d + n)

Since Time is Inversely Proportional to Speed over this Constant Distance, the Time spent driving will INCREASE by (n/d) =

+(3/4) Increase in the Usual, no accident Time

This 3/4 Increase in the Usual Time corresponds to 36 extra minutes in which the driver is Late


Let the Usual Time to travel from the spot of the accident - to - the Destination = T

After the accident, the Time taken is:

T + 36 minutes = T + (3/4)T

36 = (3/4)T

T = 36 * (4/3)

T = 48 minutes = Usual Time had the accident not occurred and the driver was able to drive @ Normal Speed

-D-

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A train meets with an accident and travels at 4/7th of its regular spe 12 Nov 2020, 08:49
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Bunuel
A train meets with an accident and travels at 4/7th of its regular speed hereafter and hence it reaches its destination 36 minutes late. How much time does the train take to reach its destination from the site of the accident had it traveled at its regular speed?

A. 40 minutes.
B. 42 minutes.
C. 46 minutes.
D. 48 minutes.
E. 50 minutes.
Show more

Solution:

We can let 60 mph be 4/7th of the regular speed of the train; thus, the regular speed of the train is 60 / (4/7) = 60 x 7/4 = 105 mph. If we let t = the time, in hours, it takes the train to reach its destination from the site of the accident at its regular speed, we can create the equation:

105t = 60(t + 36/60)

105t = 60t + 36

45t = 36

t = 36/45 = 4/5

We see that it takes the train 4/5 of an hour, or 4/5 x 60 = 48 minutes, to reach its destination from the site of the accident at its regular speed.

Answer: D
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A train meets with an accident and travels at 4/7th of its regular spe 25 Nov 2020, 12:06
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Hi Scott, how did you decide on the number 60? ScottTargetTestPrep
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Bunuel
A train meets with an accident and travels at 4/7th of its regular speed hereafter and hence it reaches its destination 36 minutes late. How much time does the train take to reach its destination from the site of the accident had it traveled at its regular speed?

A. 40 minutes.
B. 42 minutes.
C. 46 minutes.
D. 48 minutes.
E. 50 minutes.

Solution:

We can let 60 mph be 4/7th of the regular speed of the train; thus, the regular speed of the train is 60 / (4/7) = 60 x 7/4 = 105 mph. If we let t = the time, in hours, it takes the train to reach its destination from the site of the accident at its regular speed, we can create the equation:

105t = 60(t + 36/60)

105t = 60t + 36

45t = 36

t = 36/45 = 4/5

We see that it takes the train 4/5 of an hour, or 4/5 x 60 = 48 minutes, to reach its destination from the site of the accident at its regular speed.

Answer: D
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A train meets with an accident and travels at 4/7th of its regular spe 13 Dec 2020, 04:52
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Naomilobo16
Hi Scott, how did you decide on the number 60? ScottTargetTestPrep
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Bunuel
A train meets with an accident and travels at 4/7th of its regular speed hereafter and hence it reaches its destination 36 minutes late. How much time does the train take to reach its destination from the site of the accident had it traveled at its regular speed?

A. 40 minutes.
B. 42 minutes.
C. 46 minutes.
D. 48 minutes.
E. 50 minutes.

Solution:

We can let 60 mph be 4/7th of the regular speed of the train; thus, the regular speed of the train is 60 / (4/7) = 60 x 7/4 = 105 mph. If we let t = the time, in hours, it takes the train to reach its destination from the site of the accident at its regular speed, we can create the equation:

105t = 60(t + 36/60)

105t = 60t + 36

45t = 36

t = 36/45 = 4/5

We see that it takes the train 4/5 of an hour, or 4/5 x 60 = 48 minutes, to reach its destination from the site of the accident at its regular speed.

Answer: D
Show more

Explanation:

Hi Naomilobo16, here's an explanation of why I chose 60 mph for the train speed.

When it is possible to solve a question by picking numbers, we want to choose our numbers in such a way that the arithmetic that follows is easy to perform. For instance, working with fractions is generally harder than working with integers, so if the problem involves division, we choose numbers which are multiples of the divisors in the question. If the problem involves taking square or cube roots, we choose numbers which will not produce any radicals (if possible). If the problem involves percents, we choose multiples of 100 so that we get whole numbers when we take percents. We also try to keep our numbers as small as possible so that the calculations are easy.

For this problem, before deciding the number to choose, we scan the problem and determine that the number we choose will be divided by 4/7 at some point. Therefore, the only condition when choosing a value for the (reduced) speed of the train is that the number should be divisible by 4. We chose 60 mph because it is divisible by 4 and it seems a reasonable value for the speed of a train. We could actually use 4 mph and arrive at the same answer; here’s how:

If 4/7 of the regular speed of the train is 4 mph, then the regular speed of the train is 4 / (4/7) = 7 mph. Let t be the number of hours for the train to reach its destination, as in my original solution. Then:

7t = 4(t + 36/60)

7t = 4(t + 3/5)

7t = 4t + 12/5

35t = 20t + 12

15t = 12

t = 12/15 = 4/5

As we can see, we obtained the same value of t = 4/5 by choosing the reduced speed of the train to be 4 mph instead of 60 mph. Any multiple of 4 would work the same way; however, I like to choose values that agree with real life, and one doesn’t see a lot of trains traveling at 7 or 4 miles per hour. That’s why I chose 60 mph.
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A train meets with an accident and travels at 4/7th of its regular spe 13 Dec 2020, 05:55
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A train meets with an accident and travels at 4/7th of its regular speed hereafter and hence it reaches its destination 36 minutes late. How much time does the train take to reach its destination from the site of the accident had it traveled at its regular speed?

Let's assume the original speed to be R, and Time=t
New speed after the accident: 4/7 R, and new time: T+36(because it takes 36 minutes more to reach the destination after the accident).

As the distance travelled by the train will be the same, irrespective of the accident
Therefore, we can create an equation where
(distance travelled before the accident = distance travelled after the accident)
i.e RT= 4R/7 *(T+36)
RT= 4RT/7 + 4/7*36R
RT-4 RT/7 = 4/7*36 R
3RT/7 = 4/7*36R
3T= 4*36
T= 48 minutes.

IMO: D
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A train meets with an accident and travels at 4/7th of its regular spe 08 Sep 2022, 03:53
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A train meets with an accident and travels at 4/7th of its regular speed hereafter and hence it reaches its destination 36 minutes late. How much time does the train take to reach its destination from the site of the accident had it traveled at its regular speed?

A. 40 minutes.
B. 42 minutes.
C. 46 minutes.
D. 48 minutes.
E. 50 minutes.

We can use ratios, If a train travelled at 4/7 of its original speed then the time taken would be inverse of speed = (7/4) Speed is inversely propotional with time.

From Stem : 4/7th Speed will delay the destination of the train by 36 mins
This can be written as 7/4T = T + 36
7T = 4(T+36)
7T = 4T + 144
3T = 144
T = 48

D
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A train meets with an accident and travels at 4/7th of its regular spe 27 Jul 2025, 02:21
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