Speed, distance and time problem

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November 13th 2009, 06:20 AM
saberteeth

Speed, distance and time problem

Two workers X and Y staying at the same place are working in the same factory. X takes 20 mins while Y takes 30 mins to reach the factory by the same road. One day, Y started at 7.10 am from his place and X started at 7.16am. At what time would they meet each other on their way?
November 13th 2009, 06:58 AM
Wilmer

Not enough information given.
They'll never meet if both speeds are same
...unless they always leave from same spot
Note: missed "staying at same place"; apologies.
November 13th 2009, 07:43 AM
stapel

Quote:

Originally Posted by saberteeth

Two workers X and Y staying at the same place are working in the same factory. X takes 20 mins while Y takes 30 mins to reach the factory by the same road. One day, Y started at 7.10 am from his place and X started at 7.16am. At what time would they meet each other on their way?

To learn how to set up and solve this sort of "uniform rate" exercise, try here. (Wink)

Then use the standard set-up, noting that "meeting" here actually means "passing". The rate for X, given that the distance from home to the factory is "d", is d/20 units per minute; the rate for Y is d/30 units per minute. Then:

. . . . .X:
. . . . .rate: d/20
. . . . .time: t
. . . . .distance: dt/20

. . . . .Y:
. . . . .rate: d/30
. . . . .time: t - 6)
. . . . .distance: d(t - 6)/30

Since they covered the same distance (from home to when X, having started later but moving faster, caught up to Y), set the two "distance" expressions equal. Since d does not equal zero (they do not live in the factory), you can divide through.

Solve the resulting linear equation. (Cool)
November 15th 2009, 07:21 AM
saberteeth

Hey thanks but, as per your solution [dividing dt/20 by d(t - 6)/30] i get t = -12 as the answer, which is incorrect.. the correct answer is 7:28am. (Worried)

Quote:

Originally Posted by stapel

To learn how to set up and solve this sort of "uniform rate" exercise, try here. (Wink)

Then use the standard set-up, noting that "meeting" here actually means "passing". The rate for X, given that the distance from home to the factory is "d", is d/20 units per minute; the rate for Y is d/30 units per minute. Then:

. . . . .X:
. . . . .rate: d/20
. . . . .time: t
. . . . .distance: dt/20

. . . . .Y:
. . . . .rate: d/30
. . . . .time: t - 6)
. . . . .distance: d(t - 6)/30

Since they covered the same distance (from home to when X, having started later but moving faster, caught up to Y), set the two "distance" expressions equal. Since d does not equal zero (they do not live in the factory), you can divide through.

Solve the resulting linear equation. (Cool)
November 15th 2009, 08:34 AM
ukorov

suppose d is the distance they both travel, in metre for example.
then the speed of x, $v_x = \frac{d}{20}$
and the speed of y, $v_y = \frac{d}{30}$
both speeds are in metre per minutes

the difference in the speeds
= $v_x - v_y$
= $\frac{3d - 2d}{60}$
= $\frac{d}{60}$

By the time x started walking, y has already travelled a distance of:
$\frac{d}{30}$ m/min x 6min
= $\frac{d}{5}$ (metres)

therefore the time x needed to catch up y is $\frac{d}{5}$ divided by $\frac{d}{60}$
= $\frac{60}{5}$
= 12 minutes

Since x started walking at 7.16am, so they will meet at 7.28am
November 15th 2009, 08:47 AM
saberteeth

Thank you so much!

Quote:

Originally Posted by ukorov

suppose d is the distance they both travel, in metre for example.
then the speed of x, $v_x = \frac{d}{20}$
and the speed of y, $v_y = \frac{d}{30}$
both speeds are in metre per minutes

the difference in the speeds
= $v_x - v_y$
= $\frac{3d - 2d}{60}$
= $\frac{d}{60}$

By the time x started walking, y has already travelled a distance of:
$\frac{d}{30}$ m/min x 6min
= $\frac{d}{5}$ (metres)

therefore the time x needed to catch up y is $\frac{d}{5}$ divided by $\frac{d}{60}$
= $\frac{60}{5}$
= 12 minutes

Since x started walking at 7.16am, so they will meet at 7.28am