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Re: Two objects are located 390 meters away from each other on a straight [#permalink]
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First object ..............................(390 m) ...........................................Second object

First object: 6 , 12, 18, 24, 30, 36, ....

Second object: 12 , 12, 12, 12, ....

In first five minutes: First object has covered total distance of : 6 + 12 + 18 + 24 + 30 = 90.

Remaining distance is 390 - 90 = 300 m.

In next five minute distance covered :

First object: 36 + 42 + 48 + 54 + 60 = 240

Second object: 12 + 12 + 12 + 12 + 12 = 60

Total : 240 + 60 = 300

So, total 10 minutes.

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Two objects are located 390 meters away from each other on a straight [#permalink]
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sherxon wrote:
Two objects are located 390 meters away from each other on a straight path. The first object covers 6 meters in the first minute, and the distance covered in every next minute is 6 meters greater than that in the previous minute. The second object moves at a speed of 12 meters per minute. The first object starts to move toward the second one, and after 5 minutes the second object starts moving toward the first one. In how many minutes after the first object started the movement will the two objects meet?

A. 5
B. 10
C. 12
D. 15
E. 20


For any EVENLY SPACED SET:
sum = (count)(median)

For the first five minutes, only the first object travels.

Minutes 1-5:
Since the initial 6 meter-per-minute rate for the first object increases by 6 meters each minute, we get the following set of evenly spaced distances:
6, 12, 18, 24, 30
sum = (count)(median) = 5*18 = 90

From this moment on, the two objects travel toward each other.
When objects travel toward each other, they WORK TOGETHER to cover the distance between them.
As a result, we ADD THEIR RATES.

Next minute:
Distance traveled by the first object = (previous distance) + (6 additional meters) = 30+6 = 36
Distance traveled by the second object = 12
Combined rate for the two objects = 36+12 = 48

Minutes 6-10:
Since the first object increases the rate by 6 meters each minute, we get the following set of evenly spaced distances:
48, 54, 60, 66, 72
sum = (count)(median) = 5*60 = 300

In minutes 1-10, the total distance traveled = 90+300 = 390
Thus, the total time for the 390-meter distance to be traveled = 10 minutes

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Re: Two objects are located 390 meters away from each other on a straight [#permalink]
Given: Two objects are located 390 meters away from each other on a straight path. The first object covers 6 meters in the first minute, and the distance covered in every next minute is 6 meters greater than that in the previous minute. The second object moves at a speed of 12 meters per minute. The first object starts to move toward the second one, and after 5 minutes the second object starts moving toward the first one.
Asked: In how many minutes after the first object started the movement will the two objects meet?

Distance covered by first object in t th minute = 6 + (t-1)6 = 6t
Total distance covered by first object after time t = 6 + 12 + ..... + 6t = 6 t(t+1)/2 = 3t(t+1)
Total distance covered by second object after time t = 12(t-5)
Since the two objects are moving towards each other, the Toal distance covered by 2 objects after time t = 3t(t+1) + 12(t-5)
When they meet after time t, the total distance covered is 390 meters
3t(t+1) + 12(t-5) = 390
3t^2 + 3t + 12t - 60 = 390
3t^2 + 15t - 450 = 0
t^2 + 5t - 150 = 0
t^2 + 15t - 10t - 150 =0
(t+15)(t-10) = 0
t = 10 minutes

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Re: Two objects are located 390 meters away from each other on a straight [#permalink]
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Re: Two objects are located 390 meters away from each other on a straight [#permalink]
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