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Two heavily loaded sixteen-wheeler transport trucks are 770 [#permalink]

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06 Sep 2013, 03:29

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Two heavily loaded sixteen-wheeler transport trucks are 770 kilometers apart, sitting at two rest stops on opposite sides of the same highway. Driver A begins heading down the highway driving at an average speed of 90 kilometers per hour. Exactly one hour later, Driver B starts down the highway toward Driver A, maintaining an average speed of 80 kilometers per hour. How many kilometers farther than Driver B, will Driver A have driven when they meet and pass each other on the highway?

Re: Two heavily loaded sixteen-wheeler transport trucks are 770 [#permalink]

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06 Sep 2013, 03:38

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caro2789 wrote:

Two heavily loaded sixteen-wheeler transport trucks are 770 kilometers apart, sitting at two rest stops on opposite sides of the same highway. Driver A begins heading down the highway driving at an average speed of 90 kilometers per hour. Exactly one hour later, Driver B starts down the highway toward Driver A, maintaining an average speed of 80 kilometers per hour. How many kilometers farther than Driver B, will Driver A have driven when they meet and pass each other on the highway?

A) 90 B) 130 C) 150 D) 320 E) 450

In one hour A covers 90 kilometers. So, by the time B starts traveling the distance between the trucks is 770-90=680 kilometers.

To cover that distance the trucks need (total distance)/(combined rate)=680/(90+80)=4 hours.

In 4 hours B covered 4*80=320 kilometers and A covered 770-320=450 kilometers.

Re: Two heavily loaded sixteen-wheeler transport trucks are 770 [#permalink]

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06 Sep 2013, 18:19

caro2789 wrote:

Two heavily loaded sixteen-wheeler transport trucks are 770 kilometers apart, sitting at two rest stops on opposite sides of the same highway. Driver A begins heading down the highway driving at an average speed of 90 kilometers per hour. Exactly one hour later, Driver B starts down the highway toward Driver A, maintaining an average speed of 80 kilometers per hour. How many kilometers farther than Driver B, will Driver A have driven when they meet and pass each other on the highway?

A) 90 B) 130 C) 150 D) 320 E) 450

1. Driver A traveled totally for 1 hr + (770-90)/ (90+80) hrs= 5 hrs. Distance traveled = 90*5=450 km 2. Driver B traveled 1 hr less i.e., he traveled for 4 hrs. Distance traveled = 80*4=320 km 3. Difference is, 450-320=130km _________________

Re: Two heavily loaded sixteen-wheeler transport trucks are 770 [#permalink]

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09 Sep 2013, 03:46

Two heavily loaded sixteen-wheeler transport trucks are 770 kilometers apart, sitting at two rest stops on opposite sides of the same highway. Driver A begins heading down the highway driving at an average speed of 90 kilometers per hour. Exactly one hour later, Driver B starts down the highway toward Driver A, maintaining an average speed of 80 kilometers per hour. How many kilometers farther than Driver B, will Driver A have driven when they meet and pass each other on the highway?

We are looking for how long it takes the two drivers to meet at some point between both ends of the highway. We are NOT looking for how long it takes for each truck to reach the other side of the highway. If Driver B sets off one hour after A, then A has already traveled for 90km when driver B begins to move. After one hour, driver A and B have to cover 770-90 = 680 km. Their total average speed is 90+80 km/h which means it will take them 680km/170km/h = 4 hours. In this time A travels 90 + (4*90) = 450 km which means that B travels 770 - 450 = 320 km so A travels 130 km more than B.

ANSWER: B) 130

I got tripped up when I couldn't figure out how to determine how long B traveled for. I was trying to solve for how long it would take B to travel from one side of the highway to the other when the question was asking for how long it would take for them too pass one another. All this means is that A travels for one more hour than B and that from that distance (770-90 = 680km) we can determine how long it took for them to meet up. When we calculate how many km A traveled, we multiply the time it took A to meet B by its rate and add in the 90 km it traveled before B set off.

Re: Two heavily loaded sixteen-wheeler transport trucks are 770 [#permalink]

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27 Feb 2014, 18:45

Bunuel wrote:

caro2789 wrote:

Two heavily loaded sixteen-wheeler transport trucks are 770 kilometers apart, sitting at two rest stops on opposite sides of the same highway. Driver A begins heading down the highway driving at an average speed of 90 kilometers per hour. Exactly one hour later, Driver B starts down the highway toward Driver A, maintaining an average speed of 80 kilometers per hour. How many kilometers farther than Driver B, will Driver A have driven when they meet and pass each other on the highway?..

A) 90 B) 130 C) 150 D) 320 E) 450

In one hour A covers 90 kilometers. So, by the time B starts traveling the distance between the trucks is 770-90=680 kilometers.

To cover that distance the trucks need (total distance)/(combined rate)=680/(90+80)=4 hours.

In 4 hours B covered 4*80=320 kilometers and A covered 770-320=450 kilometers.

Re: Two heavily loaded sixteen-wheeler transport trucks are 770 [#permalink]

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03 May 2014, 01:59

Bunuel wrote:

caro2789 wrote:

Two heavily loaded sixteen-wheeler transport trucks are 770 kilometers apart, sitting at two rest stops on opposite sides of the same highway. Driver A begins heading down the highway driving at an average speed of 90 kilometers per hour. Exactly one hour later, Driver B starts down the highway toward Driver A, maintaining an average speed of 80 kilometers per hour. How many kilometers farther than Driver B, will Driver A have driven when they meet and pass each other on the highway?

A) 90 B) 130 C) 150 D) 320 E) 450

]In one hour A covers 90 kilometers. So, by the time B starts traveling the distance between the trucks is 770-90=680 kilometers.

To cover that distance the trucks need (total distance)/(combined rate)=680/(90+80)=4 hours.

In 4 hours B covered 4*80=320 kilometers and A covered 770-320=450 kilometers.

The difference = 450 - 320 = 130 kilometers.

Answer: B.

Bunuel bit confused on the line highlighted above. If we take it conversly ,

In 4 hrs A covered 90*4 = 360 km , B would have covered 770-360 = 410 kms. Am I missing something ? Is the rationale behind your sequencing of taking B first and subtracting for A is the fact that A has a higher speed so it would cover a larger distance ?

My thought process was -

For 1 hr only A was driving @ 90kph so distance covered = 90kph Thus remaining distance = 770-90 = 680kph when both A and B are approaching each other.

Now since time of start is same , distance covered is in ratio of speeds. Let distance covered by A = x , then B = 680-x

x/680-x = 90/80. Solving for x , x= 360 = Distance A drove , 680-x = 320 kph.

Total extra dist travelled by A = 360 + 90 ( of first 1 hr) - 320 = 130 kph

Re: Two heavily loaded sixteen-wheeler transport trucks are 770 [#permalink]

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03 May 2014, 04:30

Expert's post

himanshujovi wrote:

Bunuel wrote:

caro2789 wrote:

Two heavily loaded sixteen-wheeler transport trucks are 770 kilometers apart, sitting at two rest stops on opposite sides of the same highway. Driver A begins heading down the highway driving at an average speed of 90 kilometers per hour. Exactly one hour later, Driver B starts down the highway toward Driver A, maintaining an average speed of 80 kilometers per hour. How many kilometers farther than Driver B, will Driver A have driven when they meet and pass each other on the highway?

A) 90 B) 130 C) 150 D) 320 E) 450

]In one hour A covers 90 kilometers. So, by the time B starts traveling the distance between the trucks is 770-90=680 kilometers.

To cover that distance the trucks need (total distance)/(combined rate)=680/(90+80)=4 hours.

In 4 hours B covered 4*80=320 kilometers and A covered 770-320=450 kilometers.

The difference = 450 - 320 = 130 kilometers.

Answer: B.

Bunuel bit confused on the line highlighted above. If we take it conversly ,

In 4 hrs A covered 90*4 = 360 km , B would have covered 770-360 = 410 kms. Am I missing something ? Is the rationale behind your sequencing of taking B first and subtracting for A is the fact that A has a higher speed so it would cover a larger distance ?

My thought process was -

For 1 hr only A was driving @ 90kph so distance covered = 90kph Thus remaining distance = 770-90 = 680kph when both A and B are approaching each other.

Now since time of start is same , distance covered is in ratio of speeds. Let distance covered by A = x , then B = 680-x

x/680-x = 90/80. Solving for x , x= 360 = Distance A drove , 680-x = 320 kph.

Total extra dist travelled by A = 360 + 90 ( of first 1 hr) - 320 = 130 kph

A starts driving 1 hour before B. They meet 4 hours AFTER B starts. If you want to do the other way around it would be: A travels for 4+1=5 hours, thus covers 5*90=450 kilometers, hence B covers 770-450=320 kilometers. The difference = 450-320 = 130 kilometers.

Re: Two heavily loaded sixteen-wheeler transport trucks are 770 [#permalink]

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03 May 2014, 07:32

Hi Bunuel, first of all thank you for all the material you are posting. It is of great help and value! There is something I don't understand in this question, how come you can add their average speed? What is the reasoning behind that because in the document GMAT club math book, p. 39 it is written that you cannot add the speeds.. Thank you very much

Re: Two heavily loaded sixteen-wheeler transport trucks are 770 [#permalink]

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04 May 2014, 08:21

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Expert's post

StephOD wrote:

Hi Bunuel, first of all thank you for all the material you are posting. It is of great help and value! There is something I don't understand in this question, how come you can add their average speed? What is the reasoning behind that because in the document GMAT club math book, p. 39 it is written that you cannot add the speeds.. Thank you very much

We CAN add or subtract rates (speeds) to get relative rate.

For example if two cars are moving toward each other from A to B (AB=100 miles) with 10mph and 15mph respectively, then their relative (combined) rate is 10+15=25mph, and they'll meet in (time)=(distance)/(rate)=100/25=4 hours;

Similarly if car x is 100 miles ahead of car y and they are moving in the same direction with 10mph and 15mph respectively then their relative rate is 15-10=5mph, and y will catch up x in 100/5=20 hours.

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