Two heavily loaded sixteen-wheeler transport trucks are 770

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.

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.

Hope it's clear.

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.

Hope it's clear.

Visualizing the problem in Distance and Speed question always helps in easy understanding of the question asked.

Refer the below diagram



We are given that the initial distance between drivers A and B is 770 kms. Driver A drives for an hour towards driver B before he starts driving

Distance covered by driver A before driver B starts driving = Speed * time = 90 * 1 = 90 Kms

So, when driver B starts driving , the distance between driver A and driver B is 770 - 90 = 680 kms.

The problem can now be simplified to both drivers driving towards each other with constant speeds. Let's assume the time taken by them to meet is t.

Speed of driver A = 90 kms/hour
Speed of driver B = 80 kms/hour

Let's assume the distance traveled by driver A is \(x\).
Hence distance traveled by driver B = 680 - \(x\)

As the time taken by both of them to meet each other is same, we can write

\(\frac{x}{90} = \frac{680-x}{80}\) which gives us the value of \(x\) = 360 kms

So total distance traveled by driver A = \(x\) + Initial distance covered by driver A before driver B starts driving
= 360 + 90 = 450 kms

Distance traveled by driver B = 680 - \(x\) = 320 kms

Hence, extra distance traveled by driver A = 450 - 320 = 130 kms

Hope this helps

Regards
Harsh

Solving for a distance "gap" in a "kiss" or "crash" scenario (traveling in opposite directions, toward one another).

At first the distance gap between them is 770 km. That number matters only once - it decreases when A travels alone.

Truck A travels alone for one hour. RT=D. (90 kmh * 1 hr) = 90 km. Part of A's total distance is this 90 km.

Now the gap between them is 770 - 90 = 680 km

In a distance gap:
--if A and B travel in opposite directions, whether towards or away from one another, add their rates. (They close -- or expand -- the gap much faster when both move than when only one moves.)
--if vehicles travel the same direction (e.g. parallel train tracks, "chase"), subtract their rates

Add (or subtract) rates, and calculate time to meet, only when both travel simultaneously.

Here, add rates: 90 + 80 = 170 kmh

\(\frac{Distance}{CombinedRate} = Time\) to meet

\(\frac{680}{170} = 4\) hours to meet

A's total distance:90 km alone.
Then (RT=D): (90*4) = 360 km
90 + 360 = 450 km

B 's total distance, RT = D
(80*4) = 320 km

A travels (450 - 320) = 130 km farther than B

Answer B

Trick is to understand what is being asked.

We are given distance, D = 770 kms

Let's start with time; A started 1 hr early and traveled 90 kms before B starts

Remaining distance for both to travel = 770 - 90 = 680 kms

Effective speed of A & B = 90+80 = 170 kmph

Time = 680/170 = 4 hrs (when both were moving)

As A traveled 1 hr more, Total time of A = 1+4 = 5 hrs

Distance traveled by A = 90 x 5 = 450 kms

Distance traveled by B = 80 x 4 = 320 kms

Difference = 450 - 320 = 130 kms (B)

We let t = the time for Driver B, and so (t + 1) is the time for Driver A. We can create the distance equation:

90(t + 1) + 80t = 770

90t + 90 + 80t = 770

170t = 680

t = 4

When the two meet, Driver A will have driven 90 x 5 =450 miles, and Driver B will have driven 80 x 4 = 320 miles. Thus, Driver A will have
driven 450 - 320 = 130 miles more.

Answer: B

Since A travels at 90 kph, the remaining distance between A and B after 1 hour = 770-90 = 680 km

When A and B travel toward each other, they WORK TOGETHER to cover the remaining 680 km.
When elements work together, we add their rates:
(A's rate) + (B's rate) = = 90+80 = 170 kph

Of every 170 kilometers traveled each hour by A and B together, 80 kilometers are traveled by B.
Thus, B travels \(\frac{80}{170}\) of the remaining 680 kilometers:
\(\frac{80}{170} * 680 = 320\) km

Since B travels 320 of the 770 kilometers, the distance traveled by A = 770-320 = 450 km
Resulting difference:
(A's distance) - (B's distance) = 450-320 = 130 km

My approach:

We know the two trucks are 770 kilometers apart.

Driver A drove for one hour at 90 kilometers per hour = 770 - 90 = 680 kilometers apart.

How many kilometers farther than Driver B, will Driver A have driven when they meet and pass each other on the highway?

Lets add both rates: 80 + 90 = 170 kilometers per hour.

680 / 170 = 4 hours

We know Driver A drives 10 kilometers more per hours vs Driver B.

4 * 10 = 40 + 90 kilometers that Driver A drove in the beginning = 130

Answer is B.

Initial distance between them = 770 km
In 1 hr, driver A covers 90 km and then distance between them = 680 km

Since their speeds are in the ratio 9:8, they will cover the total 680 km distance in the ratio 9:8. Since the total distance in ratio terms is 17, the multiplier is 40 and hence the difference of 1 in distance covered on ratio scale is equal to 40 km.

So A covered 90 + 40 = 130 kms more than B.

Answer (B)

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