Two buses, A and B, are going from city P to city Q and back without a

Question

Two buses, A and B, are going from city P to city Q and back without a break. Bus A starts from city P and is moving at constant speed of 50km/h, while bus B starts from city Q and is moving at constant speed of 20km/h. If the distance between the cities is 1,400 km, how far will bus A be from city Q when the buses meet for the 22nd time?

A. 400 km
B. 800 km
C. 1,000 km
D. 1,200 km
E. 1,400 km

Are You Up For the Challenge: 700 Level Questions

A. 400 km
B. 800 km
C. 1,000 km
D. 1,200 km
E. 1,400 km

Are You Up For the Challenge: 700 Level Questions

rishi02 · Answer

Since the buses are travelling in opposite direction, they will intercept each other after 1400 km the first time and after 2800 km each time after that

Distance travelled by each bus will be ratio of their speeds i.e  \frac{Distance(A)}{Distance(B)}  =  \frac{50}{20}  =  \frac{5}{2}

Therefore Bus A will cover  \frac{5}{2}  * 1400 = 1000 km while Bus B will cover 400 km the first time they meet and Bus A will cover  \frac{5}{2}  * 2800 = 2000 km while Bus B will cover 800 km during every subsequent meet

When they meet for the 22nd time, Bus A will have travelled 1000 + 21*2000 = 43,000 km

After covering 2800 km, Bus A returns to point P. Therefore after covering 43,000 km (i.e 2800*15 + 1000), it will be 400 km from point Q (since PQ = 1400 km)

Lorak13 · Answer

It takes bus A 56hrs to travel to Q and back
(2800km / 50km/h)

It takes bus B 140hrs to travel to P and back
(2800km / 20km/h)

We note that the greatest common multiple of 56 and 140 is 280. So after 280hrs, both buses will be back in their original positions.

How many times did they meet during 280hrs? We know the faster bus (A) did 5 two-way trips in this time, so they met 10 times.

So after 560hrs they are both back in their original positions and met 20 times. So the question is where do they meet the 2nd time, equivalent to the 22nd.

First time they meet after 1400km/(20+50)km/hr =20hrs, so busA did 1000km. So 400km from Q.

It takes bus A 8hrs to get to Q, so bus B travels 160km more. Bus A is at Q and bus B 560km from Q.

Now bus B catches A with 30km/hr so in 560/30 hours. The distance travelled, and therefore 22nd meeting point is at 560/30 * 50 km from Q, but this is bot one of the answer choices. Did i go wrong somwhere?

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fireagablast · Answer

I think the answer is C

22(1400/50+20) = 22*1400/70 = 22*20 = 440

D = (440)*(50) = 22,000

22,000/1400 = 15*1400 + 1000

So departing from Q for his 16th trip, he goes 1000 miles.
So Bus A is 1000 miles from Q, and 400 miles from P

gurmukh · Answer

Check for yourself every 10th time that you are talking about they will not meet when they are comming back to their original point. So the 20th meeting that you are talking about is actually 18th meeting. But answer still is not matching.

gurmukh · Answer

Why you are assuming that evertime they have a relative speed of 70km/hr when they are in same direction they have a relative speed of 30km/hr

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chetan2u · Answer

The two buses start from opposite ends, and their speed is in ratio A:B = 5:2, so the distance covered will be in ratio 2:5.

Therefore, when A makes 5 trips one way, B makes 2 trips one way, and at the end of this A and B are at Q.
Another 5 trips by A, and both A and B will be back in original position.

In these 10 rounds, the distance covered by both combined is (10+4)*1400 = 14*1400. 
When they meet together at original point, we can say that average distance covered per meet is 1400*2=2800, so they meet 14*1400/2800=7 times..
You can prove it by finding each time they meet but not required here.

Now extrapolating this further, when they meet 21st time, that is 7*3, they both will be at original position, A at P and B at Q.

So A will meet B at a combined distance of 1400, in the ratio 5:2, So A travels 1400*5/7=1000 and is 1400-1000=400 from Q

A

monikakumar · Answer

Hi  chetan2u ,
speed is directly proportional to distance for a given amount of time...so when speed is in 5:2 ratio, distance must also be in 5:2, right, pls correct me if im wrong, and can u pls elaborate more on this, i can't get it...your insights would be mush helpful

thanks.

ScottTargetTestPrep · Answer

The two buses meet for the 22nd time when the total distance traveled by the two buses is exactly 22 x 1400 miles. The combined rates of the two buses is 50 + 20 = 70 mph; therefore, the two buses will meet for the 22nd time (22 x 1400)/70 = 22 x 20 = 440 hours after they start moving. So, the problem now becomes finding the distance of Bus A from city Q exactly 440 hours after it starts traveling.

Bus A travels 440 x 50 = 22,000 miles in 440 hours. Dividing 22,000 miles by 1,400 produces a quotient of 15 and a remainder of 1,000. This means that bus A makes 15 full trips (at which point it will be at city P) and drives further a distance of 1,000 miles. Since bus A is 1,000 miles from city P, it is 1400 - 1000 = 400 miles from city Q.

Answer: A

monikakumar · Answer

so 22*1400 is the combined distance they have covered together for 440 hours..is it right?? and one more doubt is, is it for every time they meet, they cover 1400km distance together with each covering some distance with respect to their rates..pls clarify me on this..
Thanks

Shobhit7 · Answer

For 22nd meeting:
Relative distance covered= 1400 per trip * 22 trips = 1400*22
At a Relative speed = 50+20 = 70 kmph
Time of travel = (1400*22) / 70 = 440 hrs

Now, in 440 hrs:
Distance traveled by Bus A @ 50 kmph = 440 * 50
Converting this distance into trips of 1400 km each = (440*50) / 1400 = 15 + 5/7
That means, 15 full trips and 5/7th on 16th trip.

Bus A at the end of 1st, 3rd, 5th ....15th trip will be on Point Q.
So, Distance from Point Q after 5/7th trip on return from Q towards P = 1400 * 5/7 = 1000 kms

IMO, Ans C

GMathGuy · Answer

Hi,

Since everyone's answers disagreed with each other, I calculated this problem manually for the first 10 laps of A - at this point, A goes back to P and B goes back to Q, so both are effectively in their starting positions. I found that A & B meet 9 times during these first 10 laps, they only do not meet in A's 6th lap because both A & B had started from Q for that lap. Based on this calculations, the final answer should be B - A is 800 m away from city Q as per the repetition of the cycle.

Cheers!

Lorak13 · Answer

I agree the correct answer is B. They meet the 5th time in Q and when A is back in P and B in Q they met only 9 times. Another cycle to 18. And another 5 to get both in Q to meet 23 times. So we can roll them backwords to see where they met at 22, and it is 800km from Q. 
 Bunuel  could you please clarify the answer?

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effatara · Answer

This post merits an in-depth discussion because it highlights the danger of automatically adopting an approach based on the assumption that the two vehicles will always cover a combined distance equal to twice the length of the route between each meeting and the next. Fact is, this is true only when the speed of the faster vehicle is equal to or less than twice that of the slower one. However, this is not the case here because Bus A (BA) is 2.5 times faster than Bus B (BB) which is why the sum of the distances covered by each bus between meets is not 2800 miles as shown below:
The 1st meet (M1) takes place 400 miles from Q. By the time BA covers this 400 miles to Q, BB travels 160 miles (400*(2/5) from M1 towards P so the distance separating them is 560 miles (400+160). BA turns around and catches up with BB in 560/30 hours covering (56/3)*50 miles. So the distances covered by BA and BB between the 1st and 2nd meet is (56*50/3+400) and (56*50/3-400) respectively. The combined distance covered by the two buses is 2*(56*50/3)=1866.66 miles.

SOLUTION:
This is basically the  Lorak13  approach but with a bit more detail. 
Let us break down the number of BA’s laps into 5-lap segments. We have to remember 3 things:
(a) At the end of each segment, BA will be at the opposite end of where it was at the beginning of the segment (if it was at P at the beginning of the segment it will be at Q at the end and vice versa because BA completes 5 (odd number) of laps per segment) but BB will always be at its starting point Q since it completes 2 (even number) laps in each segment;
(b) Since the buses started off from opposite ends (BA from P and BB from Q), BA will have met BB once during each of its laps. Hence, 5 meetings in the 1st segment and both buses will be at Q at the end of the segment
(c) However, there is no meeting in BA’s first lap of the 2nd segment since both the buses start simultaneously from the same end (Q) and since BA is faster it is always ahead of BB until it reaches P. So BA will meet BB once in each of the remaining 4 laps of the 2nd segment. Hence, 4 meetings in the 2nd segment and BA will be at P and BB at Q and the 5-4 cycle will continue 
So by the end of 4 segments, 18 meetings (5+4+5+4) will have taken place. Since the buses are at opposite ends (BA at P and BB at Q) at this point, there will a meeting in each of BA’s laps in the 5th segment. So the 22nd meeting will take place in BA’s 24th lap which will, of course, commence after he has completed his 23rd lap. By the time BA completes 23 laps, BB will have completed 9.2 laps (23*2/5) which means it will be on its way from P to Q at a distance of 280 miles (1400*0.2) from P or 1120 miles (1400-280) from Q. So, at this point there are 1120 miles separating the two buses and BA moves from Q to P and BB towards Q. Thus, BA will have covered 800 miles (1120*5/7) when they meet. So, BA will be 800 miles from city Q when the two buses meet for the 22nd time.

ANS:B

rsama · Answer

Can we really have such ambiguous questions in GMAT ? not to mention lengthy too. As far as I have come to know, GMAT won't contain question which may people won't be able to solve with-in 2 minutes.

JaiPavuluri · Answer

Relative Speed (Opposite direction) = Speed of A + Speed of B = 50 + 20 = 70

DUST => D = S * T => 1400 = 70 * t => t = 20

at t = 20, D A = 50*20 = 1,000 km from City P

at t = 20, D B = 20*20 = 400 km from City Q

22 laps = 22 * 1400 = 11 * 2 * 7 * 200 = 11 * 7 * 400

After every 7 laps, A & B will meet at City Q => 400 * 7 = 2800

this happens for 10 times, and the 11th time we have 400 KM away from City Q again

Tinytrotter · Answer

I am with your calculation so far where Bus A has to travel 1,000km while Bus B has to cover 400 km the first time they meet. However, for every subsequent meeting, Bus A has to travel only an additional 1,000 km from the first meet and Bus B has to travel an additional 400km for the second meet because the second meeting will be when cumulative distance travelled is 1400 km i.e. at 2800 km then at 4200 km (where A has travelled cumulative distance of 3000km and B has travelled cumulative distance of 1200km) and 5600km and so on.

Now when they meet for the 22nd time, the total cumulative distance travelled by both of them has to be 22*1400 = 30,800km where A has travelled 22,000 km (1000*22) and B has travelled 8,800 km (400*22).

A started his journey from P so he makes a total of 15 complete trips by covering a cumulative distance of 21,000 km (1400*15). Because he started at P, at the end of the 15th trip, he has reached Q. Starting from Q, he has to travel 1,000 km before the 22nd meeting point. Hence, at the 22nd meeting point, A is 1,000 km away from Q and 400 km away from P.

I do not understand how 400km can be the answer. What am I missing here?

Two buses, A and B, are going from city P to city Q and back without a

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