Two trains started simultaneously from opposite ends of a

Question

Two trains started simultaneously from opposite ends of a 100-mile route and traveled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100-mile trip in 5 hours; train Y, travelling at a constant rate, completed the 100-mile trip in 3 hours. How many miles had X traveled when it met train Y?

A/ 37.5
B/ 40.0
C/ 60.0
D/ 62.5
E/ 77.5

A. 37.5
B. 40.0
C. 60.0
D. 62.5
E. 77.5

Bunuel · Accepted Answer

As the ratio of the rates of X and Y is 3 to 5 then the distance covered at the time of the meeting (so after traveling the same time interval) would also be in that ratio, which means that X would cover 3/(3+5)=3/8 of 100 miles: 100*3/8=37.5 miles.

Answer: A.

shwetank7508 · Answer

Speed of train X= Distance/Time
 = 100/5
 = 20 mph

Speed of train Y= Distance/Time
 = 100/3
 = 33.33 mph

Relative Speed (opposite direction) = 20+33.33
 = 53.33 mph

Now, 
Time when X & Y meet:

Distance = Relative Speed x Time

Time= D/RS
 = 100/53.33
 = 1.87 hrs

After 1.87 hrs X & Y meet at some distance

In 1.87 hrs, X traveled

Distance= 20 x 1.87
 =37.50 miles (Answer)

And Y traveled,

Distance= 33.33 x 1.87
 = 62.3 (approx.)

Answer is A.

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gmat1220 · Answer

Train X : 100 m in 5 hrs 
speed = 100/5 = 20mph
Train Y : 100 m in 3 hrs 
speed = 100/3 mph

The distance shrinking at effective speed (20 + 100/3 ) mph
Time of intersect = 100 / (20+100/3) = 15/8 hrs
Distance travelled by X < Distance travelled by Y

Distance travelled by X = time * speed = 15/8 * 20 = 75/2

mymbadreamz · Answer

Bunuel, could you please explain this. It is not clear to me. thanks!

Bunuel · Answer

Since, the ratio of times of X and Y to cover the same distance of 100 miles is is 5:3, then the ratio of their rates is 3:5. Consider this, say the rates of trains X and Y are X and Y respectively, then:

Distance=Rate*Time --> X*5=Y*3 --> ratio of the rates is X:Y=3:5. At the time they meet, so after they travel the same time interval the ratio of distances covered by X and Y would also be in that ratio (for example if X=3 mph and Y=5 mph then they would meet in 100/(3+5)=100/8 hours, hence train X would cover 300/8 miles and train Y would cover 500/8 miles --> ratio of distances covered (300/8):(500/8)=3:5).

Now, since the the ratio of distances covered by X and Y is 3:5 then X covered 3/(3+5)=3/8 of the total distance.

Hope it's clear.

theGame001 · Answer

Why is that when the meet they would have covered 100 miles? This bit is little confusing.

Bunuel · Answer

When they meet one train covers some part of 100-mile distance and another covers the remaining part of 100-mile distance, so combined they cover 100 miles.

JusTLucK04 · Answer

Speed of first train: 100/5 = 20 Mph
Speed of second train: 100/3 = 33.33 Mph
 
Now in 1 hr distance covered by X is 20 & that by Y is 33.33.In the next hr it will be 40 & 66.66 resp.
40+66.66=106.6> the distance between them...So they would have met by now & the answer would be something less than 40..which leaves us with A

ScottTargetTestPrep · Answer

The combined distance traveled by the two trains was 100 miles. Each train traveled for t hours. We can create the distance equation:

100/5 * t + 100/3 * t = 100

Multiplying by 15 to clear the fractions from the equation, we have:

300t + 500t = 1500

800t = 1500

t = 15/8

Thus, train X had traveled 15/8 x 100/5 = 15/8 x 20 = 37.5 miles by the time it reached Y.

Answer: A

Kritisood · Answer

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/03 ... of-ratios/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/03 ... os-in-tsd/

Karishma explaining a similar question to this.

LaveenaPanchal · Answer

The time travelled by both the trains X & Y is equal when they meet.

If Train X travelled distance D, train Y travelled 100-D

Time travelled by Train X = D / 20. Time travelled by Y = (100- D)/(100/3) = 3(100-D)/100

Now D / 20 = 3 (100-D)/100 
5D = 300 - 3 D, 8D = 300
So the distrance travelled X i.e. D = 300/8 = 37.5 answer A

Sneha2021 · Answer

If both X and Y have same speed, they will meet in the middle. In this case, both X and Y will travel 50 miles
Now X is slower than Y, so X will travel less than 50 miles whereas Y will travel more than 50.
So we can reject C,D,E before starting the calculation because the answer must be less than 50

Now A & B are left
Let's pick B because 40 seems a nice number (no decimals)
Speed of X = 100/5=20, Speed of Y= 100/3=33.3
Now if X has travelled 40 miles, time taken is 40/20 = 2 hrs
Y has travelled 33.3*3=66 miles
So the distance doesn't add up to 100 (40+66 = 106)
Therefore A is correct

BrentGMATPrepNow · Answer

Train X completed the 100-mile trip in 5 hours  
Speed = distance/time = 100/5 =  20 mph

Train Y completed the 100-mile trip in 3 hours  
Speed = distance/time = 100/3 ≈  33 mph  (This approximation is close enough. You'll see why shortly)

How many miles had Train X traveled when it met Train Y?  
Let's start with a  word equation . 
When the two trains meet, each train will have been traveling for the  same amount of time  
So, we can write:   Train X's travel time  =  Train Y's travel time

time = distance/speed
We know each train's speed, but not the distance traveled (when they meet). So, let's assign some variables.

Let  d  = the distance train X travels
So,  100-d  = the distance train Y travels (since their COMBINED travel distance must add to 100 miles)

We can now turn our word equation into an algebraic equation.
We get:   d/20  =  (100 - d)/33  
Cross multiply to get: (33)(d) = (20)(100 - d)
Expand: 33d = 2000 - 20d
Add 20d to both sides: 53d = 2000
So, d = 2000/53

IMPORTANT : Before you start performing any long division, first notice that 2000/ 50  = 40
Since the denominator (53) is greater than 50, we can conclude that 2000/ 53  is LESS THAN 40
Since only one answer choice is less than 40, the correct answer must be A

PriyamRathor · Answer

Lets find out the Speed first of both the trains.
X - 100/5 = 20 Miles/Hour
Y - 100/3 = 33.33 Miles/Hour

Now , lets visualize the problem

0----------------------------------------------0 
X starts from here

0-----------------------------------------------0 Y Starts from here

Now after 1 hour lets understand the position of both the trains.
 See Image

After 2 hours the position will be

See Image

After 2 hours of travelling both the trains must have crossed each other.
If the distance travelled by Y after 2 hours would have been 60 Miles. Then they both have met exactly after 2 hours and X would have travelled 40 Miles.

But Since Y travelled for 66.66 Miles , then it is sure that X and Y must have met before X travelled 40 Miles.

Hence Answer is A.

Thanks.

GmatPoint · Answer

Since the train X is travelling at a speed of 100 miles per 5 hours,
Speed per hour = 100/5 = 20 miles per hour

Similarly, the speed of the train Y = 100/3 miles per hour.

Since the distance between the trains is 100 miles, and the trains are travelling toward each other,

Time to meet = 100/(20+100/3) = 300/160 = 15/8 hours.

Distance covered by X to meet Y = 15*20/8 miles = 37.5 miles.

Thus, the correct option is A.

agoyal1999 · Answer

To solve this logically, we begin by calculating the speeds of both trains:

Train X travels 100 miles in 5 hours, so its speed is 100÷5=20 miles per hour.

Train Y travels 100 miles in 3 hours, giving it a speed of 100÷3≈33.33 miles per hour.

Since the trains are moving toward each other, their relative speed is the sum of their individual speeds: 20+33.33=53.33 miles per hour.

This means that together, they cover approximately 53.33×2=106.67 miles in 2 hours. However, the total distance between them is only 100 miles, so they must have met before the 2-hour mark.

At 2 hours, Train X would have traveled 20×2=40 miles. Since the trains meet before 2 hours, Train X must have traveled less than 40 miles at the point of meeting.

Given that only option A is below 40 miles, the correct answer is A.

bumpbot · Answer

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Two trains started simultaneously from opposite ends of a

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