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02 Jun 2014, 02:41
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pretzel
When the trains pass each other A covered 100 miles and B covered x/6 miles. Together they covered the whole distance of d miles, no?
{NY}----(100 miles)----{meeting point}--(x/6 miles)--{Boston}
02 Jun 2015, 05:25
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Bunuel
Hi, if you havd substituted (i) into (ii) instead this would have created a quadratic equation to hard to solve (in such a short time frame). But making that mistake and then doing the other substitution would have eaten up precious minutes. How did you decide which equation to sub into which?
04 Sep 2015, 09:39
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rohitgoel15
Let x be the speed of train B.
Let M be the point where Train A and B meet.
Train A travels 100 miles in 1 hr before it meets B at point M.
Time taken by train B to travel 100 miles will be 100/x
Train B travels for 10 mins or 1/6 hrs before it meets train A at point M. So the distance covered by train B in 10 mins will be x/6 and total distance will be 100 + x/6
Time taken by train A to travel x/6 distance will be x/600
It is given that total time for travel for both trains is 2 hours.
So,
1 + \(\frac{x}{600}\) +\(\frac{1}{6}\) + \(\frac{100}{x}\) = 2
or \(\frac{x}{600}\) + \(\frac{100}{x}\) = \(\frac{5}{6}\)
or \(x^2 + 60000 = 500x\)
or \(x^2 - 500x + 60000 =0\)
or \((x-300)(x-200) = 0\)
or \(x= 300, 200\)
Now,
the distance between point M and boston when x= 200 is 200/6 miles and when x = 300 is 300/6 or 50 miles
Total travel time taken by train A and train B, when distance between point M and boston is 200/6 miles is
Train A:
1+(\(\frac{200}{6}\) * \(\frac{1}{100}\)) = 1 + \(\frac{1}{3}\) = 1 hour and 20 minutes.
time of arrival at boston is 4:20 pm
Train B:
\(\frac{1}{6}\) + \(\frac{100}{200}\) = \(\frac{2}{3}\) = 40 minutes
time of arrival at new york is 4:30 pm
Total travel time taken by train A and train B, when distance between point M and boston is 50 miles is
Train A:
1+ (50*\(\frac{1}{100}\)) = 1+ \(\frac{1}{2}\) = 1 hour and 30 minutes
time of arrival at boston is 4:30 pm
Train B:
\(\frac{1}{6}\) + \(\frac{100}{300}\) = \(\frac{1}{2}\) = 30 minutes
time of arrival at new york is 4:20 pm
Statement 1:-
Train B arrived in New York before Train A arrived in Boston.
This is true when speed of Train B is 300 miles per hour. Train B arrives new york at 4:20 pm
SUFFICIENT
Statement 2:-
The distance between New York and Boston is greater than 140 miles.
This is possible only when the distance between point P and boston is 50 miles.
So, at 300 miles/hr train B arrives new york at 4:20 pm
SUFFICIENT
Answer:- D
Hugoba
Joined: 27 Dec 2011
Last visit: 01 Feb 2017
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Location: Brazil
Concentration: Entrepreneurship, Strategy
Schools: Insead Sept '17 IESE '18 Haas '18 CBS '18 Wharton '18
GMAT 1: 620 Q48 V27
GMAT 2: 680 Q46 V38
GMAT 3: 750 Q50 V41
GPA: 3.5
12 Oct 2015, 12:23
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I was trying to find a pure geometric solution to this problem. I could't. However, I think I came up with an alternative solution to this problem:
Since the velocities are constants, the trajectories of the trains are represented by straight lines in a graph Distance x Time. So, this problem can be represented accordingly with the figure attached. By simmilarity of triangles we have:
A- \(\triangle ABD\)\(\approx\)\(\triangle GFD\) \(->\) \(\frac{\frac{1}{6}}{t_2}=\frac{d}{100}\) \(->\)\(t_2=\frac{100}{6d}\);
B- \(\triangle BCD\)\(\approx\)\(\triangle FED\) \(->\) \(\frac{t_1}{1}=\frac{d}{100}\) \(->\)\(t_1=\frac{d}{100}\)
Now, the problem says that the total time traveled by the trains is 2 h, or \((t_1+1)+(t_2+\frac{1}{6})=2\) \(->\) C- \(t_1+t_2=\frac{5}{6}\). Clearly, this will lead a second degree equation in d by substituting A and B in C:
\(\frac{d}{100}+\frac{100}{6d}=\frac{5}{6}\); or by multiplying the equation by 6d:
D- \(\frac{6d^2}{100}-5d+100=0\)
It is easy to calculate the discriminant of equation D: \(\triangle=25-4*\frac{6}{100}*100=1\). So \(d=\frac{5+1}{2*\frac{6}{100}}=50\) or \(d=\frac{5-1}{2*\frac{6}{100}}=\frac{100}{3}\).
We have two possible values for the distance between New York and Boston. One is 150 miles other is 133 miles. Also, this leads to two pairs of values for \((t_1,t_2)\):
If d=50 miles then \((t_1,t_2)=(0.50,0.33)\) or
If d=33 miles then \((t_1,t_2)=(0.33,0.50)\)
(1) This statement allows us to decide between the two situations above. Since the trains encounter in point D, if \(t_2<t_1\) then train B arrives at NY earlier than A arrives in Boston (just like the figure represents). Sufficient.
(2) Again, this statement allows us to decide between the two situations above (only if d=50 miles). Sufficient.
Letter D.
Since the velocities are constants, the trajectories of the trains are represented by straight lines in a graph Distance x Time. So, this problem can be represented accordingly with the figure attached. By simmilarity of triangles we have:
A- \(\triangle ABD\)\(\approx\)\(\triangle GFD\) \(->\) \(\frac{\frac{1}{6}}{t_2}=\frac{d}{100}\) \(->\)\(t_2=\frac{100}{6d}\);
B- \(\triangle BCD\)\(\approx\)\(\triangle FED\) \(->\) \(\frac{t_1}{1}=\frac{d}{100}\) \(->\)\(t_1=\frac{d}{100}\)
Now, the problem says that the total time traveled by the trains is 2 h, or \((t_1+1)+(t_2+\frac{1}{6})=2\) \(->\) C- \(t_1+t_2=\frac{5}{6}\). Clearly, this will lead a second degree equation in d by substituting A and B in C:
\(\frac{d}{100}+\frac{100}{6d}=\frac{5}{6}\); or by multiplying the equation by 6d:
D- \(\frac{6d^2}{100}-5d+100=0\)
It is easy to calculate the discriminant of equation D: \(\triangle=25-4*\frac{6}{100}*100=1\). So \(d=\frac{5+1}{2*\frac{6}{100}}=50\) or \(d=\frac{5-1}{2*\frac{6}{100}}=\frac{100}{3}\).
We have two possible values for the distance between New York and Boston. One is 150 miles other is 133 miles. Also, this leads to two pairs of values for \((t_1,t_2)\):
If d=50 miles then \((t_1,t_2)=(0.50,0.33)\) or
If d=33 miles then \((t_1,t_2)=(0.33,0.50)\)
(1) This statement allows us to decide between the two situations above. Since the trains encounter in point D, if \(t_2<t_1\) then train B arrives at NY earlier than A arrives in Boston (just like the figure represents). Sufficient.
(2) Again, this statement allows us to decide between the two situations above (only if d=50 miles). Sufficient.
Letter D.
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