Train A leaves New York for Boston at 3 PM and travels at the constant speed of 100 mph. An hour later, it passes Train B, which is making the trip from Boston to New York at a constant speed. If Train B left Boston at 3:50 PM and if the combined travel time of the two trains is 2 hours, what time did Train B arrive in New York?
(1) Train B arrived in New York before Train A arrived in Boston.
(2) The distance between New York and Boston is greater than 140 miles.
(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) Each statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.
Please help me!!!!-tel & explain the answer
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IMO D
Info given:
Train A speed = 100 mph
A leaves NY at 3:00PM
The 2 trains meet after 1 hr at 4:00PM
Train B leaves Boston at 3:50PM
Ta+Tb=2hrs
(Ta-time taken by A; Tb time taken by B)
Let speed of train B = S
Total distance between NY and Boston = D
Distance between NY and point of meeting = 100mph*1hr = 100miles (Speed of A*time taken)
Hence D = 100+X
X is the distance covered by train B from Boston to meeting point
X=S*10/60=S/6
(10/60 is the time taken by train B to reach meeting point (MP) in hours. B started at 3:50PM and reached meeting point at 4PM. Hence the total time taken to reach MP = 10mins)
D=100+S/6
D=(600+S)/6 -----equ1
Ta+Tb=2
(D/100) + (D/S) = 2-----equ2
Sub. The value of D in equ2.
(600+S)/600 + (600+S)/S = 2
Simplify to get the equ: S^2-500S+60,000=0
Solving for S we get S= 300, 200
When S=300 D=150
When S=200 D=800/6
St 1 Train B arrived in New York before Train A arrived in Boston.
Tb>Ta
When S=300 and D=150 (sub S in equ 1 to find D)
Ta = D/100 = 4:30pm
Tb = D/S = 4:20pm ---- Tb>Ta satisfies the condition
When S=200 and D=800/6
Ta=D/100=4:20pm
Tb=D/S = 4:30pm------Ta>Tb does not satisfy the condition
Hence Tb=4:20pm St 1----SUFF
St 2 The distance between New York and Boston is greater than 140 miles.
I.e. D>140miles
D=150miles, S=300 and Tb=4:20pm
St 2----SUFF
Answer is D
Info given:
Train A speed = 100 mph
A leaves NY at 3:00PM
The 2 trains meet after 1 hr at 4:00PM
Train B leaves Boston at 3:50PM
Ta+Tb=2hrs
(Ta-time taken by A; Tb time taken by B)
Let speed of train B = S
Total distance between NY and Boston = D
Distance between NY and point of meeting = 100mph*1hr = 100miles (Speed of A*time taken)
Hence D = 100+X
X is the distance covered by train B from Boston to meeting point
X=S*10/60=S/6
(10/60 is the time taken by train B to reach meeting point (MP) in hours. B started at 3:50PM and reached meeting point at 4PM. Hence the total time taken to reach MP = 10mins)
D=100+S/6
D=(600+S)/6 -----equ1
Ta+Tb=2
(D/100) + (D/S) = 2-----equ2
Sub. The value of D in equ2.
(600+S)/600 + (600+S)/S = 2
Simplify to get the equ: S^2-500S+60,000=0
Solving for S we get S= 300, 200
When S=300 D=150
When S=200 D=800/6
St 1 Train B arrived in New York before Train A arrived in Boston.
Tb>Ta
When S=300 and D=150 (sub S in equ 1 to find D)
Ta = D/100 = 4:30pm
Tb = D/S = 4:20pm ---- Tb>Ta satisfies the condition
When S=200 and D=800/6
Ta=D/100=4:20pm
Tb=D/S = 4:30pm------Ta>Tb does not satisfy the condition
Hence Tb=4:20pm St 1----SUFF
St 2 The distance between New York and Boston is greater than 140 miles.
I.e. D>140miles
D=150miles, S=300 and Tb=4:20pm
St 2----SUFF
Answer is D