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08 Jun 2021, 08:30
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
74% (02:54) correct
26%
(03:02)
wrong
based on 299
sessions
History
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Two trains start from A and B and move towards B and A, respectively. The train from A starts at 6 a.m. and reaches B at 2 p.m., whereas the train from B starts at 8 a.m. and reaches A at 2 p.m. In how much time after 8 a.m. will the two trains meet?
(A) \(1 \frac{8}{9}\) h
(B) \(\frac{15}{7}\) h
(C) \(2 \frac{1}{6}\) h
(D) \(2 \frac{4}{7}\) h
(E) \(3 \frac{1}{6}\) h
(A) \(1 \frac{8}{9}\) h
(B) \(\frac{15}{7}\) h
(C) \(2 \frac{1}{6}\) h
(D) \(2 \frac{4}{7}\) h
(E) \(3 \frac{1}{6}\) h
08 Jun 2021, 11:20
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we know that train A starts at 6 am and reaches B at 2 pm. Instead, train B starts at 8 am and reaches A at 2 pm. So, it takes 8 hours for train A and 6 hours for train B. This means that their LCM is 24 (since 8= 2x2x2 and 6= 2x3).
Now we need to find their rates.
r * T=D so
A: r*8=24 ---> r= 3
B: r*6= 24 ---> r=4
Since A travels alone for the first two hours: 3*2= 6 miles. Thus 24-6= 18 miles left.
Their relative speed is 3+4=7
so if we want to know at what time they meet: T= D/R= 18/7
answer D
Now we need to find their rates.
r * T=D so
A: r*8=24 ---> r= 3
B: r*6= 24 ---> r=4
Since A travels alone for the first two hours: 3*2= 6 miles. Thus 24-6= 18 miles left.
Their relative speed is 3+4=7
so if we want to know at what time they meet: T= D/R= 18/7
answer D
08 Jun 2021, 22:33
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Solution:
let the distance between A and B be = \(x\) km.
Train from A starts at 6 a.m. and reaches B at 2 p.m, which means this train takes 8 hours to travel \(x\) km.
Hence the speed of train from A = \(\frac{x}{8} km/hr. \)
Train from B starts at 8 a.m. and reaches A at 2 p.m, which means this train takes 6 hours to travel \(x\) km.
Hence the speed of train from B = \(\frac{x}{6} km/hr. \)
In the time of 6 am to 8am train from A travels = \(\frac{x}{8}*2 =\frac{x}{4} km\).
So, the distance between the 2 trains at 8 am = \(x-\frac{x}{4}=\frac{3x}{4}km\)
We know a formula that if 2 bodies are moving towards each other at speeds \(S_1\) and \(S_2\) and distance between them is \(D\),
then they will meet each other after \( \frac{D}{S_1+S_2}\) hours.
To get the answer let us find the sum of the speeds of 2 trains = \(\frac{x}{8}+\frac{x}{6 }= \frac{3x+4x}{24} = \frac{7x}{24} km/hr.\)
So, after 8am the 2 trains will meet in = \(\frac{3x}{4}/\frac{7x}{24} = \frac{18}{7} hour. \)
Hence the right answer is Option D.
let the distance between A and B be = \(x\) km.
Train from A starts at 6 a.m. and reaches B at 2 p.m, which means this train takes 8 hours to travel \(x\) km.
Hence the speed of train from A = \(\frac{x}{8} km/hr. \)
Train from B starts at 8 a.m. and reaches A at 2 p.m, which means this train takes 6 hours to travel \(x\) km.
Hence the speed of train from B = \(\frac{x}{6} km/hr. \)
In the time of 6 am to 8am train from A travels = \(\frac{x}{8}*2 =\frac{x}{4} km\).
So, the distance between the 2 trains at 8 am = \(x-\frac{x}{4}=\frac{3x}{4}km\)
We know a formula that if 2 bodies are moving towards each other at speeds \(S_1\) and \(S_2\) and distance between them is \(D\),
then they will meet each other after \( \frac{D}{S_1+S_2}\) hours.
To get the answer let us find the sum of the speeds of 2 trains = \(\frac{x}{8}+\frac{x}{6 }= \frac{3x+4x}{24} = \frac{7x}{24} km/hr.\)
So, after 8am the 2 trains will meet in = \(\frac{3x}{4}/\frac{7x}{24} = \frac{18}{7} hour. \)
Hence the right answer is Option D.
