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  • Typical Day of a UCLA MBA Student - Recording of Webinar with UCLA Adcom and Student

     December 14, 2018

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    Carolyn and Brett - nicely explained what is the typical day of a UCLA student. I am posting below recording of the webinar for those who could't attend this session.
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Two cites... help me :(

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Joined: 13 Aug 2018
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Two cites... help me :(  [#permalink]

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New post 13 Aug 2018, 06:30
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There are two cites P and Q. A and B start from P and Q and move on their motorcycles towards Q and P receptively. After reaching their destination, they reverse their direction and move back to where they started from. First time A and B meet at a distance of 65 km from P and second time while they were returning to the point where they have started from, they meet at a distance of 55 km from Q.Assuming that A and B started at the same time and travelled at uniform speeds, find the distance between P and Q.

A) 130
B) 140
C) 150
D) can't be determined
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Two cites... help me :(  [#permalink]

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New post 13 Aug 2018, 08:48
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charan79 wrote:
There are two cites P and Q. A and B start from P and Q and move on their motorcycles towards Q and P receptively. After reaching their destination, they reverse their direction and move back to where they started from. First time A and B meet at a distance of 65 km from P and second time while they were returning to the point where they have started from, they meet at a distance of 55 km from Q.Assuming that A and B started at the same time and travelled at uniform speeds, find the distance between P and Q.

A) 130
B) 140
C) 150
D) can't be determined


OA:B

let the Average speed of A be \(v_a\) and Average speed of B be \(v_b\)
Let the distance between P and Q be \(x\)
\(t_1\) = time after start, A and B meet first time
\(t_2\) = time interval between when they first meet and when they meet 2nd time

\(t_1=\frac{65}{v_a}=\frac{x-65}{v_b}\)
\(\frac{v_a}{v_b} = \frac{65}{x-65}\) ...........(1)

\(t_2=\frac{x-65+55}{v_a}=\frac{65+x-55}{v_b}\)
\(\frac{v_a}{v_b} =\frac{x-10}{x+10}\) ........(2)

Equating 1 and 2 , we get

\(\frac{65}{x-65}=\frac{x-10}{x+10}\)

\(65(x+10)=(x-65)(x-10)\)

\(65x+650=x^2-10x-65x+650\)

\(x^2-140x=0\)

\(x(x-140)=0\)

Rejecting \(x=0\), we get \(x =140\)
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Two cites... help me :(  [#permalink]

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New post 13 Aug 2018, 09:15
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charan79 wrote:
There are two cites P and Q. A and B start from P and Q and move on their motorcycles towards Q and P receptively. After reaching their destination, they reverse their direction and move back to where they started from. First time A and B meet at a distance of 65 km from P and second time while they were returning to the point where they have started from, they meet at a distance of 55 km from Q.Assuming that A and B started at the same time and travelled at uniform speeds, find the distance between P and Q.

A) 130
B) 140
C) 150
D) can't be determined


let d=distance between P and Q
at first meeting, combined distance=d
at second meeting, combined distance=3d
if A travels 65 km of d, he will travel 3*65=195 km of 3d
A's total distance to second meeting=d+55 km
d+55=195 km
d=140 km
B
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Two cites... help me :( &nbs [#permalink] 13 Aug 2018, 09:15
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Two cites... help me :(

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