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# Two cites... help me :(

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Intern
Joined: 13 Aug 2018
Posts: 1
Two cites... help me :(  [#permalink]

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13 Aug 2018, 06:30
00:00

Difficulty:

(N/A)

Question Stats:

60% (01:24) correct 40% (00:00) wrong based on 5 sessions

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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
Senior Manager
Joined: 22 Feb 2018
Posts: 411
Two cites... help me :(  [#permalink]

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13 Aug 2018, 08:48
1
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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VP
Joined: 07 Dec 2014
Posts: 1128
Two cites... help me :(  [#permalink]

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13 Aug 2018, 09:15
2
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
Two cites... help me :( &nbs [#permalink] 13 Aug 2018, 09:15
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