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On Sunday, Ron begins to drive due north from point T at a constant

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Aabhash777
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On Sunday, Ron begins to drive due north from point T at a constant [#permalink] New post   10 Aug 2022, 01:49
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On Sunday, Ron begins to drive due north from point T at a constant rate of x miles per hour. Two hours after Ron started, Sam begins to drive due south from point T at a constant rate of y miles per hour. If y>4x, for how many hours Ron would have driven when Sam had covered thrice as much distance as Ron?

A. x/(x - 2y)
B. 2y/(x - 3y)
C. 2y/(y - 3x)
D. 2x/(x + 4y)
E. 2x/(y - 3x)
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C
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Re: On Sunday, Ron begins to drive due north from point T at a constant [#permalink] New post   11 Aug 2022, 11:09
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Bunuel KarishmaB chetan2u

Please help me solve this ques
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Re: On Sunday, Ron begins to drive due north from point T at a constant [#permalink] New post   11 Aug 2022, 19:02
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Aabhash777 wrote:
On Sunday, Ron begins to drive due north from point T at a constant rate of x miles per hour. Two hours after Ron started, Sam begins to drive due south from point T at a constant rate of y miles per hour. If y>4x, for how many hours Ron would have driven when Sam had covered thrice as much distance as Ron?

A. x/(x - 2y)
B. 2y/(x - 3y)
C. 2y/(y - 3x)
D. 2x/(x + 4y)
E. 2x/(y - 3x)


Two ways…

(I)
Choose numbers for the variables.

Let x and y be 10 and 50…..50>4*10.
If the time is a, then 3(10*a)=50*(a-2)
30a=50a-100 or a=5.
Look for the option that gives you answer 5.
A and B will have negative denominator, so eliminate.
\(\frac{2y}{(y-3x)}=\frac{2*50}{(50-3*10)}=\frac{100}{20}=5\)

C

(II)
Algebraic
Let the time Ron has traveled be a, so Sam has traveled for a-2 hours.
Distance traveled by Ron is x*a, while Sam travels y(a-2).
Now it is given that \(3(x*a)=y(a-2)……3xa=ay-2y……..ay-3ax=2y………a(y-3x)=2y……..a=\frac{2y}{y-3x}\)

C
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Re: On Sunday, Ron begins to drive due north from point T at a constant [#permalink] New post   12 Aug 2022, 03:17
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Time taken by Ron = t
so Sam time = t-2 hours.
Distance traveled by Ron is x*t, while Sam travels y(t-2).
Now it is given that 3(x∗t)=y(t−2)
3xt=ty−2y
3xt-ty = -2y
ty-3xt = 2y
t(y-3x) = 2y
t= 2y/(y-3x)
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Re: On Sunday, Ron begins to drive due north from point T at a constant [#permalink] New post   12 Aug 2022, 03:57
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Aabhash777 wrote:
On Sunday, Ron begins to drive due north from point T at a constant rate of x miles per hour. Two hours after Ron started, Sam begins to drive due south from point T at a constant rate of y miles per hour. If y>4x, for how many hours Ron would have driven when Sam had covered thrice as much distance as Ron?

A. x/(x - 2y)
B. 2y/(x - 3y)
C. 2y/(y - 3x)
D. 2x/(x + 4y)
E. 2x/(y - 3x)


Ron travels for x mph for 2 hrs. Say x = 1 so he travels 2 miles in 2 hrs.
Sam begins and say drives for an hour. Ron covers 3 hrs in this time and Sam should cover three times that i.e. 9 miles in 1 hr. So y = 9 mph (y > 4x so this constraint is satisfied).
Ron has driven for 3 hrs at this time.

Put x = 1 and y = 9 in the options. Only option (C) gives 3 hrs. (Options A and B are negative and D and E are less than 1)

Answer (C)
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Re: On Sunday, Ron begins to drive due north from point T at a constant [#permalink] New post   30 Aug 2022, 10:04
Aabhash777 wrote:
On Sunday, Ron begins to drive due north from point T at a constant rate of x miles per hour. Two hours after Ron started, Sam begins to drive due south from point T at a constant rate of y miles per hour. If y>4x, for how many hours Ron would have driven when Sam had covered thrice as much distance as Ron?

A. x/(x - 2y)
B. 2y/(x - 3y)
C. 2y/(y - 3x)
D. 2x/(x + 4y)
E. 2x/(y - 3x)


I worked this way.

Lets y drove t hours.
so, ty= 3(2x+tx)
ty-3tx=6x
t(y-3x)=6x
t=6x/(y-3x)

so ron drove 6x/(y-3x)+2 hours
which gives 2y/(y-3x)
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Re: On Sunday, Ron begins to drive due north from point T at a constant [#permalink] New post   21 Sep 2023, 04:51
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Re: On Sunday, Ron begins to drive due north from point T at a constant [#permalink]
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