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# Radina and Russell are participating the Greystone Annual Run/Walk. Ra

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Joined: 02 Sep 2009
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23 Jan 2018, 23:45
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45% (medium)

Question Stats:

71% (03:20) correct 29% (03:02) wrong based on 69 sessions

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Radina and Russell are participating the Greystone Annual Run/Walk. Radina leaves a half hour after Russell and runs 1 mile per hour slower than twice Russell's rate. If Radina arrives at the finish line 4 hours after she started and finishes at the same time as Russell, approximately how many miles long is the race?

A. 4/3
B. 7/2
C. 9/2
D. 32/7
E. 36/7

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Re: Radina and Russell are participating the Greystone Annual Run/Walk. Ra  [#permalink]

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24 Jan 2018, 10:34
2
Speed of Russell = s
Speed of Radina = 2s - 1

Time taken by Radina = 4 hrs
Time taken by Russell = 4.5 hrs

(2s - 1)*4 = s * 4.5
s = 8/7

Distance = s * 4.5 = (8/7) * (9/2) = 36/7

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24 Jan 2018, 17:22
Bunuel wrote:
Radina and Russell are participating the Greystone Annual Run/Walk. Radina leaves a half hour after Russell and runs 1 mile per hour slower than twice Russell's rate. If Radina arrives at the finish line 4 hours after she started and finishes at the same time as Russell, approximately how many miles long is the race?

A. 4/3
B. 7/2
C. 9/2
D. 32/7
E. 36/7

Set up (R*T) = D for each runner. Set distances equal to each other, which will yield one actual rate. Use that rate to find distance.

1) TIME for each runner
Russell: 30 minutes MORE:$$4\frac{1}{2}=\frac{9}{2}$$ hours

2) RATE for each in variables
Let Russell's rate = R
Radina's rate = (2R - 1)

3) Find Russell's actual rate from D = D (they run the same distance)
$$(2R - 1)(4) = (R)(\frac{9}{2})$$
$$(2R - 1)(8) = 9R$$
$$16R - 8 = 9R$$
$$7R = 8$$
$$R = \frac{8}{7}$$
mph

4) Actual distance?
Use Russell's rate * time:
$$D = (\frac{8}{7}mph * \frac{9}{2}hrs) = \frac{36}{7}$$ miles

If D = $$\frac{36}{7}$$, then Radina's rate is D/T: $$\frac{\frac{36}{7}}{4} = \frac{36}{28} = \frac{9}{7}$$
Radina's rate is also (2R - 1)
Radina's rate: $$[(\frac{8}{7}*2) - 1] = (\frac{16}{7} - \frac{7}{7}) = \frac{9}{7}$$
That's a match. The answer is correct.

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Re: Radina and Russell are participating the Greystone Annual Run/Walk. Ra  [#permalink]

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26 Jan 2018, 10:42
Bunuel wrote:
Radina and Russell are participating the Greystone Annual Run/Walk. Radina leaves a half hour after Russell and runs 1 mile per hour slower than twice Russell's rate. If Radina arrives at the finish line 4 hours after she started and finishes at the same time as Russell, approximately how many miles long is the race?

A. 4/3
B. 7/2
C. 9/2
D. 32/7
E. 36/7

We can let Russell’s rate = r and Radina's rate = 2r -1

Radina’s time is 4 hours, and Russell’s is 4.5 hours. Thus, we have:

4(2r -1) = 4.5r

8r - 4 = 4.5r

3.5r = 4

r = 4/3.5 = 40/35 = 8/7

Thus, the distance is 8/7 x 4.5 = 36/7 = miles.

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Re: Radina and Russell are participating the Greystone Annual Run/Walk. Ra  [#permalink]

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30 Jan 2018, 01:17
1
Let Rusell start the run at 1 pm and let his speed be x mph.
So, Radina will start the run at 1.30 pm and her speed will be 2x-1 mph

Now, Radina reaches the finish line after 4 hrs. i.e at 5.30 pm will a speed of 2x-1 mph.
Therefore, the distance travelled by Radina = (2x-1)*4 miles----------------1

Rusell takes 4.5 hrs to cover the same distance with speed of x mph
so , distance travellled by Rusell = 4.5*x-----------------2

Since the total distance is same, equate eqn 1 & 2

(2x-1)*4 = 4.5x
x=8/7 mph

Now required Distance = 4.5*8/7 = 36/7

So E is correct.

Regards
Kshitij
Re: Radina and Russell are participating the Greystone Annual Run/Walk. Ra &nbs [#permalink] 30 Jan 2018, 01:17
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