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# Carl and Alvin stand at points C and A, respectively, as represented

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Re: Carl and Alvin stand at points C and A, respectively, as represented [#permalink]
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Carl and Alvin stand at points C and A, respectively, as represented in the figure above (not drawn to scale), with distances labeled in feet. Carl starts running toward point B at a rate of 400 ft/min, and Alvin starts running toward point B at a rate of 500 ft/min. When Alvin reaches point B, he turns around and runs in the opposite direction until he reaches point A, whereupon he reverses his direction again and runs toward point B. If Alvin repeats this pattern indefinitely, how far will the two runners be from point B when they meet? Assume that Alvin is able to reverse his direction instantaneously.

A. 20 feet
B. 31$$\frac{2}{5}$$ feet
C. 35 feet
D. 47$$\frac{7}{9}$$ feet
E. 59 feet

Let Sc = Speed of Carl = 400ft/min, Sa = Speed of Alvin = 500ft/min.
Time taken by Carl to reach point A = $$\frac{2200}{400} = 5.5 min$$
Now, in same time how much Alvin traveled = 500 * 5.5 = 2750 ft
Also, distance AB = 70ft,
Number of turns Alvin took while covering 2750 ft = 2750/70 = 39.28 where 0.28 equates to 20ft(as remainder)
Since Alvin starts running towards B he takes 1st turn at B, 2nd at A, then 3rd at B and so on...
Thus, every odd number denotes a turn taken by Alvin at B and every even denotes at A. So, 39 denotes Alvin's direction(opposite) is towards A.

As Alvin already covered 20ft from B towards A, distance between Carl and Alvin = 70 - 20 = 50ft(let's call this point as X) when Carl is at A.
Time taken to cover 50ft(distance AX) = $$\frac{50}{400+500} = \frac{5}{90}$$ min
Distance covered by Carl from A to meet Alvin = $$\frac{5}{90} * 400 = \frac{200}{9}$$ ft
Hence distance between point of meeting of Carl & Alvin and point B = $$70 - \frac{200}{9} = 47\frac{7}{9}$$

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Re: Carl and Alvin stand at points C and A, respectively, as represented [#permalink]
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We know that Alvin runs at 500ft/min while Carl runs at 400ft/min.
Carl will take 2200/400 = 5.5minutes to get to point A. Within this 5.5 minutes timeframe, Alvin would have covered a distance of 5.5*500=2750ft.
How many 70ft laps would Alvin have made within the 5.5 minutes? He would have made 2750/70 = 39.2 laps implying, Alvin would be moving in the direction from point B towards A. and he would be 2750-39*70 ft away from B = 20ft.
So we can see that Alvin and Carl are 50ft away from each other at time 5.5minutes and running towards each other. Hence their relative speed = 900ft/min
The time taken for Alvin and Carl to meet, t, can be determined as 50/900=5/90.
So they will meet at a distance 20+500*5/90=20+250/9 = 20+27+7/9=47+7/9.

The answer is, therefore, option D.
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Re: Carl and Alvin stand at points C and A, respectively, as represented [#permalink]
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Time it would take for C to reach A = 2200/400 ; 5.5 mins
so in 5.5 mins A would have covered ; 5.5 * 500 ; 2750 feet
i.e 2750/70 ; 39+ 2/7 feet ; and A must be now 2/7 * 70 ; 20 feet away from B and running towards A ;
so distance b/w C & A ; 50 feet and relative speed= 400+500 ; 900
time it will take them to meet ; 50/900 ; 1/18 min
Alvin would have run ; 500 * 1/18 ; 250/9 mtrs or say 27 7/9 and add 20 mts ; we get 47 7/9
IMO D

Carl and Alvin stand at points C and A, respectively, as represented in the figure above (not drawn to scale), with distances labeled in feet. Carl starts running toward point B at a rate of 400 ft/min, and Alvin starts running toward point B at a rate of 500 ft/min. When Alvin reaches point B, he turns around and runs in the opposite direction until he reaches point A, whereupon he reverses his direction again and runs toward point B. If Alvin repeats this pattern indefinitely, how far will the two runners be from point B when they meet? Assume that Alvin is able to reverse his direction instantaneously.

A. 20 feet

B. 31-2/5 feet

C. 35 feet

D. 47-7/9 feet

E. 59 feet
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Re: Carl and Alvin stand at points C and A, respectively, as represented [#permalink]
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Carl = 400 ft/min
Alvin= 500 ft/min

—> CA= 2200 ft
$$\frac{2200}{400}= 5.5 min$$ (it takes 5.5 min for Carl to get to the point A.)

—> 500* 5.5= 2750( in 5.5 minutes, Alvin can run 2750 ft)
—> 2750: 140= 19 (remainder 90)

That means the distance between Carl and Alvin is 50 ft now and Alvin is just 20 ft away from B.

In order to meet each other, they need:
—> 50= (400+500)t
$$t = \frac{50}{900}= \frac{1}{18}$$ min
—> Alan =$$( \frac{1}{18})* 500= \frac{250}{9} = 27 \frac{7}{9}$$

$$27 \frac{7}{9}+ 20 = 47 \frac{7}{9}$$

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Re: Carl and Alvin stand at points C and A, respectively, as represented [#permalink]
Image
Carl and Alvin stand at points C and A, respectively, as represented in the figure above (not drawn to scale), with distances labeled in feet. Carl starts running toward point B at a rate of 400 ft/min, and Alvin starts running toward point B at a rate of 500 ft/min. When Alvin reaches point B, he turns around and runs in the opposite direction until he reaches point A, whereupon he reverses his direction again and runs toward point B. If Alvin repeats this pattern indefinitely, how far will the two runners be from point B when they meet? Assume that Alvin is able to reverse his direction instantaneously.

A. 20 feet

B. 31253125 feet

C. 35 feet

D. 47794779 feet

E. 59 feet

To reach to point A, Carl will take = (2200/400) =$$\frac{5 1}{2}$$ minutes. In this meantime, Alvin can cover 2750 feet. Alvin will be 20 feet from point B when Carl reached point A. The remaining distance is (70-20) = 50 feet. Lets say, they will meet at point d which is between this 50 feet.

d/500 = (50-d)/400 = 27.77. I choose the closet option, B.
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Re: Carl and Alvin stand at points C and A, respectively, as represented [#permalink]
C--------------------------------A------M--------D----B (CA=2200 ft; AB=70 ft; D indicates Alvin's position when Carl reaches A and M is the point where they meet)

By the time Carl covers 2200 ft to reach A, Alvin will have run (5/4)*2200=2750 ft which is equal to 2750/70=39 and 2/7 laps between A an B. Since he has completed an odd number of laps, Alvin would be headed towards A at this point. BD=(2/7)*70=20 ft which means there is 70-20=50 ft separating Carl and Alvin now from each other. DA=50. Alvin covers (5/9)th of DA=(5/9)*50=250/9 ft. DM=250/9.
Alvin is (DM+DB)=250/9+20=[47 and (7/9)] ft from B when he meets Carl. ANS: C
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Re: Carl and Alvin stand at points C and A, respectively, as represented [#permalink]
Bunuel wrote:

Carl and Alvin stand at points C and A, respectively, as represented in the figure above (not drawn to scale), with distances labeled in feet. Carl starts running toward point B at a rate of 400 ft/min, and Alvin starts running toward point B at a rate of 500 ft/min. When Alvin reaches point B, he turns around and runs in the opposite direction until he reaches point A, whereupon he reverses his direction again and runs toward point B. If Alvin repeats this pattern indefinitely, how far will the two runners be from point B when they meet? Assume that Alvin is able to reverse his direction instantaneously.

A. 20 feet
B. $$31\frac{2}{5}$$ feet
C. 35 feet
D. $$47\frac{7}{9}$$ feet
E. 59 feet

Carl = X
Alvin = Y

Time for X to get to A: 2200/400 = 5.5 mins
In 5.5 mins Y will have traveled: 5.5*500 = 2750
Y traveled from B to A: 2750/70 = 39+2/7 cycles
In 5.5 mins Y will be 2/7 distance of B to A: 2/7*70 = 20 from B or 50 from A
Time that X and Y will meet at opposite directions: 50/(400+500)=5/90=1/18 mins
In 1/18 mins Y will have traveled: 1/18*500=500/18
Y's distance from B: 500/18+20=250+180=430/9=47+7/9 feet

Ans (D)
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Re: Carl and Alvin stand at points C and A, respectively, as represented [#permalink]
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Re: Carl and Alvin stand at points C and A, respectively, as represented [#permalink]
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