Adam, Bill, and Chris simultaneously start from City A to City B. Adam

Question

Adam, Bill, and Chris simultaneously start from City A to City B. Adam reaches City B first, then turns back and meets Bill at a distance of 9 miles from City B. When Bill reaches City B, he too turns back and meets Chris at a distance of 7 miles from City B. If three times the speed of Adam is equal to five times Chris' speed, what could be the distance between the City A and City B?

A. 40 miles
B. 35 miles
C. 63 miles
D. 90 miles
E. 120 miles

A. 40 miles
B. 35 miles
C. 63 miles
D. 90 miles
E. 120 miles

nick1816 · Accepted Answer

\frac{A}{B}=\frac{D+9}{D-9} .......(1)
 \frac{B}{C} = \frac{D+7}{D-7} .......(2)

Multiply (1) and (2)

\frac{A}{C}=\frac{(D+9)(D+7)}{(D-9)(D-7)}

\frac{(D+9)(D+7)}{(D-9)(D-7)}=\frac{5}{3}

D^2-64D+63=0 
D=1(Not possible)
or D=63

ParthSanghavi · Answer

Chetan2u VeritasKarishma Bunuel Could you please explain this problem in detail?

KarishmaB · Answer

Let distance between cities A and B be d.

Adam, Bill and Chris start together from A.

(ABC) -------------- d ---------------

A is faster so reaches the other end.

------------------(B)-------------(A)

Then A turns back and meets B after 9 miles

------------------------(BA)--- 9 ---

In the same time, A travelled (d + 9) miles while B travelled (d - 9) miles

Speed of A: Speed of B = (d + 9):(d - 9)

Thereafter, B is faster than C so reaches the other end.

------------------(C)-------------(B)

Then B turns back and meets C after 7 miles

------------------------(CB)--- 7 ---

So in the same time, B travelled (d + 7) miles while C travelled (d - 7) miles

Speed of B:Speed of C = (d+7):(d - 7)

We need to find Speed of A:Speed of C since we know it is 5:3.

We already know 
Speed of A: Speed of B = (d + 9):(d - 9)
Speed of B:Speed of C = (d+7):(d - 7)
So Speed of A:Speed of C will be found by equating the speed of B in the two ratios (When we know A:B and B:C, we get A:C by making B same in both ratios)

I might plug in values at this point. The first value of d that I will try will be 63 - the reason that it is a multiple of 7 as well as 9. It is likely that when 7 or 9 is added/subtracted, we will get some simple values. Since 5:3 are simple values, that is what we are looking for. 
Adding 7 to 40 will give 47 so we won't get a simple ratio.
Adding 9 to 35 will give 44 so we will get a multiple of 11.
Adding 7 to 90 will give 97, again a difficult number
etc

say d = 63

Speed of A: Speed of B = (d + 9):(d - 9) = 72:54 = 4:3 = 20:15
Speed of B:Speed of C = (d+7):(d - 7) = 70:56 = 5:4 = 15:12

Speed of A : Speed of C = 20:12 = 5:3 (Matches)

Answer (C)

CrackverbalGMAT · Answer

This is a moderately difficult question on Time & Distance, which is based on the concept that, when time is constant, the ratio of the speeds of two objects is the same as the ratio of the distances travelled by them.

That is, when T = constant, \(\frac{S_1}{S_2}\)= \(\frac{D_1}{D_2}\)

Let us draw some simple diagrams to represent the situations described in the question. Rather than going the variables way, let us try to use options to our advantage. The moment I look at 9 and 7 in the same question, I will try to look for a common multiple of these – 63. That then, is our starting point.

Attachment:

13th June 2019 - Reply 4.JPG [ 31.05 KiB | Viewed 3990 times ]

Let the distance between A and B be 63 miles. Then, in the first case, Adam travelled a total of 72 miles while Bill travelled 54. Therefore,

\(\frac{S_{Adam}}{S_{Bill}}\) = \(\frac{72}{54}\) = \(\frac{4}{3}\).

Similarly, \(\frac{S_{Bill}}{S_{Chris}}\) = \(\frac{70}{56}\) = \(\frac{5}{4}\).

Combining the two ratios, we have \(\frac{S_{Adam}}{S_{Chris}}\) = \(\frac{S_{Adam}}{S_{Bill}}\) * \(\frac{S_{Bill}}{S_{Chris}}\) = \(\frac{4}{3}\) * \(\frac{5}{4}\) = \(\frac{5}{3}\).

But, this is what the question says about the ratio of the speeds of Adam and Chris. 3 \(S_{Adam}\) = 5 \(S_{Chris}\) only means that \(\frac{S_{Adam}}{S_{Chris}}\) = \(\frac{5}{3}\).
Therefore, the value that we started off assuming HAS to be the right answer.

Obviously, the variable method will also proceed in a similar manner, but, dealing with numbers is always easier. Hope you’ll all agree with that!

Hope this helps!

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Adam, Bill, and Chris simultaneously start from City A to City B. Adam

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