Suppose distance between A and B is x.

Tom travelled x+9 miles in the same time in which Jerry travelled x-9 miles

\(V_T : V_J = (x+9) : (x-9)\).....(1)

Similarly

\(V_J: V_B = (x+7) : (x-7).\)...(2)

From (1) and (2)

\(\frac{V_T}{V_J }= \frac{(x+9)(x+7)}{(x-9)(x-7)} = \frac{5}{3}\)

We can clearly see that (x+9)(x+7) must be a multiple of 5, as all the options are integer. x= 5k+1 or 5k+3 . Eliminate A, C and D


Option B- \(\frac{(31+9)(31+7)}{(31-9)(31-7)} = \frac{40*38}{22*24} ≠ \frac{5}{3}\) Eliminate B

E

OR

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

\(3(x^2+16x+63) = 5(x^2-16x+63)\)

\(2x^2 - 128x + 126 = 0\)

\(x^2-64x+63 = 0\)

x= 1 or 63

E

Let total Distance be D
Speed of Tom Be St
Speed of Jerry be Sj
Speed of Bill be Sb.

Tom reaches point B first and turns back and meets Jerry at a distance of 9 miles from B.
Time taken by both Tom and Jerry to meet will be same so the equation will be
(D+9)/St=(D-9)/Sj
Sj=(D-9)*St/(D+9) -----(1)

When Jerry reaches B, he too turns back and meets Bill at a distance of 7 miles from B.

Time taken by both Jerry and Bill to meet will be same so the equation will be
(D+7)/Sj=(D-7)/Sb ----(2)

If 3 times the speed with which Tom drives his car is equal to 5 times Bill’s speed

3*St=5*Sb or
Sb=3*St/5-----(3)

Putting value in equation (2).
We get,
(D+7)/[(D-9)*St/(D+9)]=(D-7)/[3*St/5]
(D+7)*(D+9)*3=5*(D-7)(D-9)
3D^2+48D+63*3=5D^2-80D+63*5
128D=2D^2+126
D^2-64D+63
(D-63)(D-1)
D=1(not possible)
D=63 ( Answer)

Answer is E

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_________________

- Let the distance between A and B be x miles.
- Let speeds of Tom, Jerry & Bill be T, J & B, respectively.

When Tom met Jerry,
Tom travelled (x+9) miles
Jerry travelled (x–9) miles
Speed ratio: T/J=(x+9)/(x-9) ...(1)

When Jerry met Bill,
Jerry travelled (x+7) miles
Bill travelled (x–7) miles
Speed ratio: J/B=(x+7)/(x-7) ...(2)

Speed ratio of T/B:
T/B = (T/J)*(J/B) = (1)*(2)
5/3 = [(x+9)/(x-9)]*[(x+7)/(x-7)]
5*(x-9)(x-7) = 3*(x+9)(x+7)
2x^2 -128x + 63*2 = 0
x^2 -64x + 63 = 0
(x-63)(x-1)=0
x=63 or x=1 (NA)

FINAL ANSWER IS (E)

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Tom, Jerry and Bill start from point A at the same time in their cars to go to B. Tom reaches point B first and turns back and meets Jerry at a distance of 9 miles from B. When Jerry reaches B, he too turns back and meets Bill at a distance of 7 miles from B. If 3 times the speed with which Tom drives his car is equal to 5 times Bill’s speed, what could be the distance between the points A and B?

A. 24 miles
B. 31 miles
C. 40 miles
D. 45 miles
E. 63 miles
Let distance between point A and B = D; Speed of Tom = \(S_t\), Speed of Jerry = \(S_j\) and Speed of Bill = \(S_b\) (considering all drive at constant speed)
Also given \(3S_t = 5S_b\) OR \(\frac{S_b}{S_t} = \frac{3}{5}\)
Here time taken by Tom and Jerry to cover their respective distances when they meet is equal. \(T_t\) = \(T_j\)
So, when they meet:
Distance covered by Tom = D + 9
Distance covered by Jerry = D - 9
Hence \(\frac{D + 9 }{S_t} = \frac{D - 9}{S_j}\) OR \(\frac{D + 9 }{D - 9} = \frac{S_t}{S_j}\) Eqn. 1

Similarly, when Jerry and Bill meet time taken by them is equal. \(T_j\)' = \(T_b\)
When Jerry and Bill meet:
Distance covered by Jerry = D + 7
Distance covered by Bill = D - 7
Hence \(\frac{D + 7 }{S_j} = \frac{D - 7}{S_b}\) OR Hence \(\frac{D + 7 }{D - 7} = \frac{S_j}{S_b}\) Eqn. 2

Multiplying Eqn. 1 and 2
Hence \(\frac{(D + 9)(D + 7)}{(D - 9)(D - 7)} = \frac{S_t}{S_b} = \frac{5}{3}\)
\(D^2 - 64D + 63 = 0\)
(D - 63)(D - 1) = 0
D ≠ 1 So
D = 63

Answer E.
_________________

Tom, Jerry and Bill start from point A at the same time in their cars to go to B. Tom reaches point B first and turns back and meets Jerry at a distance of 9 miles from B. When Jerry reaches B, he too turns back and meets Bill at a distance of 7 miles from B. If 3 times the speed with which Tom drives his car is equal to 5 times Bill’s speed, what could be the distance between the points A and B?

A. 24 miles
B. 31 miles
C. 40 miles
D. 45 miles
E. 63 miles

3Ts=5Bs
time is same here t=d/s [for tom and jerry]
x+9/Ts=x-9/Js ------->1
time is same here t=d/s [for jerry and bill]
x+7/Js=x-7/Bs
Js=(x+7)3Ts/(x-7)5
sub this in (1)
(x+9)/Ts=(x-9)(x-7)5/(x+7)3Ts
on solving we get,
x^2-64x+63=0,
x=1,63...dist can be 1
so dist is 63
Ans E

It's a good question,but I think it can't be solved in 2 mins. So, do these types of questions really come in real gmat?

Hea234ven
Yes, you may be right but its better to practice problems like these. The more you practice the better you get at solving them.

Consider this one, if you got absolutely outflanked by it on reading the first time check the options given. On reading the questions and trying to figure out what it says and asks, you might have an intuition(don't rely on it but better confirm it) of the correct answer. While writing out the equations you may release that 63 can be the answer since they involve ±9 and ±7. This may be a fluke but it becomes lesser of a fluke as you practice these kind of questions more. The best part about having such intuition is that it lets you save some time.

Cheers..!!
_________________

Let:

Speed of Tom = T
Speed of Jerry = J
Speed of Bill = B

In each scenario, drove for same travel time. When Time is constant, ratio of distances traveled is directly proportional to the ratio of speeds.

3T = 5B
T/B = 5/3


(T/J) * (J/B) = T/B

Let d = distance from A to B


(1)
T/J = (d + 9) / (d - 9)

(2)
J/B = (d + 7) / (d - 7)


Multiplying the 2 equations:

T/B = (d^2 + 16d + 63) / (d^2 - 16d + 63)

T/B = 5/3


Setting the 2 equations equal, cross multiplying and simplifying:

(d)^2 - 64d + 63 = 0

(d - 1) (d - 63) = 0

d = 63

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