GMATPrepNow wrote:

Ann and Bea leave X-ville at the same time and travel towards Y-ville, which is 70 kilometers away. Their individual speeds are constant, but Ann’s speed is greater than Bea’s speed. Upon reaching Y-ville, Ann immediately turns around and drives toward X-ville until she meets Bea. When they meet, how far has Bea traveled?

1) Ann’s speed is 30 kilometers per hour greater than Bea’s speed

2) Ann’s speed is twice Bea’s speed

*kudos for all correct solutions

Let Speeds of Ann and Bea be\(S_{A}\) and \(S_{B}\) respectively ----- (1)

Let Time of Ann and Bea be \(T_{A}\) and \(T_{B}\) respectively ------- (2)

Let distance covered by Ann after completing 70Km is D.

Therefore,

Total distance covered by Ann = 70 + D --------- (3)

Total distance covered by Bea = 70 - D ---------- (4) (As Bea is D kms far from completing 70 kms)

As we know both meet at same time there we can equate their times, i.e.,

\(T_{A}\) = \(T_{B}\) -------- (5)

Now we know \(S = \frac{D}{T}\) ------------ (6)

Using (1), (2), (3), (4), (5), (6), we get ->

\(\frac{70 + D}{S_{A}} = \frac{70 - D}{S_{B}}\) ---------- (7)

We need to determine (70 - D)

GMATPrepNow wrote:

1) Ann’s speed is 30 kilometers per hour greater than Bea’s speed

Here we determine,

\(S_{A}\) = 30 + \(S_{B}\)

Substituting same in equation (7), we get

\(\frac{70 + D}{30 + S_{B}} = \frac{70 - D}{S_{B}}\)

This equation cannot be used to determine value of D or (70 - D).

Hence Insufficient.GMATPrepNow wrote:

2) Ann’s speed is twice Bea’s speed

Here we determine,

\(S_{A}\) = 2 * \(S_{B}\)

Substituting same in equation (7), we get

\(\frac{70 + D}{2 * S_{B}} = \frac{70 - D}{S_{B}}\)

Now here we can cancel \(S_{B}\) and determine the value of D, henceforth (70 - D).

Hence Sufficient.Hence Option B
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Regards

AD

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A Kudos is one more question and its answer understood by somebody !!!