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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

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Hi
Distance travelled by Ann on her journey beck from Y to X – c.
Total distance travelled by Ann d_1 = 70 + c
Total distance travelled by Bea d_2 = 70 – c
1) Ann’s speed is 30 kilometers per hour greater than Bea’s speed
Bea’s speed = v
Ann’s speed = v + 30
(70 + c )/ (v + 30) = (70 – c) / v
70c + cv = (70 – c)(v + 30)
c(v + 15) = 1050
We can derive many values for v and c. Insufficient.
2) Ann’s speed is twice Bea’s speed
(70 + c) / 2v = (70 – c) / v
70v + cv = 140v – 2cv
3cv = 70v
c = 70/3
Distance travelled by Bea: 70 – 70/3 = 140/3 Sufficient.
Answer B.
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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 \(a\) be speed of Ann & \(b\) be the speed of Bea.

Let the distance traveled by Bea be \(D\). Hence distance traveled by Ann is \(70 + D\)

Since the time required is same, we get \(\frac{(70+D)}{a}\) = \(\frac{D}{b}\)

Statement 1: \(a = b + 30\)

Substituting this in above equation we get an equation with 2 unknowns.

Hence Statement 1 alone is Not Sufficient.

Statement 2: \(a = 2b\)

Substituting this, we can find the value of \(D\).

Hence Statement 2 is sufficient.


Answer B.


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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
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
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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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

distance = rate * time

Clearly insufficient. Unable to solve for distance with 2 variables.

2) Ann’s speed is twice Bea’s speed

let 2x = Ann's speed
let x = Bea's speed

2x * time = 70
x * time = 70

Add the two equations together: 3x * time = 140
distance = 140/3

Statement 2 is sufficient.

Answer is B.
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When A and B finally meet, they will have together combined to cover the 70 km distance TWICE

Combined distance of A and B = 140 km

A and B both depart from the same location at the same instant of time. Therefore, A and B will have traveled for the SAME TIME when A turns around at Y-ville to come meet B.

Concept: when the Time traveled is constant, the Speed at which each travels will be DIRECTLY PROPORTIONAL to the DISTANCE each covers

Or

Ratio of Speeds = Ratio of Distance Covered

To determine how much of the 140 km Bea covers, if we can determine the relative speed of the two then we can determine how much of the total distance Bea covered with respect to Anne

S1: speed of A = 30 + speed of B

Case 1: B = 10 mph
A = 40 mph

Ratio of speeds = A : B = 40 : 10 = 4 : 1

Bea will cover (1/5) of the total combined distance of 140 while Anne will cover (4 / 5)

Distance Bea covers = 140 * (1/5) = 28 miles

Case 2: B = 20
A = 50

Ratio of Speeds = A : B = 5 : 2

Bea covers (2 /7) of the 140 combined distance = (140) (2 / 7) = 40 miles

S1 not sufficient

S2: Anne speed = 2 (Bea Speed)

Ratio of speeds = A : B = 2 : 1

Bea covers (1 / 3) of the 140 distance

S2 is sufficient alone
Bea distance = (140 / 3)

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