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Ann and Bea leave X-ville at the same time and travel towards Y-ville,

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Ann and Bea leave X-ville at the same time and travel towards Y-ville,  [#permalink]

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New post 10 Apr 2017, 07:33
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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

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Re: Ann and Bea leave X-ville at the same time and travel towards Y-ville,  [#permalink]

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New post 10 Apr 2017, 08:23
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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

Common sense: B's speed 20 kmph A s speed 50 kmph

B's speed 5kmph A's speed 35kmph

these will yield diff. result.
NS.


2) Ann’s speed is twice Bea’s speed

A-2x, A travelled 2xt.
B- x, B travelled xt.

2xt + xt = 70 + 70

xt = 140/3.

Suff.

B.

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Re: Ann and Bea leave X-ville at the same time and travel towards Y-ville,  [#permalink]

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New post 10 Apr 2017, 08:27
2
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


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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Re: Ann and Bea leave X-ville at the same time and travel towards Y-ville,  [#permalink]

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New post 11 Apr 2017, 07:16
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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


Target question: When they meet, how far has Bea traveled?

Statement 1: Ann’s speed is 30 kilometers per hour greater than Bea’s speed
We can see that is not sufficient if we examine some EXTREME CASES:
Case a: Ann's speed = 30.00000001 kilometers per hour, and Bea's speed = 0.00000001 kilometers per hour. In this case, Bea travels almost 0 kilometers
Case b: Ann's speed = 40 kilometers per hour, and Bea's speed = 10 kilometers per hour. In this case, Bea travels more than 0 kilometers
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: Ann’s speed is twice Bea’s speed
One option here is to test a bunch of cases to see what happens. If we do this, we'll find that we keep getting the same answer to the target question

Alternatively, we can use some algebra:
Let B = the distance Bea traveled
Let R = Bea's speed.


NOTE: the total distance from Townville to Villageton and then BACK TO Townville = 140 kilometers.

So, 140 - B = the distance Ann traveled
And 2R = Ann's speed (since her speed is TWICE Bea's speed)


From here, let's create a WORD EQUATION that uses distance and speed.
How about: Ann's travel time = Bea's travel time

Time = distance/rate, so we get:
(140 - B)/2R = B/R
Cross multiply to get: (B)(2R) = (R)(140 - B)
Expand: 2BR = 140R - BR
Add BR to both sides: 3BR = 140R
Divide both sides by R to get: 3B = 140
Divide both sides by 3 to get: B = 140/3
In other words, Bea traveled 140/3 kilometers
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer:

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Re: Ann and Bea leave X-ville at the same time and travel towards Y-ville,  [#permalink]

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New post 13 Aug 2018, 10:56
1
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 \(a\) 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.


Thanks,
GyM
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Re: Ann and Bea leave X-ville at the same time and travel towards Y-ville,  [#permalink]

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New post 13 Aug 2018, 11:26
1
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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Re: Ann and Bea leave X-ville at the same time and travel towards Y-ville, &nbs [#permalink] 13 Aug 2018, 11:26
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Ann and Bea leave X-ville at the same time and travel towards Y-ville,

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