Ashok and Brian are both walking east along the same path; Ashok walks at a faster constant speed than does Brian. If Brian starts 30 miles east of Ashok and both begin walking at the same time, how many miles will Brian walk before Ashok catches up with him?
(1) Brian's walking speed is twice the difference between Ashok's walking speed and his own.
(2) If Ashok's walking speed were five times as great, it would be three times the sum of his and Brian's actual walking speeds.
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
Speed Time Distance MGMAT
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Let A = A's rate and B = Brian's rate.Ashok and Brian are both walking east along the same path; Ashok walks at a faster constant speed than does Brian. If Brian starts 30 miles east of Ashok and both begin walking at the same time, how many miles will Brian walk before Ashok catches up with him?
(1) Brian's walking speed is twice the difference between Ashok's walking speed and his own.
(2) If Ashok's walking speed were five times as great, it would be three times the sum of his and Brian's actual walking speeds.
Ashok has to CATCH-UP by 30 miles.
The CATCH-UP rate is equal to the DIFFERENCE between the two rates.
If A = 3mph, while B = 2mph, then every hour A walks 1 more mile than B, with the result that every hour A catches up by 1 mile -- the DIFFERENCE between the two rates:
A-B = 3-2 = 1mph.
Statement 1: Brian's walking speed is twice the difference between Ashok's walking speed and his own.
Thus:
B = 2(A-B)
B = 2A - 2B
3B = 2A
A = (3/2)B.
Case 1: B = 10mph, A = (3/2)(10) = 15mph
Here, the catch-up rate = A-B = 15-10 = 5mph.
Time for A to catch up by 30 miles = (catch-up distance)/(catch-up rate) = 30/5 = 6 hours.
In 6 hours, the distance traveled by B at a rate of 10mph = r*t = 10*6 = 60 miles.
Case 2: B = 20mph, A = (3/2)(20) = 30mph
Here, the catch-up rate = A-B = 30-20 = 10mph.
Time for A to catch up by 30 miles = (catch-up distance)/(catch-up rate) = 30/10 = 3 hours.
In 3 hours, the distance traveled by B at a rate of 20mph = r*t = 20*3 = 60 miles.
Since B travels the SAME DISTANCE in each case, SUFFICIENT.
Statement 2: If Ashok's walking speed were five times as great, it would be three times the sum of his and Brian's actual walking speeds.
Thus:
5A = 3(A+B)
5A = 3A + 3B
2A = 3B
A = (3/2)B.
Same information as statement 1.
SUFFICIENT.
The correct answer is D.
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Since both A and B started at the same time, they both spent same time, t.
Let d be the extra distance B traveled beyond 30m.
Time spent by Ashok = Time spent by Brian
Time spent by Ashok = distance traveled / speed =
Let Speed of Ashok = a meters; distance = 30 + d
Time spent by Ashok = (30+d)/a..............(1)
Time spent by Brian = distance Brian traveled / speed of Brian
Speed of Brian = b; distance traveled by Brian = d meters
Time spent by Brian = d/b...................(2)
Equating (1) to (2), we have
(30+d)/a = d/b
30b + bd = ad
a/b = d/(30+d)...if we know the ratio of their speed(a/b) that would be sufficient
to find d.
Both statements can be reduced to give a ratio of a/b.. D.
Let d be the extra distance B traveled beyond 30m.
Time spent by Ashok = Time spent by Brian
Time spent by Ashok = distance traveled / speed =
Let Speed of Ashok = a meters; distance = 30 + d
Time spent by Ashok = (30+d)/a..............(1)
Time spent by Brian = distance Brian traveled / speed of Brian
Speed of Brian = b; distance traveled by Brian = d meters
Time spent by Brian = d/b...................(2)
Equating (1) to (2), we have
(30+d)/a = d/b
30b + bd = ad
a/b = d/(30+d)...if we know the ratio of their speed(a/b) that would be sufficient
to find d.
Both statements can be reduced to give a ratio of a/b.. D.
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We can use our standard Distance = Rate * Time equation here, but we have to be careful about how we define each of the terms.
Let's say that Brian's rate is b, and the distance he travels is d. We'll say that Ashok's rate is a, and since he has to travel 30 miles further than Brian does, Ashok's distance is (d + 30).
Since they start walking at the same time and meet each other at the same time, their times are the same, so let's call each guy's time t.
This gives us the equations
d = bt
d + 30 = at
which reduces to bt = at - 30. We want to solve for bt, since bt = Brian's Rate * Brian's Time = Brian's Distance.
S1 tells us that b = 2*(a - b), or 3b = 2a, or a = (3/2)b. Subbing this into our equation above gives us
bt = (3/2)bt - 30, or
30 = (1/2)bt, or
60 = bt
So b*t = 60, and we're done!
S2 gives us 5a = 3(a + b), or 2a = 3b, which is the same as what we got in S1, making this statement ALSO sufficient.
Let's say that Brian's rate is b, and the distance he travels is d. We'll say that Ashok's rate is a, and since he has to travel 30 miles further than Brian does, Ashok's distance is (d + 30).
Since they start walking at the same time and meet each other at the same time, their times are the same, so let's call each guy's time t.
This gives us the equations
d = bt
d + 30 = at
which reduces to bt = at - 30. We want to solve for bt, since bt = Brian's Rate * Brian's Time = Brian's Distance.
S1 tells us that b = 2*(a - b), or 3b = 2a, or a = (3/2)b. Subbing this into our equation above gives us
bt = (3/2)bt - 30, or
30 = (1/2)bt, or
60 = bt
So b*t = 60, and we're done!
S2 gives us 5a = 3(a + b), or 2a = 3b, which is the same as what we got in S1, making this statement ALSO sufficient.
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GIVEN: When the men start walking, Brian has a 30-mile leadrommysingh wrote:Ashok and Brian are both walking east along the same path; Ashok walks at a faster constant speed than does Brian. If Brian starts 30 miles east of Ashok and both begin walking at the same time, how many miles will Brian walk before Ashok catches up with him?
(1) Brian's walking speed is twice the difference between Ashok's walking speed and his own.
(2) If Ashok's walking speed were five times as great, it would be three times the sum of his and Brian's actual walking speeds.
Let B = Brian's walking speed (in miles per hour)
Let A = Ashok's walking speed (in miles per hour)
Since Ashok's speed is greater than Brian's speed, the rate at which the gap shrinks = (A - B) miles per hour
For example, if A = 5 and B = 2, then the 30-mile gap will shrink at a rate of (5 - 2) mph.
time = distance/speed
So, time for 30-mile gap to shrink to zero = 30/(A - B)
Target question: How many miles will Brian walk before Ashok catches up with him?
This is a good candidate for rephrasing the target question.
Aside: Here's a video with tips on rephrasing the target question: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100
distance = (speed)(time)
So, the distance Brian travels = (B)(30/(A - B))
Simplify to get: 30B/(A - B)
REPHRASED target question: What is the value of 30B/(A - B)?
Statement 1: Brian's walking speed is twice the difference between Ashok's walking speed and his own.
We can write: B = 2(A - B)
Expand: B = 2A - 2B
This means: 3B = 2A
So: 3B/2 = A
Or we can say: 1.5B = A
Now take 30B/(A - B) and replace A with 1.5B to get: 30B/(1.5B - B)
Simplify: 30B/(0.5B)
Simplify: 30/0.5
Evaluate 60 (miles)
Perfect!! The answer to the REPHRASED target question is Brian will travel 60 miles
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT
Statement 2: If Ashok's walking speed were five times as great, it would be three times the sum of his and Brian's actual walking speeds.
We can write: 5A = 3(A + B)
Expand: 5A = 3A + 3B
Rewrite as: 2A = 3B
We get: A = 3B/2 = 1.5B
At this point, we're at the same place we got to for statement 1.
So, since statement 1 is sufficient, we know that statement 2 is also sufficient.
Answer: D
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This is a catching up problem. Recall that the time needed for the slower person (Brian) to catch up with the faster person (Ashok) is (difference in distances)/(difference in speeds). Here the difference in their distances is 30, and the difference in their speeds is (a - b), where a is Ashok's speed and b is Brian's speed. So the time for Brian to catch up with Ashok is 30/(a - b). If we can determine that, then we can determine the distance walked by Brian.rommysingh wrote:Ashok and Brian are both walking east along the same path; Ashok walks at a faster constant speed than does Brian. If Brian starts 30 miles east of Ashok and both begin walking at the same time, how many miles will Brian walk before Ashok catches up with him?
(1) Brian's walking speed is twice the difference between Ashok's walking speed and his own.
(2) If Ashok's walking speed were five times as great, it would be three times the sum of his and Brian's actual walking speeds.
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
Statement One Alone:
Brian's walking speed is twice the difference between Ashok's walking speed and his own.
We are given that b = 2(a - b). That is,:
b = 2a - 2b
3b = 2a
a = 3b/2.
Therefore, the time for Brian to catch up with Ashok is 30/(a - b) = 30/(3b/2 - b) = 30/(b/2) = 60/b and the distance walked by Brian is b x 60/b = 60 miles. Statement one alone is sufficient.
Statement Two Alone:
If Ashok's walking speed were five times as great, it would be three times the sum of his and Brian's actual walking speeds.
We are given that 5a = 3(a + b). That is:
5a = 3a + 3b
2a = 3b
a = 3b/2.
We can see that this statement provides the same information as statement one. So statement two is also sufficient.
Answer: D
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