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Re M1608 [#permalink]
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16 Sep 2014, 00:58
Official Solution: Statement (1) by itself is insufficient. The bus could have gone slowly or fast during the last 2 hours of the journey. Statement (2) by itself is insufficient. We know nothing about the value of the bus's speed for either first or second half. Statements (1) and (2) combined are insufficient. S2 tells us that the bus spent \(\frac{4}{3}\) hours on the first half of the distance and \(\frac{8}{3}\) hours on the second half. To answer the question, we have to know the distance between \(A\) and \(B\). Whatever the distance between \(A\) and \(B\)(as long as it's not too big or too small), the bus could spend \(\frac{4}{3}\) hours on the first half, then adjust its speed to arrive at 100 mile mark after 2 hours of the journey, and then adjust again to finish the rest of the distance in 2 hours. Answer: E
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Re: M1608 [#permalink]
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24 Apr 2015, 21:10
Bunuel wrote: Official Solution:
Statement (1) by itself is insufficient. The bus could have gone slowly or fast during the last 2 hours of the journey. Statement (2) by itself is insufficient. We know nothing about the value of the bus's speed for either first or second half. Statements (1) and (2) combined are insufficient. S2 tells us that the bus spent \(\frac{4}{3}\) hours on the first half of the distance and \(\frac{8}{3}\) hours on the second half. To answer the question, we have to know the distance between \(A\) and \(B\). Whatever the distance between \(A\) and \(B\)(as long as it's not too big or too small), the bus could spend \(\frac{4}{3}\) hours on the first half, then adjust its speed to arrive at 100 mile mark after 2 hours of the journey, and then adjust again to finish the rest of the distance in 2 hours.
Answer: E Hi Bunnel, Don't we get the speed from the statement 1 for the first half of the trip. When combining this information with statement 2, cant we get the speed for second trip. From this we can find the distance for 1st half and 2nd half. and hence can calculate average speed. please let me know. Thanks.



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Re: M1608 [#permalink]
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25 Apr 2015, 05:07
CKG wrote: Bunuel wrote: Official Solution:
Statement (1) by itself is insufficient. The bus could have gone slowly or fast during the last 2 hours of the journey. Statement (2) by itself is insufficient. We know nothing about the value of the bus's speed for either first or second half. Statements (1) and (2) combined are insufficient. S2 tells us that the bus spent \(\frac{4}{3}\) hours on the first half of the distance and \(\frac{8}{3}\) hours on the second half. To answer the question, we have to know the distance between \(A\) and \(B\). Whatever the distance between \(A\) and \(B\)(as long as it's not too big or too small), the bus could spend \(\frac{4}{3}\) hours on the first half, then adjust its speed to arrive at 100 mile mark after 2 hours of the journey, and then adjust again to finish the rest of the distance in 2 hours.
Answer: E Hi Bunnel, Don't we get the speed from the statement 1 for the first half of the trip. When combining this information with statement 2, cant we get the speed for second trip. From this we can find the distance for 1st half and 2nd half. and hence can calculate average speed. please let me know. Thanks. I think your doubt is addressed here: ifittookabus4hourstogetfromtownatotownbwhat102361.htmlHope it helps.
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Re: M1608 [#permalink]
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15 Nov 2015, 03:26
Dear Bunuel,
Thanks for drafting such a good question.
I have a doubt regarding explanation of statement 2.
Can we say that "the bus spent 4/3 hours on the first half of the distance and 8/3 hours on the second half". I think to draw this conclusion we must have a finite known distance.
I only able to deduce that time spent by bus is first half is 1/2 of time spend in the second half.
Please correct me if my analysis is wrong.



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Re: M1608 [#permalink]
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15 Jan 2016, 07:32
From statement 1: We can know the avg speed value in first half.
From statement 2: Consider the total distance as D. As per the statement, 2s is the speed of first half i.e. D/2 and s is the speed of second half . Hence avg speed of total journey is total distance/time taken, D/ ((D/2)/2s+(D/2)/s)..whose value will come in value of s (D will be cancelled). Hence, statement 2 we will get a value in variables of s.
Combing 1 and 2 statement: avg value in first half=s and value of s from statement 2 gives me the answer.
Please let me know anything wrong in this reasoning



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Re: M1608 [#permalink]
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03 Aug 2016, 09:31
Hi Bunuel, The trick is that the 1st half of the distance doesn't mean the distance covered in 1st two hours. If the question has addressed that the bus traveled 100 KM distance in 1st half. Then it would be sufficient. Right? Thanks, Sathish



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Re: M1608 [#permalink]
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27 Dec 2016, 10:36
Bunuel wrote: Official Solution:
Statement (1) by itself is insufficient. The bus could have gone slowly or fast during the last 2 hours of the journey. Statement (2) by itself is insufficient. We know nothing about the value of the bus's speed for either first or second half. Statements (1) and (2) combined are insufficient. S2 tells us that the bus spent \(\frac{4}{3}\) hours on the first half of the distance and \(\frac{8}{3}\) hours on the second half. To answer the question, we have to know the distance between \(A\) and \(B\). Whatever the distance between \(A\) and \(B\)(as long as it's not too big or too small), the bus could spend \(\frac{4}{3}\) hours on the first half, then adjust its speed to arrive at 100 mile mark after 2 hours of the journey, and then adjust again to finish the rest of the distance in 2 hours.
Answer: E Hi Bunuel, I pick C as the answer. Combining two statements, I find that • Total travel time 4 hours • Bus travelled 2 hours to cover 100 mils. Hence, average speed is 100/2 = 50MPH • Average speed in the next 2 (Total travel time 4 hours and travel time for the first half is 2 hours) hours = (50/2) = 25MPH • Bus travelled next 2 hours at a speed of 25MPH. Hence, the bus covered = (2*25) = 50 miles in next 2 hours. • Therefore, bus travelled (100+(2*25)) = 150 miles in 4 hours. So, average speed for the entire trip is (150/4) = 37.5MPH. Please let me where I miss the trick.



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Re: M1608 [#permalink]
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28 Dec 2016, 06:46
jahidhassan wrote: Bunuel wrote: Official Solution:
Statement (1) by itself is insufficient. The bus could have gone slowly or fast during the last 2 hours of the journey. Statement (2) by itself is insufficient. We know nothing about the value of the bus's speed for either first or second half. Statements (1) and (2) combined are insufficient. S2 tells us that the bus spent \(\frac{4}{3}\) hours on the first half of the distance and \(\frac{8}{3}\) hours on the second half. To answer the question, we have to know the distance between \(A\) and \(B\). Whatever the distance between \(A\) and \(B\)(as long as it's not too big or too small), the bus could spend \(\frac{4}{3}\) hours on the first half, then adjust its speed to arrive at 100 mile mark after 2 hours of the journey, and then adjust again to finish the rest of the distance in 2 hours.
Answer: E Hi Bunuel, I pick C as the answer. Combining two statements, I find that • Total travel time 4 hours • Bus travelled 2 hours to cover 100 mils. Hence, average speed is 100/2 = 50MPH • Average speed in the next 2 (Total travel time 4 hours and travel time for the first half is 2 hours) hours = (50/2) = 25MPH • Bus travelled next 2 hours at a speed of 25MPH. Hence, the bus covered = (2*25) = 50 miles in next 2 hours. • Therefore, bus travelled (100+(2*25)) = 150 miles in 4 hours. So, average speed for the entire trip is (150/4) = 37.5MPH. Please let me where I miss the trick. I think your doubt is addressed here: ifittookabus4hourstogetfromtownatotownbwhat102361.htmlHope it helps.
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Re: M1608 [#permalink]
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29 Dec 2016, 14:18
Statements (1) and (2) combined are insufficient. S2 tells us that the bus spent \(\frac{4}{3}\) hours on the first half of the distance and \(\frac{8}{3}\) hours on the second half.
Answer: E[/quote]
How did you arrive to number presented above? it is unclear.



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Re: M1608 [#permalink]
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02 Feb 2017, 17:50
Subramonian wrote: jahidhassanYour point is not very clear To me, the answer is C. Average speed of the entire trip can be found using two statements. Combining two statements, I find that • Total travel time is 4 hours • Bus traveled first 2 hours to cover 100 mils. Hence, average speed is 100/2 = 50MPH • Average speed in the next 2 hours (Total travel time is 4 hours and travel time for the first half is 2 hours) = (50/2) = 25MPH • Bus traveled next 2 hours at a speed of 25MPH. Hence, the bus covered = (2*25) = 50 miles in next 2 hours. • Therefore, bus traveled (100+(2*25)) = 150 miles in 4 hours. So, average speed for the entire trip is (150/4) = 37.5MPH. I couldn't understand how the OA is E. Thanks.



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My 2 cents, Wow these gmatclub test never disappoint in making me realize how little i know and i NEED to be more thorough with my approach!!! kudos to question...
A2ssB d1Od2
Avg Speed of entire trip = D/T = d1+d2/(t1 + t2) = d1+d2/( d1/s1 + d2/s2 )
(1) In the first 2 hours the bus covered 100 miles. some portion between AB is 100 miles. D > 100 but we don't know what exact value of D is. NOT SUFFICITENT
(2) The average speed of the bus for the first half of the distance was twice its average speed for the second half. Avg s1 = d1/t1 and Avg s2 = d2/t2 d1/t1 = 2*(d2/t2); NOTE we don't have a relationship between d1/s1 and d2/s2 BUT between d1/t1 and d2/t2. Here we don't know what are the actual values of s1,d1,t1 and s2,d2,t2. So we can't say: s1/s2 = 2 = d1/d2 = t2/t1 we can only get s1 = 2*s2 => d1/t1 = 2*(d2/t2)  NOT SUFFICITENT
(1 + 2): D > 100 miles D = 100+x and Avg speed = (100+x)/(d1/s1 + d2/s2). Thus E



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Question: mikemcgarry, Bunuel, Engr2012, VeritasPrepKarishma, Abhishek009, msk0657, abhimahna, Skywalker18, VyshakHow can anyone break 4 into 4/3 and 8/3 as per 2nd statement? What i want to ask is that what is the though process behind it because i can understand what you have done but i don't know what to do so that i can myself learn to break a integer in fraction as that in other cases. So if possible PLEASE elaborate. Thanks



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Re: M1608 [#permalink]
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31 Mar 2017, 08:14
is there any other short cut to this problem ?



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manishtank1988 wrote: How can anyone break 4 into 4/3 and 8/3 as per 2nd statement? What i want to ask is that what is the though process behind it because i can understand what you have done but i don't know what to do so that i can myself learn to break a integer in fraction as that in other cases. So if possible PLEASE elaborate. Thanks Hi, Consider this as follows: Statement 2 says the average speed in the 1st half(Say S1) was twice the average speed in the 2nd half(say S2) Or S1 = 2(S2) Now, Time taken for 1st half = Distance(d)/Speed(S1) = d/S1 = d/2(S2)  (1) Time taken for 2nd half = Distance(d)/Speed(S2) = d/S2  (2) Total Time Taken : (1) + (2) = d/S2 (3/2) (I hope you are clear on this.) Total Distance = 2d Average Speed of Entire Journey = Total Distance/Total Time Taken = 2d/(d/S2) (3/2) = 4/3S2. Since we have an unknown variable S2 in the equation, we can say B is not sufficient. We need the value of S2, which cannot be derived using A. Hence Answer is E. We have got 4/3 as per above solution and as we know that Speed is inversely proportional to time. So, for a given distance, when speed is doubled, it means time would be half and vice versa. Let me know in case of any confusion.
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Re: M1608 [#permalink]
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31 Mar 2017, 15:02
abhimahna wrote: manishtank1988 wrote: How can anyone break 4 into 4/3 and 8/3 as per 2nd statement? What i want to ask is that what is the though process behind it because i can understand what you have done but i don't know what to do so that i can myself learn to break a integer in fraction as that in other cases. So if possible PLEASE elaborate. Thanks Hi, Consider this as follows: Statement 2 says the average speed in the 1st half(Say S1) was twice the average speed in the 2nd half(say S2) Or S1 = 2(S2) Now, Time taken for 1st half = Distance(d)/Speed(S1) = d/S1 = d/2(S2)  (1) Time taken for 2nd half = Distance(d)/Speed(S2) = d/S2  (2) Total Time Taken : (1) + (2) = d/S2 (3/2) (I hope you are clear on this.) Total Distance = 2d Average Speed of Entire Journey = Total Distance/Total Time Taken = 2d/(d/S2) (3/2) = 4/3S2. Since we have an unknown variable S2 in the equation, we can B is not sufficient. We need the value of S2, which cannot be derived using A even. Hence Answer is E. We have got 4/3 as per above solution and as we know that Speed is inversely proportional to time. So, for a given distance, when speed is doubled, it means time would be half and vice versa. Let me know in case of any confusion. Absolutely clear, thanks a lot abhimahna



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Re: M1608 [#permalink]
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06 Apr 2017, 09:32
Guys this questions is ridiculously easy.
100 miles does NOT necessarily equal half the distance of the trip. You have no idea what the total distance is. 100 miles is an arbitrary number.
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