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M16-08

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M16-08  [#permalink]

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New post 15 Sep 2014, 23:58
1
8
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A
B
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D
E

Difficulty:

  75% (hard)

Question Stats:

44% (01:21) correct 56% (01:20) wrong based on 90 sessions

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If it took a bus 4 hours to get from town \(A\) to town \(B\), what was the average speed of the bus for the trip?


(1) In the first 2 hours the bus covered 100 miles.

(2) The average speed of the bus for the first half of the distance was twice its average speed for the second half.

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Re M16-08  [#permalink]

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New post 15 Sep 2014, 23:58
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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M16-08  [#permalink]

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New post 31 Mar 2017, 06:59
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...

A|-----2s-----|-----s-----|B
--------d1----O----d2

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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New post 31 Mar 2017, 07:14
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
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M16-08  [#permalink]

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New post 31 Mar 2017, 07:17
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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: M16-08  [#permalink]

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New post 31 Mar 2017, 14:02
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: M16-08  [#permalink]

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New post 06 Apr 2017, 08:32
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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Re: M16-08  [#permalink]

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New post 02 Jul 2018, 21:40
Hi Bunuel

Can you please expand on the official solution? The OA skips a couple of steps and so it's not very clear.
There is no alternate explanation available by other expert; nor do the other posts elaborate on the OA; the other posts end up in questions or address some parts of the solution.

Specifically, why is statement 2 not sufficient? I believe some parts of the algebraic explanation of statement 2 is used again when both statements are combined together resulting in those 4/3, 8/3 which also I don't understand. If you could expand on the algebra, it would be really helpful.

Thanks.
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Re: M16-08  [#permalink]

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New post 02 Jul 2018, 22:05
sanjay1810 wrote:
Hi Bunuel

Can you please expand on the official solution? The OA skips a couple of steps and so it's not very clear.
There is no alternate explanation available by other expert; nor do the other posts elaborate on the OA; the other posts end up in questions or address some parts of the solution.

Specifically, why is statement 2 not sufficient? I believe some parts of the algebraic explanation of statement 2 is used again when both statements are combined together resulting in those 4/3, 8/3 which also I don't understand. If you could expand on the algebra, it would be really helpful.

Thanks.


I think THIS POST addresses exactly your doubt.

Hope it helps.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M16-08  [#permalink]

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New post 06 Oct 2018, 03:18
As per option -2
Let, total distance = 2D
Speed for 1st half distance = 2S and speed for second half distance = S;
Time for 1st half = D/2S = x hr (let x = D/2S)
So time for 2nd half distance = D/S = 2x hr
Total time = 3x = 4 hr
or x= 4/3 hrs
As per option 1 and 2 together
4/3*2S + (2-4/3*S) = 100;
S = 30
So, Total distance = 4/3*60 + (4-3/4)*30
Total time = 4hrs
Average speed can be determined.
Answer choice C is the correct one.
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Re: M16-08 &nbs [#permalink] 06 Oct 2018, 03:18
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