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nk_81: Junior | Next Rank: 30 Posts; Posts: 24; Joined: Thu Dec 23, 2010 11:13 pm; Location: India; GMAT Score:640

Rate Problem. DS question.

by nk_81 » Thu Dec 30, 2010 2:23 am

Bill and Sally see each other across the field, They are 500 feet apart. If at the same instant of time they start running towards each other in a direct line, how far will Bill have travelled when they meet?

1) Bill ran at an average speed 50% greater than Sally's average speed

2) Bill ran at an average speed 4 feet per second faster than Sally's average speed.

I have not understood the solution. Can some one please explain?

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Source: Beat The GMAT — Problem Solving | Thu Dec 30, 2010 2:23 am

GMAT/MBA Expert

Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pm; Location: Milpitas, CA; Thanked: 1854 times; Followed by:523 members; GMAT Score:770

by Anurag@Gurome » Thu Dec 30, 2010 2:33 am

nk_81 wrote:Bill and Sally see each other across the field, They are 500 feet apart. If at the same instant of time they start running towards each other in a direct line, how far will Bill have travelled when they meet?

1) Bill ran at an average speed 50% greater than Sally's average speed
2) Bill ran at an average speed 4 feet per second faster than Sally's average speed.

Statement 1: Bill's speed = 50% greater than Sally's speed
Say Sally's speed = x feet/second
=> Bill's speed = Sally's speed + 50% of Sally's speed = 3x/2 feet/second

As they are traveling towards each other, in each second they reduces the distance between them by (x + 3x/2) feet = 5x/2 feet

Therefore they will cover the 500 feet distance in 500/(5x/2) second = 200/x second. In that time bill will travel a distance of (3x/2)*(200/x) = 300 feet

Sufficient

Statement 2: Bill's speed = Sally's speed + 4 feet/second
Say Sally's speed = x feet/second
=> Bill's speed = Sally's speed + 4 feet/second = (x + 4) feet/second

As they are traveling towards each other, in each second they reduces the distance between them by (x + x + 4) feet = (2x + 4) feet

Therefore they will cover the 500 feet distance in 500/(2x + 4) second = 250/(x + 2) second. In that time bill will travel a distance of (x + 4)*(250/(x + 2)) --> This quantity depends upon x.

Not sufficient

The correct answer is A.

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Geva@EconomistGMAT: GMAT Instructor; Posts: 905; Joined: Sun Sep 12, 2010 1:38 am; Thanked: 378 times; Followed by:123 members; GMAT Score:760

by Geva@EconomistGMAT » Thu Dec 30, 2010 2:35 am

nk_81 wrote:Bill and Sally see each other across the field, They are 500 feet apart. If at the same instant of time they start running towards each other in a direct line, how far will Bill have travelled when they meet?

1) Bill ran at an average speed 50% greater than Sally's average speed

2) Bill ran at an average speed 4 feet per second faster than Sally's average speed.

I have not understood the solution. Can some one please explain?

Stat. (1) is sufficient because of the direct correlation between speed and distance: If you speed increases times 3/2, the distance you cover (in the same amount of time) increases by the same times 3/2.

Thus, whether Bill and Sally's speeds are 30 ft/sec and 20 ft/sec respectively, or 300 ft/sec and 200 ft/sec, Bill will always cover 3/2 times the distance that Sally covers in the time it takes them to run: the 500 feet are divided at a ratio of 3:2, or 300 feet Bill : 200 feet Sally, just because you know the ratio of the speeds.

Stat. (2) is insufficient because the same principle does not apply to addition/subtraction. +4 fee/sec speed can make different ratios, depending on the initial speed: If Sally runs at 1 ft/sc, then Bill runs at 5 ft/sec, and the speed ratio is 5:1; If Sally runs at 20 ft/sec then bill runs at 24 ft/sec, and the speed ratio is a different 24:20 = 6:5. We've established that the speed ratio will determine the distance ratio, but stat. (2) does not lock the speed ratio, so the distance ratio cannot be determined.

Geva
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nk_81: Junior | Next Rank: 30 Posts; Posts: 24; Joined: Thu Dec 23, 2010 11:13 pm; Location: India; GMAT Score:640

by nk_81 » Thu Dec 30, 2010 3:17 am

Dear Anurag, Dear Geva,

Thank You.... I just wanted to know, was this a 600 level or a 700 level question?

What is perplexing me during my prep is that I am able to solve some 700 level questions mentally even though some questions, like this one, I find it hard to comprehend.

Can you throw some light on that? Is that normal?

Regards
Naveen

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Geva@EconomistGMAT: GMAT Instructor; Posts: 905; Joined: Sun Sep 12, 2010 1:38 am; Thanked: 378 times; Followed by:123 members; GMAT Score:760

by Geva@EconomistGMAT » Thu Dec 30, 2010 3:46 am

nk_81 wrote:Dear Anurag, Dear Geva,

Thank You.... I just wanted to know, was this a 600 level or a 700 level question?

What is perplexing me during my prep is that I am able to solve some 700 level questions mentally even though some questions, like this one, I find it hard to comprehend.

Can you throw some light on that? Is that normal?

Regards
Naveen

This question is indeed advanced, as it tests an obscure math concept (direct relationship between speed and distance) in a tricky way (reflected in the direct relationship between speed ratio and distance ratio of the two runners), and offers an obvious trap answer choice of C.

I would hesitate to say whether it's a 600 or a 700 - such decrees are pure speculation without data. Your score is determined by the percentile of your performance in the population of test takers, calculated by comparing your performance on the test with the performance of test-takers 5 years back. to know whether this is a 600 or a 700, or an 800 with any degree of accuracy, I'll need to present it to the same test taker population and see how many get it wrong, and what is THEIR score compared to the population.

This is also the reason why I wouldn't be concerned with your seemingly erratic performance. There's just no real way to determine that what you consider a 700 question is really a 700 question, so the whole discussion is sort of pointless. The difference between questions may also be in areas of strengths and weaknesses: you may be very good with the tough algebra concepts, but weaker in these abstract word problems where the concept tested is hidden. Without additional information, it is very difficult to make any sort of judgement, but I wouldn't worry about it too much either. At the end of the day, your score is not based on whether you manage to solve a particularly tough question or not, but on your overall performance in either section.

Geva
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HPengineer: Master | Next Rank: 500 Posts; Posts: 165; Joined: Wed Mar 24, 2010 8:02 am; Thanked: 2 times; Followed by:1 members

by HPengineer » Fri Dec 31, 2010 10:08 am

@ experts

In a similar problem if they gave us the ratio of say time instead of speed in the first statement would we be able to solve?

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Geva@EconomistGMAT: GMAT Instructor; Posts: 905; Joined: Sun Sep 12, 2010 1:38 am; Thanked: 378 times; Followed by:123 members; GMAT Score:760

by Geva@EconomistGMAT » Sat Jan 01, 2011 11:01 pm

HPengineer wrote:@ experts

In a similar problem if they gave us the ratio of say time instead of speed in the first statement would we be able to solve?

Indeed, as time also has a direct correlation with work: If you run for twice the time (at the same constant rate), you cover double the distance. In the problem above, the time is fixed (they both run the same amount of time until they meet), so the speed ratio determines the distance ratio. If the problerm were constructed so that Bill and Sandy somehow run at the same rate, only Bill runs for 50% longer (for example, they both run at the same rate, but Bill begins earlier and runs for 50% seconds more then Sandy), the the distance ratio would divide by the same ratio.

The 2nd statement would also be insufficient, for the same reason: If Bill were to run 4 seconds more than sandy, we would be unable to discern the time ratio (and from that the distance ratio) because we don't know how long they run together: we'd be unable to determine whether those 4 additional seconds compose 50%, 100%, or any other percent of the time they run together, unless they actually tell us the speed they run at.

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HPengineer: Master | Next Rank: 500 Posts; Posts: 165; Joined: Wed Mar 24, 2010 8:02 am; Thanked: 2 times; Followed by:1 members

by HPengineer » Sat Jan 01, 2011 11:08 pm

Do you got another beer problem u can create for me to try regarding this concept?

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Geva@EconomistGMAT: GMAT Instructor; Posts: 905; Joined: Sun Sep 12, 2010 1:38 am; Thanked: 378 times; Followed by:123 members; GMAT Score:760

by Geva@EconomistGMAT » Sat Jan 01, 2011 11:42 pm

HPengineer wrote:Do you got another beer problem u can create for me to try regarding this concept?

Sure

This is made up on the spot, so it's probably got ambigous holes in it. But here goes:

Geva sits at the pub and drink beers at a constant rate of x beers and hour. A while later, he is joined by HPengineer who matches him pint for pint (i.e. orders a beer only whenever Geva orders one). If a Beer costs $5 each, and the beer tab when they both pass out simultaneously ends up at $150, how much did Geva pay, assuming they both bought nothing but beers?

(1) If their pub stay is measured from their first respective beer to the time they pass out, Geva's pub stay is 50% longer than HPengineer's.

(2) Geva spent 40 minutes drinking beer before he was joined by HP engineer.

Geva
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HPengineer: Master | Next Rank: 500 Posts; Posts: 165; Joined: Wed Mar 24, 2010 8:02 am; Thanked: 2 times; Followed by:1 members

by HPengineer » Sun Jan 02, 2011 12:18 am

im stumped...

sadly best i can come up with is a total time which might not even be right

geva = 3/2T

hpengineer = T

total time at pub = 5T/2

really stuck as what to do now... X is our respective rates so could we calculate total time multiplied by rate to = 150 or would we take total time and multiply by 5 dollars to get 150... im not really sure where to proceed..

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Geva@EconomistGMAT: GMAT Instructor; Posts: 905; Joined: Sun Sep 12, 2010 1:38 am; Thanked: 378 times; Followed by:123 members; GMAT Score:760

by Geva@EconomistGMAT » Sun Jan 02, 2011 12:24 am

HPengineer wrote:im stumped...

sadly best i can come up with is a total time which might not even be right

geva = 3/2T

hpengineer = T

total time at pub = 5T/2

really stuck as what to do now... X is our respective rates so could we calculate total time multiplied by rate to = 150 or would we take total time and multiply by 5 dollars to get 150... im not really sure where to proceed..

I'll hold off on the explanation to let other people take a whack at it, if you don't mind. ping me at the end of the day if you don't see an explanation you like, or can't figure it out for yourself?

PS take another look at the concept we'd discussed above.

Geva
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nk_81: Junior | Next Rank: 30 Posts; Posts: 24; Joined: Thu Dec 23, 2010 11:13 pm; Location: India; GMAT Score:640

by nk_81 » Sun Jan 02, 2011 1:48 am

Geva@MasterGMAT wrote:
HPengineer wrote:Do you got another beer problem u can create for me to try regarding this concept?
Sure

This is made up on the spot, so it's probably got ambigous holes in it. But here goes:

Geva sits at the pub and drink beers at a constant rate of x beers and hour. A while later, he is joined by HPengineer who matches him pint for pint (i.e. orders a beer only whenever Geva orders one). If a Beer costs $5 each, and the beer tab when they both pass out simultaneously ends up at $150, how much did Geva pay, assuming they both bought nothing but beers?

(1) If their pub stay is measured from their first respective beer to the time they pass out, Geva's pub stay is 50% longer than HPengineer's.

(2) Geva spent 40 minutes drinking beer before he was joined by HP engineer.

Statement 2 - INSUFFICIENT - 40 mins earlier does not tell us how many beers Geva consumed, hence can't say what he paid.

Statement 1 - ?????

Combine 1&2-
Let HPengineer stay be n/60 hrs, then Geva would have stayed for n/40 hrs (3n/2 converted in to hrs)
Hence
n/40 - n/60 = 2/3 ==> 3n-2n/120 = 2/3 ==> 3n= 240 ==> n = 80
Therefore HP Engineer was in the Pub for 80 mins and Geva in the Pub for 120 mins
We also know that 30 cans of beers were consumed (150$/5$)

Now i am confused........!!!!!!! I need a beer :@

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pm; Thanked: 147 times; Followed by:3 members

by anshumishra » Sun Jan 02, 2011 7:12 am

Geva@MasterGMAT wrote:
HPengineer wrote:Do you got another beer problem u can create for me to try regarding this concept?
Sure

This is made up on the spot, so it's probably got ambigous holes in it. But here goes:

Geva sits at the pub and drink beers at a constant rate of x beers and hour. A while later, he is joined by HPengineer who matches him pint for pint (i.e. orders a beer only whenever Geva orders one). If a Beer costs $5 each, and the beer tab when they both pass out simultaneously ends up at $150, how much did Geva pay, assuming they both bought nothing but beers?

(1) If their pub stay is measured from their first respective beer to the time they pass out, Geva's pub stay is 50% longer than HPengineer's.

(2) Geva spent 40 minutes drinking beer before he was joined by HP engineer.

Statement 1 :
Since Geva spent 50% more time and beer consumption rate is constant, Hence Geva will be paying 50% more than HPengineer (90$ Geva, 60$ HPengineer) -> Sufficient

Statement 2 :
We can't determine what % of total drinks Geva has already taken based on the time "40" minutes (Depends how fast Geva drinks

) -> Insufficient.

Hence A.

Thanks
Anshu

(Every mistake is a lesson learned )

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pm; Location: Gandhinagar; Thanked: 41 times; Followed by:2 members

by shashank.ism » Sun Jan 02, 2011 7:29 am

nk_81 wrote:Bill and Sally see each other across the field, They are 500 feet apart. If at the same instant of time they start running towards each other in a direct line, how far will Bill have travelled when they meet?

1) Bill ran at an average speed 50% greater than Sally's average speed

2) Bill ran at an average speed 4 feet per second faster than Sally's average speed.

I have not understood the solution. Can some one please explain?

It seems to be a love game so they should meet soon...well here Bill is moving 50% faster than sally...Its nature's rule Boys have to do this....All is fair in love and war.
Well coming to the question..

Distance to crack for them to meet = 500 feet (seems to be very long distance between two lovers

)
Statement 1: Let avg speed of sally be v feet/s.... so avg speed of bill = 1.5v feet/s.
let they take t time to meet.
so distance covered by bill 1.5vt and sally = vt.
so 1.5vt +vt = 500 => 2.5vt = 500 => vt = 200
so distance covered by bill = 1.5vt = 1.5 x 200 = 300 feet

Sufficient

Statement 2: Let avg speed of sally be v feet/s.... so avg speed of bill = (v + 4) feet/s.
let they take t time to meet.
so distance covered by bill (v+4)t and sally = vt.
(v+4)t + vt = 500 => (2v+4)t = 500

here we can't calculate vt as in single . so we can't calculate the distance travelled by bill.
Insufficient

Hence ans A

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