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Bill and Sally see each other across a field. They are 500 feet apart. [#permalink]
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05 Oct 2017, 04:36
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Re: Bill and Sally see each other across a field. They are 500 feet apart. [#permalink]
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05 Oct 2017, 05:17
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Distance to be covered is 500 feet to find: distance travelled by bill when bill meet sally to find this, we need to know the speeds of both bill and sally
statement1: Bill's average speed is 1.5times (or 50% higher than) the average speed of sally there can be many combinations of speeds of bill and sally such as (75,50), (150,100), (300,200) ...etc. but because both the speeds are relative to each other i.e. x and 1.5x, the individual distance travelled by bill and sally will remain fixed, Bill will cover 300 feet and sally will cover 200 feet Sufficient Statement2: bill ran at an average speed of 4feet faster than sally's average speed let sally's average speed be "a" then bill's average speed is "a+4" as we have no other information given, we are left with constant "a", and hence distance covered by bill will be calculated in terms of constant "a" and we will not get a value Hence insufficient Option A



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Re: Bill and Sally see each other across a field. They are 500 feet apart. [#permalink]
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06 Oct 2017, 06:02
niks18 can you explain how (A) is sufficient ?



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Bill and Sally see each other across a field. They are 500 feet apart. [#permalink]
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06 Oct 2017, 06:34
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kunalsinghNS wrote: niks18 can you explain how (A) is sufficient ? Hi kunalsinghNSAs a first step you need to analyse the stem and rewrite it in the form of an equation. Given the distance between B & S\(=500\). Let speed of B be \(x\) and that of S be \(y\) As they are in opposite direction and move towards each other, their Relative speed will be \(= x+y\) When they meet together they would have covered the entire \(500\) ft. So time taken to meet each other \(t= \frac{500}{(x+y)}\) The question is asking the distance covered by B when they meet \(= x*t\). substitute the value of \(t\) here to get \(x*\frac{500}{(x+y)}\) Now to know the value we need to know the relation between \(x\) & \(y\) and nothing else. Note that till this process I have not looked at the statements. Statement 1: Gives me a relation between \(x\) & \(y\) i.e. their speeds, hence Sufficient. No need to calculate the value as it is a DS question. Just for clarity, given that \(x=1.5y\), substitute this in equation \(x*\frac{500}{(x+y)} = 300\). Statement 2: this implies \(x=4+y\), here we have two variable and one equation, hence cannot be solved. InsufficientOption A



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Re: Bill and Sally see each other across a field. They are 500 feet apart. [#permalink]
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07 Oct 2017, 00:35
niks18 thank you for the explanation !



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Bill and Sally see each other across a field. They are 500 feet apart. [#permalink]
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07 Oct 2017, 00:48
I got this wrong at first attempt, but then i gave it a second thought w/o seeing solutions. Chose a no. which satisfies a+b = 500 and a=1.5 b. Took it like, distance traveled by Bill will be 1.5 times of Sally (as speed of Bill is 1.5 times that of Sally > 1.5 times Distance covered in same time as of Sally). So, got the ans as combination of 300, 200 very quick, which satisfies the above conditions.
Thanks.



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Re: Bill and Sally see each other across a field. They are 500 feet apart. [#permalink]
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07 Oct 2017, 07:04
Bunuel wrote: Bill and Sally see each other across a field. They are 500 feet apart. If at the same instant of time they start running toward each other in a direct line how far will Bill have traveled when they meet?
(1) Bill ran at an average speed that was 50% greater than Sally's average speed.
(2) Bill ran at an average speed 4 feet per second faster than Sally's average speed. Hi.. Distance is given, so what should tell us the distance at which they meet  speed individually or relative speed..lets see the distance: (1) Bill ran at an average speed that was 50% greater than Sally's average speed. relative speed is given so should be sufficient..but let's solve it as in PS.. speed of B = 1.5S you can use weighted average method = \(\frac{1.5S}{S+1.5S}*500=\frac{1.5}{2.5}*500=\frac{3}{5}*500=300\) Sufficient (2) Bill ran at an average speed 4 feet per second faster than Sally's average speed. answer will depend on the speeds insuff A
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Bill and Sally see each other across a field. They are 500 feet apart. [#permalink]
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09 Oct 2017, 00:50
As both are moving towards each other, the time taken by both of them will be same. So lets assume the total distance traveled by Bill is x. So distance by Bill will be (500x). Bill>X<Sally I) According to 1:  Speed  Distance  Time Sally  S  X  T1 Bill 1.5 S  500X  T2 As T1= T2 Dist/Speed of Sally = dist/speed of Bill. X/S = 500X/1.5S S gets cancel out. We have X= 200. Hence sufficient. 2) Not sufficient as ratio of relative speed is not known.
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Bill and Sally see each other across a field. They are 500 feet apart. [#permalink]
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21 Feb 2018, 21:38
Hi chetan2u Bunuel niks18 I have one question. The question doesn't say they ran at constant speed throughout. Maybe Sally ran fast till they met and slowed down after meet, but Bill ran slowly till they met and picked up his speed after they met and made his average speed greater than Sally's. Similar question, I was tricked into for assuming constant speed. I will share the link of that question in the next post. If we assume constant speed, answer is definite A Please help. I am confused Posted from my mobile device



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Re: Bill and Sally see each other across a field. They are 500 feet apart. [#permalink]
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21 Feb 2018, 21:50
hellosanthosh2k2 wrote: Hi chetan2u Bunuel niks18 I have one question. The question doesn't say they ran at constant speed throughout. Maybe Sally ran fast till they met and slowed down after meet, but Bill ran slowly till they met and picked up his speed after they met and made his average speed greater than Sally's. Similar question, I was tricked into for assuming constant speed. I will share the link of that question in the next post. If we assume constant speed, answer is definite A Please help. I am confused Posted from my mobile deviceHi hellosanthosh2k2the statements mention Average speed. so even if their speeds varied during the distance covered, we have a relation between their average speeds, which takes care of any variability.



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Bill and Sally see each other across a field. They are 500 feet apart. [#permalink]
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21 Feb 2018, 22:18
Hi niks18, Thanks for reply. https://gmatclub.com/forum/stationsxa ... 00088.htmlThis is the question, I was mentioning. In the link question, statement 2 is not sufficient, can we apply the same logic for this question. Posted from my mobile device



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Re: Bill and Sally see each other across a field. They are 500 feet apart. [#permalink]
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21 Feb 2018, 23:18
hellosanthosh2k2 wrote: Hi niks18, Thanks for reply. https://gmatclub.com/forum/stationsxa ... 00088.htmlThis is the question, I was mentioning. In the link question, statement 2 is not sufficient, can we apply the same logic for this question. Posted from my mobile deviceHi hellosanthosh2k2There is a basic difference between two questions. In this question, we can use average speed because when the two meet then relatively they have traversed the entire distance. and we only need to know the distance covered till that point. In the other question, trains meet and then the trains continue to run till they reach their destination. So you can use average speed concept only till the point they meet, post that to know the time taken to reach their destination you will have to know their individual speeds for the remaining distance. the average speed of both train P & Q is not mentioned in statement 2 when they meet. let us assume the average speed of train P was a and that of train Q was y when they crossed each other, so you will have 2a+2b=250 =>a+b=125. Now we don't know what is a or b here. So the statement 2 is insufficient



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Bill and Sally see each other across a field. They are 500 feet apart. [#permalink]
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22 Feb 2018, 03:02
niks18 wrote: hellosanthosh2k2 wrote: Hi niks18, Thanks for reply. https://gmatclub.com/forum/stationsxa ... 00088.htmlThis is the question, I was mentioning. In the link question, statement 2 is not sufficient, can we apply the same logic for this question. Posted from my mobile deviceHi hellosanthosh2k2There is a basic difference between two questions. In this question, we can use average speed because when the two meet then relatively they have traversed the entire distance. and we only need to know the distance covered till that point. In the other question, trains meet and then the trains continue to run till they reach their destination. So you can use average speed concept only till the point they meet, post that to know the time taken to reach their destination you will have to know their individual speeds for the remaining distance. the average speed of both train P & Q is not mentioned in statement 2 when they meet. let us assume the average speed of train P was a and that of train Q was y when they crossed each other, so you will have 2a+2b=250 =>a+b=125. Now we don't know what is a or b here. So the statement 2 is insufficient Hi niks18I was trying to prove Statement 1 as not sufficient with two cases Say average speed of Bill = 300 ft/hr (1.5 times Sally's as mentioned in statement 1), Sally = 200 ft/hr Case 1: Until the point they meet, Let average speed of Bill = 100ft/hr, Sally = 400 ft/hr so time to meet = 500/(100+400) = 1 hr so they meet at 100ft distance from where Bill start. After meet, say average speed of Bill is B for remaining distance of 400 ft, then 500/(1 + (400/B)) = 300 (his overall average speed), solving we get B = 600ft/hr, so after meet, Bill speeds up to 600 ft/hr Similarly after meet, say average speed of Sally S for remaining distance of 100ft, then 500/(1 + 100/S) = 200 (her overall average speed), solving we get S = 100/1.5 ft/hr, So after meet, Sally slows down to 100/1.5 ft/hr so in this case, Bill average speed = 300 ft/hr and Sally average speed = 200 ft/hr, but they meet at 100 ft from where Bill started. Case 2: Bill and Sally travel at same constant speed of 300 ft/hr and 200 ft/hr respectively, in this case, they meet at 300 ft from where Bill started. So two cases, different point of meet. I think unless and until the statement says, Bill and Sally ran at constant speed, statement 1 won't be sufficient. Am i missing anything? Thanks



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Re: Bill and Sally see each other across a field. They are 500 feet apart. [#permalink]
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22 Feb 2018, 08:25
hellosanthosh2k2 wrote: niks18 wrote: hellosanthosh2k2 wrote: Hi niks18, Thanks for reply. https://gmatclub.com/forum/stationsxa ... 00088.htmlThis is the question, I was mentioning. In the link question, statement 2 is not sufficient, can we apply the same logic for this question. Posted from my mobile deviceHi hellosanthosh2k2There is a basic difference between two questions. In this question, we can use average speed because when the two meet then relatively they have traversed the entire distance. and we only need to know the distance covered till that point. In the other question, trains meet and then the trains continue to run till they reach their destination. So you can use average speed concept only till the point they meet, post that to know the time taken to reach their destination you will have to know their individual speeds for the remaining distance. the average speed of both train P & Q is not mentioned in statement 2 when they meet. let us assume the average speed of train P was a and that of train Q was y when they crossed each other, so you will have 2a+2b=250 =>a+b=125. Now we don't know what is a or b here. So the statement 2 is insufficient Hi niks18I was trying to prove Statement 1 as not sufficient with two cases Say average speed of Bill = 300 ft/hr (1.5 times Sally's as mentioned in statement 1), Sally = 200 ft/hrCase 1: Until the point they meet, Let average speed of Bill = 100ft/hr, Sally = 400 ft/hrso time to meet = 500/(100+400) = 1 hr so they meet at 100ft distance from where Bill start. After meet, say average speed of Bill is B for remaining distance of 400 ft, then 500/(1 + (400/B)) = 300 (his overall average speed), solving we get B = 600ft/hr, so after meet, Bill speeds up to 600 ft/hr Similarly after meet, say average speed of Sally S for remaining distance of 100ft, then 500/(1 + 100/S) = 200 (her overall average speed), solving we get S = 100/1.5 ft/hr, So after meet, Sally slows down to 100/1.5 ft/hr so in this case, Bill average speed = 300 ft/hr and Sally average speed = 200 ft/hr, but they meet at 100 ft from where Bill started. Case 2: Bill and Sally travel at same constant speed of 300 ft/hr and 200 ft/hr respectively, in this case, they meet at 300 ft from where Bill started. So two cases, different point of meet. I think unless and until the statement says, Bill and Sally ran at constant speed, statement 1 won't be sufficient. Am i missing anything? Thanks Hi hellosanthosh2k2For this question which is Case 1, you have considered one average speed and then again considered a different average speed. Representing the question again here for clarity  Quote: Bill and Sally see each other across a field. They are 500 feet apart. If at the same instant of time they start running toward each other in a direct line how far will Bill have traveled when they meet?
(1) Bill ran at an average speed that was 50% greater than Sally's average speed.
(2) Bill ran at an average speed 4 feet per second faster than Sally's average speed. here when Bill & Sally meet, then their average speed hold the relation of 1.5:1. This average is in totality till they meet. So let the two meet at point P Bill______________P(Bill meets Sally)___________Sally. Bill's average speed till P is x & Sally's average speed till P is y, so we have x=1.5y. Within this average speed we are not going to again assume some different average speed. And for the second question Quote: Stations X and Y are connected by two separate, straight, parallel rail lines that are 250 miles long. Train P and train Q simultaneously left Station X and Station Y, respectively, and each train traveled to the other’s point of departure. The two trains passed each other after traveling for 2 hours. When the two trains passed, which train was nearer to its destination?
(1) At the time when the two trains passed, train P had averaged a speed of 70 miles per hour. (2) Train Q averaged a speed of 55 miles per hour for the entire trip. We know XY=250 X_____________Y. Let the trains meet at point A i.e X___________A_______Y Average speed of train P till point A=70 and time taken 2 hours. So distance XA=70*2=140 we know XY=250, so AY=XYXA=250140=110 so train Q traveled 110 Km. Now train P has to reach Y which and train Q has to reach X, so you can see which train is closer to its destination AT THE TIME WHEN THEY PASSED each other. So in both these cases Statement 1 is sufficient.



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Bill and Sally see each other across a field. They are 500 feet apart. [#permalink]
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22 Feb 2018, 09:24
Hi niks18 Thanks again for reply. you mentioned as "when Bill & Sally meet, then their average speed hold the relation of 1.5:1" this is exactly my point of confusion is. I think Bill's overall average speed and Sally's overall average speed hold 1.5:1, but the average speeds till the point of meet need not hold 1.5:1. is this correct?. So i gave an example different speeds of Bill and Sally in case 1 of my example, but overall average speed will hold 1.5:1. Because unlike in statement 1 of link question, it is explicitly mentioned as at point of meet, Train X average speed as 70 m/hr making it sufficient Thanks



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Bill and Sally see each other across a field. They are 500 feet apart. [#permalink]
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22 Feb 2018, 09:33
hellosanthosh2k2 wrote: Hi niks18 Thanks again for reply. you mentioned as "when Bill & Sally meet, then their average speed hold the relation of 1.5:1" this is exactly my point of confusion is. I think Bill's overall average speed and Sally's overall average speed hold 1.5:1, but the average speeds till the point of meet need not hold 1.5:1. is this correct?. So i gave an example different speeds of Bill and Sally in case 1 of my example, but overall average speed will hold 1.5:1. Because unlike in statement 1 of link question, it is explicitly mentioned as at point of meet, Train X average speed as 70 m/hr making it sufficient Thanks Hi hellosanthosh2k2The question stem ends at a point when they meet and does not provide any information that they continue to run thereafter. In all possibility they met and then stopped there, so we will have to assume here that whatever information is provided is till the time they meet. The question does not provide any information regarding any other kind of journey. So the average speed relation given has to be till the point they meet.



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Re: Bill and Sally see each other across a field. They are 500 feet apart. [#permalink]
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22 Feb 2018, 09:39
niks18 wrote: hellosanthosh2k2 wrote: Hi niks18 Thanks again for reply. you mentioned as "when Bill & Sally meet, then their average speed hold the relation of 1.5:1" this is exactly my point of confusion is. I think Bill's overall average speed and Sally's overall average speed hold 1.5:1, but the average speeds till the point of meet need not hold 1.5:1. is this correct?. So i gave an example different speeds of Bill and Sally in case 1 of my example, but overall average speed will hold 1.5:1. Because unlike in statement 1 of link question, it is explicitly mentioned as at point of meet, Train X average speed as 70 m/hr making it sufficient Thanks Hi hellosanthosh2k2The question stem ends at a point when they meet and does not provide any information that they continue to run thereafter. In all possibility they met and then stopped there, so we will have to assume here that whatever information is provided is till the time they meet. The question does not provide any information regarding any other kind of journey. So the average speed relation given has to be till the point they meet. Hi niks18, Thanks i missed that part of the question, they end up meeting and the question didnt provide enough information about what happens after they meet. In which case, Statement 1 is clearly sufficient. Yes, I run into these sort of confusions with the words in the question. Hope i don't run into these confusions on the DDay. Need to keep my eyes wide open Thanks again




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