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# Bill and Sally see each other across a field. They are 500 feet apart.

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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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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.
[Reveal] Spoiler: OA

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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
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 kunalsinghNS

As a first step you need to analyse the stem and re-write 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. 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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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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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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 (500-x).

Bill|>-----------------------|X|--------------------------------------------<|Sally

I) According to 1:
| Speed | Distance | Time
Sally | S | X | T1
Bill |1.5 S | 500-X | T2

As T1= T2
Dist/Speed of Sally = dist/speed of Bill.
X/S = 500-X/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] 09 Oct 2017, 00:50
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