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# Starting from the same point, a sparrow and a hawk flew in

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Manager
Joined: 05 Mar 2011
Posts: 95
Starting from the same point, a sparrow and a hawk flew in  [#permalink]

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Updated on: 10 Jul 2013, 14:06
1
19
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Difficulty:

65% (hard)

Question Stats:

58% (01:52) correct 42% (02:07) wrong based on 309 sessions

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Starting from the same point, a sparrow and a hawk flew in opposite directions. Each traveled at a constant speed until they were 200 feet apart. How far did the sparrow travel?

(1) The ratio of the sparrow's speed to the hawk's speed was 3 to 2.
(2) The average speed of the sparrow was 5 feet per second faster than the average speed of the hawk.

Originally posted by ashiima on 10 Dec 2011, 17:54.
Last edited by Bunuel on 10 Jul 2013, 14:06, edited 1 time in total.
RENAMED THE TOPIC.
Manager
Joined: 30 Aug 2009
Posts: 222
Location: India
Concentration: General Management

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12 Dec 2011, 08:55
1
ashiima wrote:
Starting from the same point, a sparrow and a hawk flew in opposite directions. Each traveled at a constant speed until they were 200 feet apart. How far did the sparrow travel?

(1) The ratio of the sparrow's speed to the hawk's speed was 3 to 2.
(2) The average speed of the sparrow was 5 feet per second faster than the average speed of the hawk.

time for which sparrow and hawk will fly will be equal , lets say T

stmnt1 : ratio of speed is 3:2 (sparrow:hawk)
let speed be 2S for hawk and sparrow be 3S

now we have total distance = 200 = 3S*T + 2S*T = 5ST

and T = 200/5S = 40/S

Distance traveled by Sparrow is 3S* 40/S = 120 feet.

Hence suff

stmnt 2: let speed of hawk be S and speed of sparrow is S+ 5
Time will remain same

so we get 200 = S* T + (S+5)*T = (2S + 5)T

=> T = 200/(2S + 5)

Speed of sparrow = (S+5) * 200/(2S + 5)
This can give us more than 1 value

e.g S = 5/7.5/10

so insuff

Hence A
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Joined: 25 Oct 2011
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14 Dec 2011, 09:20
Let's test Option A -

Let speed of sparrow = 3 feet/sec
therefore, speed of hawk = 2 feet/sec

Relative speed = 5 feet/sec
Therefore, time taken to travel 200 feet = 40 sec.
So, distance travelled by sparrow = 3*40 = 120 feet -- SUFFICIENT

Lets test Option B -

This introduces 2 variables - time and speed of hawk - hence can't be solved. - INSUFFICIENT

If you look carefully, the ratio in Option A is sufficient to tell us that we can solve the problem, no need to go through with the complete solution.

Manager
Joined: 12 Oct 2011
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14 Dec 2011, 11:33
1
ashiima wrote:
Starting from the same point, a sparrow and a hawk flew in opposite directions. Each traveled at a constant speed until they were 200 feet apart. How far did the sparrow travel?

(1) The ratio of the sparrow's speed to the hawk's speed was 3 to 2.
(2) The average speed of the sparrow was 5 feet per second faster than the average speed of the hawk.

Statement 1:
Let the speed of sparrow be 3.
Let the speed of hawk be 2.
Let the distance covered by sparrow be d.
Hence the distance covered by hawk = 200 - d

Hence the time taken by sparrow = d/3
Hence the time taken by hawk = (200 - d)/2

The times the birds flew in opposite directions is equal.

Hence we can equate the two time equations as follows:
d/3 = (200 - d)/2

This will give you a unique value for d (the distance covered by sparrow. Even if the speeds are assumed to be different (6 & 4), the distance traveled by the sparrow will be the same because the ratio of the speeds is given.

Statement 2:
Let the average speed of hawk be h.
Hence the average speed of sparrow is h + 5.

Using the same logic as above, we can set up an equation for the times taken by each bird.

Time taken by sparrow = Time taken by hawk
i.e. d/(h + 5) = (200 - d)/h

This equation cannot be solved because we have two unknowns and only a single equation.

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14 Dec 2011, 16:49
1
1
ashiima wrote:
Starting from the same point, a sparrow and a hawk flew in opposite directions. Each traveled at a constant speed until they were 200 feet apart. How far did the sparrow travel?

(1) The ratio of the sparrow's speed to the hawk's speed was 3 to 2.
(2) The average speed of the sparrow was 5 feet per second faster than the average speed of the hawk.

There's a general principle you can use here to answer this question without algebra. If, say, two cars travel for the *same* amount of time, and the first car travels twice as fast as the second, then the first car will travel exactly twice as far as the second.

So here, from Statement 1, the two birds travel for the same time, and we know the sparrow travels 50% faster than the hawk. So the sparrow must travel 50% further than the hawk. Since we know they covered 200 feet combined, we can find the distance each traveled, and Statement 1 is sufficient.

For Statement 2, again we do not need any algebra. If you understand why Statement 1 is sufficient, then you'll know that the ratio of the two speeds determines the ratio of the distances traveled. From Statement 2, the speeds of the sparrow and the hawk could be in many different ratios, so Statement 2 cannot be sufficient.

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15 Dec 2011, 03:07
IanStewart wrote:
There's a general principle you can use here to answer this question without algebra. If, say, two cars travel for the *same* amount of time, and the first car travels twice as fast as the second, then the first car will travel exactly twice as far as the second.

So here, from Statement 1, the two birds travel for the same time, and we know the sparrow travels 50% faster than the hawk. So the sparrow must travel 50% further than the hawk. Since we know they covered 200 feet combined, we can find the distance each traveled, and Statement 1 is sufficient.

For Statement 2, again we do not need any algebra. If you understand why Statement 1 is sufficient, then you'll know that the ratio of the two speeds determines the ratio of the distances traveled. From Statement 2, the speeds of the sparrow and the hawk could be in many different ratios, so Statement 2 cannot be sufficient.

Thanks a lot for showing that there is always a better method of getting at the answer. Such approaches are needed to get to the answer quickly on test day. Thanks again.
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Joined: 13 May 2013
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Re: Starting from the same point, a sparrow and a hawk flew in  [#permalink]

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01 Aug 2013, 16:23
1
Starting from the same point, a sparrow and a hawk flew in opposite directions. Each traveled at a constant speed until they were 200 feet apart. How far did the sparrow travel?

(1) The ratio of the sparrow's speed to the hawk's speed was 3 to 2.
The rate of speed of the sparrow to the rate of speed of the hawk was 3:2 so for every three feet the sparrow travels the hawk travels two.
3s+2s=200
5s=200
s=40

sparrow traveled 120 ft, hawk traveled 80 ft.
SUFFICIENT

(2) The average speed of the sparrow was 5 feet per second faster than the average speed of the hawk.
Average speed = distance/time

(y/t) = ([200-y]/t) + 5
Not enough information to solve.
INSUFFICIENT

(A)
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Re: Starting from the same point, a sparrow and a hawk flew in  [#permalink]

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02 Aug 2013, 00:35
Statement 1) Suppose Sparrow travels 3 feet/min and hawk travel 2 feet /min, so in 1 minute they are 5 feet apart, after 40 minutes they must be 200 feet apart, and the sparrow at the rate of 3 feet/min must have traveled (40*3=120 feet):: Sufficient (you can take any value of rates for sparrow and hawk (3,2 or 6,4...you will get the same answer)

Statement 2) From this statement it can be deducted that ratio of their speed is not constant (Sparrow=15, Hawk=10 or Sparrow =105 Hawk=100 ..etc), Not sufficient

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Re: Starting from the same point, a sparrow and a hawk flew in  [#permalink]

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31 Dec 2013, 07:23
1
ashiima wrote:
Starting from the same point, a sparrow and a hawk flew in opposite directions. Each traveled at a constant speed until they were 200 feet apart. How far did the sparrow travel?

(1) The ratio of the sparrow's speed to the hawk's speed was 3 to 2.
(2) The average speed of the sparrow was 5 feet per second faster than the average speed of the hawk.

This is similar to those weighted average problems in that the only thing that we needed was the actual ratio of the weights, anyways let's apply some of these concepts to Distance/Rate problems

So we have that they both traveled the same time and covered a total distance of 200 feet. How much did the sparrow covered?

Statement 1

If the sparrow speed is 3x and the hawk's speed is 2x then total is 5x

We have that they covered 200 feet in 5X (Remember time is same for both so its a constant)

So x=40 and the distance the sparrow covered was of course 120 feet

Statement 2

The difference between their speeds is 5 ft/second

We can't solve with this. We only need the ratio between the speeds not the actual amount

Hence A is the correct answer

Hope it helps!
Cheers!
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Starting from the same point, a sparrow and a hawk flew in  [#permalink]

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12 Mar 2019, 10:37
The two birds flew in opposite directions, so we add their rates*time to find total distance (200)
rate S*t + rate H*t = 200 --> t*(rate S + rate H) = 200

We are looking for rate S*t = ?

1) rate S/rate H = 3/2
So now we can use 1 variable, R... t*(3R+2R) = 200
t*R = 40
Although we have 2 variables, it doesn't matter what the actual breakdown of the 40 is, i.e. 4*10 or 2*20 because it will always be 120 for the sparrow. Sufficient.

2) rate S-5 = rate H
t*(rate S + rate S-5) = 120
t*(2rate S-5) = 120
There isn't a unique solution here because we don't have a ratio or another equation for rate S to rate H like in statement 1), Insufficient.
Starting from the same point, a sparrow and a hawk flew in   [#permalink] 12 Mar 2019, 10:37
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