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Starting from the same point, a sparrow and a hawk flew in opposite di 02 Jan 2010, 07:44
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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.
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Starting from the same point, a sparrow and a hawk flew in opposite di 02 Jan 2010, 11:21
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Can anyone help me better understand the logic behind this one?

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.
Show more

(1) \(\frac{s}{h}=\frac{3}{2}\), where \(s\) is speed of sparrow and \(h\) is speed of hawk --> \(h=s\frac{2}{3}\), as sparrow and hawk flew for same time and the speed of hawk was \(\frac{2}{3}\) of sparrow's, hence hawk would cover \(\frac{2}{3}\) of the distance as would cover sparrow --> \(Ds+\frac{2}{3}Ds=200\), where \(Ds\) the distance of sparrow --> \(Ds=120\). Sufficient.

(2) \(s=h-5\), clearly insufficient to calculate \(Ds\).

Answer: A.
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Starting from the same point, a sparrow and a hawk flew in opposite di 12 Dec 2011, 08:55
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ashiima
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.
Show more

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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Starting from the same point, a sparrow and a hawk flew in opposite di 14 Dec 2011, 11:33
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ashiima
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.
Show more
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.

Hence, the answer is A.
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Starting from the same point, a sparrow and a hawk flew in opposite di 14 Dec 2011, 16:49
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ashiima
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.
Show more

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.

The answer is A.
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Starting from the same point, a sparrow and a hawk flew in opposite di 01 Aug 2013, 16:23
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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.
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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Starting from the same point, a sparrow and a hawk flew in opposite di 31 Dec 2013, 07:23
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ashiima
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.
Show more


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!
J :)
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Starting from the same point, a sparrow and a hawk flew in opposite di 18 May 2014, 06:02
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x13069
Can anyone help me better understand the logic behind this one?

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.

(1) \(\frac{s}{h}=\frac{3}{2}\), where \(s\) is speed of sparrow and \(h\) is speed of hawk --> \(h=s\frac{2}{3}\), as sparrow and hawk flew for same time and the speed of hawk was \(\frac{2}{3}\) of sparrow's, hence hawk would cover \(\frac{2}{3}\) of the distance as would cover sparrow --> \(Ds+\frac{2}{3}Ds=200\), where \(Ds\) the distance of sparrow --> \(Ds=120\). Sufficient.

(2) \(s=h-5\), clearly insufficient to calculate \(Ds\).

Answer: A.
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Hi Bunuel, i am not able to understand how the sum of 2 distance is 200 ? Should't it be total dist - (distance covered by sparrow + dist covered by hawk), since 200 is the distance between them.
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Starting from the same point, a sparrow and a hawk flew in opposite di 19 May 2014, 02:33
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x13069
Can anyone help me better understand the logic behind this one?

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.

(1) \(\frac{s}{h}=\frac{3}{2}\), where \(s\) is speed of sparrow and \(h\) is speed of hawk --> \(h=s\frac{2}{3}\), as sparrow and hawk flew for same time and the speed of hawk was \(\frac{2}{3}\) of sparrow's, hence hawk would cover \(\frac{2}{3}\) of the distance as would cover sparrow --> \(Ds+\frac{2}{3}Ds=200\), where \(Ds\) the distance of sparrow --> \(Ds=120\). Sufficient.

(2) \(s=h-5\), clearly insufficient to calculate \(Ds\).

Answer: A.

Hi Bunuel, i am not able to understand how the sum of 2 distance is 200 ? Should't it be total dist - (distance covered by sparrow + dist covered by hawk), since 200 is the distance between them.
Show more

We are told that 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. Hence 200 feet is the distance they covered together.

Does this make sense?
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Starting from the same point, a sparrow and a hawk flew in opposite di 19 May 2014, 04:17
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x13069
Can anyone help me better understand the logic behind this one?

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.

(1) \(\frac{s}{h}=\frac{3}{2}\), where \(s\) is speed of sparrow and \(h\) is speed of hawk --> \(h=s\frac{2}{3}\), as sparrow and hawk flew for same time and the speed of hawk was \(\frac{2}{3}\) of sparrow's, hence hawk would cover \(\frac{2}{3}\) of the distance as would cover sparrow --> \(Ds+\frac{2}{3}Ds=200\), where \(Ds\) the distance of sparrow --> \(Ds=120\). Sufficient.

(2) \(s=h-5\), clearly insufficient to calculate \(Ds\).

Answer: A.

Hi Bunuel, i am not able to understand how the sum of 2 distance is 200 ? Should't it be total dist - (distance covered by sparrow + dist covered by hawk), since 200 is the distance between them.
Show more

We are told that 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. Hence 200 feet is the distance they covered together.

Does this make sense?[/quote]


Ahh ok , did not read the question properly. It thought they are coming towards each other.. Thanks
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Starting from the same point, a sparrow and a hawk flew in opposite di 28 May 2015, 11:26
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In Distance and Speed questions, it's helpful to represent the given information visually.



Both the Sparrow and the Hawk start from the same starting point, at the same time, in the opposite directions.

From the image, it is clear that:

(Time taken by Sparrow to cover x feet) = (Time taken by Hawk to cover 200 - x feet) . . . (1)

Now, we know that Time = Distance/Speed

So, if we assume the Sparrow's speed to be S feet/min and the Hawk's speed to be H feet/min, we can write Equation (1) as:


\(\frac{x}{S}= \frac{(200-x)}{H}\)

This equation can also be rewritten as:

\(\frac{x}{(200-x)} = \frac{S}{H}\) . . . (2)

The question asks us to find the value of x. From Equation (2), it's clear that to find the value of x, we need to know the value of S:H

Statement 1 straightaway gives us the value of S:H. So, St. 1 is sufficient.

Statement 2 gives us the equation S = H + 5. However, since this equation is not sufficient to get us the value of S:H, St. 2 is not sufficient.

So, the correct answer is Option A.

Hope this helped! :)

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Starting from the same point, a sparrow and a hawk flew in opposite di 12 Mar 2019, 10:37
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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.
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Starting from the same point, a sparrow and a hawk flew in opposite di 03 Dec 2023, 12:05
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