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Princeton Review Q

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Princeton Review Q

by nirupshetty » Thu Feb 28, 2008 5:22 pm
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.


The answer was A.

I find it hard to agree with the explanation given by them. any help is appreciated.
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Source: — Data Sufficiency |
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by codesnooker » Thu Feb 28, 2008 10:12 pm
I think the answer should be C.

Lets try to solve the problem.

According to the question, total distance traveled = D = 200 feet.
Lets say speed of Hawk = H and speed of Sparrow = S and both have traveled for t seconds.

Therefore, distance traveled by Hawk = d1 = Ht and distance traveled by Sparrow = d2 = St

and D = d1 + d2 => D = St + Ht = (S + H)t = 200 -----> equation 1

Now according to 1st statement,
S:H::3:2 => S/H = 3/2 => H = 2S/3 -----> equation 2

Now put the value of H from equation 2 in the equation 1

=> (S + 2S/3)t = 200 ------> equation 3

Now as in equation 3, there are two unknown variables, therefore it is impossible to solve until and unless we guess any of the variable, as it is a linear equation.

Now consider statement 2

=> S = H + 5 --------> equation 4

If we try to make equation from this statement also, we will left with again 2 variables in a linear equation, that again requires guessing.

Now from equation 2 and 4, we can reduce S.

i.e S = 2S/3 + 5 => 3S = 2S + 15 => S = 15 feet/second ---> equation 5

Now from equation 5, place the value of S from 3 to get t (however, no need of doing this step for this question).

(15 + 10)t = 200 => t = 8 seconds.

Therefore, both statement can together sufficient the deduce the speed of sparrow but none of them alone is sufficient.
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Re: Princeton Review Q

by Stuart@KaplanGMAT » Thu Feb 28, 2008 10:34 pm
nirupshetty 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.


The answer was A.

I find it hard to agree with the explanation given by them. any help is appreciated.
We know that the total distance is 200, that they started together and that they flew in opposite directions.

(1) if we know the ratio of the speeds, and we know that time is the same for both of them, we also know the ratio of the distances.

Since the sparrow accounts for 3/5 of the total rate (3:2 ratio), the sparrow will also account for 3/5 of the total distance. Therefore, the sparrow traveled 3/5(200) = 120 feet: sufficient.

(2) doesn't give us the ratios of the speeds (for determining new ratios, multiplication/division good, addition/subtraction bad!): insufficient.

(1) is sufficient, (2) isn't: choose (a).
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by codesnooker » Thu Feb 28, 2008 10:40 pm
Thanks again Stuart :D
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by nirupshetty » Fri Feb 29, 2008 9:37 am
thanks stuart...

i for some reason just ignored the line 'they started at the same point'.... was wondering actually how can i determine this without the time spent :)).. thanks again.

cheers!
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