Starting from the Same Point, a Sparrow and a Hawk Flew in Opposite Directions GMAT Data Sufficiency

Question: 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. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
2. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
3. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
4. EACH statement ALONE is sufficient.
5. Statements (1) and (2) TOGETHER are NOT sufficient.

“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?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book "GMAT Quantitative Review". GMAT Quant section consists of a total of 31 questions. GMAT Data Sufficiency questions consist of a problem statement followed by two factual statements. GMAT data sufficiency comprises 15 questions which are two-fifths of the total 31 GMAT quant questions.

Answer:

Approach Solution 1:

(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 = 120

Hence sufficient.

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

Correct option: A

Approach Solution (2):

Time for which sparrow and hawk will fly will be equal, let’s 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 = \(\frac{200}{5S}=\frac{40}{S}\)

Distance traveled by Sparrow is 3S* \(\frac{40}{S}\) = 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

Correct option: A

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