In 1979 approximately 1/3 of the 37.3 million airline passengers trave

Question

In 1979 approximately 1/3 of the 37.3 million airline passengers traveling to or from the United States used Kennedy Airport. If the number of such passengers that used Miami Airport was 1/2 the number that used Kennedy Airport and 4 times the number that used Logan Airport, approximately how many millions of these passengers used Logan Airport that year?

(A) 18.6
(B) 9.3
(C) 6.2
(D) 3.1
(E) 1.6

(A) 18.6 
(B) 9.3 
(C) 6.2 
(D) 3.1 
(E) 1.6

BrentGMATPrepNow · Accepted Answer

This question illustrates the importance of   ALWAYS checking the answer choices BEFORE   getting bogged down with time-consuming calculations. 
 For more on this important strategy, you can watch the video below

Since the answer choices are VERY spread apart, we can be quite aggressive in our rounding and estimation.

In 1979 approximately 1/3 of the 37.3 million airline passengers traveling to or from the United States used Kennedy Airport.   
Let's say that "In 1979 approximately 1/3 of the   36   million airline passengers traveling to or from the United States used Kennedy Airport. "
1/3 of 36 =  12 
So,  12  million passengers used Kennedy Airport

The number of such passengers that used Miami Airport was 1/2 the number that used Kennedy Airport   
 12  million passengers used Kennedy Airport
1/2 of  12  million =  6 million 
So,  6 million  passengers used Miami Airport

The number of such passengers that used Miami Airport was 4 times the number that used Logan Airport  
In other words, the number of such passengers that used Logan Airport was  1/4  the number that used Miami Airport 
 1/4  of  6 million  =   1.5 million

So,   1.5 million    passengers used Logan Airport

Check the answer choices . . . only answer choice E is close to   1.5

Answer: E

RELATED VIDEO
 https://www.youtube.com/watch?v=8GqcJGUVIk4

CounterSniper · Answer

number of
passengers using Kennedy airport = 37/3 =~12.43
Passengers using Miami airport = 12.43/2=~6.2
passengers using logan airport = 6.2/4 =~1.54

so E

Pritishd · Answer

Hi Brent,

In general when we have statements like  'Miami Airport was 1/2 the number that used Kennedy Airport and 4 times the number that used Logan Airport' , we interpret them as  M = \frac{1}{2}K + 4L

Any specific reason for spliting into two equations here?

Warm Regards,
 Pritishd

aggvipul · Answer

Lets not do calculations rather look at the kind of formula that emerges finally- \frac{37.3}{4*3*2}
Since, 4*3*2 = 24, therefore the division would result in 1 < number < 2, hence answer E

BrentGMATPrepNow · Answer

EDIT: Someone just pointed out that my earlier response was incorrect. I should have read it better. 
Here's my new response to the above question:

I think you may have misinterpreted information in the question.

It tells us that:  the number of such passengers that used Miami Airport was 1/2 the number that used Kennedy Airport   and   4 times the number that used Logan Airport 
In other words: M = 0.5K and M = 4L

I believe you are reading the information as:  the number of such passengers that used Miami Airport was 1/2 the number that used Kennedy Airport   PLUS   4 times the number that used Logan Airport

Kimberly77 · Answer

Hi  BrentGMATPrepNow , If I use M=1/2K+4L
 M = 1/2 (12.4) + 4L
 M = 6 + 4L or L = M-6/4 
Not sure how to proceed from here in order to get 1/4 of 6 million = 1.5 million or have I missed something here? Thanks Brent.

BrentGMATPrepNow · Answer

I think you may have misinterpreted information in the question.

It tells us that:  the number of such passengers that used Miami Airport was 1/2 the number that used Kennedy Airport   and   4 times the number that used Logan Airport 
In other words: M = 0.5K and M = 4L

I believe you are reading the information as:  the number of such passengers that used Miami Airport was 1/2 the number that used Kennedy Airport   PLUS   4 times the number that used Logan Airport

Kimberly77 · Answer

Thanks Brent  BrentGMATPrepNow  . so M = 0.5K and M = 4L is not equal to  M = \frac{1}{2}K + 4L  ? 
I followed calculation from this post therefore got stuck in M = 1/2 (12.4) + 4L
 M = 6 + 4L or L = M-6/4 ?

So this is not correct then?

BrentGMATPrepNow · Answer

Sorry,  Kimberly77 . 
One of my earlier responses to a student's question was incorrect. Please see above.

The answer to your question is no, the pair of equations  M = 0.5K  and  M = 4L  cannot be combined to get  M = \frac{1}{2}K + 4L

Consider these values: M = 20, K = 40 and L = 5
Notice that these values satisfy the equations  M = 0.5K  and  M = 4L

If we plug the same values into the equation  M = \frac{1}{2}K + 4L , we get:  20 = (\frac{1}{2})40 + 4(5)  
Simplify to get:  20 = 20 + 20 , which is not true.

Kimberly77 · Answer

Brilliant thanks  BrentGMATPrepNow . Crystal clear now. 
So "and" & "Plus" do not mean the same thing in Maths generally?

BrentGMATPrepNow · Answer

Tricky question!
It really depends on the context. 
If I say I have 3 cats and 5 dogs, and then the total number of pets I have = 3 + 5

Kimberly77 · Answer

Noted thanks BrentGMATPrepNow

bumpbot · Answer

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In 1979 approximately 1/3 of the 37.3 million airline passengers trave

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