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kevincan
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120 business executives were asked which of three leading Updated on: 24 Aug 2006, 13:35
Originally posted by kevincan on 24 Aug 2006, 03:38.
Last edited by kevincan on 24 Aug 2006, 13:35, edited 1 time in total.
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120 business executives were asked which of three leading business newspapers they read in the previous week. There were three times as many executives who reported reading only one newspaper as those who reported reading none. 45% of the executives surveyed reported reading two of the three newspapers. How many elements are in the set of possible values for the number of executives who read all three?

(A) 5 (B) 6 (C) 14 (D) 15 (E) 16

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Ruby
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120 business executives were asked which of three leading 24 Aug 2006, 08:04
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I am getting B and C...

If someone knows the correct answer please let me know.

120 = 3 read none + 1 read none + 54 + Read all 3
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prashrash
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120 business executives were asked which of three leading 24 Aug 2006, 09:56
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all options (A-E) seem to be working for me!
all three = 120 - (read none) - (read one) - 54
if read none = x,
all three = 120 - 4x - 54
= 66 - 4x
Even if x = 16, 4x = 64.
Hence all answers are valid!

Pls let me know what I am doing wrong.
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120 business executives were asked which of three leading 24 Aug 2006, 09:58
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Let reading two be D and reading none be N.
-> D=3N

Let reading three be T.

120-54=66=4N+T
In order that T>0, N could be 1~16. Thus choose (E).
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120 business executives were asked which of three leading 24 Aug 2006, 10:09
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Both B and C seems to be working.
x = None
A = All

120 = 3x+54 + A + x
x = (66-A)/4

B gives x = 15 and C gives x = 13. Both are possible. All others give fractions.
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120 business executives were asked which of three leading 24 Aug 2006, 10:28
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ps_dahiya,

I think the question asks how many As there could possibly be,
rather than what value A could possibly be. :roll:
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ps_dahiya
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120 business executives were asked which of three leading Updated on: 25 Aug 2006, 07:49
Originally posted by ps_dahiya on 24 Aug 2006, 10:33.
Last edited by ps_dahiya on 25 Aug 2006, 07:49, edited 1 time in total.
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Beegeez2
ps_dahiya,

I think the question asks how many As there could possibly be,
rather than what value A could possibly be. :roll:
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Oh I got it. E is the answer.

x = (66-A)/4
x is an integer for values of A = (2,6,10,14.........62)

62 = 2 + (n-1) * 4
n = 16
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gk3.14
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120 business executives were asked which of three leading 24 Aug 2006, 10:53
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beegeez2 , you are absolutely right.. It is 16..
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ps_dahiya
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120 business executives were asked which of three leading Updated on: 25 Aug 2006, 07:48
Originally posted by ps_dahiya on 25 Aug 2006, 07:08.
Last edited by ps_dahiya on 25 Aug 2006, 07:48, edited 1 time in total.
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prashrash
ps_dahiya

x is an integer for values of A = (2,6,10,14.........64) agreed

64 = 2*2 + (n-1) * 4 could you pls explain the logic behind this split pls ?
n = 16

Pls see my comments in the quote (in red).
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This is an A.P. (Arithmatic Progression)

A[n] = A[1] + (n-1) * d
where d is the difference in terms and A[n] is nth term and A[1] is first term.
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