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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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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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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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Beegeez2 wrote:
ps_dahiya,

I think the question asks how many As there could possibly be,
rather than what value A could possibly be. :roll:

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

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 , you are absolutely right.. It is 16..
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prashrash wrote:
ps_dahiya wrote:
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).

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.

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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