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Number properties question [#permalink]
 26 Jan 2010, 13:02
Question Stats:
100% (02:14) correct
0% (00:00) wrong based on 1 sessions
Which of the following could be the sum of the reciprocals of two different prime numbers?
(a) 7/13
(b) 10/21
(c)11/30
(d)23/50
(e)19/77
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Re: Number properties question [#permalink]
26 Jan 2010, 13:15
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roceeet wrote: Which of the following could be the sum of the reciprocals of two different prime numbers?
(a) 7/13
(b) 10/21
(c)11/30
(d)23/50
(e)19/77 B..sum of reciprocals of 7 and 3..
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Re: Number properties question [#permalink]
26 Jan 2010, 13:19
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roceeet wrote: Which of the following could be the sum of the reciprocals of two different prime numbers?
(a) 7/13
(b) 10/21
(c)11/30
(d)23/50
(e)19/77 Two primes - a and b. [m]1/a + 1/b = (b+a)/(ab)[m] So: (a) - not possible, because 13 is a prime - ab cannot equal 13. (b) - 7,3 work 7*3=31, 7+3=10, the answer is B... (c) 30 -- 15,2 (not prime), 10,3 (not prime), 5,6 (not prime), 30,1 (not prime)... 30 does not factor into only two numbers (i.e. primes), so not possible. (d) Like 30, 50 will not factor into solely two primes. (e) 77 = 7 * 11, so two primes. 7+11 = 18, NOT 19, so not E.
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Re: Number properties question [#permalink]
26 Jan 2010, 15:13
Thanks, you guys are awesome.
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Re: Number properties question [#permalink]
26 Jan 2010, 15:18
If x<12, then it must be true that
(a) -x<-12
(b)-x - 2<14
(c)-x + 2< -10
(d) x + 2< 10
(e) x - 2 < 11
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Re: Number properties question [#permalink]
26 Jan 2010, 17:30
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IMO E
given x<12,
-x > -12. ( inequality changes when multiplied by -1)
-x+2 > -12+2 ( inequality preserved when adding or subtracting numbers)
-x+2 > -10
x-2 < 10 (inequality changed when multiplied by -1)
if x-2 < 10 is true then x-2 < 11 must definitely be true. Hence E
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Re: Number properties question [#permalink]
26 Jan 2010, 20:21
awesome thanks for the reply.
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Re: Number properties question [#permalink]
27 Jan 2010, 03:20
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Re: Number properties question [#permalink]
28 Jan 2010, 07:33
If \frac{2}{x} + \frac{3}{y} = 4 and xy = 5 then 3x + 2y =
(a) \frac{1}{5}
(b) \frac{1}{4}
(c) \frac{4}{5}
(d) 4
(e) 20
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Re: Number properties question [#permalink]
28 Jan 2010, 07:44
roceeet wrote: If \frac{2}{x} + \frac{3}{y} = 4 and xy = 5 then 3x + 2y =
(a) \frac{1}{5}
(b) \frac{1}{4}
(c) \frac{4}{5}
(d) 4
(e) 20  E.Add the 2 fractions and replace the value of xy.
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Re: Number properties question [#permalink]
28 Jan 2010, 07:52
=> [(2/x)+(3/y)]=4 =>[3x+2y]=4xy putting the value of xy =>3x+2y=4*5=20 answer (e)
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Re: Number properties question [#permalink]
28 Jan 2010, 15:33
For any numbers a and b, a * b = a + b - ab
If a * b = 0 which of the following CANNOT be a value of b?
(a) 2
(b) 1
(c) 0
(d) -1
(e) \frac{-3}{2}
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Re: Number properties question [#permalink]
28 Jan 2010, 16:20
a*b = a + b - ab
given a*b = 0, then 0 = a + b -ab
-b = a(1-b)
b = a(b-1)
a = b / (b-1)
if b=1, then a becomes undefined. So b=1 cannot be valid. IMO B
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Re: Number properties question [#permalink]
29 Jan 2010, 00:08
It was a confussing question. thank for answering it guy.
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Re: Number properties question [#permalink]
29 Jan 2010, 14:35
Thanks for the reply, I found this question confusing myself.
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Re: Number properties question [#permalink]
29 Jan 2010, 14:41
\frac{(0.3)^5}{(0.3)^3}=
(a) 0.001
(b) 0.01
(c) 0.09
(d) 0.9
(e) 1.0
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Re: Number properties question [#permalink]
29 Jan 2010, 15:50
==>0.3 ^ 2 = 0.09 C.. Seems straightforward?
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Re: Number properties question [#permalink]
29 Jan 2010, 20:56
b; prime number reciprocal 10/21
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Re: Number properties question [#permalink]
29 Jan 2010, 20:56
(e) x - 2 < 11 must be true
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Re: Number properties question [#permalink]
29 Jan 2010, 20:57
(c) 0 CANNOT be a value of b
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Re: Number properties question
[#permalink]
29 Jan 2010, 20:57
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