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# Number properties question

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26 Jan 2010, 12:02
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100% (02:14) correct 0% (00:00) wrong based on 3 sessions

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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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26 Jan 2010, 12:15
1
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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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26 Jan 2010, 12:19
1
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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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26 Jan 2010, 14:18
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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26 Jan 2010, 16:30
1
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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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27 Jan 2010, 02:20
2
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Expert's post
If $$x<12$$, then it must be true that:

(a) $$-x<-12$$ --> $$x>12$$ not true;
(b) $$-x-2<14$$ --> $$x>-16$$ not always true;
(c) $$-x+2<-10$$ --> $$x>12$$ not true;
(d) $$x+2<10$$ --> $$x<8$$, not always true;
(e) $$x-2<11$$ --> $$x<13$$ now if $$x<12$$ (given), $$x$$ will definitely be less than $$13$$. Always true.

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28 Jan 2010, 06:33
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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28 Jan 2010, 06:44
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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28 Jan 2010, 06:52
=> [(2/x)+(3/y)]=4
=>[3x+2y]=4xy
putting the value of xy
=>3x+2y=4*5=20

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28 Jan 2010, 14:33
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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28 Jan 2010, 15:20
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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29 Jan 2010, 13:41
$$\frac{(0.3)^5}{(0.3)^3}=$$

(a) 0.001
(b) 0.01
(c) 0.09
(d) 0.9
(e) 1.0
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29 Jan 2010, 14:50
==>0.3 ^ 2 = 0.09 C.. Seems straightforward?
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29 Jan 2010, 19:56
b; prime number reciprocal 10/21
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29 Jan 2010, 19:56
(e) x - 2 < 11 must be true
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29 Jan 2010, 19:57
(c) 0 CANNOT be a value of b
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29 Jan 2010, 23:25
=.09

=(.3)^2 =.09
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30 Jan 2010, 07:01
$$\frac{(0.3)^5}{(0.3)^3}=$$

(a) 0.001
(b) 0.01
(c) 0.09
(d) 0.9
(e) 1.0
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31 Jan 2010, 15:05
ya that is correct 0.09, had to double check. Thanks for your replies
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31 Jan 2010, 15:07
if x is an integer and $$y = 3x + 2$$, which of the following CANNOT be a divisor of y?

(a) 4
(b) 5
(c) 6
(d) 7
(e) 8
Re: Number properties question   [#permalink] 31 Jan 2010, 15:07

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