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Good ps questions from a Turkish exam (kpss 2009) - I

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maliyeci
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Good ps questions from a Turkish exam (kpss 2009) - I [#permalink] New post   06 Aug 2009, 02:30
1
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A
B
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D
E

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(N/A)

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I solved this exam yesterday. I solved all of the problems. Got the right answers. I liked some of the problems so much that I wanted you to see these. I will post each in its own topic.

if one of the roots of the equation \(x^2+4x-7=0\) is \(a\), then what is the result of the multiplication below?
\((a-1).(a+1).(a+3).(a+5)=?\)

a)12 b)14 c)15 d)18 e)20



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Re: Good ps questions from a Turkish exam (kpss 2009) - I [#permalink] New post   06 Aug 2009, 06:30
The answer is in this spoiler.
Show Spoiler
\(x^2+4x-7= x^2+4x+4-11=(x+2)^2-11^=0\)
Then
\((x+2)^2=11\)
this makes
\((a+2)^2=11\)(since a is a root of this equation)
Say \(b=(a+2)\)
Then the equation below becomes :
\((b-3).(b-1).(b+1).(b+3)=(b^2-9).(b^2-1)\) since \((a+2)^2=11\) this makes
\(b^2=11\)
So
\((b^2-9).(b^2-1)=(11-9).(11-1)=20\) Answer is E
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Re: Good ps questions from a Turkish exam (kpss 2009) - I [#permalink] New post   Updated on: 13 Aug 2009, 02:12
Another question:

What is the equal of \(sqrt{5-sqrt{21}}\)
Answer is
Show Spoiler
A

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Originally posted by maliyeci on 13 Aug 2009, 01:43.
Last edited by maliyeci on 13 Aug 2009, 02:12, edited 2 times in total.
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Re: Good ps questions from a Turkish exam (kpss 2009) - I [#permalink] New post   13 Aug 2009, 02:08
2
Kudos
if one of the roots of the equation \(x^2+4x-7=0\) is \(a\), then what is the result of the multiplication below?
\((a-1).(a+1).(a+3).(a+5)=?\)

a)12 b)14 c)15 d)18 e)20[/quote]

rephrase question: (a-1).(a+5).(a+3).(a+1) =(a^2+4a+3)(a^2+4a-5)

from stem

a^2+4a = 7

thus subst:
(7+3)(7-5) = 20...E
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Re: Good ps questions from a Turkish exam (kpss 2009) - I [#permalink] New post   13 Aug 2009, 02:11
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Good solution yezz +1 ;)
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Re: Good ps questions from a Turkish exam (kpss 2009) - I [#permalink] New post   13 Aug 2009, 02:17
maliyeci wrote:
Good solution yezz +1 ;)


excellent GMAT-type question.. thanks for sharing.
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Re: Good ps questions from a Turkish exam (kpss 2009) - I [#permalink] New post   14 Aug 2009, 05:54
maliyeci wrote:
Another question:

What is the equal of \(sqrt{5-sqrt{21}}\)
Answer is
Show Spoiler
A


Does not anybody have any solution to this problem???
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Re: Good ps questions from a Turkish exam (kpss 2009) - I [#permalink] New post   14 Aug 2009, 06:38
maliyeci wrote:
maliyeci wrote:
Another question:

What is the equal of \(sqrt{5-sqrt{21}}\)
Answer is
Show Spoiler
A


Does not anybody have any solution to this problem???


we can brute force it but i am sure there is an eithier way...that i dont know of :)
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maliyeci
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Re: Good ps questions from a Turkish exam (kpss 2009) - I [#permalink] New post   15 Aug 2009, 13:35
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\(sqrt{5-sqrt{21}}\)???

\(5-sqrt{21} = \\
1/2*(10-2*sqrt{21})=\\
1/2*(7+3-2*sqrt{21})=\\
1/2*(sqrt{7}-sqrt{3})^2\)
Thus
\(sqrt{5-sqrt{21}}\)=

\(sqrt{1/2*(sqrt{7}-sqrt{3})^2}=\\
(sqrt{7}-sqrt{3})/sqrt{2}\)
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Re: Good ps questions from a Turkish exam (kpss 2009) - I [#permalink] New post   15 Aug 2009, 13:38
1
Kudos
maliyeci wrote:
\(sqrt{5-sqrt{21}}\)???

\(5-sqrt{21} = \\
1/2*(10-2*sqrt{21})=\\
1/2*(7+3-2*sqrt{21})=\\
1/2*(sqrt{7}-sqrt{3})^2\)
Thus
\(sqrt{5-sqrt{21}}\)=

\(sqrt{1/2*(sqrt{7}-sqrt{3})^2}=\\
(sqrt{7}-sqrt{3})/sqrt{2}\)



to be honest , it wouldnt have even crossed my mind in a million light years .:) ...great sol, Kudos+1 from me
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Re: Good ps questions from a Turkish exam (kpss 2009) - I [#permalink] New post   15 Aug 2009, 13:44
This forum showed me a lot of this type good solutions. Thanks for all.
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Re: Good ps questions from a Turkish exam (kpss 2009) - I [#permalink] New post   17 Aug 2009, 01:14
An easy one:
a and b are positive numbers.
LCM(a,b)=\(2.3^2\)
How many different (a,b) numbers are there?
a)6
b)12
c)13
d)14
e)15
Note:(a,b) is different from (b,a)

Answer is
Show Spoiler
A




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Re: Good ps questions from a Turkish exam (kpss 2009) - I [#permalink]
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