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If a, b, c, d are each positive, a+b+c+d=8, a^2+b^2+c^2+d^2=

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xcusemeplz2009
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If a, b, c, d are each positive, a+b+c+d=8, a^2+b^2+c^2+d^2= [#permalink] New post   Updated on: 09 Jul 2013, 10:13
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If a, b, c, d are each positive, a+b+c+d=8, a^2+b^2+c^2+d^2=25 and c=d, then what is the greatest value of c?

(A) 1/2
(B) 3/2
(C) 5/2
(D) 7/2

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D

Originally posted by xcusemeplz2009 on 07 Dec 2009, 11:57.
Last edited by Bunuel on 09 Jul 2013, 10:13, edited 2 times in total.
Added the OA.
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Re: If a, b, c, d are each positive, a+b+c+d=8, a^2+b^2+c^2+d^2= [#permalink] New post   07 Dec 2010, 05:01
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xcusemeplz2009 wrote:
If a,b,c,d are each +ve, a+b+c+d=8, a^2+b^2+c^2+d^2=25 and c=d ..Then what is the greatest value of c..?
(A)1/2 (B) 3/2 (C) 5/2 (D) 7/2

oa soon



This is not a GMAT question so I wouldn't worry about it at all. Anyway below is an algebraic solution for it:

Given: \(a+b+2c=8\) and \(a^2+b^2+2c^2=25\).

\(a+b+2c=8\) --> \(b=8-a-2c\) --> \(a^2+(8-a-2c)^2+2c^2=25\) --> \(2a^2+4a(c-4)+(6c^2-32c+39)=0\). Now, this quadratic equation to have real solutions for \(a\) its discriminat must be more than or equal to zero: \(d=4^2*(c-4)^2-4*2(6c^2-32c+39)\geq{0}\) --> \(4c^2-16c+7\leq{0}\) --> \({\frac{1}{2}}\leq{c}\leq{\frac{7}{2}}\) --> \(c_{max}=\frac{7}{2}\).

Answer: D.
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Re: If a, b, c, d are each positive, a+b+c+d=8, a^2+b^2+c^2+d^2= [#permalink] New post   07 Dec 2010, 19:43
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nikhilpoddar wrote:
can this be a gmat prob ???


GMAT problems sometimes look tricky. Nevertheless, all of them can be easily solved. Something like this isn't fun to do.. just tedious... so very slim chance that it will appear on GMAT... If you do get stuck on something, don't fret. Use options.
Here try 7/2 since it is the greatest value. Make d also 7/2 and split the leftover 1 from the sum of 8 evenly between a and b. It will work. It is not a neat little problem but still intuitive.
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Re: If a, b, c, d are each positive, a+b+c+d=8, a^2+b^2+c^2+d^2= [#permalink] New post   07 Dec 2009, 12:11
xcusemeplz2009 wrote:
If a,b,c,d are each +ve, a+b+c+d=8,a^2+b^2+c^2+d^2=25 and c=d ..Then what is the greatest value of c..?
(A)1/2 (B) 3/2 (C) 5/2 (D) 7/2

oa soon


c -7/2

d = 7/2 ,a and b = 1/2

sum of these will be 8 and sum of there squares will be 25
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Re: If a, b, c, d are each positive, a+b+c+d=8, a^2+b^2+c^2+d^2= [#permalink] New post   08 Dec 2009, 09:20
can this be a gmat prob ???
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Re: If a, b, c, d are each positive, a+b+c+d=8, a^2+b^2+c^2+d^2= [#permalink] New post   08 Dec 2009, 09:58
nikhilpoddar wrote:
can this be a gmat prob ???


don't think so...but you never know :lol:
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Re: If a, b, c, d are each positive, a+b+c+d=8, a^2+b^2+c^2+d^2= [#permalink] New post   07 Dec 2010, 02:55
Can someone please take a look at this one? I think there should be some shortcut or simplification.
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Re: If a, b, c, d are each positive, a+b+c+d=8, a^2+b^2+c^2+d^2= [#permalink] New post   17 Aug 2013, 13:32
I tried hit and trial method.

let a= 1
b=2
c=3

1+2+6=9
a^2 + b^2 + 2 c^2= 1 + 4 + 18 = 23

c should be near to 3 and 7/2 is the nearest value.
Its not a foolproof way, but good to guess a possible answer :P

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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Re: If a, b, c, d are each positive, a+b+c+d=8, a^2+b^2+c^2+d^2= [#permalink]
17 Aug 2013, 13:32
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