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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: \(d=4^2*(c-4)^2-4*2(6c^2-32c+39)\geq{0}\) --> \(4c^2-16c+7\leq{0}\) --> \({\frac{1}{2}}\leq{c}\leq{\frac{7}{2}}\) --> \(c_{max}=\frac{7}{2}\).

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. _________________

Re: If a, b, c, d are each positive, a+b+c+d=8, a^2+b^2+c^2+d^2= [#permalink]

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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 _________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

gmatclubot

Re: 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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17 Aug 2013, 13:32

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