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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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Manager
Joined: 09 May 2009
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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]

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07 Dec 2009, 11:57
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95% (hard)

Question Stats:

43% (02:54) correct 57% (02:06) wrong based on 126 sessions

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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 09 Jul 2013, 10:13, edited 2 times in total.

Kudos [?]: 257 [0], given: 13

Senior Manager
Joined: 30 Aug 2009
Posts: 283

Kudos [?]: 188 [0], given: 5

Location: India
Concentration: General Management
Re: easy one but cant solve it [#permalink]

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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

Kudos [?]: 188 [0], given: 5

Manager
Joined: 14 Jan 2006
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Schools: HKUST
Re: easy one but cant solve it [#permalink]

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08 Dec 2009, 09:20
can this be a gmat prob ???

Kudos [?]: 40 [0], given: 2

Senior Manager
Joined: 30 Aug 2009
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Re: easy one but cant solve it [#permalink]

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08 Dec 2009, 09:58
nikhilpoddar wrote:
can this be a gmat prob ???

don't think so...but you never know

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Director
Joined: 23 Apr 2010
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Kudos [?]: 95 [0], given: 7

Re: easy one but cant solve it [#permalink]

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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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Math Expert
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Re: easy one but cant solve it [#permalink]

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07 Dec 2010, 05:01
3
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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: $$d=4^2*(c-4)^2-4*2(6c^2-32c+39)\geq{0}$$ --> $$4c^2-16c+7\leq{0}$$ --> $${\frac{1}{2}}\leq{c}\leq{\frac{7}{2}}$$ --> $$c_{max}=\frac{7}{2}$$.

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Veritas Prep GMAT Instructor
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Re: easy one but cant solve it [#permalink]

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07 Dec 2010, 19:43
1
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Expert's post
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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Kudos [?]: 17335 [1], given: 232

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Status: Everyone is a leader. Just stop listening to others.
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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]

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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
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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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