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Tough DS

by gmatmachoman » Sun May 09, 2010 11:59 am
If square roots of numbers are considered positive and sqrt( a) +sqrt( b) =sqrt (c )+ sqrt( d )+sqrt( e) , then is a < c?

(1) c = d


(2) sqrt (b )+ sqrt (d )< sqrt ( e)


This is areal tooughie!!
Last edited by gmatmachoman on Sun May 09, 2010 11:11 pm, edited 1 time in total.
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by Stuart@KaplanGMAT » Sun May 09, 2010 12:58 pm
gmatmachoman wrote:If square roots of numbers are considered positive and sqrt( a) +sqrt( b) =sqrt (c )+ sqrt( d )+sqrt( e) , then is a < c?

(1) c = d


(2) sqrt (b )+ sqrt (d ) sqrt (< e)


This is areal tooughie!!
Is that the full wording of the question? Nowhere does it say that a, b, c, d and e are positive; if there's additional language, please share! Statement (2) definitely has the inequality in the wrong spot.

Also, what's the source?

(1) no info about the left side of the equation, so clearly insufficient.

(2) without the inequality in the right spot, and without knowing if the variables could be 0, impossible to evaluate.
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by kuntalkkc » Sun May 09, 2010 2:21 pm
Nothing tough....no need to know whether a,b,c,d,e positive or not.cant be negetive thats for sure coz imaginary is not to be considered here to bring argund plane.

but a,b,c,d,e can be 0 or not thts not given...Fine.

take the statement 1: c=d..cant conclude that much because those quantities can be 0 also.

Take statement 2 only:

sqrt(b)+sqrt(d)<sqrt(e)[it can be seen it's a printing mistake in the post,so the actual inequality should be
like this]
=> sqrt(b)+sqrt(d)+m=sqrt(e) where m is a positive quantity>0 thats make the difference.

now take the given statement:

sqrt(a)+sqrt(b)=sqrt(c)+sqrt(d)+sqrt(e)
=>sqrt(a)+sqrt(b)=sqrt(c)+sqrt(d)+sqrt(b)+sqrt(d)+m...rite?

sqrt(b) gets cancelled from both sides.

=>sqrt(a)=sqrt(c)+sqrt(d)+sqrt(d)+m

now great thing is i know m is positive....so i dont bother about d where it +ve or 0.

gives,sqrt(a)>sqrt(c) as m is +ve

which gives a>c

which gives the ans that by statement 2 only we can ans this....any doubt?please post.
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by ansumania » Tue May 11, 2010 9:10 am
hi,

you mentioned :

sqrt(a)=sqrt(c)+sqrt(d)+sqrt(d)+m

if we m is + ve , then

sqrt(a)=sqrt(c)+sqrt(d)+sqrt(d)

=> sqrt(a)>=sqrt(c) or sqrt(a)<=sqrt(c)

as we know sqrt(d)>0 , but we don't know what the value is.

pl. suggest
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by analyst218 » Tue May 11, 2010 10:05 am
gmatmachoman wrote:If square roots of numbers are considered positive and sqrt( a) +sqrt( b) =sqrt (c )+ sqrt( d )+sqrt( e) , then is a < c?

(1) c = d


(2) sqrt (b )+ sqrt (d )< sqrt ( e)


This is areal tooughie!!
I agreet with stuart. sqr roots is positive is not enough info.
however, if had the question been 'is sqr a< sqr c?'
then
well statement 1 is clearly insuff.
in 2, you need sqr A bigger than sqr C since sqr d+sqr e will be greater than sqr b
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by kuntalkkc » Tue May 11, 2010 7:47 pm
It is clearly mentioned in the question that "sqrt of numbers are positive",

now as it is true that,

sqrt(a)=sqrt(c)+sqrt(d)+sqrt(d)+m where m is a +ve quantity>0

now worst case scenario is sqrt(d)=0,please note it cant be negetive as per question says.

so,for this worst case also,sqrt(a)=sqrt(c)+0+0+m(something +ve)

which means,after adding something +ve to sqrt(c) we are getting sqrt(a)

so even if sqrt(c) is 0,still we have sqrt(a) is positive.

so it is for sure that sqrt(a) is always greater than sqrt(c)[it is simply greater than(>),not greater than equal to(>=)]

now again by question as sqrt of numbers are mentioned as positive so definitely a has to be greater than b(because a,c cannot be negetive but c has always a chance to be 0.that is why in this question +ve is not mentioned)

so by statement 2 only we can reach the conclusion for sure...

Hope this explanation clears your doubt.
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