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Source: — Data Sufficiency |

by ajith » Thu Jan 21, 2010 2:46 am
rahul.s wrote:Kindly provide a detailed explanation. Will provide the OA after some discussion
x^2+3x+c = x^2+( a+b) x + ab


=> a+b =3
also, ab = c

Now 1 or 2 is independently sufficient to find out c
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by GhassanMBA » Thu Jan 21, 2010 2:55 am
ajith's explanation is good. I got D as well.
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by rahul.s » Thu Jan 21, 2010 3:48 am
ajith wrote:
rahul.s wrote:Kindly provide a detailed explanation. Will provide the OA after some discussion
x^2+3x+c = x^2+( a+b) x + ab


=> a+b =3
also, ab = c

Now 1 or 2 is independently sufficient to find out c
Ajith,

Could you please elaborate?
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by ajith » Thu Jan 21, 2010 4:10 am
rahul.s wrote:
ajith wrote:
rahul.s wrote:Kindly provide a detailed explanation. Will provide the OA after some discussion
x^2+3x+c = x^2+( a+b) x + ab


=> a+b =3
also, ab = c

Now 1 or 2 is independently sufficient to find out c
Ajith,

Could you please elaborate?
Comparing the coefficients of x and and the constant term we get,

=> a+b =3
also, ab = c

now if a+b =3

using (1) b = 2

c= ab = 2*1 = 2

Using (2)

a= 1
c= ab = 2*1 = 2

So (2) or (1) can be used to find out the value of c
Always borrow money from a pessimist, he doesn't expect to be paid back.
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by bhumika.k.shah » Thu Jan 21, 2010 5:24 am
Ajith,


I understand one needs to solve questions correctly at a super fast speed. But while providing answer explanations it would be better if you could elaborate in depth because the person who posted this question or someone like me are not quite as fast in understanding concepts @ math as you are.

Would appreciate a step wise answer solution.

Regards,
Bhumika Shah

"Some people are born geniuses, but most of us have to work hard, but ultimately we all get there. I respect the latter ones more."
ajith wrote:
rahul.s wrote:
ajith wrote:
rahul.s wrote:Kindly provide a detailed explanation. Will provide the OA after some discussion
x^2+3x+c = x^2+( a+b) x + ab


=> a+b =3
also, ab = c

Now 1 or 2 is independently sufficient to find out c
Ajith,

Could you please elaborate?
Comparing the coefficients of x and and the constant term we get,

=> a+b =3
also, ab = c

now if a+b =3

using (1) b = 2

c= ab = 2*1 = 2

Using (2)

a= 1
c= ab = 2*1 = 2

So (2) or (1) can be used to find out the value of c
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by ajith » Thu Jan 21, 2010 5:33 am
bhumika.k.shah wrote:Ajith,


I understand one needs to solve questions correctly at a super fast speed. But while providing answer explanations it would be better if you could elaborate in depth because the person who posted this question or someone like me are not quite as fast in understanding concepts @ math as you are.

Would appreciate a step wise answer solution.

Regards,
Bhumika Shah

Hey Bhumika,

I really appreciate your feedback will try to explain the answer in great detail in future posts.

As for this problem, please tell me where exactly does it get complicated so that I can explain that part in detail.
Always borrow money from a pessimist, he doesn't expect to be paid back.
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by bhumika.k.shah » Thu Jan 21, 2010 5:37 am
Please can u break it down step by step.

That would be of great help

Thanking you in advance.

Regards,
ajith wrote:
bhumika.k.shah wrote:Ajith,


I understand one needs to solve questions correctly at a super fast speed. But while providing answer explanations it would be better if you could elaborate in depth because the person who posted this question or someone like me are not quite as fast in understanding concepts @ math as you are.

Would appreciate a step wise answer solution.

Regards,
Bhumika Shah

Hey Bhumika,

I really appreciate your feedback will try to explain the answer in great detail in future posts.

As for this problem, please tell me where exactly does it get complicated so that I can explain that part in detail.
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by bhumika.k.shah » Thu Jan 21, 2010 6:51 am
x^2 + 3x + c = x^2 + (a+b)x + ab

a+b=3 ; ab=??

1. a=1 so b=2

ab= c= 2

Sufficient

1. b=2 so a=1

ab=c =2

Sufficient

Answer D

Hope this solves ur doubt rahul.s
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by rahul.s » Thu Jan 21, 2010 6:54 am
Yeah, it does.

Thank you :)

I hate Quadratics!
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