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Today's Problems

by alnasser973 » Wed May 23, 2012 3:24 pm
Is p + q = 0 ?

(1) p = 1/(1+ q)

(2) 2q =1/(1− p)

----------------

Is x = 1, y = 2, and z = 3?

(1) 5x + 2z + 3 = 3x + 4y = y + 2z + 3

(2) 5x + z + 3 = 3x + 4y and 3x + 4y = y + 2z + 3

-----------------

Is (p + q)/(q + s)
equal to 1?

(1) (p+q)/(r+s)=1

(2) (p+s)/(r+q)=1

------------------

Is (ps + qr)/(pq + rs)= 1?

(1) (p + q)/(r + s)= 1

(2) (p + s)/(r + q)= 1
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Source: — Data Sufficiency |
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by aneesh.kg » Sat May 26, 2012 8:33 am
alnasser973 wrote:Is p + q = 0 ?

(1) p = 1/(1+ q)

(2) 2q =1/(1− p)
Statement(1):
p(q + 1) = 1
In order to test if p + q = 0, let q = -p
p(1 - p) = 1
p^2 - p + 1 = 0
No Real Roots from this equation for p (because b^2 - 4ac = 1 - 4 < 0), so
p CANNOT be equal to -q.
SUFFICIENT

Statement(2):
2q(1 - p) = 1
In order to test if p + q = 0, let p = -q
2q(1 + q) = 1
2q^2 - 2q + 1 = 0
No Real Roots from this equation for q(because b^2 - 4ac = 4 - 4*2 < 0), so
p CANNOT be equal to -q.
SUFFICIENT

[spoiler](D)[/spoiler] is correct.
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by Ashujain » Sat May 26, 2012 10:21 am
alnasser973 wrote:Is p + q = 0 ?

(1) p = 1/(1+ q)

(2) 2q =1/(1− p)

----------------

Is x = 1, y = 2, and z = 3?

(1) 5x + 2z + 3 = 3x + 4y = y + 2z + 3

(2) 5x + z + 3 = 3x + 4y and 3x + 4y = y + 2z + 3

-----------------

Is (p + q)/(q + s)
equal to 1?

(1) (p+q)/(r+s)=1

(2) (p+s)/(r+q)=1

------------------

Is (ps + qr)/(pq + rs)= 1?

(1) (p + q)/(r + s)= 1

(2) (p + s)/(r + q)= 1

Solution1:
St1: (1) pq+p=1 it is insufficient
St2: (2) 2q-2pq=1 it is insufficient

Merge both the statements: (1)---> pq+p=1, 2*(1)--->2pq+2p=2, now add (1) and (2)-->2p+2q=3
therefore, p+q=3/2 (p+q not equal to 0) We are getting a unique value of p+q
Hence, the answer is C


Solution2:
St1: 5x+2z+3=3x+4y=y+2z+3, now if we substitute the values of x,y, z in this equation it will become 14=11=11 which is incorrect and hence x, y and z can not be equal to the values given. And hence this statement is sufficient as it gives a unique answer - NO
St2: It is same as statement1 and hence it is also sufficient.

Hence the answer is D


Solution3:
we have to find whether (p+q)/(q+s)=1--->p+q=q+s--->p=s?

St1: p+q=r+s--->not sufficient
St2: p+s=r+q--->not sufficient
Merge statements: it is also insufficient
Hence answer is E


Solution4:
Similar to solution3 we have to find whether p=r?
St1: it is insufficient
St2: it is insufficient
Merger Statements: Add ststements 1 and 2 and u will get p=r
Hence, the answer is C

Kindly correct me if I am wrong...
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by alnasser973 » Mon May 28, 2012 5:23 pm
Thank you all for your answers and i'm sorry for the late reply.

The OA are as follows:

Q1 : A

but I'm don't know why i'm getting as you aneesh D

Q2 : E

Don't know why!!!!

Q3 : E

Q4 : C

Thanks agian...
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by aneesh.kg » Mon May 28, 2012 7:45 pm
I had made a small mistake.
Corrected Post:

Statement(1):
p(q + 1) = 1
In order to test if p + q = 0, let q = -p
p(1 - p) = 1
p^2 - p + 1 = 0
No Real Roots from this equation for p (because b^2 - 4ac = 1 - 4 < 0), so
p CANNOT be equal to -q.
SUFFICIENT

Statement(2):
2q(1 - p) = 1
In order to test if p + q = 0, let p = -q
2q(1 + q) = 1
2q^2 + 2q - 1 = 0
Real Roots exist.(because b^2 - 4ac = 4 + 4*2 > 0), so
p = -q for two values, but not for the other values.
INSUFFICIENT

[spoiler](A)[/spoiler] is correct.
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by neelgandham » Tue May 29, 2012 12:34 am
Ashujain wrote: Solution4:
Similar to solution3 we have to find whether p=r?
A small correction.

Is (ps + qr)/(pq + rs)= 1? Let me try and rephrase the question.
Is (ps + qr)=(pq + rs) ?
Is (ps - pq)=(rs - rq) ?
Is (ps - pq)-(rs - rq) = 0 ?
Is p(s - q)- r(s - q) = 0 ?
Is (p - r)(s - q)= 0 ?
So the question is now Is p = r or Is s = q?
(1) (p + q)/(r + s)= 1
(p + q)=(r + s)
p - r = s - q
We can't determine if the value of p=r or s=q. Hence, Statement 1 is insufficient to answer the question.
(2) (p + s)/(r + q)= 1
(p + s)=(r + q).
p - r = q - s.
We can't determine if the value of p=r or s=q. Hence, Statement 1 is insufficient to answer the question.
From 1 and 2
p - r = s - q = q - s
s - q = q - s
2s = 2q
s = q.
Bingo! Statement 1 + 2 is sufficient to answer the question.

IMO C
Anil Gandham
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