Is p + q = 0 ?

Question

Is p + q = 0 ?

(1) p = 1/1+q 
(2) 2q = 1/1- p

Chapter name: Algebraic expressions, page: 296, problem set M, number :2

Now by using statement (2), we can find 1-p = 1/2q . then i added this equation to statement (1). Then by calculating we can have easily the values of q ,which are 1 and -1/2 .

By using these two values into either equations we can have the values of p.

when q=1 then p =1/2 and when q = -1/2 then p = 2 . 
so clearly , p ≠ q (just found by using two statements together);
so p + q is not equal to zero at any cases. 
Consequently (c) must be the Answer.

But the book says the answer is (A) ……….!!!

Who is right, book or me ?

mau5 · Accepted Answer

From F.S 1, we know that p = \frac{1}{1+q} . Adding 'q' on both sides , we have  p+q = \frac{1}{1+q}+q = \frac{q^2+q+1}{1+q} .
Thus, now the question stands at : Is  \frac{q^2+q+1}{1+q} = 0?

The only way, the above expression can be zero is if  q^2+q+1 = 0 .

Note that:  q^2+q+1  can be written as :  q^2+q+\frac{1}{4} + \frac{3}{4} = (q+\frac{1}{2})^2+\frac{3}{4}  This can never be zero. Thus, p+q can never equal zero. Sufficient.

Hope this helps.

Asifpirlo · Answer

yes man...thats awesome..now got it.. thanks for help... :lol:

fameatop · Answer

Statement 1 p = \frac{1}{(1+q)} Adding q on both sides, we get p + q= \frac{1}{(1+q)} + q = \frac{(q^2+ q + 1)}{(1+q)} So the question becomes IS \frac{(q^2+ q + 1)}{(1+q)} = 0 or IS (q^2+ q + 1) = 0 ---------This is a quadratic equation In other words the question is asking whether any solution exists for the above quadratic equation or not? Solution does not exist if b^2 - 4ac <0 1^2 - 4.1.1 <0 -3<0 It means no solution exists, Thus \frac{(q^2+ q + 1)}{(1+q)} can never equal 0 Sufficient Statement 2 2q = \frac{1}{(1-p)} -------> q = \frac{1}{2(1-p)} Adding p on both sides, we get p + q= \frac{1}{2(1-p)} + p = \frac{(1+ 2p - 2p^2)}{(2-2p)} So the question becomes IS \frac{(1+ 2p - 2p^2)}{(2-2p)} = 0 or IS (2p^2 - 2p -1) = 0 ---------This is a quadratic equation In other words the question is asking whether any solution exists for the above quadratic equation or not? Solution does not exist if b^2 - 4ac <0 (-2)^2 - 4.2.(-1)1 <0 12<0 ----> Incorrect It means for two values of "P" \frac{(1+ 2p - 2p^2)}{(2-2p)} will be 0. But as per question we don't know what value "P" can take. Insufficient Answer A

gmatprav · Answer

Same approach as fametop. +1 A.

KarishmaB · Answer

There is another approach to such questions.
Each statement gives you an equation in p and q. You need to find whether p+q = 0 holds or not. Take each statement at a time.

(1) p = 1/1+q 
p(1+q) = 1
Is p+q = 0? 
We would like to know two things: (i) Is it necessary that p+q = 0 for all p and q? If yes, we answer the question with 'yes' and the statement alone is sufficient. (ii) If not, is it possible that p+q = 0? If no, we answer the question with 'No' and the statement alone is sufficient. If p+q can be 0 but is not necessarily so, then this statement alone is not sufficient to answer the question. (Ensure you understand this)

So we can try and solve these two as simultaneous equations. If we get that they hold for all values of p and q, then the answer is yes. If we get that they hold for no real values of p and q, then the answer is no. If we get some specific values of p and q, the answer is may be and the statement alone is not sufficient.

p(1+q) = 1 ......(1)
p+q = 0
Substitute q = -p in (1)
p^2 - p + 1 = 0
This equation has no real values for p (Discriminant = b^2 - 4ac is negative). Hence there are no values of p and q for which p+q can be 0 and p(1+q) can be 1. If p(1+q) is 1, p+q cannot be 0. Hence this statement alone is sufficient to answer the question with 'No'.

Similarly, analyze statement 2
(2) 2q = 1/1- p
2q(1-p) = 1 ....(1)
p+q = 0
Substitute q = -p in (1)
-2p(1-p) -1 = 0
2p^2 - 2p -1 = 0
Here the discriminant will be positive and the equation will have two values for p. Hence p+q may be 0 (if p takes one of these particular values) and p+q may not be 0 (if p doesn't take these values)
Hence this statement alone is not sufficient.

Answer (A)

bumpbot · Answer

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Is p + q = 0 ?

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