Ritwik wrote:Is pq = 1?
(1) pqp = p

(2) qpq = q


OA is E

I have found lot many posts on this topic.

solution of (1) is p = 0 or pq = 1
solution of (2) is q = 0 or pq = 1

Combining, MGMAT says the answer is E.

Explanation from experts:
we could certainly pick p=q=1 to get a "yes" answer to the question.

However, we could also pick p=q=0 to get a "no" answer to the question.

Therefore, even after combining we don't have enough info to solve: choose E.

I disagree here:

Lets take another similar example:
is x=2 ?
1) (x-1)(x-2)=0
2) (x-2)(x-3)=0

I will mark answer of the above question as C.

as 1) gets x=1 or x=2
2) gets x=2 or x=3

Combining we get x=2 , hence answer is C.
yes in this case answer will be c
but for previous question c in DS question if you have yes and no both for the given statements .this means it cannot be answer as answer can be 1 only

In the original pq question the answer should be C as after combining we will get pq=1 solution.

c the bold

As the questions stands the answer should be D. Howver, if the question is asking whether P=1 or Q=1, then the answer would be E. Experts please intervene. Thanks for the question.

Ritwik wrote:Is pq = 1?
(1) pqp = p

(2) qpq = q


OA is E

Statement 1: pqp = p

We could use p=0, q=0, because then pqp=0*0*0=0. Does pq=1? No, because 0*0=0.
We could use p=1, q=1, because then pqp=1*1*1=1. Does pq=1? Yes, because 1*1=1.
Since the answer can be both yes and no, INSUFFICIENT.

Statement 2: qpq = q

We could use p=0, q=0, because then qpq=0*0*0=0. Does pq=1? No, because 0*0=0.
We could use p=1, q=1, because then qpq=1*1*1=1. Does pq=1? Yes, because 1*1=1.
Since the answer can be both yes and no, INSUFFICIENT.

Statement 1 and 2 together:

Since p=0, q=0 works in both statements, making pq=0, and p=1, q=1 works in both statements, making pq=1, we can't tell whether pq=1. INSUFFICIENT.

The correct answer is E.

Hope this helps!

GMATGuruNY and others,

We can reduce the equations in stmts 1 and 2 to pq = 1 by dividing both part of the equations by whats common on each side of the equation thus simplifying it. Why is there an exception to the rule in this case.

Thanks!

Haaress wrote:GMATGuruNY and others,

We can reduce the equations in stmts 1 and 2 to pq = 1 by dividing both part of the equations by whats common on each side of the equation thus simplifying it. Why is there an exception to the rule in this case.

Thanks!

Statement 1: pqp=p

If p=q=0, we can't simply statement 1 by dividing each side of the equation by p. Why? Because then we'd be dividing by 0, and division by 0 is undefined.

Statement 2: qpq=q

If p=q=0, we can't simply statement 2 by dividing each side of the equation by q. Why? Because then we'd be dividing by 0, and division by 0 is undefined.

The situation above illustrates one danger of approaching a DS question algebraically: you could break a rule of math even without realizing it! Plugging in real values is much safer.

truly said i don't get this quesion.

Is pq = 1?
(1) pqp = p
dividing each side to "p" and getting pq=1 , sufficient

(2) qpq = q
dividing each side to "q", and getting pq=1 , sufficient

i know i'm missing something, but i don't know what exactly.

blaster wrote:truly said i don't get this quesion.

Is pq = 1?
(1) pqp = p
dividing each side to "p" and getting pq=1 , sufficient

(2) qpq = q
dividing each side to "q", and getting pq=1 , sufficient

i know i'm missing something, but i don't know what exactly.

Hi Blaster,

Let me try to put it in simple terms for you.
pqp = p will mean pq=1 only if p is not equal to zero. Let me explain how.
bring the p on to the LHS.

you will see pqp-p=0 which is same as p(pq-1)=0.
So either p has to be 0 or pq-1 has to be 0.
This is what is being said in all the above posts.

If you dont get this, let me give an example, where cancelling out zero on both sides is a danger.
let me try to prove that 1=2
what is 1*0 ?
0
What is 2*0 ?
0

So 1*0 = 2*0 . cancel out zero on both sides. You get 1=2.

We know that this cannot be true. So 0 cannot be cancelled out on both sides.

Hope this helps!!

blaster wrote:truly said i don't get this quesion.

Is pq = 1?
(1) pqp = p
dividing each side to "p" and getting pq=1 , sufficient

(2) qpq = q
dividing each side to "q", and getting pq=1 , sufficient

i know i'm missing something, but i don't know what exactly.

Let me try to help.

What if we were asked to solve the following for p?

5p = p.

If we divide each side of the equation by p, we get:

5 = 1. Huh?

How can 5=1? It can't. How did we got such a nonsensical answer? Because when we simplified the equation by dividing each side by p, we broke a fundamental rule of math. The only value of p that works in 5p=p is 0; when we divided each side of the equation by p, we were dividing by 0. And division by 0 is undefined.

Before we use algebra, we have to understand how the problem is restricted; otherwise, we could break a rule or miss something.

Here's another example:

y^2 = y.

If we divide each side by y, we get:

y = 1.

But this isn't the only possible of y. y=0 also works, because 0^2=0. Using algebra alone could cause us to miss a possible value of y.

It's important to recognize when a value could equal 0.

Getting back to our DS question above. Since p=q=0 works in both statements, we can't just divide by p or q. Because then we'd be dividing by 0.

Does this help?

thanks, i got it ))