pbanavara wrote:If p,q,r, and s are nonzero numbers, is
(p – 1)(q - 2)2(r – 3)3(s – 4)4 ≧ 0?
(1) q > 2 and s > 4
(2) p > 1 and r > 3


How do you guys go about solving such problems ? Plugging in actual values seemed to take a lot of time ..

- pradeep

Pradeep,

if you look at the statement you will see how they match with the question stem

q>2 hmmm so it means q-2 > 0 right ?

Please apply this hint to the rest of the questions and see what you got. If you get stucked somewhere post your solution and I will be more than happy to help you.

Cramya will proud of me today!

(q-2)^2 and (s-4)^4 will always be greater than 0
so we can ignore this part

we need to test
(p – 1)(r – 3)3 ≧ 0

A
no info for p and r
so insuff

B
p>1
and r>3
so
(p – 1)(r – 3)3 >0
hence suff

soa and is B

logitech wrote:

pbanavara wrote:If p,q,r, and s are nonzero numbers, is
(p – 1)(q - 2)2(r – 3)3(s – 4)4 ≧ 0?
(1) q > 2 and s > 4
(2) p > 1 and r > 3


How do you guys go about solving such problems ? Plugging in actual values seemed to take a lot of time ..

- pradeep

Pradeep,

if you look at the statement you will see how they match with the question stem

q>2 hmmm so it means q-2 > 0 right ?

Please apply this hint to the rest of the questions and see what you got. If you get stucked somewhere post your solution and I will be more than happy to help you.

Cramya will proud of me today!

Thanks - That helps : The question is actually
(p – 1)(q - 2)^2(r – 3)^3(s – 4)^4 ≧ 0?

Stmnt 1 : q-2 > 0 and s-4 > 0 => p -1 could be < 0 and r-3 < 0
so the value of the equation could be >0 <0 or =0

Stmnt 2 : r-1 >0 and s-3 > 0 implies the product has to be >= 0 because no matter what value q and s have they will be positive.

So the answer is B - which is the OA btw

stop@800 wrote:(q-2)^2 and (s-4)^4 will always be greater than 0
so we can ignore this part

we need to test
(p – 1)(r – 3)3 &#8807; 0

A
no info for p and r
so insuff

B
p>1
and r>3
so
(p – 1)(r – 3)3 >0
hence suff

soa and is B

Stop my man:

1) You ruined the poster's homework
2) How did you come up with (q-2)^2 and (s-4)^4 ?

:roll:

Cramya will proud of me today!

U bet I am. I am almost in tears after seeing your efforts to explain.....

pbanavara wrote:

logitech wrote:

pbanavara wrote:If p,q,r, and s are nonzero numbers, is
(p – 1)(q - 2)2(r – 3)3(s – 4)4 &#8807; 0?
(1) q > 2 and s > 4
(2) p > 1 and r > 3


How do you guys go about solving such problems ? Plugging in actual values seemed to take a lot of time ..

- pradeep

Pradeep,

if you look at the statement you will see how they match with the question stem

q>2 hmmm so it means q-2 > 0 right ?

Please apply this hint to the rest of the questions and see what you got. If you get stucked somewhere post your solution and I will be more than happy to help you.

Cramya will proud of me today!

Thanks - That helps : The question is actually
(p – 1)(q - 2)^2(r – 3)^3(s – 4)^4 &#8807; 0?

Stmnt 1 : q-2 > 0 and s-4 > 0 => p -1 could be < 0 and r-3 < 0
so the value of the equation could be >0 <0 or =0

Stmnt 2 : r-1 >0 and s-3 > 0 implies the product has to be >= 0 because no matter what value q and s have they will be positive.

So the answer is B - which is the OA btw

That's right my man!! YOU GOT IT!

I am sure you will apply this take-away to different questions down the hill.

logitech wrote:

stop@800 wrote:(q-2)^2 and (s-4)^4 will always be greater than 0
so we can ignore this part

we need to test
(p – 1)(r – 3)3 &#8807; 0

A
no info for p and r
so insuff

B
p>1
and r>3
so
(p – 1)(r – 3)3 >0
hence suff

soa and is B

Stop my man:

1) You ruined the poster's homework

Sorry, Actually when I started replying your reply was not there

2) How did you come up with (q-2)^2 and (s-4)^4 ?

I think its clear now...

:roll: