Advanced Math

[morrowgsm]

I read the explanation - however, I don't believe that 'A' could be the answer because what if the consecutive numbers were:
-2,-1,0,1.. then 0 > -1, however, if the numbers were 1,2,3,4 then 3<8.

06/22/09
Question
If p, q, r, and s are consecutive integers, with p < q < r < s, is pr < qs?

(1) pq < rs
(2) ps < qr

[lalitkc]

Take -4,-3,-2-,-1 as the 4 consecutive integers. The answer is A

[morrowgsm]

the numbers can be -4,-3,-2,-1 as well but do we know that those are the numbers?

-2,-1,0,1.. then 0 > -1, however, if the numbers were 1,2,3,4 then 3<8.

[shaji]

If p, q, r, and s are consecutive integers, with p < q < r < s and pq < rs;then r>0.5 and cannot be 0. In fact pr<qs making statement 1 sufficient.
Statement 2 give no information other than the universal truth 2>0.
The correct answer is indeed A.

[Kweku.Amoako]

P<q<r<s
If they are consecutive
Then q = p+1 , r = p+2 and s = p+3
The question is pr <qs let plug in
Is p(p+2) < (p+1)(p+3) lets simplify
Is p^2 + 2p < P^2 +4p +3 --> -2p < 3 --> p>-1.5

So we can rephrase the question to --> is p >-1.5

(1) pq <rs
p(p+1) < (p+2)(p+3)
p^2 +p < p^2 +5p+6
-4p <6
p>-1.5 SUFFICIENT

(2) ps <qr
P(P+3) < (P+1)(P+2)
P^2+3P < P^2+3P+2
0 <2
This tells us nothing. Insufficient

Answer = A

[RonPurewal]

P<q<r<s
If they are consecutive
Then q = p+1 , r = p+2 and s = p+3
The question is pr <qs let plug in
Is p(p+2) < (p+1)(p+3) lets simplify
Is p^2 + 2p < P^2 +4p +3 --> -2p < 3 --> p>-1.5

So we can rephrase the question to --> is p >-1.5

(1) pq <rs
p(p+1) < (p+2)(p+3)
p^2 +p < p^2 +5p+6
-4p <6
p>-1.5 SUFFICIENT

(2) ps <qr
P(P+3) < (P+1)(P+2)
P^2+3P < P^2+3P+2
0 <2
This tells us nothing. Insufficient

Answer = A

well done.

[RonPurewal]

I read the explanation - however, I don't believe that 'A' could be the answer because what if the consecutive numbers were:
-2,-1,0,1.. then 0 > -1, however, if the numbers were 1,2,3,4 then 3<8.

your first example is out-of-bounds for statement (1), because it does not satisfy the condition pq < rs. (using those numbers, that would be 2 < 0.)

[jyothi h]

P<q<r<s
If they are consecutive
Then q = p+1 , r = p+2 and s = p+3
The question is pr <qs let plug in
Is p(p+2) < (p+1)(p+3) lets simplify
Is p^2 + 2p < P^2 +4p +3 --> -2p < 3 --> p>-1.5

So we can rephrase the question to --> is p >-1.5

(1) pq <rs
p(p+1) < (p+2)(p+3)
p^2 +p < p^2 +5p+6
-4p <6
p>-1.5 SUFFICIENT

(2) ps <qr
P(P+3) < (P+1)(P+2)
P^2+3P < P^2+3P+2
0 <2
This tells us nothing. Insufficient

Answer = A

well done.

I had a doubt on this. If they state that all 4 numbers are positive integers , then even though the method mentioned above doesn't prove statement 2 to be sufficient , logically statement 2 should be sufficient , right ?
since statement 2 says : p*s < q*r
so it is obvious that p*r ( since r is less than s) is < q*s ( since s is greater than r)
Could you please clarify on this ?

[tim]

sounds right. so if we change the question to one where we have only positive integers, then indeed 2 would be sufficient. this is a good way to study math problems by the way - take the original question and add a twist to make it a slightly different challenge..