S99-18

Question

If  p ,  q ,  r , and  s  are consecutive integers, with  p \lt q \lt r \lt s , is  pr \lt qs ?

(1)  pq \lt rs

(2)  ps \lt qr

Bunuel · Answer

Official Solution:

We can solve this problem either by algebra or by number-plugging. Let's use algebra. All four variables can be expressed in terms of just one variable, since they are consecutive integers and we know their order. If we keep  p  as the basic variable, then  q = p + 1 ,  r = p + 2 , and  s = p + 3 .
 
Now we can rephrase the question:
 
Is  pr \lt qs ? 
 
Is  p(p + 2) \lt (p + 1)(p + 3) ?
 
Is  p^2 + 2p \lt p^2 + 4p + 3 ? 
 
Is  2p \lt 4p + 3 ?
 
Is  0 \lt 2p + 3 ?
 
Is  -3 \lt 2p ?
 
Is  -\frac{3}{2} \lt p ?
 
Since  p  is an integer, the question is answered "yes" if  p = -1  or greater, and "no" if  p = -2  or less. 
 
Statement (1): SUFFICIENT.
 
We rephrase the statement similarly.
  pq \lt rs 
  p(p + 1) \lt (p + 2)(p + 3) 
  p^2 + p \lt p^2 + 5p + 6 
  0 \lt 4p + 6 
  0 \lt 2p + 3 
  -\frac{3}{2} \lt p 
 
This is precisely the same condition as asked in the question. Thus, we can answer the question definitively.
 
Statement (2): INSUFFICIENT.
 
Again, we rephrase the statement similarly.
  ps \lt qr 
  p(p + 3) \lt (p + 1)(p + 2) 
  p^2 + 3p \lt p^2 + 3p + 2 
  0 \lt 2 
 
Since 0 is always less than 2, no matter the value of  p , the statement is always true. Thus, we do not gain any information that would help us answer the question.

Answer: A

Yadnesh · Answer

We can solve this problem either by algebra or by number-plugging. Let's use algebra. All four variables can be expressed in terms of just one variable, since they are consecutive integers and we know their order. If we keep p as the basic variable,

Bunuel - how do you suggest to solve this by number plugging.

Also im not great at using number plugging to solve questions and generally take the algaebric ways which costs me more time and is very unfeasibe in some questions.

Are you aware of a bank of questions for me to practice number plugging and also some tips and tricks about the same.
Thanks in advance . You are a saviour.

Arvind90 · Answer

Shoudn't the question have specified that the integers can't be 0?

pratistha29 · Answer

if we use number plugging method the ans will be B 
1st statement: pq<rs
let the values be 1,2,3,4
then 2<12 its true.
assumen values be -2,-1,0,1
then 2>0 the statement is not sufficient.

2nd statement: ps<qr
values be 1,2,3,4
then 4<6 true.
values be -2,-1,0,1
then -2<0 hence statement is sufficient.
please help me where i am wrong

Bunuel · Answer

With 1,2,3,4 you have and YES answer to the question ( pr \lt qs ) while with -2,-1,0,1 you get a NO answer to the question ( pr \gt qs ). This means that the second statement is not sufficient.

Tusmat · Answer

If we wish to use number plugging, then we have to consider 4 different cases for values for p q r s.

Case 1 : 1, 2, 3, 4

Case 2 : -4, -3, -2, -1

Case 3 : -2, -1, 0, 1

Case 4 : -1, 0, 1, 2

Now, 
check for statement 1 : pq<rs

For all the cases for which pq<rs holds true, pr<qs should give the same answer  . 
Here, it does. Hence, SUFFICIENT.

Now, 
check for statement 2 : ps<qr

Here, all the cases for which ps<qr holds true donot give the same answer to 'Is pr<qs ?'. 
Hence, INSUFFICIENT.

Therefore, the answer ====> A

Sarjaria84 · Answer

Hi  Bunuel

How were you able to subtract 2p from both the sides in the highlighted portion without knowing the sign of P. The question does not specify that it is a positive intiger.

Thanks
Saurabh

Bunuel · Answer

We can safely add or subtract any value to both sides of inequality. For example, 3 > 1. We can add any x to both sides to get 3 + x > 1 + x. I think you are mixing adding/subtracting with dividing/multiplying.

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