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# S99-18

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Math Expert
Joined: 02 Sep 2009
Posts: 59590

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16 Sep 2014, 01:53
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Difficulty:

95% (hard)

Question Stats:

41% (02:28) correct 59% (01:39) wrong based on 86 sessions

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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) $$ps \lt qr$$

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16 Sep 2014, 01:53
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.

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01 Oct 2014, 02:41
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.
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Joined: 27 Jan 2016
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17 Oct 2016, 08:13
Shoudn't the question have specified that the integers can't be 0?
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25 Nov 2016, 02:53
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.
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25 Nov 2016, 06:17
pratistha29 wrote:
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.

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.
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02 Oct 2018, 12:53
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

[NOTICE the difference between case 3 and 4 carefully]

Now,
check for statement 1 : pq<rs

For all the cases for which pq<rs holds true, pr<qs should give the same answer [either YES or NO].
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.

Re: S99-18   [#permalink] 02 Oct 2018, 12:53
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# S99-18

Moderators: chetan2u, Bunuel