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Math Expert V
Joined: 02 Sep 2009
Posts: 58390
If p, q, r, and s are consecutive integers, with p < q < r < s, is pr  [#permalink]

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11 00:00

Difficulty:   85% (hard)

Question Stats: 56% (02:43) correct 44% (02:30) wrong based on 271 sessions

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Tough and Tricky questions: Inequalities.

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) $$ps \lt qr$$

Kudos for a correct solution.

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Math Expert V
Joined: 02 Sep 2009
Posts: 58390
Re: If p, q, r, and s are consecutive integers, with p < q < r < s, is pr  [#permalink]

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Bunuel wrote:

Tough and Tricky questions: Inequalities.

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) $$ps \lt qr$$

Kudos for a correct solution.

Official Solution:

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

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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Senior Manager  Joined: 13 Jun 2013
Posts: 266
Re: If p, q, r, and s are consecutive integers, with p < q < r < s, is pr  [#permalink]

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Bunuel wrote:

Tough and Tricky questions: Inequalities.

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) $$ps \lt qr$$

Kudos for a correct solution.

let p=k-1, q=k, r=k+1, s=k+2

now question states is pr<qs

(k-1)(k+1)<k(k+2)
k^2 - 1 < k^2 + 2k
or k>-1/2
or the question becomes is k>-1/2

st.1
pq<rs
(k-1)k<(k+1)((k+2)
k^2 -k < k^2 + 3k + 2
or 4k>-2
or k>-1/2

hence statement 1 alone is sufficient

st.2

ps<qr

actually this statement doesn't provide us any new information. because for any 4 set of consecutive integers the above inequality will always hold true. for e.g. let 4 consecutive integers be -5,-4,-3,-2.
as can be seen (-5)*(-2)<(-4)*(-3).

hence statement 2 alone is not sufficient.

hence answer should be A
Manager  Joined: 22 Sep 2012
Posts: 123
Concentration: Strategy, Technology
WE: Information Technology (Computer Software)
Re: If p, q, r, and s are consecutive integers, with p < q < r < s, is pr  [#permalink]

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For the set of consecutive integers, p,q,r,s, we have find out whether pr < qs is true
Statement 1: pq < rs
This statement is true if
a) All the integers are positive. In that case, pr < qs is true
b) When the numbers are p= -1, q= 0, r= 1, s= 2. In this case also, pr < qs is true

There are not other sets of consecutive integers that fulfills the condition of pq<rs. Since, from this condition we always get pr < qs as true, this statement is sufficient.

Statement 2 : ps < qr
Here are the cases:
1. p= -1, q= 0, r= 1, s= 2 ( pr < qs is true )
2. p= -2, q= -1, r= 0, s= 1 ( pr < qs is false)
3. p= -3, q= -2, r= -1, s= 0 ( pr < qs is false)
.......
Since, we get both true and false from this condition. This condition is insufficient.

Hence A) is the answer
Intern  Joined: 10 Mar 2014
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Concentration: General Management, Technology
GMAT 1: 650 Q47 V32 GPA: 4
Re: If p, q, r, and s are consecutive integers, with p < q < r < s, is pr  [#permalink]

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lets have consecutive numbers be: p, p+1,p+2,p+3

now pr < qs => p(p+2) < (p+1) (p+3) => p > -3/2

Statement1: p(p+1) < (p+2) (p+3) => p> -3/2 Sufficient

Statement2: p(p+3) < (p+1) (P+2) => which eliminates all P, so we can not conclude anything.

Hence answer is A.
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Re: If p, q, r, and s are consecutive integers, with p < q < r < s, is pr  [#permalink]

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Given that, p , q , r & s are consecutive integers & is pr<qs ?

SO solving pr<qs
p(q+s)/2<s(p+r)/c ; since they are consecutive hence can be simplified to their mean.
pq+ps<ps+sr
hence, pq<rs.............A

1) pq<rs,
This statement proves equation A. SUFFICIENT.

2) ps<qr
There is no fruitful link with equation A.INSUFFICIENT.

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Re: If p, q, r, and s are consecutive integers, with p < q < r < s, is pr  [#permalink]

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Bunuel wrote:

Tough and Tricky questions: Inequalities.

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) $$ps \lt qr$$

Kudos for a correct solution.

picked A...didn't think that much..
if the consecutive numbers start with negative values and reach positive values, then we can't know for sure anything...
1. clearly states that all the numbers are positive - sufficient.
2. not much offered.

A
Intern  B
Joined: 10 Aug 2017
Posts: 17
GMAT 1: 720 Q48 V41 Re: If p, q, r, and s are consecutive integers, with p < q < r < s, is pr  [#permalink]

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I have a question about how to approach it at an actual exam.

The above explanations are very mathematical and accurate, but I don't think I will have time to really think it through like that.

My method of getting through this problem was
1. noticing that they could be negative or positive (all or in part)
2. plug in two sets of numbers to explore sufficiency of each option.

Am I putting myself at a risk of missing something crucial by plugging it in?
Math Expert V
Joined: 02 Aug 2009
Posts: 7961
Re: If p, q, r, and s are consecutive integers, with p < q < r < s, is pr  [#permalink]

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HappyQuakka wrote:
I have a question about how to approach it at an actual exam.

The above explanations are very mathematical and accurate, but I don't think I will have time to really think it through like that.

My method of getting through this problem was
1. noticing that they could be negative or positive (all or in part)
2. plug in two sets of numbers to explore sufficiency of each option.

Am I putting myself at a risk of missing something crucial by plugging it in?

Hi..

for saving time, one should exactly do the way you have mentioned...

the MAIN point of these Q where consecutive integers are involved is they could be positive or negative and plugging 2-3 sets of number should tell you..
the center TWo would always be greater than product of !st and 4th, so statement II would not tell us whether the numbers are positive or negative..

a set of number in each positive integers and negative integers would tell you that statement I will give you a clearcut answer.

Having said that, it is not always that such method will work. If these were not given as consecutive integers, there would have been many options possible. there may have been some other info then in statements to get to your answer
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Re: If p, q, r, and s are consecutive integers, with p < q < r < s, is pr  [#permalink]

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_________________ Re: If p, q, r, and s are consecutive integers, with p < q < r < s, is pr   [#permalink] 05 Oct 2018, 15:59
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