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p and q are positive integers such that 2p -10 > -q, and 3q -20 < -p.

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p and q are positive integers such that 2p -10 > -q, and 3q -20 < -p.  [#permalink]

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New post 27 Nov 2018, 08:06
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Question Stats:

35% (01:43) correct 65% (02:30) wrong based on 93 sessions

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p and q are positive integers such that 2p -10 > -q, and 3q -20 < -p. If m is the minimum possible value of p and n is the maximum possible value of q, then which of the following pairs accurately represents (m, n)


A (1, 6)
B (6, 2)
C (3, 6)
D (3, 5)
E No such values exist.
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Re: p and q are positive integers such that 2p -10 > -q, and 3q -20 < -p.  [#permalink]

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New post 29 Nov 2018, 05:33
rencsee wrote:
p and q are positive integers such that 2p -10 > -q, and 3q -20 < -p. If m is the minimum possible value of p and n is the maximum possible value of q, then which of the following pairs accurately represents (m, n)


A (1, 6)
B (6, 2)
C (3, 6)
D (3, 5)
E No such values exist.



Could some pls post the solution?
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p and q are positive integers such that 2p -10 > -q, and 3q -20 < -p.  [#permalink]

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New post 30 Nov 2018, 01:50
1
rencsee wrote:
p and q are positive integers such that 2p -10 > -q, and 3q -20 < -p. If m is the minimum possible value of p and n is the maximum possible value of q, then which of the following pairs accurately represents (m, n)


A (1, 6)
B (6, 2)
C (3, 6)
D (3, 5)
E No such values exist.



GMATPrepNow Brent, i applied your concept


KEY CONCEPT: If the inequality signs of two inequalities are facing the SAME DIRECTION, then we can ADD those inequalities to create a new inequality.

3q -20 < -p
2p -10 > -q multiply both sides by -1 so we get -2p+10<q

so now we have

3q -20 < -p
-2p+10<q ---- > rearrange -q+10<-2p

add both inequelities

3q -20 < -p
-q+10<-2p

after addition we have

2q-10<-3p


so if i plug in any of the options below, where p is first term and q is second term i.e. 1 = p, q=6 , so non of the options is TRUE

A (1, 6) 1 = p, q=6
B (6, 2) 6 = p, q=2
C (3, 6) 3 = p, q=6
D (3, 5) 3 = p, q=5


whats wrong with my reasoning now :?
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Re: p and q are positive integers such that 2p -10 > -q, and 3q -20 < -p.  [#permalink]

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New post 30 Nov 2018, 07:42
3
2p -10 > -q
=> 2p + q > 10 --- 1

3q -20 < -p
=> -p - 3q > -20 -- 2

Make coefficient of 1 & 2 same for a single variable

2p+ q > 10
-2p - 6q > - 40

-5q > - 30
q < 6 ----- 3

Now ,
m = minimum possible value of p
n = maximum possible value of q

Maximum possible value of q is 5 = n

Put 5 in the equation 1
=> 2p + 5 > 10
p >2.5

Minimum possible value of p is 3 = m

(3,5) is the required answer.

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Re: p and q are positive integers such that 2p -10 > -q, and 3q -20 < -p.  [#permalink]

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New post 30 Nov 2018, 09:51
1
rencsee wrote:
p and q are positive integers such that 2p -10 > -q, and 3q -20 < -p. If m is the minimum possible value of p and n is the maximum possible value of q, then which of the following pairs accurately represents (m, n)


A (1, 6)
B (6, 2)
C (3, 6)
D (3, 5)
E No such values exist.



----
2p-10 > -q => 2p + q > 10 (*)
3q-20 <-p => -3q -p > -20 => -2p - 6q > -40 (**)
Add (*) and (**) we have: -5q > -30 => q<6 => MAX q = 5 (n=5)

2p + q > 10 & q<6 => 2p + q - q > 10-6 = 4 => 2p > 4 then p > 2


FINALLY we have q<6 (MAX q = 5 =n) & p>2 then D satisfies this condition.

***********If P>Q and R<S then (P-R)>(Q-S)
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Re: p and q are positive integers such that 2p -10 > -q, and 3q -20 < -p.  [#permalink]

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New post 30 Nov 2018, 10:11
1
Top Contributor
dave13 wrote:
rencsee wrote:
p and q are positive integers such that 2p -10 > -q, and 3q -20 < -p. If m is the minimum possible value of p and n is the maximum possible value of q, then which of the following pairs accurately represents (m, n)


A (1, 6)
B (6, 2)
C (3, 6)
D (3, 5)
E No such values exist.



GMATPrepNow Brent, i applied your concept


KEY CONCEPT: If the inequality signs of two inequalities are facing the SAME DIRECTION, then we can ADD those inequalities to create a new inequality.

3q -20 < -p
2p -10 > -q multiply both sides by -1 so we get -2p+10<q

so now we have

3q -20 < -p
-2p+10<q ---- > rearrange -q+10<-2p

add both inequelities

3q -20 < -p
-q+10<-2p

after addition we have

2q-10<-3p


so if i plug in any of the options below, where p is first term and q is second term i.e. 1 = p, q=6 , so non of the options is TRUE

A (1, 6) 1 = p, q=6
B (6, 2) 6 = p, q=2
C (3, 6) 3 = p, q=6
D (3, 5) 3 = p, q=5


whats wrong with my reasoning now :?


I have highlighted the error above, in green.

If we rearrange -2p+10<q, we don't get -q+10<-2p

Cheers,
Brent
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Re: p and q are positive integers such that 2p -10 > -q, and 3q -20 < -p.  [#permalink]

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New post 30 Nov 2018, 13:53
thanks a lot Brent :). have an awesome weekend ! :)
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Re: p and q are positive integers such that 2p -10 > -q, and 3q -20 < -p.   [#permalink] 30 Nov 2018, 13:53
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p and q are positive integers such that 2p -10 > -q, and 3q -20 < -p.

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