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Manager
Joined: 13 May 2017
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p and q are positive integers such that 2p 10 > q, and 3q 20 < p.
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27 Nov 2018, 07:06
Question Stats:
35% (01:40) correct 65% (02:31) wrong based on 91 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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Joined: 02 Oct 2018
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Re: p and q are positive integers such that 2p 10 > q, and 3q 20 < p.
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29 Nov 2018, 04:33
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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Joined: 09 Mar 2016
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p and q are positive integers such that 2p 10 > q, and 3q 20 < p.
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30 Nov 2018, 00:50
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 2q10<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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Joined: 14 Jun 2018
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Re: p and q are positive integers such that 2p 10 > q, and 3q 20 < p.
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30 Nov 2018, 06:42
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.
D



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Re: p and q are positive integers such that 2p 10 > q, and 3q 20 < p.
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30 Nov 2018, 08:51
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.  2p10 > q => 2p + q > 10 (*) 3q20 <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 > 106 = 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 (PR)>(QS)



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Re: p and q are positive integers such that 2p 10 > q, and 3q 20 < p.
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30 Nov 2018, 09:11
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 2q10<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.
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30 Nov 2018, 12:53
thanks a lot Brent . have an awesome weekend !




Re: p and q are positive integers such that 2p 10 > q, and 3q 20 < p. &nbs
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30 Nov 2018, 12:53






