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If j, k, m, and n are consecutive integers and j < k < m < n, then j +
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07 Sep 2022, 00:16
07 Sep 2022, 00:16
Expert Reply
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Question Stats:
85% (00:43) correct
14% (00:27) wrong based on 7 sessions
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If j, k, m, and n are consecutive integers and j < k < m < n, then j + m is how much less than k + n?
(A) 8
(B) 4
(C) 2
(D) 1
(E) 0
(A) 8
(B) 4
(C) 2
(D) 1
(E) 0
Part of the project: GRE QCQ & PS 3 Questions Every One DAILY Challenge - (2022) - Gain 20 Kudos & Get FREE Access to GRE Prep Club TESTS
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GRE Math Essentials (2022)
GRE Geometry Formulas
GRE - Math Book
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GRE Math Essentials (2022)
GRE Geometry Formulas
GRE - Math Book
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
Re: If j, k, m, and n are consecutive integers and j < k < m < n, then j +
[#permalink]
11 Sep 2022, 01:55
11 Sep 2022, 01:55
1
Expert Reply
Since \(j, k, m, n\) are consecutive integers and \(j<k<m<n\)
assume that j=a then
\(k=a+1\)
\(m=a+2\)
\(n=a+3\)
We need to find the difference
\((k+n)-(j+m)\)
= \((a+1+a+3) - (a+a+2)\) {by substituting values}
= \((2a+4)-(2a+2)\)
= \(2a+4-2a-2\)
= \(2\)
So the answer is 2
Answer: C
assume that j=a then
\(k=a+1\)
\(m=a+2\)
\(n=a+3\)
We need to find the difference
\((k+n)-(j+m)\)
= \((a+1+a+3) - (a+a+2)\) {by substituting values}
= \((2a+4)-(2a+2)\)
= \(2a+4-2a-2\)
= \(2\)
So the answer is 2
Answer: C
gmatclubot
Re: If j, k, m, and n are consecutive integers and j < k < m < n, then j + [#permalink]
11 Sep 2022, 01:55
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