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25 Oct 2010, 02:15
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
66% (01:19) correct
34%
(01:37)
wrong
based on 140
sessions
History
Date
Time
Result
Not Attempted Yet
Three consecutive even numbers are such that thrice the first number exceeds double the third by two, the third number is
a) 12
b) 6
c) 8
d) 10
e) 14
a) 12
b) 6
c) 8
d) 10
e) 14
Ellipse
Joined: 16 Oct 2010
Last visit: 07 Jul 2013
Posts: 64
Given Kudos: 3
Location: United States
Concentration: Finance, Entrepreneurship
Schools: INSEAD Jan '13 IE '15 Kellogg 1YR '15 Sloan '16
GMAT 1: 700 Q49 V35
WE:Information Technology (Finance: Investment Banking)
25 Oct 2010, 03:16
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feruz77
Here you go:
Let's say the first even number is P. So the next two numbers would be P+2 and P+4.
The question says-
3P = 2(P+4) + 2
P = 10
Hence the third number would be - P + 4 => 14
You can also solve this by plugging-in the answers.
22 Mar 2017, 12:41
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feruz77
Let the first even # be \(x\) (so the 3 consecutive integers are \(x\), \(x+2\), and \(x+4\)), then: \(3x=2(x+4)+2\) --> \(x=10\) --> third # is \(x+4=10+4=14\).
OR:
If we take the first integer to be \(x-2\) (so the 3 consecutive integers are \(x-2\), \(x\), and \(x+2\)) then: \(3(x-2)=2(x+2)+2\) --> \(x=12\) --> third # is \(x+2=12+2=14\)
Answer: E.
Hope it's clear.
