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01 Apr 2012, 12:33
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
60% (01:58) correct
40%
(02:13)
wrong
based on 926
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If y = 0.jkmn, where j, k, m, and n each represent a nonzero digit of y, what is the value of y ?
(1) j < k < m < n
(2) j + a = k, k + a = m, and m + a = n, where j > a > 1.
(1) j < k < m < n
(2) j + a = k, k + a = m, and m + a = n, where j > a > 1.
01 Apr 2012, 12:56
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If y = 0.jkmn, where j, k, m, and n each represent a nonzero digit of y, what is the value of y ?
(1) j < k < m < n. Many combinations are possible: 1<2<3<4, 2<3<4<5, 1<3<4<5, ... Not sufficient.
(2) j + a = k, k + a = m, and m + a = n, where j > a > 1.
Since \(j+a=k\) and \(j\) and \(k\) represent digits then \(a\) must be an integer.
Next, since \(j > a > 1\) then the least value of \(a\) is 2 and the least value of \(j\) is 3.
So, from \(j+a=k\) the least value of \(k\) is 3+2=5, from \(k + a = m\) the least value of \(m\) is 5+2=7 and from \(m + a = n\) the least value of \(n\) is 7+2=9.
Now, if our initial number, \(a\), is more than 2, 3 for example, then the values of all other variables will increase and \(n\) will become more than 9, which is not possible since each variable represents a single nonzero digit.
Hence: \(j=3\), \(k=5\), \(m=7\), and \(n=9\). Sufficient.
Answer: B.
(1) j < k < m < n. Many combinations are possible: 1<2<3<4, 2<3<4<5, 1<3<4<5, ... Not sufficient.
(2) j + a = k, k + a = m, and m + a = n, where j > a > 1.
Since \(j+a=k\) and \(j\) and \(k\) represent digits then \(a\) must be an integer.
Next, since \(j > a > 1\) then the least value of \(a\) is 2 and the least value of \(j\) is 3.
So, from \(j+a=k\) the least value of \(k\) is 3+2=5, from \(k + a = m\) the least value of \(m\) is 5+2=7 and from \(m + a = n\) the least value of \(n\) is 7+2=9.
Now, if our initial number, \(a\), is more than 2, 3 for example, then the values of all other variables will increase and \(n\) will become more than 9, which is not possible since each variable represents a single nonzero digit.
Hence: \(j=3\), \(k=5\), \(m=7\), and \(n=9\). Sufficient.
Answer: B.
General Discussion
unceldolan
Joined: 21 Oct 2013
Last visit: 03 Jun 2015
Posts: 151
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Location: Germany
Schools: Owen '17 (A$) Sauder '17 (A$) Schulich '17 (WD) Anderson FEMBA '17 (S) McDonough '17 (A) Desautels '17 (S)
GMAT 1: 660 Q45 V36
GPA: 3.51
Schools: Owen '17 (A$) Sauder '17 (A$) Schulich '17 (WD) Anderson FEMBA '17 (S) McDonough '17 (A) Desautels '17 (S)
GMAT 1: 660 Q45 V36
Posts: 151
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16 Jan 2014, 04:36
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I use this pattern:
AD
BCE
cross out what's incorrect.
(1) j<k<m<n --> could be 1,2,3,4 or 2,3,4,5 hence IS --> cross out AD
possible answers left: BCE
(2) j + a = k, k + a = m, and m + a = n, where j > a > 1
Since we're talking about units digit, every variable here is an integer. hence a is at least 2 and j at least 3. If you simplify the given expressions you get j+3a = n. If j = 3 and a = 2 we get n = 9, which is the highest possible value (0 is not an option). Hence you can calculate the other numbers. SUFFICIENT. Cross out C and E.
Left with answer B.
I basically followed the same approach as bunuel, but simplified the expressions first. Hope it's clear!
AD
BCE
cross out what's incorrect.
(1) j<k<m<n --> could be 1,2,3,4 or 2,3,4,5 hence IS --> cross out AD
possible answers left: BCE
(2) j + a = k, k + a = m, and m + a = n, where j > a > 1
Since we're talking about units digit, every variable here is an integer. hence a is at least 2 and j at least 3. If you simplify the given expressions you get j+3a = n. If j = 3 and a = 2 we get n = 9, which is the highest possible value (0 is not an option). Hence you can calculate the other numbers. SUFFICIENT. Cross out C and E.
Left with answer B.
I basically followed the same approach as bunuel, but simplified the expressions first. Hope it's clear!
