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.
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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!

Great Question
Here y=0.jklm
To get y we need j,k,l,m
Statement 1
using test cases
y=0.9876
y=0.8765
Clearly not sufficient
Statement 2
This statement looks interesting
Here j>1
and a=> (1,j)
so least possible value of j is 3
a=2
=> y=0.3579
Hmmm
let j=4
a=2
=> y=0.468[10]=> not posible as m is a single digit integer.
hence the only possible value of y is 0.3579

Hence Sufficient
Hence B
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1) Insufficient as it could be 2<3<4<5 or 3<4<5<6 or 4<5<6<7 etc
2)
j+a=k
k+a=m
m+a=n
j>a>1

a=2 ;j=3
3+2=k -> k=5
5+2=m -> m=7
7+2=n -> n=9

jkmn=3759 -> y=0.3579 -> Statement 2 is sufficient -> B

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

Clearly insufficient.
j = 1, k = 2, m = 3, n = 4
j = 2, k = 3, m = 4, n = 5

(2) j + a = k, k + a = m, and m + a = n, where j > a > 1.

Sufficient.
j = 3, a = 2, m = 7, n = 9 YES
j = 4, a = 2, m = 8, n = 10 NO (n cannot be a two-digit integer)

B.
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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
j = 6 k = 7, m = 8, n = 9
j = 5, k = 6, m = 7, n = 8 , so this is not sufficient.

(2) j + a = k, k + a = m, and m + a = n, where j > a > 1.

j = 3, a = 2, m = 7, n = 9 YES
j = 4, a = 3, m = 7, n = 10 NO (n cannot be a two-digit integer)
so the only possible values will be x=0.3579 , Sufficient.

Ans B.