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When k, l, and m are different positive integers greater than 1, k+l+m

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When k, l, and m are different positive integers greater than 1, k+l+m  [#permalink]

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New post 11 Jan 2016, 16:48
2
10
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

57% (01:14) correct 43% (01:13) wrong based on 449 sessions

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When k, l, and m are different positive integers greater than 1, k+l+m=?

1) km=15
2) kml=30


*A solution will be posted in two days.

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Re: When k, l, and m are different positive integers greater than 1, k+l+m  [#permalink]

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New post 11 Jan 2016, 19:50
k, l and m are different positive integers greater than

from statement 1.
km=15,
possible solution are 3 and 5 or 5 and 3. we don't know value of l.

from statement 2

klm=30

possible values are 2,3 and 5.
all are possible different integers and greater than 1.

we can take any possible varibles as 2,3 and 5 or 3,2 and 5 . so sum equals=10.

so option B is correct.
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Re: When k, l, and m are different positive integers greater than 1, k+l+m  [#permalink]

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New post 12 Jan 2016, 03:33
IMO E is correct
A. km =15
k can be 3 or 5 according to which m will be 5 or 3
insufficient

B, kml=30
k can be 2,3,5 according to which m nd l can also have multiple values

A+B
l =2
but again kor m can be anyone from 3 or 5


SO E
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Re: When k, l, and m are different positive integers greater than 1, k+l+m  [#permalink]

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New post 12 Jan 2016, 20:02
k, l and m are different positive integers greater than 1.

from st.1:
km=15
which means --> km = 3*5 or 5*3 (order doesn't matter as finally they will add up)
But no information on l..
So, NOT SUFFICIENT.

from st.2:
kml=30
which means --> kml = 2*3*5 or 5*3*2 or any other way with 3 different positive integers (greater than 1) 2,3 and 5 (again, order doesn't matter as finally they will add up)
this is the only way in which 3 factors are available for kml=30
so k+m+l = 2+3+5 = 10
Hence, SUFFICIENT.

Answer : B
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Re: When k, l, and m are different positive integers greater than 1, k+l+m  [#permalink]

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New post 13 Jan 2016, 18:58
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.


When k, l, and m are different positive integers greater than 1, k+l+m=?
1) km=15
2) kml=30

In the original condition, there are 3 variables, which should match with the number of equations. So you need 3 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make E the answer. When 1) & 2), k+l+m=10 si derived from (k,l,m)=(3,2,5),(5,2,3), which is unique and sufficient. Then the answer is C. Since this is an integer question which is one of the key questions, apply the mistake type 4(A).
In case of 1), you don’t know l, which is not unique and not sufficient.
In case of 2), in (k,l,m)=(2,3,5),(2,5,3),(3,2,5),(3,5,2),(5,2,3),(5,3,2), k+l+m=10 is derived, which is unique and sufficient. Therefore, the answer is B.


 For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
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Re: When k, l, and m are different positive integers greater than 1, k+l+m  [#permalink]

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New post 26 Aug 2018, 21:49
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Re: When k, l, and m are different positive integers greater than 1, k+l+m   [#permalink] 26 Aug 2018, 21:49
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