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Source: — Data Sufficiency |
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by kstv » Thu May 06, 2010 5:13 pm
1) x = y-1 , z = y+1 since x,y & z are cons. nos.
x+y = 2y is divisible by 10 i.e. y is multiple of 5
let it be 5a , then x+y+z = (5a-1)+5a+(5a+1) = 15a
not sufficient.

2) let y a multiple of 10 = 10 a
x+y+z = (10a-1)+10a+(10a+1) = 30a
which is divisible by 10
Sufficient

IMO B
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by outreach » Fri May 07, 2010 12:14 am
i took the number substiution approach
1)
x=4,y=5,z=6
then X+Y+Z is not divisible by 10

x=9,y=10,z=11
then X+Y+Z is divisible by 10

insufficient

2)

x=9,y=10,z=11
then X+Y+Z is divisible by 10

x=19,y=20,z=21
then X+Y+Z is divisible by 10


hence option 2 is enough
answer is B
colakumarfanta wrote:X,Y,Z are consecutive inetegers . Is X + Y + Z divisible by 10?
1) X + Z divisible by 10
2) Y is a multiple of 10
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by liferocks » Fri May 07, 2010 2:43 am
from question three integers are a-1,a,a+1

from 1 X+Z is divisible by 10
so a-1+a+1 =10n
or a=5n

n can be even or odd so a may or may not be divisible by 10...insufficient

from2 a=10n
hence X+Y+Z=10n-1+10n+10n+1=30n..divisible by 10...sufficient

Ans option B
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by tryingtocrack » Sun May 09, 2010 9:34 am
X, Y , Z

y=x+1
z=x+2

x+y+z = 3(x+1) Q is x+1 having 2, 5 as factors ?

statement a
x+z = 2(x+1) / 10 ( means x+1 ) surely has one 5 in it but not necessarily 2 , so insufficent

statement b , x+1/ 10 , means x+1 has 2, 5 as factors , so b is sufficient
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by sk818020 » Sun May 09, 2010 10:04 am
I think this would actually be C.

The reason I would say B is not suffient is because the question stem does not state that x<y<z. Without that (2) only tells us that y is a multiple of ten meaning;

If x=9, y=10, z=11, then

x+y+z=30

It could also be true that,

y=10, x=11, z=12

x+y+z=33.

Using (1) and (2) you can then deduce that x<y<z or z<y<x, because that is the only way x+z and y could be multiples of ten.

Thus, the answer is C.

Could you please confirm the official answer?
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