The way

[dungtruong]

I. How many ways are there to split a group of 6 boys into two groups of 3 boys each? (The order of the groups does not matter)

1. 8

2. 10

3. 16

4. 20

5. 24

II. X, Y and Z are possitive integers. Is X% of Y bigger than Y% of Z?

1. X - Y = Y - Z

2. X=2

Pls discuss, everyone! Thanks a lot

[12rk34]

QI :
No. of ways = 6C3 = 20
Hence D
QII. :
If X% of Y > Y% of Z => XY > YZ
=> X > Z
1. X - Y = Y - Z
INSUFFICIENT
2. X=2
INSUFFICIENT
Combining,
INSUFFICIENT
Hence E.

[dungtruong]

Thanks 12rk34

[cliq]

Originally Posted by 12rk34

QII. :
If X% of Y > Y% of Z => XY > YZ
=> X > Z
1. X - Y = Y - Z
INSUFFICIENT
2. X=2
INSUFFICIENT
Combining,
INSUFFICIENT
Hence E.

I have doubts related to QII.

from opt 1 we can say X = 2Y -Z

Combining opt1 and 2,
we get 2 = 2Y - Z ................ ( since x = 2)

case 1 -
if Y = 1,
then 2 = 2 - Z ..
z = 0... but z is a positive integer, case 1 not possible

case 2,
if y = 2,
then 2 = 4 - z
z = 2 ... x = z and therefore the question (X > Z) is FALSE

case 3
if y = 3,
then 2 = 6 - z
z = 4.. x < z .. and therefore the question (X > Z) is FALSE

... so for all values of positive integers x, y and z , where x is 2 .we can determine that x > Z is false...
so the answer should be C.

[clock60]

Originally Posted by 12rk34

QI :
No. of ways = 6C3 = 20
Hence D
QII. :
If X% of Y > Y% of Z => XY > YZ
=> X > Z
1. X - Y = Y - Z
INSUFFICIENT
2. X=2
INSUFFICIENT
Combining,
INSUFFICIENT
Hence E.

dear 12rk34
sorry for my doubt to problem 1
but i think that the answer is 10 not 20
it is (6C3)/2 since we have two teams and the order of forming of teams does not matter
if we have six persons
x1.x2,x3,x4,x5.x6 the division
x1,x2,x3 first team and x4,x5,x6 second team
is the same as
x4.x5.x6 first team and x1.x2.x3.second team
please correct me ....

[12rk34]

Originally Posted by clock60

dear 12rk34
sorry for my doubt to problem 1
but i think that the answer is 10 not 20
it is (6C3)/2 since we have two teams and the order of forming of teams does not matter
if we have six persons
x1.x2,x3,x4,x5.x6 the division
x1,x2,x3 first team and x4,x5,x6 second team
is the same as
x4.x5.x6 first team and x1.x2.x3.second team
please correct me ....

Yes, it will be 10 only.

[cliq]

Originally Posted by cliq

I have doubts related to QII.

from opt 1 we can say X = 2Y -Z

Combining opt1 and 2,
we get 2 = 2Y - Z ................ ( since x = 2)

case 1 -
if Y = 1,
then 2 = 2 - Z ..
z = 0... but z is a positive integer, case 1 not possible

case 2,
if y = 2,
then 2 = 4 - z
z = 2 ... x = z and therefore the question (X > Z) is FALSE

case 3
if y = 3,
then 2 = 6 - z
z = 4.. x < z .. and therefore the question (X > Z) is FALSE

... so for all values of positive integers x, y and z , where x is 2 .we can determine that x > Z is false...
so the answer should be C.

Anybody ?

[Lock]

Here are my thoughts on problem 2:
We are given that X, Y and Z are all positive integers.
Statement 1: X-Y=Y-Z (including the provision that X, Y and Z are all positive integers having different values) means that X>Z and that Y lies at the middle of X and Z. But, in this example values X, Y and Z can't be different as that would contradict Statement 2 and the question stem. So, we may assume that X=Y=Z.
Statement 2: X=2 is insufficient for judging about values of Y and Z.
Hence, IMO C.

[clock60]

i think that the answer on problem 2-E.. my reasoning
is (x/100)y>(y/100)z
is xy-yz>0, y(x-z)>0 as y>0 the problem asks
is x>z???
(1) x-y=y-z, 2y=x+z
y=1. 2=1+1 and x=z
y=2 4=1+3 or 2+2.. insuffisient
(2)x=2 but no information about z insufficient
together also insufficient so E

[cliq]

Thanks Lock and Clock60

Originally Posted by clock60

(1) x-y=y-z, 2y=x+z
y=1. 2=1+1 and x=z
y=2 4=1+3 or 2+2.. insuffisient

together also insufficient so E

i think together it is sufficient.
your equation from opt 1 is 2y = x + z
combining 1 and 2

2y = 2 + z
y cannot be 1..because z cannot be zero

for rest of the values of y i.e 2 , 3 , 4, 5
z is 2,4,6,8

2 NEVER > z
x NEVER > z