Data Sufficiency Questions

[Anamaiy]

Guys it would be great if you can help me out with the solutions of these Data Sufficiency questions:

Are x and y both positive ?
(1). 2x-2y = 1
(2). x/y > 1

What fraction of this year graduation students at a certain college are males ?
(1). Of this year graduating students, 33% of the males and 20% of the females transferred from another college/
(2). Of this year's graduating students, 25% transferred from another college.

Is m+z > 0 ?
(1). m-3z > 0
(2). 4z-m > 0

Are at least 10% of the people in country X who are 65 years or older and employed ?
(1). In country X, 11.3% of the population is 65 years or older.
(2). In country X, of the population 65 years or older, 20% of the men and 10% of the woman are employed.

On the number line, distance between x and y is greater than the distance between x and z. Does z lie between x and y on the number line ?
(1). xyz < 0
(2). xy < 0

[Night reader]

Are x,y>0
st(1) x=1/2+y Not Sufficient, since we don't know the signs of x and y, they could be both -ve OR +ve;
st(2) x/y>1, |x|>|y|; x and y can be both -ve OR +ve, hence Not Sufficient;
Combined st(1&2): x=1/2+y, since |x|>|y| and x,y can be both +ve OR -ve --> If x,y<0 then x lies LHS from y, i.e. x<y which is impossible for x=1/2+y, hence x,y>0 and x>y. Our answer is yes x,y>0 Sufficient

IOM C

Guys it would be great if you can help me out with the solutions of these Data Sufficiency questions:

Are x and y both positive ?
(1). 2x-2y = 1
(2). x/y > 1

[Night reader]

Find ratio, males/graduating students-? OR M/S-?
st(1) 0.33M+0.2F=TS(transferred from another college), no information about males in this college -Not Sufficient;
st(2) 0.25S=TS, no information about males -Not Sufficient;
Combined st(1&2): 0.25(M+F)=0.33M+0.2F and M+F=100, by solving for M we can find ratio M/S -Sufficient

IOM C

What fraction of this year graduation students at a certain college are males ?
(1). Of this year graduating students, 33% of the males and 20% of the females transferred from another college/
(2). Of this year's graduating students, 25% transferred from another college.

[Night reader]

Is m>-z ?
st(1) m>3z, Not Sufficient as we have no information about signs of m and z (-ve OR +ve);
st(2) 4z>m, Not Sufficient, as two cases are possible m=<4z when (m+z>0) and m>4z when (m+z<0)
Combined st(1&2): [m-3z] + [4z-m]>0, z>0; If z is +ve, i.e. z>0 then m-3z>0 means that m is also +ve! Our answer is Yes (m+z>0) Sufficient

IOM C

Is m+z > 0 ?
(1). m-3z > 0
(2). 4z-m > 0

[force5]

Question(1) IMO -E
statement1= x-y=1/2 ( x and y can be both +ve or -ve, but cant have diff signs)

Statement 2= x/y>1 ( again same sign)
combining= we cant say if both are +ive or -ve.

Question (2) IMO- c
0.33m+0.20f=m+f/4 ----(1)
m+f=100-------(2)
hence m and f can be obtained and hence m/m+f

Question (3)IMO- c

stmnt1- m-3z>0
m/3>z (insufficient since m and z can be +ve or -ve)

stmnt2- 4z-m>0
z>m/4 (Insufficient since m and z can be +ve or -ve)

combining.
add ineqalities. z>0 ( +ve) as per stmnt (1) m>3z ( if z is +ve them m will be positive too) hence m+n will be +ve

Question (4) IMO- B
stmnt1- insufficient
stmnt2- sufficient
say men be M and women be W. hence the total figure is

( 20 % of M + 10 % of W )
------------------------------
( M + W )

this gives a weighted average whose value will always be greater than 10 and less than 20.

Question (5)IMO- E

stmnt1- XYZ < 0.
1 negative or
all 3 negatives.

all can be on the left (-ve x axis)

So A is insufficient.
stmnt 2
XY < 0.
only one -ve
any one can be on the -ve axis.but z can lie anywhere. eg- x 0 z y or z x 0 y. (insufficient)

combining-
from stmnt 2 only one can be -ve hence when we combine both statements one of x, y are -ve and rest two have to be +ve to maintain (xyz<0)
hence if x is -ve then x 0 z y
if y -ve then y 0 x z
hence E

[srcc25anu]

Q5. IMO it should be C
stat 2 says xy <0 i.e. x and y have opp signs no indication of Z (Not Sufficient)

stat 1 says xyz < 0 that is either one or all three are -ve. (not sufficient)

using both statements:
xy < 0, that is x and y have diff signs. and xyz = -ve that is z = positive

.__y________x___Z_____

and zero lies between y and x. therefore we can answer using both statements.

[force5]

yes anu you are correct that zero will lie between x and y. however as per the original poster- the question is does Z lie between x and y. you can see there are two scenarios possible here

if x is -ve then x 0 z y
if y -ve then y 0 x z

hope that helps

[Night reader]

you marked E.

Combined st(1) x=1/2+y AND st(2) x/y>1

from st(1&2) if (1/2+y)/y>1 then (1/2y)>0 Hence 2y>0 AND y>0 //agree?
from st(1) x=1/2 +y Hence with positive y ---> x=1/2 + y {positive value of y} We can answer Not only x and y are positive, BUT even x>y // agree?

answer is C

Question(1) IMO -E
statement1= x-y=1/2 ( x and y can be both +ve or -ve, but cant have diff signs)

Statement 2= x/y>1 ( again same sign)
combining= we cant say if both are +ive or -ve.

[Anamaiy]

Thanks guys

[force5]

Question(1) IMO -E
statement1= x-y=1/2 ( x and y can be both +ve or -ve, but cant have diff signs)

Statement 2= x/y>1 ( again same sign)
combining= we cant say if both are +ive or -ve.

Guys please correct the 1st question- the answer should be C and not E.

[HSPA]

Are x and y both positive ?
(1). 2x-2y = 1
(2). x/y > 1


Let us first use the line equation first : 2x-2y=1 => y= x-0.5.. So we have line in quadrants 3,1,4
Now let us use option 2... x/y is positive and greater than 1.. this is giving quadrants 1 and 3

But the question is if both x and y are from quadrant 1...
Here answer shall be E...

[ankur.agrawal]

Combined st(1) x=1/2+y AND st(2) x/y>1

from st(1&2) if (1/2+y)/y>1 then (1/2y)>0 Hence 2y>0 AND y>0 //agree?
from st(1) x=1/2 +y Hence with positive y ---> x=1/2 + y {positive value of y} We can answer Not only x and y are positive, BUT even x>y // agree?

answer is C

How did (1/2y)>0 convert to 2y>0. { Are we taking reciprocal) ? Plz explain.

[Night reader]

Hi Ankur, since 1>0, then in order for 1/2y to be greater than 0 OR to be positive, 2y must be greater than 0 OR must be positive. Hence 2y>0
thanks

Combined st(1) x=1/2+y AND st(2) x/y>1

from st(1&2) if (1/2+y)/y>1 then (1/2y)>0 Hence 2y>0 AND y>0 //agree?
from st(1) x=1/2 +y Hence with positive y ---> x=1/2 + y {positive value of y} We can answer Not only x and y are positive, BUT even x>y // agree?

answer is C

How did (1/2y)>0 convert to 2y>0. { Are we taking reciprocal) ? Plz explain.

[force5]

Are x and y both positive ?
(1). 2x-2y = 1
(2). x/y > 1


Let us first use the line equation first : 2x-2y=1 => y= x-0.5.. So we have line in quadrants 3,1,4
Now let us use option 2... x/y is positive and greater than 1.. this is giving quadrants 1 and 3

But the question is if both x and y are from quadrant 1...
Here answer shall be E...

Hi HSPA this is an excellent doubt. ......

you are correct till eliminating A and B . This is what it is when we think about C

as per stmnt 1 your line y=x-0.5 has a positive slope and -ve y intercept. ( its in 1, 3, 4 th quadrant)
how ever also see that for this line x is always greater than y ( when x=0, y =-0.5, when y =0, x=0.5)

From stmnt 2 x and y have same sign.
when both are positive ( x>y)
when both are negative ( x<y)

now combining both. x >y can only happen when both are positive.


Hope that helps dear ...............

[HSPA]

Super force5... I sometimes forget to do the complete calculation... this is what is haunting me from days
i did the same in some other question which says by how much 12 should be reduced to get 4.5.. i kept 37.5%