Set 10 Q13

If x and y are positive, is 3x > 7y?

(1) x > y + 4

here if y =2 then x =6 & 3x>7y

y =3 then x =7 so 3x =7y

but if y is greater than 4 then 3x<4y

Hnece this cannot be sufficeint

2) -5x < -14y

i.e 5x > 14y

so we have 2.5x>7y

so suff

Ans should be B

arocks wrote:If x and y are positive, is 3x > 7y?
(1) x > y + 4
(2) -5x < -14y

I think the answer is B. The OA is D.

Stmt2
-5x<-14y or 5x>14y or 2.5x>7y
So stmt2 is suff.

stmt1 - Not sure how statement1 is useful?
Pls explain. Thanks.

x > y + 4
Put Y = 0,1,2...
x>4, x>5 , x>6..so on
Put y=0.5
x>4.5

We need to find out : (y/n)
is 3x > 7y?
x>(7/3)y ---- x>2.3 y

for y=0
x>0,
Stmt 1 says x>4 and querstion : Is x > 0... Ans YES

for y=0.5
x> (23/5)y
x> 4.6 y

Stmt 1 is x>4.5 and Question is X > 4.6 y

Answer NO

Therfore B

is the OA

mmslf75 wrote:

arocks wrote:If x and y are positive, is 3x > 7y?
(1) x > y + 4
(2) -5x < -14y

I think the answer is B. The OA is D.

Stmt2
-5x<-14y or 5x>14y or 2.5x>7y
So stmt2 is suff.

stmt1 - Not sure how statement1 is useful?
Pls explain. Thanks.

x > y + 4
Put Y = 0,1,2...
x>4, x>5 , x>6..so on
Put y=0.5
x>4.5

We need to find out : (y/n)
is 3x > 7y?
x>(7/3)y ---- x>2.3 y

for y=0
x>0,
Stmt 1 says x>4 and querstion : Is x > 0... Ans YES

for y=0.5
x> (23/5)y
x> 4.6 y

Stmt 1 is x>4.5 and Question is X > 4.6 y

Answer NO

Therfore B

is the OA

Some creative math here!

Everyone is ok with (2), so let's just focus on (1).

We know that x and y are both positive, so we can only select positive values (i.e. not y=0).
We also know that x > y + 4

We want to know whether 3x > 7y.

Let's try a really small y and a really big y and see if it matters.

If y = 1, then x > 5. Let's make x as small as possible, 5.000001.

Is 3(5.0000001) > 7(1)? YES.

If y = 1000, then x > 1004. Let's make x as small as possible, 1004.0000001.

Is 3(1004.000001) > 7(1000)? NO.

Since (1) can produce both a YES and a NO, it's insufficient.

Stuart Kovinsky wrote:

mmslf75 wrote:

arocks wrote:If x and y are positive, is 3x > 7y?
(1) x > y + 4
(2) -5x < -14y

I think the answer is B. The OA is D.

Stmt2
-5x<-14y or 5x>14y or 2.5x>7y
So stmt2 is suff.

stmt1 - Not sure how statement1 is useful?
Pls explain. Thanks.

x > y + 4
Put Y = 0,1,2...
x>4, x>5 , x>6..so on
Put y=0.5
x>4.5

We need to find out : (y/n)
is 3x > 7y?
x>(7/3)y ---- x>2.3 y

for y=0
x>0,
Stmt 1 says x>4 and querstion : Is x > 0... Ans YES

for y=0.5
x> (23/5)y
x> 4.6 y

Stmt 1 is x>4.5 and Question is X > 4.6 y

Answer NO

Therfore B

is the OA

Some creative math here!

Everyone is ok with (2), so let's just focus on (1).

We know that x and y are both positive, so we can only select positive values (i.e. not y=0).
We also know that x > y + 4

We want to know whether 3x > 7y.

Let's try a really small y and a really big y and see if it matters.

If y = 1, then x > 5. Let's make x as small as possible, 5.000001.

Is 3(5.0000001) > 7(1)? YES.

If y = 1000, then x > 1004. Let's make x as small as possible, 1004.0000001.

Is 3(1004.000001) > 7(1000)? NO.

Since (1) can produce both a YES and a NO, it's insufficient.

That was great...!! Thanks stuart..

I like the way u used 5.00000001 and 1004.000001 ..wow !

I do it this way:

stmt # 1: x > y + 4
=> x-y > 4

lets take x=9, y=4 3x = 27 < 7y = 28
so, 9-4 > 4

lets take x=9, y=2 3x = 27 > 7y = 14
so, 9-2 > 4

Thus stmt # 1 is INSUFFICIENT. (Q.E.D - Quadrat Errat Demonstrandum