If y 0, is x/y

Question

If  y > 0 , is  \frac{x}{y} < 3 ?

(1)  x(x + y) − 4y(x + y) < 0

(2)  5(y + 8) < 20y – 3x + 40

Fdambro294 · Answer

Since y > 0 we can first rephrase the question stem:

Is X < 3Y?

(1) factoring the (x + y) common factor:

(X + Y) (X - 4Y) < 0

Since the product is less than 0, the factors must have opposite signs. So we have two possible cases:

Case 1:

(X + Y) < 0
And
(X - 4Y) > 0

Which means:

X < -Y
And
X > 4Y

Since Y is a positive value, it means that if Y were +1, for instance, these inequalities would be impossible to coexist

X <-1 ——-and——— X > 4

X can not simultaneously be less than a (-)negative number and greater than a (+)positive number. So it must be the other case

Case 2:

(X + Y) > 0
and
(X - 4Y) < 0

X > -Y —- and —— X < 4Y

Combining the inequalities:

-Y < X < 4Y

*Q* is: X < 3Y?

Not sure:

Can have

-Y < 3Y < X < 4Y ——> NO

OR

-Y < X < 3Y < 4Y ——> YES

S1 not sufficient

(2)

5Y + 40 < 20Y - 3X + 40

3X < 15Y

X < 5Y

*Q* is X < 3Y?

Not sure —- S2 is not sufficient alone

Together: statement 2 does not add any more of a restrictive condition on X

All we know for sure is that: X < 4Y

We do not know whether: X < 3Y

Answer
E

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sanjitscorps18 · Answer

To verify whether \frac{x}{y} < 3 => x < 3y since y > 0 (1) x(x + y) − 4y(x + y) < 0 => (x - 4y)(x + y) < 0 Given y > 0, let's take the below cases Note - x = 0 case is already true since it gives a negative number Case 1) x > 0 (x + y) > 0 hence (x - 4y) < 0 => x - 4y < 0 => x < 4y Case 2) x < 0 => (x - 4y) < 0 hence (x + y) < 0 => (x + y) < 0 => x < -y Range of x becomes => -y < x < 4y This includes x = 0 as well Insufficient (2) 5(y + 8) < 20y – 3x + 40 => 5y + 40 < 20y – 3x + 40 => -15y < -3x => 15y > 3x => 5y > x Insufficient Combining we get -y < x < 4y and x < 5y Still insufficient Option E

arushi118 · Answer

1st equation gives: (x-2y)^2 +y(x-8y)<0
--> as (x-2y)^2 is always more than 0: y(x-8y)<0
--> as y>0: x-8y<0
--> x/y<8
So insufficient.

2nd equation gives: 15y-3x>0
--> 5y-x>0
--> x/y<5
So insufficient

Combining both equations will also give x/y<5 so insufficient.
Hence, answer is E.

If y > 0, is x/y < 3? (1) x(x + y) − 4y(x + y) < 0 (2) 5(y + 8) < 20y

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