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Manager  Joined: 05 Jun 2009
Posts: 65
Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 69% (01:44) correct 31% (01:55) wrong based on 742 sessions

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Is $$xy < 6$$ ?

(1) $$x < 3$$ and $$y < 2$$

(2) $$\frac{1}{2} < x < \frac{2}{3}$$ and $$y^2 < 64$$

DS76602.01

Originally posted by sacmanitin on 24 Oct 2009, 04:36.
Last edited by Bunuel on 21 Sep 2019, 16:08, edited 2 times in total.
Edited the question and added the OA
Math Expert V
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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64  [#permalink]

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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64  [#permalink]

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6
2
IMO B.

stmt1: if both x and y are negative such that the product of their absolute values is > 6 then xy>6. eg. x=-8, y=-3

stmt2: x is +ve, -8<y<8.
Let us take max value of x =2/3 and max value of y=8, the product is <6.
Now, let us take max value of x=2/3 and a negative value of y=-8, the product is -ve and hence < 6.
One more try, take x=2/3 and y=5, the product is <6.
##### General Discussion
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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64  [#permalink]

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Geronimo wrote:

Is xy<6 ?
(1) x<3 and y<2
(2) 1/2 < x < 2/3 and y^2<64

Is $$xy<6$$?

(1) $$x<3$$ and $$y<2$$ --> now, if both $$x$$ and $$y$$ are equal to zero then $$xy=0<6$$ and the answer will be YES but if both $$x$$ and $$y$$ are small enough negative numbers, for example -10 and -10 then $$xy=100>6$$ and the answer will be YES. Not sufficient.

(2) $$\frac{1}{2}<x<\frac{2}{3}$$ and $$y^2<64$$, which is equivalent to $$-8<y<8$$ (not y<8, y>-8. as written above) --> even if we take the boundary values of $$x$$ and $$y$$ to maixmize their product we'll get: $$xy=\frac{2}{3}*8\approx{5.3}<6$$, so the answer to the question "is $$xy<6$$" will always be YES. Sufficient.

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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64  [#permalink]

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Clearly 1) is sufficient. It says x is less than 3 and y is less than 2 so product can not be greater than 6.

Lets examine 2) x lies between 1/2 and 2/3 and y^2 is less than 64, or y lies within (-8,8). On the extreme, xy can be just below 2/3*8 or ~5.33 so it will always be less than 6. Sufficient.

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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64  [#permalink]

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beyondgmatscore wrote:
Clearly 1) is sufficient. It says x is less than 3 and y is less than 2 so product can not be greater than 6.

Lets examine 2) x lies between 1/2 and 2/3 and y^2 is less than 64, or y lies within (-8,8). On the extreme, xy can be just below 2/3*8 or ~5.33 so it will always be less than 6. Sufficient.

Statement (1) is not sufficient: if x and y are small enough negative numbers, for example -10 and -10 then xy=100>6. So answer is B, not D.
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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64  [#permalink]

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How does y^2 <64 become y<8 and y>-8?
Retired Moderator Joined: 20 Dec 2010
Posts: 1518
Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64  [#permalink]

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1
naveenhv wrote:
How does y^2 <64 become y<8 and y>-8?

1/2 < x < 2/3 and y^2<64

It is a general rule;

$$y^2 < 64$$
$$|y| < 8$$ (Watch the less than (<) symbol)
Means;
-8<y<8

$$y^2 > 64$$
$$|y| > 8$$(Watch the greater than (>) symbol)
Means;
y>8 or y<-8

so;
-8<y<8
1/2<x<2/3

Extreme values of xy
-8*1/2 = -4
8*1/2 = 4
-8*2/3 = -16/3 > -6
8*2/3 = 16/3 < 6

Thus;
xy>-6
xy<6

We found that; yes indeed xy<6.
Sufficient.
Manager  Joined: 15 Apr 2011
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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64  [#permalink]

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Bunuel, your explanations are so easy to understand. Thanks!
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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64  [#permalink]

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Bunuel wrote:
I didn't understand this one. could someone please explain why C is not the answer? thanks.

Is $$xy<6$$?

(1) $$x<3$$ and $$y<2$$ --> now, if both $$x$$ and $$y$$ are equal to zero then $$xy=0<6$$ and the answer will be YES but if both $$x$$ and $$y$$ are small enough negative numbers, for example -10 and -10 then $$xy=100>6$$ and the answer will be NO. Not sufficient.

(2) $$\frac{1}{2}<x<\frac{2}{3}$$ and $$y^2<64$$, which is equivalent to $$-8<y<8$$ --> even if we take the boundary values of $$x$$ and $$y$$ to maixmize their product we'll get: $$xy=\frac{2}{3}*8\approx{5.3}<6$$, so the answer to the question "is $$xy<6$$?" will always be YES. Sufficient.

i dont understand i put A....i could answer the question with that information. N why are we letting x times y = to 0 why cant we let it be equal to 1 or 2?

if a number times a number is less than 6......cant we just say use 1?
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Joined: 02 Sep 2009
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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64  [#permalink]

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1
Bunuel wrote:
I didn't understand this one. could someone please explain why C is not the answer? thanks.

Is $$xy<6$$?

(1) $$x<3$$ and $$y<2$$ --> now, if both $$x$$ and $$y$$ are equal to zero then $$xy=0<6$$ and the answer will be YES but if both $$x$$ and $$y$$ are small enough negative numbers, for example -10 and -10 then $$xy=100>6$$ and the answer will be NO. Not sufficient.

(2) $$\frac{1}{2}<x<\frac{2}{3}$$ and $$y^2<64$$, which is equivalent to $$-8<y<8$$ --> even if we take the boundary values of $$x$$ and $$y$$ to maixmize their product we'll get: $$xy=\frac{2}{3}*8\approx{5.3}<6$$, so the answer to the question "is $$xy<6$$?" will always be YES. Sufficient.

i dont understand i put A....i could answer the question with that information. N why are we letting x times y = to 0 why cant we let it be equal to 1 or 2?

if a number times a number is less than 6......cant we just say use 1?

On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

Now, for x=y=0 we got an YES answer and for x=y=-10 we got a NO answer, thus the statement is NOT sufficient.

Of course we could use some other numbers to get an YES and a NO answers to prove that the statement is not sufficient: x=y=0 and x=y=-10 are just examples of many possible sets.

Hope it's clear.
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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64  [#permalink]

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4

Statement (1)

We are given x < 3 and y < 2 ; no lower bound specified for either of the variables.
x and y could be x=-3 and y=-2 we get xy=6 or they could be very small negative numbers then xy would be much greater than 6.
On the other hand x=1 y=1 which results in xy=1 which is smaller than 6 so Statement (1) is not sufficient

Statement (2)

y^2 < 64 can be rewritten as -8< y <8 ,
Since x is positive, we can test the extremes without worrying about changing the direction of the inequality sign
-8*(2/3) < xy < 8*(2/3)

-5.3333 < xy < 5.333
So we can answer the question "Is xy<6" with the Statement (2) ALONE.

Originally posted by code19 on 30 Jan 2014, 23:46.
Last edited by code19 on 01 Feb 2014, 13:04, edited 1 time in total.
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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64  [#permalink]

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2
1
Geronimo wrote:
Is xy < 6

(1) x < 3 and y < 2

(2) 1/2 < x < 2/3 and y^2 < 64

We need to determine whether the product of x and y is less than 6.

Statement One Alone

x < 3 and y < 2

Using the information in statement one we do not have enough information to determine whether the product of x and y is less than 6. For example, if x = -4 and y = -2, the product of x and y is 8, which is greater than 6. However, if x = 0 and y = 0, the product of x and y is 0, which is less than 6. We can eliminate answer choices A and D.

Statement Two Alone

1/2 < x < 2/3 and y^2 < 64

Using the information in statement two, we see that x is less than 2/3 and that y is less than 8. Thus, the maximum product of x and y is less than (2/3)(8) = 16/3, which is less 5.33 and thus less than 6. Since the maximum product of x and y is less than 6, statement two is sufficient to answer the question. Note that we did not even consider the case when -8 < y < 0 because in that case, xy will be negative and thus will be less than 6.

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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64  [#permalink]

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Test values

Statement 1
x<3 and y<2
x=-6 y=-1 xy <6? No
x=2 y=1 xy <6? yes

Statement 2
Convert to decimal for ease
0.5 <x < 0.66 and -8<y<8

We don't need to bother with negative values for y since xy with a negative y will always be less than 6 since x is known to be >0

Maximise the value of y as a positive
y=8 (not allowed but maximise to see)
x=0.6
xy=4.8
Even when maximised xy<6

Sufficient
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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64  [#permalink]

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Top Contributor
sacmanitin wrote:
Is $$xy < 6$$ ?

(1) $$x < 3$$ and $$y < 2$$

(2) $$\frac{1}{2} < x < \frac{2}{3}$$ and $$y^2 < 64$$

Target question: Is xy < 6?

Statement 1: x < 3 and y < 2
Let's TEST some values.
There are several values of x and y that satisfy statement 1. Here are two:
Case a: x = 0 and y = 0. In this case, xy = (0)(0) = 0. So, the answer to the target question is YES, xy IS less than 6
Case b: x = -5 and y = -5. In this case, xy = (-5)(-5) = 25. So, the answer to the target question is NO, xy is NOT less than 6
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 1/2 < x < 2/3 and y² < 64
Let's see if we can find the greatest possible value of xy.
Since we can see that x is POSITIVE, the maximum value of xy will be achieved when y is also POSITIVE
From the inequality y² < 64, we can conclude that y < 8
He also know that x < 2/3
(8)(2/3) = 16/3 = 5 1/3 (which is less than 6)

Since x < 2/3 and y < 8, we can be certain that xy is less than 6
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Cheers,
Brent
_________________ Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64   [#permalink] 20 Dec 2019, 05:54
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