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

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.

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.

Answer should be D.

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.

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.

Bunuel, your explanations are so easy to understand. Thanks!

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.

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.

Answer B;

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.

Answer: B

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

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

Answer: B

Cheers,
Brent

What is the concept of multiplying two inequalities? WHy can we not mltiply the two here? And if we do, how to go about it? Bunuel JeffTargetTestPrep ScottTargetTestPrep nick1816

What is the concept of multiplying two inequalities? WHy can we not mltiply the two here? And if we do, how to go about it? Bunuel JeffTargetTestPrep ScottTargetTestPrep nick1816 KarishmaB avigutman ThatDudeKnows

One of the most common things that the GMAT tests on inequalities is whether you know/remember to reverse the sign when you multiply by a negative number. If everything in both inequalities is greater than 0, you're safe. But if anything is less than 0 (or could be less than 0), there's no clean way to handle reversing the sign(s) (or even knowing whether to reverse the signs.

a>b>0 and p>r>0 --> ap*br>0 Yes

0>a>b and p>0>r (or any other set up whereby we have a negative or possible negative) --> Can't do it

I sense that you're looking for a rule here, Elite097, but, well, it's complicated.
Unlike additive reasoning, with which we can safely infer that the sum of two bigger terms will be greater than the sum of two smaller terms (a.k.a. adding inequalities), multiplicative reasoning behaves differently depending on the positioning of the terms relative to -1, 0, and 1.
The inequalities in this problem allow x and y to be located on either side of those key tick marks on the number line, so we should probably avoid multiplying them.

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