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505-555 Level|   Algebra|                           
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If x = 2t and y = t/3, what is the value of x^2 – y^2?

(1) t^2 – 3 = 6
(2) t^3 = −27

Sol. x^2 = 4t^2 and y^2 = t^2/9
1) t^2 = 9 so 4t^2 - t^2/9 so 36-1 = 35
2) t^3 = −27 so t = -3 and t^2=9 so 4t^2 - t^2/9 so 36-1 = 35

D Both Sufficient
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Bunuel
If x = 2t and y = t/3, what is the value of x^2 – y^2?

(1) t^2 – 3 = 6
(2) t^3 = −27

Kudos for a correct solution.

Given : x = 2t and y = t/3

Question : x^2 – y^2=(x+y)*(x-y) = ?

Statement 1: t^2 – 3 = 6
i.e. t^2 = 9
i.e. t = +3
i.e. y =+1 respectively for respective values of t=+3
i.e. x =+6 respectively for respective values of t=+3
i.e. x^2 – y^2 = (+6)^2 - (+1)^2 = 36-1 = 35
SUFFICIENT

Statement 2: t^3 = −27
i.e. i.e. t = -3
i.e. y =-1 respectively for respective values of t=+3
i.e. x =-6 respectively for respective values of t=+3
i.e. x^2 – y^2 = (-6)^2 - (-1)^2 = 36-1 = 35
SUFFICIENT

Answer: Option D
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If the question prompt wanted (x^3 - y^3) or (x - y) rather than (X^2 - y^2)

then statement 1 would not be sufficient, correct?
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If the question prompt wanted (x^3 - y^3) or (x - y) rather than (X^2 - y^2)

then statement 1 would not be sufficient, correct?

Yes. In this case the answer would be B.
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Hi All,

Certain DS questions are really just a 'test' of your 'thoroughness', so it's important to do all of the necessary work (on the pad; NOT 'in your head') to prove what the correct answer is. Without having that 'proof', you might inadvertently choose the wrong answer (and miss out on some easy points on Test Day).

We're told that X = 2T and Y = T/3. We're asked for the value of X^2 - Y^2

1) T^2 - 3 = 6.

With the equation in Fact 1, we can determine that T has 2 values...
T^2 - 3 = 6
T^2 = 9
T = +3 or -3
Many Test Takers would assume that these two values would lead to two different answers to the question (and assume that Fact 1 was insufficient).... but where is your PROOF that it's insufficient...?

IF....
T = 3, X=6, Y=1, then the answer to the question is 6^2 - 1^2 = 35
IF....
T = -3, X= -6, Y= -1, then the answer to the question is (-6)^2 - (-1)^2 = 35
It turns out that both values of T lead to the SAME answer - and the answer to the question is ALWAYS 35.
Fact 1 is SUFFICIENT

2) t^3 = -27

The equation in Fact 2 has just one solution: -3. With the work that we did in Fact 1, we know that there is only one answer to the question (35).
Fact 2 is SUFFICIENT

Final Answer:

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Ok. Was just wondering. Some irrational's irrational power can equal -27. In this case t may not be -3

Please guide.
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ShankSouljaBoi
Ok. Was just wondering. Some irrational's irrational power can equal -27. In this case t may not be -3

Please guide.

Hi ShankSouljaBoi,

The information in Fact 1 and Fact 2 defines specific 'powers' (a square and a cube), so there's no possibility of an 'irrational power.'

Assuming that you are focusing on Fact 2, we know that there is a CUBED number that equals - 27. By definition, that number MUST be NEGATIVE...

(positive)^3 = positive
(0)^3 = 0
(negative)^3 = negative

In addition, increasing or decreasing the value of the negative will change the value of the cubed term...

(-2)^3 = -8
(-3)^3 = -27
(-4)^3 = -64
Etc.

Thus, even if you are changing the number in really small increments (for example -3 vs -2.999 or -3.1), the end cubed value will change (meaning that there's ONLLY ONE way to get to a -27 result.... when we use -3).

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Bunuel
If x = 2t and y = t/3, what is the value of x^2 – y^2?

(1) t^2 – 3 = 6
(2) t^3 = −27

Answer: Option D

Video solution by GMATinsight

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To make the solution easy lets solve the equation in the question first:
putting the values of x and y in the equation we will get:
x^2-y^2= (35t^2)/9

Statement 1 : SUFFICIENT as it gives the value for t which is enough to find the value for x^2-y^2
Statement 2 : SUFFICIENT as it gives the value for t which is enough to find the value for x^2-y^2

therefore, D is the answer
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I had put down statement (1) as insufficient because, given t = +/- 3, I had possible values of X as +/- 6 and for Y +/- 1. But when I plug this different values in the equation that we are trying to prove (x+y)(x-y), I got different answers. Sometimes I would get 35, other times I would get -35. Therefore I had selected (1) as sufficient.

For example:
Y = 1 or -1
X = 6 or -6

Test into equation (x+y)(x-y) to make sure we get the same outcomes.

Test 1: X = 6, Y = 1
(7)(5) = 35

Test 2: X = 6, Y = -1
(5)(7) = 35

Test 3: X = -6, Y = -1
(-7)(-5) = 35

Test 4: X = -6, Y = 1
(-5)(-7) = 35

Oh, wait.... nvm - they all result in 35 - must've been a calculation error... Lol!
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Bunuel
If x = 2t and y = t/3, what is the value of x^2 – y^2?

(1) t^2 – 3 = 6
(2) t^3 = −27

Kudos for a correct solution.
Solution:

We need to determine the value of x^2 - y^2 given that x = 2t and y = t/3. Substituting 2t for x and t/3 for y, we have:

(2t)^2 - (t/3)^2 = 4t^2 - t^2/9 = 35t^2 / 9

Therefore, if we can determine the value of t, we can determine the value of x^2 - y^2 since it’s equal to 35t^2 / 9.

Statement One Alone:

Solving the equation, we have:

t^2 = 9

t = ±√9

t = ±3

Although the value of t is not unique (it can be either 3 or -3), we see that t^2 = 9, regardless of whether t is 3 or -3. Thus, the value of x^2 - y^2 is 35(9) / 9 = 35. Statement one alone is sufficient.

Statement Two Alone:

Solving the equation, we have:

t^3 = -27

t = ∛(-27)

t = -3

Since t = -3, the value of x^2 - y^2 is [35(-3)^2]/9 = 35(9)/35 = 35. Statement two alone is sufficient.

Answer: D
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answer is D
4t^2- t^2/9
[t^2(36-1)]/9
35t^2/9

Now using st 1 and st 2 individually w can find the value
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