What is the value of t?

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(1) Clearly t=6,because we can eliminate s from both side of the equation,Sufficient

(2) \(t^3\)=216,here t must be positive because other side of the equation(216) is positive.So t=\(\sqrt[3]{216}\)=\(\sqrt[3]{6^3}\)=6,Sufficient

Correct Answer D

from 1) value of t=6 so sufficient
from 2) t^3=216 so t=6

both statements are independently sufficient to answer.
so answer is D.

OG 2017 - Book Question 252:

What is the value of t?

(1) s + t = 6 + s
(2) t^3 = 216

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are NOT sufficient.



Correct Answer: D

So I picked A here instead of D because:

(1) Shows T = 6 + s - s = 6
(2) -6^3 = 216 and 6^3 = 216

The explanation for D was given as take the cube root such that you get t = 6

Am I mistaken because the question says t^3 and not (t)^3?

I just don't want to make a stupid mistake on the exam for something as easy as this.

Appreciate any insight

You missed a small part.

-6 ^ 3 will not be 216 infact it will be -216. ( -6 * -6 * -6 ).
So only possible value is 6 ^ 3.

Ahhh I can't believe I missed that, I feel so stupid! Thanks though

What is the value of t?

(1) s + t = 6 + s

S is canceled out

t =6

Sufficient

(2) t^3 = 216

Taking cubic root fro both sides

t =6

Sufficient

Answer: D

(1) \(s + t = 6 + s\)

\(s + t - s = 6\)

\(t = 6\)

Hence I is Sufficient.

(2) \(t^3 = 216\)

\(\sqrt[3]{t^3} = \sqrt[3]{216}\)

\(t = 6\)

Hence II is Sufficient.

Answer (D)...

1. rewriting t=6 sufficient
2. t^3 = 216 so t=+6 sufficient
D

We need to determine the value of t.

Statement One Alone:

s + t = 6 + s

We can subtract s from both sides of the equation and obtain:

t = 6

Statement one is sufficient to answer the question.

Statement Two Alone:

t^3 = 216

We can take the cube root of t^3 and 216 and we have:

t = 6

Statement two is sufficient to answer the question.

Answer: D
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Target question: What is the value of t?

Statement 1: s + t = 6 + s
Subtract s from both sides to get: t = 6
DONE!
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: t³ = 216
Only ONE value of t satisfies this equation. That value is 6 (6³ = 216)
So, it must be the case that t = 6
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

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1) cancel 's' from each side, then

t = 6

Sufficient

2) We can take 3rd root of number, so we will get exact one answer

Sufficient

Answer: D

What is the value of t?

(1) s + t = 6 + s
or,t = 6 sufficient
(2) t^3 = 216
or,t = 6 sufficient

correct answer D

(1) s + t = 6 + s
t=6
(2) t^3=216
t=6
D
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Silly question, but we dont have to find the value of t right. For case 2, whether or not its a perfect cubed root, doesnt really matter right? You can look and know that you could find a value of t, whether or not its rational

Answer: Option D

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does a odd root ALWAYS have one solution whereas an even root has multiple solutions?

If you're asking about equations like

x^3 = 216
or
t^5 = -32

where we're raising an unknown to an odd power, and that's equal to some number, then yes, there is always exactly one solution when the exponent is odd. If instead we have the same situation, but with an even power, there are three possibilities. If the result is equal to a positive number, e.g.

x^2 = 16
or
t^4 = 625

then there will be two solutions, one positive and one negative (4 and -4 to the first equation, 5 and -5 to the second), unless some restriction in the question obligates the solution to be positive (e.g. if x is a length or a number of cans in a warehouse, it cannot be negative).

If the result is equal to zero, there is just one solution, namely zero, so:

x^2 = 0

has only one solution, x = 0.

And if the result is equal to a negative number, which you shouldn't see on the GMAT, then there are no real solutions, so the equations:

x^4 = -64
or
t^2 = -9

have no real number solutions.
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