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From A) The answer could be 1 , 0, or -1. However, the question prompt dictates that y must be positive, eliminating -1 and 0. A is sufficient.

From B) If a<1, a could be zero, in which case y could be any positive integer. If a does not equal 0, then y = 1. There are multiple solutions, Not Sufficient.
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Official Solution




Steps 1 & 2: Understand Question and Draw Inferences


Given:

    • y is a positive integer

    o y > 0

To find:

    • Value of y

Step 3: Analyze Statement 1 independently

    • \(y^3 = y^{1/3}\)

To get rid of the cube root on the RHS, we take cube on both sides

    • \(y^9 = y\)

Per our conceptual understanding, we know that this is possible only when \(y =0, -1\) or \(1\)

As \(y > 0\) is given, we get

    • \(y = 1\)

As we got a unique value for ‘y’

Statement 1 alone is sufficient to answer this question.

Step 4: Analyze Statement 2 independently

    • \(ay = a\)

    • \(ay = a\) is satisfied in two cases

      o Case 1: If \(a = 0\), y could be any value

      o Case 2: If \(a ≠ 0, y = 1\)

As we don’t get a unique value for ‘y’

Statement 2 alone is NOT sufficient to answer this question.

Correct Answer: Option A

Note:

For Statement 2, many test-takers commit the mistake of dividing both sides of ay = a by ‘a’ thereby getting the answer as y = 1.
Please note that you are allowed to do this division only if a ≠0 is given as division by zero is not defined.


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If y is a positive integer, what is the value of y?

(1) \(y^3 = y^{1/3}\)

From this statement, since it has been given that y is positive,
the only integer which satisfies this equation is 1. Hence, sufficient.

(2) \(ay = a\), and \(a < 1\)

Since a<1, if a = 0,
we can have any value for y, since all values of y
will satisfy this equation, statement 2 alone is not sufficient. (Option A)
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Hi,

I am having difficulty with the second statement:

2) ay=a
I take the a to the LHS
ay-a=0
taking a as common
a(y-1)=0
this is where I start to get confused-I'm not sure if this is even correct or I've got it completely messed up in my head but-
a=0 or y-1=0 ---> y=1

If this is wrong could someone please help me with why and it would also be great if someone could confirm when to use the
a(y-1)=0
a=0 or y-1=0 ---> y=1?

Many thanks!
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Statement-1: y³=y^(1/3)
Since y is a positive integer, if y=1 then this satisfies this equation which results in 1³=1^(1/3)
⇒1=1
Therefore, S-1 is sufficient.

Statement-2: ay=a where a<1
So a=0,–1,–2,...

If a<0 then y=1 but if a=0 y=0
Therefore, S-2 is not sufficient.

Ans. is A
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1) y3=y1/3

whenever you are given this only values possible are 0,1,-1

here given that y is a positive integer, -1 is not possible ; infact even 0 is not allowed

hence A is sufficient
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