If r and s are positive numbers and θ is one of the operations, +, −,

We know that \(r\) and \(s\) are positive numbers and θ is one of the 4 basic operations. The original question: Which operation is θ ?

1) If \(r=s\), then

\(r+s=r+r>0\)

\(r-s=r-r=0\)

\(r\cdot s=r\cdot r>0\)

\(\frac{r}{s}=\frac{r}{r}=1\neq 0\)

θ must be subtraction. Thus, we can get a definite answer to the original question. \(\implies\) Sufficient

2) If \(r\neq s\), then

\(r+s=s+r\)

\(r-s\neq s-r\) since the two sides have different signs.

\(r\cdot s=s\cdot r\)

\(\frac{r}{s}\neq \frac{s}{r}\) since one of the sides has a value greater than 1, while the other side has a value less than 1.

θ can be either subtraction or division. Thus, we can't get a definite answer to the original question. \(\implies\) Insufficient

Answer: A

Given: r and s are positive numbers and θ is one of the operations, +, −, ×, or ÷

Target question: Which operation is θ ?

Statement 1: If r = s, then r θ s = 0
Keep in mind that r and s are POSITIVE
So, we have POSITIVE θ POSITIVE = 0

POSITIVE + POSITIVE ≠ 0 So, θ cannot represent addition
POSITIVE - POSITIVE = 0 So, θ COULD represent subtraction
POSITIVE × POSITIVE ≠ 0 So, θ cannot represent multiplication
POSITIVE ÷ POSITIVE ≠ 0 So, θ cannot represent division
The answer to the target question is θ represents subtraction
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: If r ≠ s, then r θ s ≠ s θ r
Let's test some values for each possible operation.
1 ≠ 2, yet 1 + 2 = 2 + 1. So, θ cannot be addition
1 ≠ 2, yet 1 x 2 = 2 x 1. So, θ cannot be multiplication

If r and s are DIFFERENT positive numbers, then r - s will never equal s - r. So, θ COULD represent subtraction
If r and s are DIFFERENT positive numbers, then r ÷ s will never equal s ÷ r. So, θ COULD represent division
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent

Statement One Alone:

If r = s, then r θ s = 0.

If r = s, then the only way for r θ s = 0 is for θ to represent subtraction.

Statement one alone is sufficient to answer the question.

Statement Two Alone:

If r ≠ s, then r θ s ≠ s θ r.

We can let r = 2 and s = 3.

We see that 2 - 3 does not equal 3 - 2 and 2/3 does not equal 3/2, thus θ could represent subtraction or division.

Statement two alone is not sufficient to answer the question.

Answer: A

Hi All,

We're told that R and S are POSITIVE NUMBERS and θ is one of the operations add, subtract, multiply or divide. We're asked which operation is θ. This question comes down to some basic Arithmetic and TESTing VALUES.

(1) If R =S, then R θ S = 0.

With the information in Fact 1, we have to check the 4 possible operations to see which one(s) fit the given equation.
IF....
R=S=1....
1 + 1 = 2 this is NOT a match
1 - 1 = 0 this IS a match
(1)(1) = 1 this is NOT a match
1/1 = 1 this is NOT a match

The ONLY operation that fits is SUBTRACTION, so that is the answer to the question.
Fact 1 is SUFFICIENT

(2) If R ≠ S, then R θ S ≠ S θ R

Just as we did with Fact 1, we have to check the 4 possible operations to see which one(s) fit the given equation (meaning that the two calculations do NOT equal one another).
IF....
R=1 and S=2....
1 + 2 = 3 IS equal to 2 + 1 = 3 this is NOT a match
1 - 2 = - 1 DOES NOT equal 2 - 1 = 1 this IS a match for the given information
(1)(2) = 2 IS equal to (2)(1) = 2 this is NOT a match
1/2 = 1/2 DOES NOT equal 2/1 = 2 this IS a match for the given information.

Thus, the operation could be either SUBTRACTION OR DIVISION.
Fact 2 is INSUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Hi


But it can be either r-s or s-r or |s-r|. It is still insufficient

A alone is sufficient.

For B, - and / two signs are possible.

********************
r and s are positive numbers (Remember that)

(1) If r = s, then r θ s = 0.
only when r-s=0 Sufficient

(2) If r ≠ s, then r θ s ≠ s θ r.

It can be - or / (try few numbers)

Not sufficient.

Answer is A

Video solution from Quant Reasoning:
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Forget the conventional way to solve DS questions.

We will solve this DS question using the variable approach.

DS question with 1 variable: Let the original condition in a DS question contain 1 variable. Now, 1 variable would generally require 1 equation for us to be able to solve for the value of the variable.

We know that each condition would usually give us an equation, and Since we need 1 equation to match the numbers of variables and equations in the original condition, the logical answer is D.

To master the Variable Approach, visit https://www.mathrevolution.com and check our lessons and proven techniques to score high in DS questions.

Let’s apply the 3 steps suggested previously. [Watch lessons on our website to master these 3 steps]

Step 1 of the Variable Approach: Modifying and rechecking the original condition and the question.

We have to find which operation is θ ? where operations are +, −, ×, or ÷, and 'r' and 's' are positive numbers.

Second and the third step of Variable Approach: From the original condition, we have 1 variable (θ).To match the number of variables with the number of equations, we need 1 equation. Since conditions (1) and (2) will provide 1 equation each, D would most likely be the answer.

Let’s take look at each condition separately.

Condition(1) tells us that If r = s , then r θ s = 0.

=> If r = s = 2, then r - s = 0 and hence θ = -

Since the answer is a unique value , condition(1) alone is sufficient by CMT 2.

Condition(2) tells us that r ≠ s, then r θ s ≠ s θ r .

=> If θ = - and r = 2 and s = 3 , then 2 - 3 ≠ 3 - 2

=> But If θ = ÷ and r = 2 and s = 3 , then 2 ÷ 3 ≠ 3 ÷ 2

Since the answer is not a unique value , condition(2) alone is not sufficient by CMT 2.

Condition (1) alone is sufficient.

So, A is the correct answer.

Answer: A


SAVE TIME: By Variable Approach, when you know that we need 1 equation, we will directly check each conditions to be sufficient. We will save time in checking the conditions individually.

In the -1- statement, if r and s = 0, all operations are possible.
0 + 0 = 0
0-0 = 0
0 * 0 = 0

I think only 0/0 = undefined

statement -1- would be insufficient.
0 is considered a positive number?
can someone explain this to me please?

Hi aduuuu,

The prompt tells us that R and S are POSITIVE numbers, so you are NOT allowed to use R = 0 or S = 0 when working through this question. The number '0' is neither positive nor negative (in math terms, it's actually referred to as a "null" value; in the realm of Number Properties, it is an even number though).

GMAT assassins aren't born, they're made,
Rich

Answer: Option A

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Nice to remember- 0 (zero) is not a positive number.

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