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Math Expert V
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If r and s are positive numbers and θ is one of the operations, +, −,  [#permalink]

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Difficulty:   15% (low)

Question Stats: 78% (01:15) correct 22% (01:26) wrong based on 393 sessions

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If r and s are positive numbers and θ is one of the operations, +, −, ×, or ÷, which operation is θ ?

(1) If r = s, then r θ s = 0.
(2) If r ≠ s, then r θ s ≠ s θ r.

DS37502.01
OG2020 NEW QUESTION

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Re: If r and s are positive numbers and θ is one of the operations, +, −,  [#permalink]

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From S1:

r = s (both positive numbers)
r θ s = 0, then θ can only be subtraction.
Sufficient.

From S2:

r ≠ s, then r θ s ≠ s θ r
Let r = 3
s = 4
addition and multiplication will be the same.
but, 3-4 ≠ 4-3
also, 3/4 ≠ 4/3
Hence θ can be - or /
Insufficient.

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Re: If r and s are positive numbers and θ is one of the operations, +, −,  [#permalink]

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Bunuel wrote:
If r and s are positive numbers and θ is one of the operations, +, −, ×, or ÷, which operation is θ ?

(1) If r = s, then r θ s = 0.
(2) If r ≠ s, then r θ s ≠ s θ r.

DS37502.01
OG2020 NEW QUESTION

#1
If r = s, then r θ s = 0
only subtraction possible ; sufficeint
#2
f r ≠ s, then r θ s ≠ s θ r.
insufficient
IMO A
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Re: If r and s are positive numbers and θ is one of the operations, +, −,  [#permalink]

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Bunuel wrote:
If r and s are positive numbers and θ is one of the operations, +, −, ×, or ÷, which operation is θ ?

(1) If r = s, then r θ s = 0.
(2) If r ≠ s, then r θ s ≠ s θ r.

DS37502.01
OG2020 NEW QUESTION

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

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Re: If r and s are positive numbers and θ is one of the operations, +, −,  [#permalink]

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1
Top Contributor
Bunuel wrote:
If r and s are positive numbers and θ is one of the operations, +, −, ×, or ÷, which operation is θ ?

(1) If r = s, then r θ s = 0.
(2) If r ≠ s, then r θ s ≠ s θ r.

DS37502.01
OG2020 NEW QUESTION

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

Cheers,
Brent
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Re: If r and s are positive numbers and θ is one of the operations, +, −,  [#permalink]

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Bunuel wrote:
If r and s are positive numbers and θ is one of the operations, +, −, ×, or ÷, which operation is θ ?

(1) If r = s, then r θ s = 0.
(2) If r ≠ s, then r θ s ≠ s θ r.

DS37502.01
OG2020 NEW QUESTION

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.

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Re: If r and s are positive numbers and θ is one of the operations, +, −,  [#permalink]

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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

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Re: If r and s are positive numbers and θ is one of the operations, +, −,  [#permalink]

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Hi

But it can be either r-s or s-r or |s-r|. It is still insufficient
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Re: If r and s are positive numbers and θ is one of the operations, +, −,  [#permalink]

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A alone is sufficient.

For B, - and / two signs are possible.
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" I CAN AND I WILL" Re: If r and s are positive numbers and θ is one of the operations, +, −,   [#permalink] 15 Jul 2019, 00:48
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