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Re: The symbol Ω represents one of the following operations: add [#permalink]
Thanks Bunnel, I forgot that the expression always needs to satisfy 1Ω1=1. :oops:


Bunuel wrote:
rgyanani wrote:
The symbol Ω represents one of the following operations: addition, subtraction, multiplication or division. What is the value of 1 Ω 1?

(1) 2 Ω 2 = 4

(2) 0 Ω 1 = 0

Taken from Magoosh Practice question

Can anyone tell me why it isn't E?


The symbol Ω represents one of the following operations: addition, subtraction, multiplication or division. What is the value of 1 Ω 1?

(1) 2 Ω 2 = 4. This implies that Ω represents addition or multiplication (2+2=4 and 2*2=4). If its addition, then 1 Ω 1 = 1 + 1 = 2 but if its multiplication, then 1 Ω 1 = 1*1 = 1. Not sufficient.

(2) 0 Ω 1 = 0. This implies that Ω represents multiplication or division. In either case 1 Ω 1 = 1. Sufficient.

Answer: B.

Hope it's clear.

Originally posted by ROckHIsT on 29 Dec 2013, 03:44.
Last edited by ROckHIsT on 29 Dec 2013, 03:54, edited 1 time in total.
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Re: The symbol Ω represents one of the following operations: add [#permalink]
Expert Reply
rgyanani wrote:
Thanks Bunnel, I didn't knew the concept where the expression always needs to satisfy 1Ω1=1.


Bunuel wrote:
rgyanani wrote:
The symbol Ω represents one of the following operations: addition, subtraction, multiplication or division. What is the value of 1 Ω 1?

(1) 2 Ω 2 = 4

(2) 0 Ω 1 = 0

Taken from Magoosh Practice question

Can anyone tell me why it isn't E?


The symbol Ω represents one of the following operations: addition, subtraction, multiplication or division. What is the value of 1 Ω 1?

(1) 2 Ω 2 = 4. This implies that Ω represents addition or multiplication (2+2=4 and 2*2=4). If its addition, then 1 Ω 1 = 1 + 1 = 2 but if its multiplication, then 1 Ω 1 = 1*1 = 1. Not sufficient.

(2) 0 Ω 1 = 0. This implies that Ω represents multiplication or division. In either case 1 Ω 1 = 1. Sufficient.

Answer: B.

Hope it's clear.


The question asks to find the value of 1 Ω 1.

From (1) we have that Ω represents addition or multiplication, which gives two different values of 1 Ω 1. Therefore (1) is NOT sufficient.
From (2) we have that Ω represents multiplication or division, which gives single numerical value of 1 Ω 1. Therefore (2) IS sufficient.

Hope it's clear.
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Re: The symbol Ω represents one of the following operations: add [#permalink]
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ROckHIsT wrote:
The symbol Ω represents one of the following operations: addition, subtraction, multiplication or division. What is the value of 1 Ω 1?

(1) 2 Ω 2 = 4

(2) 0 Ω 1 = 0


Target question: What is the value of 1 Ω 1?

Given: Ω represents one of the following operations: addition, subtraction, multiplication or division.

Statement 1: 2 Ω 2 = 4
2 - 2 = 0, so Ω CANNOT represent subtraction.
2 ÷ 2 = 1, so Ω CANNOT represent division.
2 + 2 = 4, so Ω COULD represent addition. In this case, the answer to the target question is 1 Ω 1 = 1 + 1 = 2
2 x 2 = 4, so Ω COULD represent multiplication. In this case, the answer to the target question is 1 Ω 1 = 1 x 1 = 1
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 0 Ω 1 = 0
0 - 1 = -1, so Ω CANNOT represent subtraction.
0 ÷ 1 = 0, so Ω COULD represent division. In this case, the answer to the target question is 1 Ω 1 = 1 ÷ 1 = 1
0 + 1 = 1, so Ω CANNOT represent addition.
0 x 1 = 0, so Ω COULD represent multiplication. In this case, the answer to the target question is 1 Ω 1 = 1 x 1 = 1

IMPORTANT: from statement 2, we can conclude that Ω represents EITHER division OR multiplication. Because Ω could represent two different operations, some students will incorrectly conclude that statement 2 is not sufficient.
However, the target question is NOT asking "What operation does Ω represent?" The target question is asking What is the value of 1 Ω 1?

In BOTH of the above cases (for statement 2), 1 Ω 1 = 1.
Since we get only one answer to the target question, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent
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Re: The symbol Ω represents one of the following operations: add [#permalink]
Bunuel wrote:
rgyanani wrote:
The symbol Ω represents one of the following operations: addition, subtraction, multiplication or division. What is the value of 1 Ω 1?

(1) 2 Ω 2 = 4

(2) 0 Ω 1 = 0

Taken from Magoosh Practice question

Can anyone tell me why it isn't E?


The symbol Ω represents one of the following operations: addition, subtraction, multiplication or division. What is the value of 1 Ω 1?

(1) 2 Ω 2 = 4. This implies that Ω represents addition or multiplication (2+2=4 and 2*2=4). If its addition, then 1 Ω 1 = 1 + 1 = 2 but if its multiplication, then 1 Ω 1 = 1*1 = 1. Not sufficient.

(2) 0 Ω 1 = 0. This implies that Ω represents multiplication or division. In either case 1 Ω 1 = 1. Sufficient.

Answer: B.

Hope it's clear.


Hi Bunuel,

Isn't \(\frac{0}{1}\) supposed to be undefined rather than be equal to zero?
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Re: The symbol Ω represents one of the following operations: add [#permalink]
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kwesibabban wrote:
Bunuel wrote:
rgyanani wrote:
The symbol Ω represents one of the following operations: addition, subtraction, multiplication or division. What is the value of 1 Ω 1?

(1) 2 Ω 2 = 4

(2) 0 Ω 1 = 0

Taken from Magoosh Practice question

Can anyone tell me why it isn't E?


The symbol Ω represents one of the following operations: addition, subtraction, multiplication or division. What is the value of 1 Ω 1?

(1) 2 Ω 2 = 4. This implies that Ω represents addition or multiplication (2+2=4 and 2*2=4). If its addition, then 1 Ω 1 = 1 + 1 = 2 but if its multiplication, then 1 Ω 1 = 1*1 = 1. Not sufficient.

(2) 0 Ω 1 = 0. This implies that Ω represents multiplication or division. In either case 1 Ω 1 = 1. Sufficient.

Answer: B.

Hope it's clear.


Hi Bunuel,

Isn't \(\frac{0}{1}\) supposed to be undefined rather than be equal to zero?


0 divided by anything but 0, is 0: 0/(anything) = 0

anything/0 is undefined. Division by 0 is not allowed.
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Re: The symbol represents one of the following operations: add [#permalink]
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Re: The symbol represents one of the following operations: add [#permalink]
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