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a*b=a^2-b

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a*b=a^2-b [#permalink] New post   07 Jun 2022, 02:00
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\(a \oplus b=a^2-b\)

\(a \odot b = a+b^2\)

Quantity A
Quantity B
\([3 \oplus 4 ] \odot 5\)
\(3 \oplus [ 4 \odot 5 ]\)



A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

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A
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B
alsafaj
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a*b=a^2-b [#permalink] New post   Updated on: 08 Jun 2022, 04:45
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Using the concept of PEMDAS, we get:
quantity A: 30
quantity B: -20

30 > -20

We always evaluate what is inside paranthesis before proceeding.

The answer is (A)

Originally posted by alsafaj on 07 Jun 2022, 22:49.
Last edited by alsafaj on 08 Jun 2022, 04:45, edited 2 times in total.
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Re: a*b=a^2-b [#permalink] New post   08 Jun 2022, 01:11
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alsafaj, could you plz explain how you get value of Quantity A as 0
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alsafaj
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a*b=a^2-b [#permalink] New post   Updated on: 08 Jun 2022, 04:45
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AkkuJi wrote:
alsafaj, could you plz explain how you get value of Quantity A as 0


Thanks for the correction. I have edited the original answer

a⊕b=a^2−b

a⊙b=a+b^2

Quantity A
[3⊕4]⊙5
[9-4]⊙5
5⊙5=5+5^2 = 30

Quantity A: 30

Originally posted by alsafaj on 08 Jun 2022, 01:48.
Last edited by alsafaj on 08 Jun 2022, 04:45, edited 1 time in total.
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Re: a*b=a^2-b [#permalink] New post   08 Jun 2022, 04:40
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Thanks alsafaj, I believe A should be 30 :angel: :angel: :angel:
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Re: a*b=a^2-b [#permalink] New post   08 Jun 2022, 04:44
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AkkuJi wrote:
Thanks alsafaj, I believe A should be 30 :angel: :angel: :angel:


haha, right! I am not sure how I was calculating all of this inside my head. Thanks for the correction again!

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Re: a*b=a^2-b [#permalink]
08 Jun 2022, 04:44
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