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If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a an

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Bunuel
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If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a an [#permalink] If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a an   14 Feb 2020, 05:01
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If the operation ∆ is defined by \(a∆b = \frac{(b-a)^2}{a^2}\) for all number a and b, and a ≠ 0, then -1∆(1∆ -1) =

A. -1
B. 0
C. 1
D. 9
E. 25
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E
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BrentGMATPrepNow
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Re: If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a an [#permalink] If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a an   14 Feb 2020, 08:39
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Bunuel
If the operation ∆ is defined by \(a∆b = \frac{(b-a)^2}{a^2}\) for all number a and b, and a ≠ 0, then -1∆(1∆ -1) =

A. -1
B. 0
C. 1
D. 9
E. 25

We want to find the value of -1∆(1∆ -1), so let's first deal with the part in brackets...

\(1∆-1 = \frac{((-1)-1 )^2}{ 1^2}= \frac{(-2)^2}{ 1^2}= \frac{4}{ 1}=4\)

So, -1∆(1∆ -1) = -1∆4
Now we need to find the value of -1∆4

\(-1∆4 = \frac{(4-(-1))^2}{(-1)^2}= \frac{5^2}{1}= \frac{25}{1}=25\)

Answer: E

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Brent
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Re: If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a an [#permalink] If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a an   14 Feb 2020, 09:43
Bunuel
If the operation ∆ is defined by \(a∆b = \frac{(b-a)^2}{a^2}\) for all number a and b, and a ≠ 0, then -1∆(1∆ -1) =

A. -1
B. 0
C. 1
D. 9
E. 25

Asked: If the operation ∆ is defined by \(a∆b = \frac{(b-a)^2}{a^2}\) for all number a and b, and a ≠ 0, then -1∆(1∆ -1) =

-1∆ 4 = 25

IMO E

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Re: If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a an [#permalink] If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a an   23 Feb 2020, 04:09
Bunuel
If the operation ∆ is defined by \(a∆b = \frac{(b-a)^2}{a^2}\) for all number a and b, and a ≠ 0, then -1∆(1∆ -1) =

A. -1
B. 0
C. 1
D. 9
E. 25

First, we simplify: a∆b = (b-a)^2/a^2 = (b^2 - 2ab + a^2)/a^2 = (b/a)^2 - 2(b/a) + 1 = (b/a - 1)^2

Working from the inside out: 1∆-1 =( -1/1 - 1)^2 = (-2)^2 = 4.
--> -1∆4 = (4/-1 -1)^2 = (-5)^2 = 25 (E)
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Re: If the operation is defined by ab = (b-a)^2/a^2 for all number a an [#permalink] If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a an   08 Aug 2022, 09:30
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Given that a ∆ b = \(\frac{(b - a)^2}{a^2}\) and we need to find the value of − 1 ∆ (1 ∆ − 1)

Lets start by finding the value of 1 ∆ − 1

To find 1 ∆ − 1 we need to compare what is before and after ∆ in 1 ∆ − 1 and a ∆ b

=> We need to substitute a with 1 and b with -1 in a ∆ b = \(\frac{(b - a)^2}{a^2}\) to get the value of 1 ∆ − 1

=> 1 ∆ -1 = \(\frac{(-1 - 1)^2}{1^2}\) = \(\frac{(-2)^2}{1}\) = 4

=> − 1 ∆ (1 ∆ − 1) = − 1 ∆ 4

Similarly, -1 ∆ 4 = \(\frac{(4 - (-1))^2}{(-1)^2}\) = \(\frac{(5)^2}{1}\) = 25

So, Answer will be E
Hope it helps!

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Re: If the operation is defined by ab = (b-a)^2/a^2 for all number a an [#permalink]
08 Aug 2022, 09:30
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