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

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Intern
Joined: 26 May 2015
Posts: 7
If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a  [#permalink]

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17 Jul 2017, 12:57
00:00

Difficulty:

15% (low)

Question Stats:

81% (01:08) correct 19% (01:27) wrong based on 50 sessions

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If the operation ∆ is defined by a∆b = (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
Senior Manager
Joined: 24 Apr 2016
Posts: 331
Re: If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a  [#permalink]

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17 Jul 2017, 15:20
1
-1∆(1∆ -1) = -1∆[(-1-1)^2]/1 = -1∆4 = [(4+1)^2]/1 = 25
Senior SC Moderator
Joined: 22 May 2016
Posts: 2632
Re: If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a  [#permalink]

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17 Jul 2017, 18:55
1
Esguitar wrote:
If the operation ∆ is defined by a∆b = $$(b-a)^2/a^2$$ for all number a and b, and a ≠ 0, then -1∆(1∆ -1) =
a. -1
b. 0
c. 1
d. 9
e. 25

Or, the longish way (not too bad - 54 seconds)

If a∆b = $$\frac{(b-a)^2}{a^2}$$, then

-1∆(1∆ -1) =

1. Parentheses first:

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

4 becomes b -->

2. New expression is -1 ∆ 4

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

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Intern
Joined: 26 May 2015
Posts: 7
Re: If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a  [#permalink]

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18 Jul 2017, 02:13
Thanks a lot, I got it.
I could not see how the expression (1∆ -1) becomes (−1−1)^2/(1)^2, I assume that it is plugin number where a=1 and b=-1.
Thanks again.
Re: If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a   [#permalink] 18 Jul 2017, 02:13
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