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Sub 505 (Easy)|   Arithmetic|   Functions and Custom Characters|               
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gmatt1476
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For all real numbers a, b, c, d, e, and f, the operation Θ is defined 20 Apr 2020, 11:30
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E
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For all real numbers a, b, c, d, e, and f, the operation Θ is defined by the equation \((a, b, c) Θ (d, e, f) = ad + be + cf\). What is the value of \((1,−2, 3) Θ (1, −\frac{1}{2}, \frac{1}{3})\) ?

A. -1

B. 5/6

C. 1

D. 5/2

E. 3

PS37631.02
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E
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For all real numbers a, b, c, d, e, and f, the operation Θ is defined 20 Apr 2020, 11:35
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\((1,−2, 3) Θ (1, −\frac{1}{2}, \frac{1}{3})\)
= \([1*1] + [(-2)*(\frac{-1}{2})] + [3*\frac{1}{3}]\)
= 1+1+1
=3
gmatt1476
For all real numbers a, b, c, d, e, and f, the operation Θ is defined by the equation \((a, b, c) Θ (d, e, f) = ad + be + cf\). What is the value of \((1,−2, 3) Θ (1, −\frac{1}{2}, \frac{1}{3})\) ?

A. -1

B. 5/6

C. 1

D. 5/2

E. 3

PS37631.02
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For all real numbers a, b, c, d, e, and f, the operation Θ is defined 20 Apr 2020, 11:35
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Answer = (1 x 1) + (-2 * - 1/2) + (3 * 1/3)
= 1 + 1 + 1
= 3

(E)
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For all real numbers a, b, c, d, e, and f, the operation Θ is defined 20 Apr 2020, 11:36
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gmatt1476
For all real numbers a, b, c, d, e, and f, the operation Θ is defined by the equation \((a, b, c) Θ (d, e, f) = ad + be + cf\). What is the value of \((1,−2, 3) Θ (1, −\frac{1}{2}, \frac{1}{3})\) ?

A. -1

B. 5/6

C. 1

D. 5/2

E. 3

PS37631.02
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As per the given conditions: \((1,−2, 3) Θ (1, −\frac{1}{2}, \frac{1}{3})\)

1*1=1 + (-2*-1/2=1) + (3*1/3=1) = 3 Ans E
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For all real numbers a, b, c, d, e, and f, the operation Θ is defined 30 Apr 2020, 09:51
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gmatt1476
For all real numbers a, b, c, d, e, and f, the operation Θ is defined by the equation \((a, b, c) Θ (d, e, f) = ad + be + cf\). What is the value of \((1,−2, 3) Θ (1, −\frac{1}{2}, \frac{1}{3})\) ?

A. -1
B. 5/6
C. 1
D. 5/2
E. 3

PS37631.02
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Given: (a, b, c) Θ (d, e, f) = ad + be + cf
So, (1,−2, 3) Θ (1, −1/2, 1/3) = (1)(1) + (-2)(-1/2) + (3)(1/3)
= 1 + 1 + 1
= 3

Answer: E

Cheers,
Brent
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For all real numbers a, b, c, d, e, and f, the operation Θ is defined 03 May 2020, 06:15
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gmatt1476
For all real numbers a, b, c, d, e, and f, the operation Θ is defined by the equation \((a, b, c) Θ (d, e, f) = ad + be + cf\). What is the value of \((1,−2, 3) Θ (1, −\frac{1}{2}, \frac{1}{3})\) ?

A. -1

B. 5/6

C. 1

D. 5/2

E. 3

PS37631.02
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Plugging into the equation, we have:

1(1) + (-2)(-1/2) + 3(1/3) = 1 + 1 + 1 = 3

Answer: E
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For all real numbers a, b, c, d, e, and f, the operation Θ is defined 31 May 2020, 00:21
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Given (a, b, c)Θ(d, e, f) = ad + be + cf and

\((1,−2,3)Θ(1,\frac{−1}{2},\frac{1}{3})\) = \((1 * 1) + (-2 * \frac{−1}{2}) + (3 * \frac{1}{3})\)

\((1,−2,3)Θ(1,\frac{−1}{2},\frac{1}{3})\) = 1 + 1 + 1 = 3

Option E
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For all real numbers a, b, c, d, e, and f, the operation Θ is defined 11 Feb 2021, 08:10
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SOLUTION:

(a,b,c)Θ(d ,e, f)=ad+ be +cf

In the expression (1,−2,3)Θ(1,−1/2,1/3) ,
a=1 d=1
b=-2 e=-1/2
c=3 f= 1/3

Using this operation, we have, ad+ be +c f

= (1.1) + (-2. -1/2) + (3.1/3)

= 1 + 1 +1

=3 (OPTION E)

Hope this helps :thumbsup:
Devmitra Sen(Math)
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For all real numbers a, b, c, d, e, and f, the operation Θ is defined 15 May 2021, 04:19
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gmatt1476
For all real numbers a, b, c, d, e, and f, the operation Θ is defined by the equation \((a, b, c) Θ (d, e, f) = ad + be + cf\). What is the value of \((1,−2, 3) Θ (1, −\frac{1}{2}, \frac{1}{3})\) ?

A. -1

B. 5/6

C. 1

D. 5/2

E. 3

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Answer: Option E

Video solution by GMATinsight

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For all real numbers a, b, c, d, e, and f, the operation Θ is defined 27 Aug 2025, 04:33
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As per the OG 2024-2025, it is positive 1/2
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For all real numbers a, b, c, d, e, and f, the operation Θ is defined 27 Aug 2025, 07:49
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thekingslay
As per the OG 2024-2025, it is positive 1/2
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For all real numbers a, b, c, d, e, and f, the operation Θ is defined 25 Oct 2025, 19:28
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